BusyBeaverWiki - User contributions [en]

Sequences

2026-05-09T03:37:40Z

Int-y1: add fractran

This page lists sequences related to the Busy Beaver functions.

These tables are incomplete, you can help by adding missing items. If you add a value, please add a reference to a paper or code with which it was computed/proved if possible.

If the "canonical" values of a sequence are maintained on another Wiki page, please link to that, instead of replicating them here.

=== Computable Sequences ===
{| class="wikitable"
!Sequence Name
!Description
!Values
![[oeis:|OEIS]] sequence
|-
|2-symbol TM count
|Number of n-state, 2-symbol, d+ in {LEFT, RIGHT}, 5-tuple (q, s, q+, s+, d+) (halting or not) Turing machines.
|
|[[oeis:A052200|A052200]]
|-
|Number of n-state 2-symbol halt-free TMs
|A Turing machine is halt-free if none of its instructions lead to the halt state.
|
|[[oeis:A337025|A337025]]
|-
|[[Lazy Beaver]]
|The smallest positive number of steps a(n) such that no n-state, m-symbol Turing machine halts in exactly a(n) steps on an initially blank tape.
|see [[Lazy Beaver#Computed Values]]
|[[oeis:A337805|A337805 (for m=2)]]
|-
|Configs A(a, b) reached in [[Antihydra]]
|
|
|[[oeis:A386792|A386792]] (for a), [[oeis:A385902|A385902]] (for b)
|}

=== Noncomputable Sequences ===
The following sequences depend on the specific behavior of programs and are grouped by their position in the [[wikipedia:Arithmetical_hierarchy|arithmetical hierarchy]].

Note that when the bbchallenge community refers to BB(n, m), we mean the Max Shift function S(n, m) defined below (if m is omitted, it is set to 2 by default). Some literature may refer to the Max Score function Σ(n, m) by BB(n, m) instead.

==== Π1 ====
{| class="wikitable"
!Sequence Name
!Symbol
!Description
!Values
![[oeis:|OEIS]] sequence
|-
|[[Busy Beaver Functions#Max Shift Function S(n, m)|Max Shift Function]]
|S(n, m)
|The maximal number of steps that an n-state, m-symbol Turing machine can make on an initially blank tape before eventually halting.
|[[Main Page|see the Main Page]]
|[[oeis:A060843|A060843]]
|-
|[[Busy Beaver Functions#Max Score Function Σ(n, m)|Max Score Function]]
|Σ(n, m)
|Maximal number of 1's that an n-state, m-symbol Turing machine can print on an initially blank tape before halting.
|
|[[oeis:A028444|A028444]]
|-
|[[Maximum Space Function]]
|BB_SPACE(n,m)
|Maximum number of memory cells visited by a halting Turing machine with n states and m symbols starting from all-0 memory tape
|see [[Maximum Space Function#Champions]]
| -
|-
|
|
|Number of n-state Turing machines which halt.
|
|[[oeis:A004147|A004147]]
|-
|[[Blanking Busy Beaver]]
|BLB(n,m)
|The maximum number of steps that an n-state m-symbol Turing machine can make on an initially blank tape until it is blank again (halting or not)
|see [[Blanking Busy Beaver Function#Champions]]
| -
|-
|BB_clean
|
|The maximum number of steps that an n-state 2-symbol Turing machine can make on an initially blank tape until it halts on a blank tape
|(see comments #75 and #77 [https://scottaaronson.blog/?p=5661 here])
|[[oeis:A119683|A119683]]
|-
|BB_ones
|
|The maximum number of 1's that an n-state 2-symbol Turing machine can make in a row, before halting on a 0 next to it
|
|
|-
|[[Maximum Consecutive Ones Function]]
|num(n)
|The maximum amount of consecutive 1's that an n-state 2-symbol Turing machine can print before eventually halting
|see [[Maximum Consecutive Ones Function]]
|
|-
|Size of the Runtime Spectrum
|<math>R(n)</math>
|The number of distinct runtimes for a machine with a given number of symbols, for increasing number of states
|see "The Spectrum of Runtimes" in "[https://www.scottaaronson.com/papers/bb.pdf The Busy Beaver Frontier]"
|
|-
|
|
|The number of non-halting programs with n states which reach infinitely many tape cells
|
|-
|[[Instruction-Limited Busy Beaver]]
|BBi(n)
|Maximum number of steps that an n-instruction Turing machine (allowing any number of states and symbols) can take on an initially blank tape before eventually halting.
|see [[Instruction-Limited Busy Beaver#Champions]]
|[[oeis:A384629|A384629]]
|-
|[[Instruction-Limited Busy Beaver|Instruction-Limited Symbol Busy Beaver]]
|Σi(n)
|Maximum number of non-blank symbols that an n-instruction Turing machine (allowing any number of states and symbols) can leave on an initially blank tape before eventually halting.
|see [[Instruction-Limited Busy Beaver#Champions]]
|[[oeis:A384766|A384766]]
|-
|[[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants|Instruction-Limited Greedy Busy Beaver]]
|gBBi(n)
|Maximum number of steps that an n-instruction Turing machine can take from a blank tape before halting, where the Turing machines first n-1 instructions are a machine which runs for gBBi(n-1) steps.
|see [[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants]]
| -
|-
|[[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants|Instruction-Limited Blanking Busy Beaver]]
|BLBi(n)
|Maximum number of steps that an n-instruction Turing machine (allowing any number of states and symbols) can take on an initially blank tape before blanking the tape again.
|see [[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants]]
| -
|-
|[[Busy Beaver for lambda calculus]]
|BBλ(n)
|Maximum beta normal form size of any closed lambda term of size n.
|see [[Busy Beaver for lambda calculus#Champions]]
|[[oeis:A333479|A333479]]
|-
|[[Fractran]]
|BBf(n)
|The maximal number of steps that a Fractran program of size n can make on the integer 2 before eventually halting.
|see [[Fractran#Champions]]
|[[oeis:A395424|A395424]]
|-
|}

==== Π2 ====
{| class="wikitable"
!Sequence Name
!Symbol
!Description
!Values
![[oeis:|OEIS]] sequence
|-
|[[Beeping Busy Beaver]]
|BBB(n)
|The latest possible step that any 2-symbol TM with n states exits a chosen state finitely many times
|see [[Beeping Busy Beaver#Results]]
|-
|[[Busy Beaver for lambda calculus#Oracle Busy Beaver|Busy Beaver for lambda calculus with a BBλ oracle]]
|BBλ<sub>1</sub>
|Maximum beta/oracle normal form size of any 1-closed lambda term of size n.
|see [[Busy Beaver for lambda calculus#Oracle Busy Beaver]]
|[[oeis:A385712|A385712]]
|}

==== Π3 ====
{| class="wikitable"
!Sequence Name
!Symbol
!Description
!Values
![[oeis:|OEIS]] sequence
|-
|[[Beeping Busy Beaver#Beeping Booping Busy Beavers|Beeping Booping busy beaver]]
|BBBB(n)
|
|see [[Beeping Busy Beaver#Beeping Booping Busy Beavers]]
|}

==== Yet ungrouped ====
{| class="wikitable"
!Sequence Name
!Symbol
!Description
!Values
![[oeis:|OEIS]] sequence
|-
|
|#S(n, m)
|The number of programs that halt after exactly S(n,m) steps ([[Busy Beaver Functions#Max Shift Function S(n, m)|Max Shift]]) for each n of a given m (including all equivalent transformations)
|#S(1,2)=32, #S(2,2)=40, #S(3,2)=16
| -
|-
|
|#Σ(n, m)
|The number of programs that halt with Σ(n, m) 1's on the tape ([[Busy Beaver Functions#Max Score Function Σ(n, m)|Max Score]]) for each n of a given m (including all equivalent transformations)
|#Σ(1,2)=16, #Σ(2,2)=4, #Σ(3,2)=40
| -
|-
|Maximum space
|#BB_SPACE(n,m)
|The number of programs that visited the most number of tape cells for a given (n,m) (including all equivalent transformations)
|#BB_SPACE(1,2)=32, #BB_SPACE(2,2)=24, #BB_SPACE(3,2)=48
| -
|-
|
|
|The average number of states that are reached infinitely many times, among all non-halting turing machines with n states
|
| -
|}

=== More possibilities ===

* The number of distinct final tape states of halting machines with n states and m symbols, for some definition of "distinct"
* Any of the above for machines with more than one tape, or tapes with more dimensions (2d grid, 3d, n-d...)
* Machines with a finite tape, or a circular one of a certain length
* Any of the above functions bounded by the number of instructions rather than states and symbols.

=== Further information ===
For more information on sequences, see the [[oeis:wiki/Busy_Beaver_numbers|OEIS Wiki: Busy Beaver Numbers]], [https://oeis.org/search?q=busy+beaver OEIS search: "busy beaver"] and [[oeis:wiki/Index_to_OEIS:_Section_Br#beaver|OEIS Wiki: "related to busy beaver"]]
[[Category:Functions]]

Fractran

2026-04-12T09:09:13Z

Int-y1: style

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A Fractran program is a list of rational numbers <math>[q_0, q_1, \dots, q_{k-1}]</math> called rules and a Fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t</math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b}</math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b)</math>. And the size of a Fractran program <math>[q_0, q_1, \dots, q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i)</math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting Fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the Fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a Fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about Fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, Fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to Fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Visualizing Fractran Programs' Space-Time Diagrams ==
Katelyn Doucette's Fractran space-time diagram visualizer produces the following space-time diagrams for some notable Fractran Programs, under the following principle: Each color represents a prime factor. Left -> right colors indicating the index of that register, and how wide the color is representing how big the value is at that step. Source code: https://github.com/Laturas/FractranVisualizer
{| class="wikitable"
|+
|[[File:Fractran_22_Cryptid.webp|alt=The space-time diagram of the BBf(22) Cryptid|460x460px]]
The space-time diagram of the BBf(22) Cryptid
|[[File:Hydra.webp|alt=The space-time diagram of Hydra.|460x460px]]
The space-time diagram of Hydra.
|[[File:Bbf21 champ full.png|alt=The space-time diagram of the BBf(21) champion.|400x400px]]

The space-time diagram of the BBf(21) champion. The width & height of the diagram can be set in the visualizer.
|[[File:Space_Needle.webp|alt=The space-time diagram of Space Needle.|460x460px]]
The space-time diagram of Space Needle.
|}

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove Fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

-d released a new decider on 25 Jan 2026: [https://discord.com/channels/960643023006490684/1438019511155691521/1464873923647639703 Beeping Permutation].

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.
All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|140 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1464873923647639703 25 Jan 2026]
Claude Opus 4.6 proof of nonhalting of all 140: [https://discord.com/channels/960643023006490684/1438019511155691521/1485168251997786173 28 March 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|2003 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1464873923647639703 25 Jan 2026]

Known [[Cryptid|Cryptids]]:

# Fenrir
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, \dots, \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|'''''17'''''
|'''''2'''''
|'''''3'''''
|'''''107'''''
|-
|'''''18'''''
|'''''2'''''
|'''''4'''''
|'''''211'''''
|-
|'''''19'''''
|'''''2'''''
|'''''5'''''
|'''''370'''''
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

<br>
==== BBf(20) ====
[[File:Screenshot 2026-04-01 104704.png|alt=Full space-time diagram of the BBf(20) champion.|left|507x507px]]
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
[[File:Bbf21 champ full.png|alt=The full space-time diagram of the BBf(21) champion until halting.|thumb|The full space-time diagram of the BBf(21) champion until halting.]]

The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased pseudo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Fenrir ===
[[File:Fractran 22 Cryptid.webp|alt=The space-time diagram of Fenrir.|thumb|Partial space-time diagram of Fenrir.]]
"Fenrir" is a family of 3 size 22 [[Cryptids]] discovered by Jason Yuen (@-d) and Claude Opus 4.6 on 22 Mar 2026. Out of 500 holdouts of size 22, Claude Opus 4.6 used Lean to prove that 497 holdouts were non-halting. The remaining 3 holdouts are the Fenrir family.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1485415054475268179]</sup> Discord user @ZTS439 shared [https://discord.com/channels/960643023006490684/1438019511155691521/1487251919444508723 some analysis] and a [https://discord.com/channels/960643023006490684/1438019511155691521/1487252789158613002 Python program] for it. Its name comes from [[wikipedia:Norse_mythology|nordic mythology]]; [[wikipedia:Fenrir|Fenrir]] is the wolf that helps destroy the world during [[wikipedia:Ragnarök|Ragnarök]].

{| class="wikitable"
|+
!Holdout number
!Holdout
!Vector Representation
|-
| 29/2003
| <code>[1/15, 27/77, 49/3, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 41/2003
| <code>[1/15, 49/3, 27/77, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 2 & 0 \\
0 & 3 & 0 & -1 & -1 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 430/2003
| <code>[27/35, 1/33, 25/3, 22/25, 21/2]</code>
|<math>\begin{bmatrix}
0 & 3 & -1 & -1 & 0 \\
0 & -1 & 0 & 0 & -1 \\
0 & -1 & 2 & 0 & 0 \\
1 & 0 & -2 & 0 & 1 \\
-1 & 1 & 0 & 1 & 0
\end{bmatrix}</math>
|}

All 3 holdouts follow a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math> (for 29/2003 and 41/2003) or <math>S(x,y) = [x, 0, 2, y, 0]</math> (for 430/2003), then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

=== Frankenstein's Monster ===
[[File:Frankenstein's Monster.webp|alt=Partial space-time diagram of Frankenstein's Monster.|thumb|Partial space-time diagram of Frankenstein's Monster.]]
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
[[File:Hydra.webp|alt=Partial space-time diagram of Hydra.|thumb|300x300px|Partial space-time diagram of Hydra.]]
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
[[File:Ftran bmo1.png|alt=Partial space-time diagram of BMO 1.|thumb|Partial space-time diagram of BMO 1.]]
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
[[File:Space Needle.webp|alt=Partial space-time diagram of Space Needle.|thumb|Partial space-time diagram of Space Needle.]]
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

Cryptids

2026-04-04T06:34:53Z

Int-y1: add fenrir

[[File:Lovecraft beaver.png|alt=A monstrous beaver in the style of HP Lovecraft with pink tentacles coming out of its mouth, 5 red spider eyes, horns on its head, elbows and tail, moss colored fur, sharp purple claws and webbed feet.|thumb|Lovecraftian Beaver fan art made by Lauren]]
'''Cryptids''' are Turing Machines whose behavior (when started on a blank tape) can be described completely by a relatively simple mathematical rule, but where that rule falls into a class of unsolved (and presumed hard) mathematical problems. This definition is somewhat subjective (What counts as a simple rule? What counts as a hard problem?). In practice, most currently known small Cryptids have [[Collatz-like]] behavior. In other words, the halting problem from blank tape of Cryptids is mathematically-hard.

If there exists a Cryptid with n states and m symbols, then BB(n, m) cannot be solved without solving this hard math problem.

The name Cryptid was proposed by Shawn Ligocki in an Oct 2023 [https://www.sligocki.com/2023/10/16/bb-3-3-is-hard.html blog post] announcing the discovery of [[Bigfoot]].

== Cryptids at the Edge ==

This is a list of notable Minimal Cryptids (Cryptids in a [[:Category:BB_Domains|domain]] with no strictly smaller known Cryptid). All of these Cryptids were "discovered in the wild" rather than "constructed".

{| class="wikitable"
|-
! Name !! BB domain !! Machine !! Date !! Discoverer !! Note
|-
|[[Bigfoot]]
|[[BB(3,3)]]
|{{TM|1RB2RA1LC_2LC1RB2RB_---2LA1LA|undecided}}
|Nov 2023
|[[User:Sligocki|Shawn Ligocki]]
|
|-
|[[Hydra]]
|[[BB(2,5)]]
|{{TM|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA|undecided}}
|May 2024
|Daniel Yuan
|
|-
|[[Bonus cryptid]]
|[[BB(2,5)]]
|{{TM|1RB3RB---3LA1RA_2LA3RA4LB0LB1LB}}
|May 2024
|Daniel Yuan
|Probviously non-halting.
|-
|[[Antihydra]]
|[[BB(6)]]
|{{TM|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA|undecided}}
|June 2024
|<code>@mxdys</code>, shown to be a Cryptid by <code>@racheline</code>.
|Same as '''Hydra''' but starting iteration from 8 instead of 3 and with termination condition <code>O > 2E</code> instead of <code>E > 2O</code>, hence the name '''Antihydra'''.
|-
|[[Lucy's Moonlight]]
|[[BB(6)]]
|{{TM|1RB0RD_0RC1RE_1RD0LA_1LE1LC_1RF0LD_---0RA}}
|Mar 2025
|Racheline
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB1RC_1LC1LE_1RA1RD_0RF0RE_1LA0LB_---1RA|undecided}}
|Jul 2024
|<code>mxdys</code>
|Variant of Hydra and Antihydra. Probviously non-halting.
|-
|
|[[BB(6)]]
|{{TM|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC|undecided}}
|Aug 2024
|<code>mxdys</code>
|Similar random walk mechanism to Hydra, Antihydra. Probviously non-halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---|undecided}}
|Sep 2024
|Daniel Yuan
|Similar random walk mechanism to Hydra, Antihydra. Probviously non-halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0LB_1LC0RE_1LA1LD_0LC---_0RB0RF_1RE1RB|undecided}}
|Nov 2024
|Racheline
|Similar random walk mechanism to Hydra, Antihydra. Probviously non-halting.
|-
|[[1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD|Space Needle]]
|[[BB(6)]]
|{{TM|1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD|undecided}}
|Jan 2025
|<code>mxdys</code>
|Probviously non-halting.
|-
|
|[[BB(6)]]
|{{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC|undecided}}
|Jul 2024
|<code>mxdys</code>
|Has near-identical behavior to 16 related BB(6) holdouts. Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB1RE_1LC1LD_---1LA_1LB1LE_0RF0RA_1LD1RF}}
|Jul 2024
|Racheline
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0RE_1LC1LD_0RA0LD_1LB0LA_1RF1RA_---1LB}}
|Jul 2024
|Racheline
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0LC_0LC0RF_1RD1LC_0RA1LE_---0LD_1LF1LA}}
|Jul 2024
|Racheline
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0LC_1LC0RD_1LF1LA_1LB1RE_1RB1LE_---0LE}}
|Nov 2024
|Racheline
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB---_0RC0RE_1RD1RF_1LE0LB_1RC0LD_1RC1RA}}
|Nov 2024
|Racheline
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE}}
|Jul 2025
|<code>mxdys</code>
|Probviously halting.
|-
|
|[[BB(6)]]
|{{TM|1RB1LE_0LC0LB_1RD1LC_1RD1RA_1RF0LA_---1RE}}
|Jul 2024
|Racheline
|Probviously decidable. Estimated to have a 3/5 chance of becoming a [[translated cycler]] and a 2/5 chance of halting.
|-
|
|[[BB(6)]]
|{{TM|1RB0RB_1LC1RE_1LF0LD_1RA1LD_1RC1RB_---1LC|undecided}}
|Aug 2024
|mxdys, shown to be a Cryptid by DrDisentangle
|Similar to Space Needle, probviously nonhalting
|-
|[[Fractran#Fenrir|Fenrir]]
|[[Fractran|BBf(22)]]
|<code>[1/15, 27/77, 49/3, 10/49, 33/2]</code> and 2 others
|Mar 2026
|Jason Yuen (@-d) and Claude Opus 4.6
|Probviously non-halting.
|}
The following machines have chaotic behavior, but have not been classified as Cryptids due to seemingly lacking a connection to any known open mathematical problems, such as Collatz-like problems.

* {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}
* {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}}
* {{TM|1RB---0RB0LA2RA_2LB2LA3RA4LB0LB|undecided}}
* {{TM|1RB3LA1LA1RA3RA_2LB2RA---4RB1LB|undecided}}
* {{TM|1RB3LA1LA1RA1RA_2LB2RA---4RB1LB|undecided}}
* {{TM|1RB3LB---4LA1RB_2LA4LA4LB3RB1RA|undecided}} [https://discord.com/channels/960643023006490684/1375584513777995957 Analysis by @mxdys]

== Larger Cryptids ==

A more complete list of notable known Cryptids over a wider range of states and symbols. These Cryptids were all "constructed" rather than "discovered".

{| class="wikitable"
|-
! Name !! BB domain !! Machine !! Announcement !! Date !! Discoverer !! Note
|-
|[[Logical independence|ZF]]
|BB(432)
|style="width:30%;word-break:break-word"|Wade's machine: https://codeberg.org/ajwade/turing_machine_explorer/src/commit/33b30300054242201a95679aacf7e74312bd8803b0df9a85d2314095efd6804e
|
|2025
|Wade, based on work by CatIsFluffy and O'Rear
|The machine halts if and only if [[wikipedia:Zermelo–Fraenkel_set_theory|Zermelo–Fraenkel set theory]] is inconsistent.
|-
|PA
|BB(372)
|https://github.com/LegionMammal978/turing_machine_explorer/blob/main/pa.py
|[https://discord.com/channels/960643023006490684/1466652214247559198/1471186212743155856 Discord message]
|2026
|LegionMammal
|The machines halts if and only if Peano-Arithmetic is inconsistent.
|-
|RH
|BB(744)
|style="width:30%;word-break:break-word"|https://github.com/sorear/metamath-turing-machines/blob/master/riemann-matiyasevich-aaronson.nql
|
|2016
|Matiyasevich and O’Rear
|The machine halts if and only if [https://en.wikipedia.org/wiki/Riemann_hypothesis Riemann Hypothesis] is false.
|-
|Goldbach
|BB(25)
|style="width:30%;word-break:break-word"|https://gist.github.com/anonymous/a64213f391339236c2fe31f8749a0df6
<div class="toccolours mw-collapsible mw-collapsed">Machine code:<div class="mw-collapsible-content"><pre style="word-break:break-all">1RB1RD_1LC1RB_0RA1LC_0LQ1RE_0LF1RG_0LC1LF_0LF0LH_1LQ1LI_0RJ0LI_1RK0LJ_0RL0RS_1RL0RM_1RN1RM_0LO0LU_0LP1LO_1RH1LX_1LR1LQ_0RK0LT_1LR1RS_---1RC_1LV1LU_0LW0LJ_0RK0LW_1RY1LX_1RE1RY</pre></div></div>
|
|2016
|anonymous
|The machine halts if and only if [[wikipedia:Goldbach's_conjecture|Goldbach's conjecture]] is false. Its behavior has been verified in Lean.<ref>https://github.com/lengyijun/goldbach_tm</ref>
|-
| Erdős
| BB(5,4) and BB(15)
|style="width:30%;word-break:break-word"|
https://docs.bbchallenge.org/other/powers_of_two_5_4.txt
<div class="toccolours mw-collapsible mw-collapsed">Machine code:<div class="mw-collapsible-content"><pre style="word-break:break-all">1RB3RA2RA1RB_0LC2RB1RA3RB_0LD1LC2LE3LC_3RE2RE---1RE_0RB1LE2LE3LE</pre></div></div>
https://docs.bbchallenge.org/other/powers_of_two_15_2.txt
<div class="toccolours mw-collapsible mw-collapsed">Machine code:<div class="mw-collapsible-content"><pre style="word-break:break-all">1RB1RO_0RC0RC_0RD1RJ_0LE1RC_0LF1LK_0LG1LE_0LH1LF_1RI0LL_0RB1LK_1RC0RA_0LI1LN_1RM---_0RI0RO_0LK1LK_1LM1RA</pre></div></div>
|| [https://arxiv.org/abs/2107.12475 arxiv preprint] || Jul 2021 || [[User:Cosmo|Tristan Stérin]] (<code>@cosmo</code>) and Damien Woods || The machine halts if and only if the following conjecture by Erdős is false: "For all n > 8, there is at least one 2 in the base-3 representation of 2^n"
|-
|Weak Collatz
|BB(124) and BB(43,4)
|style="width:30%;word-break:break-word"|https://docs.bbchallenge.org/other/weak_Collatz_conjecture_124_2.txt (unverified)
https://docs.bbchallenge.org/other/weak_Collatz_conjecture_43_4.txt (unverified)
|
|Jul 2021
|[[User:Cosmo|Tristan Stérin]]
|The machine halts if and only if the "weak Collatz conjecture" is false. The weak Collatz conjecture states that the iterated Collatz map (3x+1) has only one cycle on the positive integers.
Not independently verified, and probably easy to further optimise.
|-
| Bigfoot - compiled|| [[BB(7)]]||style="width:30%;word-break:break-word"| <code>0RB1RB_1LC0RA_1RE1LF_1LF1RE_0RD1RD_1LG0LG_---1LB</code>|| [https://github.com/sligocki/sligocki.github.io/issues/8#issuecomment-2140887228 Bigfoot Comment] || June 2024 || <code>@Iijil1</code>|| Compilation of Bigfoot into 2 symbols, there was a previous compilation [https://github.com/sligocki/sligocki.github.io/issues/8#issuecomment-1774200442 with 8 states]
|-
| Hydra - compiled
|BB(9)
|style="width:30%;word-break:break-word"|<pre>
0RB0LD_1LC0LI_1LD1LB_0LE0RG_1RF0RH_1RA---_0RD0LB_0RA---_0RF1RZ
</pre>[[File:Hydra_9_states.txt]]
|[https://discord.com/channels/960643023006490684/1084047886494470185/1251572501578780782 Discord message]
|June 2024
|<code>@Iijil1</code>
|Compilation of Hydra into 2 symbols, all [https://discord.com/channels/960643023006490684/1084047886494470185/1253193750486974464 confirmed by Shawn Ligocki]. <code>@Iijil1</code> provided 24 TMs which all emulate the same behavior.
<small>[https://discord.com/channels/960643023006490684/1084047886494470185/1247560072427474955 Previous compilation had 10 states], by Daniel Yuan, also [https://discord.com/channels/960643023006490684/1084047886494470185/1247579473042346136 confirmed by Shawn Ligocki].</small>
|}

== Beeping Busy Beaver ==

Cryptids were actually noticed in the [[Beeping Busy Beaver]] problem before they were in the classic Busy Beaver. See [[Mother of Giants]] describing a "family" of Turing machines which "[[probviously]]" [[quasihalt]], but requires solving a Collatz-like problem in order to actually prove it. They are all TMs formed by filling in the missing transition in <code>1RB1LE_0LC0LB_0LD1LC_1RD1RA_---0LA</code> with different values.
[[Category:Zoology]]
[[Category:Cryptids]]

Fractran

2026-03-28T20:15:05Z

Int-y1: /* Size 22 Cryptid */ link to formal proof

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Visualizing Fractran Programs' Space-Time Diagrams ==
A program, in development by Katelyn Doucette, produces the following space-time diagrams for some notable Fractran Programs, under the following principle: Each color represents a prime factor. Left -> right colors indicating the index of that register, and how wide the color is representing how big the value is at that step.
<table><td>[[File:Fractran 22 Cryptid.webp|alt=The space-time diagram of the BBf(22) Cryptid|thumb|The space-time diagram of the BBf(22) Cryptid|none]]</td>
<td>[[File:Hydra.webp|alt=The space-time diagram of Hydra.|none|thumb|The space-time diagram of Hydra.]]</td>
<td>[[File:Frankenstein's Monster.webp|alt=The space-time diagram of Frankenstein's Monster.|thumb|The space-time diagram of Frankenstein's Monster.|none]]</td>
<td>[[File:Space Needle.webp|alt=The space-time diagram of Space Needle.|thumb|The space-time diagram of Space Needle.|none]]</td></table>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

-d released a new decider on 25 Jan 2026: [https://discord.com/channels/960643023006490684/1438019511155691521/1464873923647639703 Beeping Permutation].

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.
All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|140 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1464873923647639703 25 Jan 2026]
Claude Opus 4.6 proof of nonhalting of all 140: [https://discord.com/channels/960643023006490684/1438019511155691521/1485168251997786173 28 March 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|2003 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1464873923647639703 25 Jan 2026]

Known [[Cryptid|Cryptids]]:

# Size 22 Cryptid
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Size 22 Cryptid ===
The size 22 [[Cryptids|Cryptid]] is a group of 3 holdouts that was discovered by Jason Yuen (@-d) and Claude Opus 4.6. Out of 500 holdouts of size 22, Claude Opus 4.6 used Lean to prove that 497 holdouts were non-halting. The remaining 3 holdouts are equivalent to this Cryptid.<sup>[https://github.com/int-y1/proofs/blob/82ffbb9dd15e9b2cb67124741b2a262db8a62625/BBfLean/Size22Summary.lean#L659-L667]</sup> Discord user @ZTS439 shared [https://discord.com/channels/960643023006490684/1438019511155691521/1487251919444508723 some analysis] and a [https://discord.com/channels/960643023006490684/1438019511155691521/1487252789158613002 Python program] for it.

{| class="wikitable"
|+
!Holdout number
!Holdout
!Vector Representation
|-
| 29/2003
| <code>[1/15, 27/77, 49/3, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 41/2003
| <code>[1/15, 49/3, 27/77, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 2 & 0 \\
0 & 3 & 0 & -1 & -1 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 430/2003
| <code>[27/35, 1/33, 25/3, 22/25, 21/2]</code>
|<math>\begin{bmatrix}
0 & 3 & -1 & -1 & 0 \\
0 & -1 & 0 & 0 & -1 \\
0 & -1 & 2 & 0 & 0 \\
1 & 0 & -2 & 0 & 1 \\
-1 & 1 & 0 & 1 & 0
\end{bmatrix}</math>
|}

All 3 holdouts follow a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math> (for 29/2003 and 41/2003) or <math>S(x,y) = [x, 0, 2, y, 0]</math> (for 430/2003), then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

TMBR

2026-03-28T08:48:30Z

Int-y1: Redirected page to Category:This Month in Beaver Research

#REDIRECT [[Category:This_Month_in_Beaver_Research]]

Fractran

2026-03-28T05:18:32Z

Int-y1: /* Size 22 Cryptid */

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|345 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|5682 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]

Known [[Cryptid|Cryptids]]:

# Size 22 Cryptid
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Size 22 Cryptid ===
The size 22 [[Cryptids|Cryptid]] is a group of 3 holdouts that was discovered by Jason Yuen (@-d) and Claude Opus 4.6. Out of 500 holdouts of size 22, Claude Opus 4.6 used Lean to prove that 497 holdouts were non-halting. The remaining 3 holdouts are this Cryptid.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1485415054475268179]</sup>

{| class="wikitable"
|+
!Holdout number
!Holdout
!Vector Representation
|-
| 29/2003
| <code>[1/15, 27/77, 49/3, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 41/2003
| <code>[1/15, 49/3, 27/77, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 2 & 0 \\
0 & 3 & 0 & -1 & -1 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 430/2003
| <code>[27/35, 1/33, 25/3, 22/25, 21/2]</code>
|<math>\begin{bmatrix}
0 & 3 & -1 & -1 & 0 \\
0 & -1 & 0 & 0 & -1 \\
0 & -1 & 2 & 0 & 0 \\
1 & 0 & -2 & 0 & 1 \\
-1 & 1 & 0 & 1 & 0
\end{bmatrix}</math>
|}

All 3 holdouts follow a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math> (for 29/2003 and 41/2003) or <math>S(x,y) = [x, 0, 2, y, 0]</math> (for 430/2003), then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

Fractran

2026-03-28T04:48:29Z

Int-y1: update summary table

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|345 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|5682 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]

Known [[Cryptid|Cryptids]]:

# Size 22 Cryptid
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Size 22 Cryptid ===
The size 22 [[Cryptid|Cryptids]] is a group of 3 holdouts that was discovered by Jason Yuen (@-d) and Claude Opus 4.6. Out of 500 holdouts of size 22, Claude Opus 4.6 used Lean to prove that 497 holdouts were non-halting. The remaining 3 holdouts are this Cryptid.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1485415054475268179]</sup>

{| class="wikitable"
|+
!Holdout number
!Holdout
!Vector Representation
|-
| 29/2003
| <code>[1/15, 27/77, 49/3, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 41/2003
| <code>[1/15, 49/3, 27/77, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 2 & 0 \\
0 & 3 & 0 & -1 & -1 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 430/2003
| <code>[27/35, 1/33, 25/3, 22/25, 21/2]</code>
|<math>\begin{bmatrix}
0 & 3 & -1 & -1 & 0 \\
0 & -1 & 0 & 0 & -1 \\
0 & -1 & 2 & 0 & 0 \\
1 & 0 & -2 & 0 & 1 \\
-1 & 1 & 0 & 1 & 0
\end{bmatrix}</math>
|}

All 3 holdouts follow a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math> (for 29/2003 and 41/2003) or <math>S(x,y) = [x, 0, 2, y, 0]</math> (for 430/2003), then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

Fractran

2026-03-28T04:33:59Z

Int-y1: /* Size 22 Cryptid */

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|345 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|5682 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Size 22 Cryptid ===
The size 22 [[Cryptid]] is a group of 3 holdouts that was discovered by Jason Yuen (@-d) and Claude Opus 4.6. Out of 500 holdouts of size 22, Claude Opus 4.6 used Lean to prove that 497 holdouts were non-halting. The remaining 3 holdouts are this Cryptid.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1485415054475268179]</sup>

{| class="wikitable"
|+
!Holdout number
!Holdout
!Vector Representation
|-
| 29/2003
| <code>[1/15, 27/77, 49/3, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 41/2003
| <code>[1/15, 49/3, 27/77, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 2 & 0 \\
0 & 3 & 0 & -1 & -1 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 430/2003
| <code>[27/35, 1/33, 25/3, 22/25, 21/2]</code>
|<math>\begin{bmatrix}
0 & 3 & -1 & -1 & 0 \\
0 & -1 & 0 & 0 & -1 \\
0 & -1 & 2 & 0 & 0 \\
1 & 0 & -2 & 0 & 1 \\
-1 & 1 & 0 & 1 & 0
\end{bmatrix}</math>
|}

All 3 holdouts follow a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math> (for 29/2003 and 41/2003) or <math>S(x,y) = [x, 0, 2, y, 0]</math> (for 430/2003), then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

Fractran

2026-03-28T04:33:28Z

Int-y1: /* Size 22 Cryptid */ group of 3 holdouts

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|345 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|5682 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Size 22 Cryptid ===
The size 22 [[Cryptid]] was discovered by Jason Yuen (@-d) and Claude Opus 4.6. Out of 500 holdouts of size 22, Claude Opus 4.6 used Lean to prove that 497 holdouts were non-halting. The remaining 3 holdouts are this Cryptid.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1485415054475268179]</sup>

{| class="wikitable"
|+
!Holdout number
!Holdout
!Vector Representation
|-
| 29/2003
| <code>[1/15, 27/77, 49/3, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 41/2003
| <code>[1/15, 49/3, 27/77, 10/49, 33/2]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 2 & 0 \\
0 & 3 & 0 & -1 & -1 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>
|-
| 430/2003
| <code>[27/35, 1/33, 25/3, 22/25, 21/2]</code>
|<math>\begin{bmatrix}
0 & 3 & -1 & -1 & 0 \\
0 & -1 & 0 & 0 & -1 \\
0 & -1 & 2 & 0 & 0 \\
1 & 0 & -2 & 0 & 1 \\
-1 & 1 & 0 & 1 & 0
\end{bmatrix}</math>
|}

All 3 holdouts follow a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math> (for 29/2003 and 41/2003) or <math>S(x,y) = [x, 0, 2, y, 0]</math> (for 430/2003), then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

Fractran

2026-03-28T04:13:58Z

Int-y1: /* Cryptids */ add size 22 cryptid

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|345 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|5682 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Size 22 Cryptid ===
The size 22 [[Cryptid]] was discovered by Jason Yuen (@-d) and Claude Opus 4.6. It was found after Claude Opus 4.6 proved many holdouts to be non-halting, but couldn't finish the proof for this Cryptid.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1485415054475268179]</sup>

<code>[1/15, 27/77, 49/3, 10/49, 33/2]</code> <math display="block">\begin{bmatrix}
0 & -1 & -1 & 0 & 0 \\
0 & 3 & 0 & -1 & -1 \\
0 & -1 & 0 & 2 & 0 \\
1 & 0 & 1 & -2 & 0 \\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}</math>

Its behavior is a biased random walk that somewhat resembles [[Hydra]]. Let <math>S(x,y) = [x, 0, 0, 2, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \to & S(0,1) \\
S(0, 2y) & = & \text{halt} \\
S(x, 2y) & \to & S(x-1, 5y+2) \\
S(x, 2y+1) & \to & S(x+2, 5y)
\end{array}</math>

The first few visited states are $$S(0, 1) \to S(2, 0) \to S(1, 2) \to S(0, 7) \to S(2, 15) \to S(4, 35)$$

TODO: Also talk about holdout 41/2003 and 430/2003.

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

Its behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y)
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><pre>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</pre>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2)
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><pre>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</pre>

== References ==
<references />

[[Category:Functions]]

TMBR: March 2026

2026-03-23T04:31:49Z

Int-y1: fractran

{{TMBRnav|February 2026|April 2026}}

''This edition of TMBR is in progress and has not yet been released. Please add any notes you think may be relevant (including in the form a of a TODO with a link to any relevant Discord discussion).''

[[:Category:This Month in Beaver Research|This Month in Beaver Research]] for March 2026. We celebrated bbchallenge's fourth birthday on 8 March.

TODO: Write a proper introductory paragraph.

== Champions ==

* Discord user 50_ft_lock [https://discord.com/channels/960643023006490684/1331570843829932063/1481871400640839691 found] a new BB(13) [[Champions|champion]] which surpasses [[Graham's number]], reducing the upper bound of Graham-beating TMs to 13 states.

== Meta ==

* Famous math youtuber [https://www.youtube.com/@twoswap 2swap] made a [https://discord.com/channels/960643023006490684/1362008236118511758/1478973587653136456 couple of videos about Turing Machines] arranged into grids and colored based on their halting status for BB(2,2), BB(3,2), BB(2,3), BB(4,2) and BB(5,2) respectively, then made a [https://www.youtube.com/watch?v=1BI3qItCJ2M two-hour long Youtube video] about the same topic on [https://www.youtube.com/@3cycle their second channel].

== BB Adjacent ==
* [[Fractran]]: In BBf(21), Claude Opus 4.6 gave a proof that all 140 holdouts do not halt. This tentatively proves that BBf(21) = 31,957,632.
* [[Fractran]]: A Cryptid was discovered in BBf(22) with the help of Claude Opus 4.6.

== Holdouts ==

* [[BB(6)]]: 1e14 machines: '''171'''. 1e15 machines: '''237'''. '''12 solved machines'''.
** Holdouts lists for machines not simulated to '''1e14 and 1e15''' steps were created. The holdouts list counts were '''178 and 252''' respectively. See [https://docs.google.com/spreadsheets/d/1mMp8bAcTFT91j7azn72liX8NSTwc2E_ozKnOGTfRCfw/edit?gid=806905077#gid=806905077 Spreadsheet for BB6].
** Later, [https://discord.com/channels/960643023006490684/1477591686514212894/1478401591307407554 prurq found 10 more machines] in the holdouts list that had previously been simulated to 1e15: thus the new 1e15 holdout count was '''242.'''
** Andrew Ducharme [https://discord.com/channels/960643023006490684/1239205785913790465/1478626634729914450 solved two machines] using FAR.
** prurq found two machines [https://discord.com/channels/960643023006490684/1239205785913790465/1478518151393312822][https://discord.com/channels/960643023006490684/1239205785913790465/1478525451826888775] to be [[Translated Cycler|Translated Cyclers]]. Shawn Ligocki [https://discord.com/channels/960643023006490684/1239205785913790465/1478781268664778989 verified one of them] and discovered its preperiod to be over <math>10^{12}</math> steps. mxdys [https://discord.com/channels/960643023006490684/1239205785913790465/1478827653636296850 verified the other]. Both were 1e14 and 1e15 holdouts, thus reducing those holdout counts by 2.
** mxdys found four more [https://discord.com/channels/960643023006490684/1239205785913790465/1478734570672095333][https://discord.com/channels/960643023006490684/1239205785913790465/1479045001089384600] Translated Cyclers in the remaining holdouts and solved [https://discord.com/channels/960643023006490684/1239205785913790465/1478828100707025151 one more TM] using FAR.
** Andrew Ducharme [https://discord.com/channels/960643023006490684/1239205785913790465/1479313219003748484 solved a machine] using FAR. This machine was a 1e14 and 1e15 holdout, thus reducing those holdout counts by 1.
** [https://discord.com/channels/960643023006490684/1477591686514212894/1480544451033170060 prurq found 3 more machines previously simulated so far] in the 1e14 list, and 2 more in the 1e15 list (both were also 1e14). This means 1e14 holdout count was reduced by 3, and 1e15 holdout count was reduced by 2.
** Discord user mammillaria [https://discord.com/channels/960643023006490684/1239205785913790465/1480686400067342346 simulated a 1e14 holdout thus far], therefore reducing that holdout count by 1.
** prurq [https://discord.com/channels/960643023006490684/1471178503235043493/1481022684975337586 found a machine to be a Translated Cycler].
** mxdys [https://discord.com/channels/960643023006490684/1239205785913790465/1481557945346035814 decided a machine] using FAR.
*[[BB(7)]]:
**Andrew Ducharme [https://discord.com/channels/960643023006490684/1369339127652159509/1481889113232904295 reduced] the number of holdouts from 18,195,192 to 18,036,852 (a 0.87% reduction) via the mxdys FAR decider.
*[[BB(2,5)]]:
**[https://discord.com/channels/960643023006490684/1259770421046411285/1481197573611061311 Peacemaker II solved a machine using FAR] and mxdys confirmed two[https://discord.com/channels/960643023006490684/1259770421046411285/1483043448855461989][https://discord.com/channels/960643023006490684/1259770421046411285/1483043657778069564] machines to be non-halting. Thus, the new holdout count is 69, or 60 considering informal proofs.
*[[BB(3,3)]]:
**mxdys [https://discord.com/channels/960643023006490684/1259770474897080380/1482680295357677651 formalised] three remaining informal results (650, 412, 279) into Rocq.

[[Category:This Month in Beaver Research|2026-03]]

Code repositories

2026-02-14T07:40:56Z

Int-y1:

{| class="wikitable sortable"
|+
!Creator
!Link
!Main language
!Notes
|-
|bbchallenge
|[https://github.com/bbchallenge/bbchallenge-deciders Link]
|?
|Official bbchallenge deciders
|-
|@mxdys
|[https://github.com/ccz181078/Coq-BB5 Link]
|Rocq
|Rocq proof of BB(5) = 47,176,870. See [[Coq-BB5]].
|-
|@mei
|[https://github.com/meithecatte/busycoq Link]
|Rocq, Rust, OCaml?
|A hybrid Rocq/Rust/OCaml repo implementing very fast deciders plus verified proofs
|-
|Georgi Georgiev ([[Skelet]])
|[https://skelet.ludost.net/bb/ Link]
|Pascal
|See [[bbfind]]. Wrapped via [https://gist.github.com/m1el/d514a353cccde531c298b725043404af bbfind-stdin by @wizord]
|-
|@mxdys
|[https://github.com/ccz181078/busycoq/tree/BB6/verify Link]
|Rocq
|Rocq proof of some deciders and some solved machines in BB(6)
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/6cdad07e15f6acf992b79dc2baf0492c Link]
|Python 3
|Accelerated TM simulators in Python utilizing memoization (like HashLife)
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/7f5e10169abbb50d1537165c6e71733b Link]
|Python 3
|Forward segment TM decider by @Mateon1 (variant of [[Halting Segment]])
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/b63eabc371ac35e2a14a9c5ce37413bc Link]
|Python 3
|[[Closed Position Set]] TM decider implementation
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/c801565e499be605cea1283a5984b4c3 Link]
|Python 3
|[[Closed Tape Language]] SAT Solver; '''TODO dependencies still Discord posts'''
|-
|@savask
|[https://github.com/savask/turing Link]
|Haskell
|Collection of deciders
|-
|@savask
|[https://gist.github.com/savask/1c43a0e5cdd81229f236dcf2b0611c3f Link]
|Haskell
|[[Closed Position Set]], reverse-engineered from Skelet's [[bbfind]] program
|-
|@savask
|[https://gist.github.com/savask/888aa5e058559c972413790c29d7ad72 Link]
|Haskell
|Decider for [[Bouncers]]
|-
|@savask
|[https://gist.github.com/savask/c7546bb6384984b2fb3cb90fc7925697 Link]
|Haskell
|Reproduction of @mxdys' [[RepWL_ES]] decider (repeated block decider)
|-
|@djmati1111
|[https://github.com/colette-b/bbchallenge Link]
|Python 3
|The first SAT-based [[Finite Automata Reduction]] decider
|-
|Frans Faase
|[https://github.com/FransFaase/SymbolicTM Link]
|C++
|An early [[Closed Tape Language]] verifier
|-
|@Iijil
|[https://github.com/Iijil1/Bouncers Link]
|Go
|Decider for [[Bouncers]]
|-
|@Iijil
|[https://github.com/Iijil1/Bruteforce-CTL Link]
|Go
|Bruteforce [[Closed Tape Language]] decider
|-
|@Iijil
|[https://github.com/Iijil1/MITMWFAR Link]
|Go
|[[Meet-in-the-Middle Weighted Finite Automata Reduction (MITMWFAR)]] decider
|-
|@Iijil
|[https://gist.github.com/Iijil1/d325da33be74a86ac2399c161a57166a Link]
|PHP
|4-symbol to 2-symbol TM compiler
|-
|Jason Yuen @-d
|[https://github.com/int-y1/proofs/tree/master/BusyLean Link]
|Lean 4
|Some progress toward a FAR verifier checked by Lean
|-
|Jason Yuen @-d
|[https://github.com/int-y1/busy-beaver-cpp Link]
|C++
|C++ version of Shawn Ligocki's Quick_Sim.py
|-
|Matthew House @LegionMammal976
|[https://github.com/LegionMammal978/bigfoot-sim Link]
|Rust
|An accelerated [[Bigfoot]] simulator
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bbchallenge-dafny-deciders Link]
|Dafny, Rust?
|Formally verified deciders using Dafny
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/busy-beaver-dafny-regex-verifier Link]
|Dafny
|Formally verified [[Closed Tape Language]] verifier
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bbchallenge-regexy-decider Link]
|Rust
|[[Closed Tape Language]] decider
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bb-simple-n-gram-cps Link]
|Rust
|[[Closed Position Set]] simplification: uses less resources and is less strong
|-
|@nickdrozd
|[https://github.com/nickdrozd/busy-beaver-stuff Link]
|?
|
|-
|@star
|[https://github.com/phinanix/busy-beavers Link]
|?
|
|-
|Shawn Ligocki @sligocki
|[https://github.com/sligocki/busy-beaver Link]
|Python 3, Rust, C++
|
|-
|Tony Guilfoyle
|[https://github.com/TonyGuil/bbchallenge Link]
|C++
|
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/uncombedcoconut/bbchallenge Link]
|Python 3
|
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/UncombedCoconut/bbchallenge-deciders/tree/FARther/decider-finite-automata-reduction Link]
|Rust
|[[Finite Automata Reduction]] decider
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/UncombedCoconut/bbchallenge-nfa-verification Link]
|Python 3
|[[Finite Automata Reduction]] verifier
|-
|Pavel Kropitz @uni
|[https://github.com/univerz/bbc Link]
|Rust
|
|-
|Dan Briggs
|[https://github.com/danbriggs/Turing Link]
|Java
|TM proofs/writing
|-
|Thomas Vigouroux @Vigoux
|[https://git.sr.ht/~vigoux/busybeaver/tree/master Link]
|Lean 4
|An attempt at formalising results regarding Busy Beavers. Contains deciders and their proof of correctness.
|-
|@mxdys
|[https://github.com/ccz181078/TM Link]
|C++
|Enumerating 7x2 TMs
|}

Code repositories

2026-02-14T07:39:54Z

Int-y1: add C++ Quick_Sim.py

{| class="wikitable sortable"
|+
!Creator
!Link
!Main language
!Notes
|-
|bbchallenge
|[https://github.com/bbchallenge/bbchallenge-deciders Link]
|?
|Official bbchallenge deciders
|-
|@mxdys
|[https://github.com/ccz181078/Coq-BB5 Link]
|Rocq
|Rocq proof of BB(5) = 47,176,870. See [[Coq-BB5]].
|-
|@mei
|[https://github.com/meithecatte/busycoq Link]
|Rocq, Rust, OCaml?
|A hybrid Rocq/Rust/OCaml repo implementing very fast deciders plus verified proofs
|-
|Georgi Georgiev ([[Skelet]])
|[https://skelet.ludost.net/bb/ Link]
|Pascal
|See [[bbfind]]. Wrapped via [https://gist.github.com/m1el/d514a353cccde531c298b725043404af bbfind-stdin by @wizord]
|-
|@mxdys
|[https://github.com/ccz181078/busycoq/tree/BB6/verify Link]
|Rocq
|Rocq proof of some deciders and some solved machines in BB(6)
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/6cdad07e15f6acf992b79dc2baf0492c Link]
|Python 3
|Accelerated TM simulators in Python utilizing memoization (like HashLife)
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/7f5e10169abbb50d1537165c6e71733b Link]
|Python 3
|Forward segment TM decider by @Mateon1 (variant of [[Halting Segment]])
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/b63eabc371ac35e2a14a9c5ce37413bc Link]
|Python 3
|[[Closed Position Set]] TM decider implementation
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/c801565e499be605cea1283a5984b4c3 Link]
|Python 3
|[[Closed Tape Language]] SAT Solver; '''TODO dependencies still Discord posts'''
|-
|@savask
|[https://github.com/savask/turing Link]
|Haskell
|Collection of deciders
|-
|@savask
|[https://gist.github.com/savask/1c43a0e5cdd81229f236dcf2b0611c3f Link]
|Haskell
|[[Closed Position Set]], reverse-engineered from Skelet's [[bbfind]] program
|-
|@savask
|[https://gist.github.com/savask/888aa5e058559c972413790c29d7ad72 Link]
|Haskell
|Decider for [[Bouncers]]
|-
|@savask
|[https://gist.github.com/savask/c7546bb6384984b2fb3cb90fc7925697 Link]
|Haskell
|Reproduction of @mxdys' [[RepWL_ES]] decider (repeated block decider)
|-
|@djmati1111
|[https://github.com/colette-b/bbchallenge Link]
|Python 3
|The first SAT-based [[Finite Automata Reduction]] decider
|-
|Frans Faase
|[https://github.com/FransFaase/SymbolicTM Link]
|C++
|An early [[Closed Tape Language]] verifier
|-
|@Iijil
|[https://github.com/Iijil1/Bouncers Link]
|Go
|Decider for [[Bouncers]]
|-
|@Iijil
|[https://github.com/Iijil1/Bruteforce-CTL Link]
|Go
|Bruteforce [[Closed Tape Language]] decider
|-
|@Iijil
|[https://github.com/Iijil1/MITMWFAR Link]
|Go
|[[Meet-in-the-Middle Weighted Finite Automata Reduction (MITMWFAR)]] decider
|-
|@Iijil
|[https://gist.github.com/Iijil1/d325da33be74a86ac2399c161a57166a Link]
|PHP
|4-symbol to 2-symbol TM compiler
|-
|Jason Yuen @-d
|[https://github.com/int-y1/proofs/tree/master/BusyLean Link]
|Lean 4
|Some progress toward a FAR verifier checked by Lean
|-
|Jason Yuen @-d
|[https://github.com/int-y1/busy-beaver-cpp Link]
|C++
|C++ version of Quick_Sim.py
|-
|Matthew House @LegionMammal976
|[https://github.com/LegionMammal978/bigfoot-sim Link]
|Rust
|An accelerated [[Bigfoot]] simulator
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bbchallenge-dafny-deciders Link]
|Dafny, Rust?
|Formally verified deciders using Dafny
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/busy-beaver-dafny-regex-verifier Link]
|Dafny
|Formally verified [[Closed Tape Language]] verifier
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bbchallenge-regexy-decider Link]
|Rust
|[[Closed Tape Language]] decider
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bb-simple-n-gram-cps Link]
|Rust
|[[Closed Position Set]] simplification: uses less resources and is less strong
|-
|@nickdrozd
|[https://github.com/nickdrozd/busy-beaver-stuff Link]
|?
|
|-
|@star
|[https://github.com/phinanix/busy-beavers Link]
|?
|
|-
|Shawn Ligocki @sligocki
|[https://github.com/sligocki/busy-beaver Link]
|Python 3, Rust, C++
|
|-
|Tony Guilfoyle
|[https://github.com/TonyGuil/bbchallenge Link]
|C++
|
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/uncombedcoconut/bbchallenge Link]
|Python 3
|
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/UncombedCoconut/bbchallenge-deciders/tree/FARther/decider-finite-automata-reduction Link]
|Rust
|[[Finite Automata Reduction]] decider
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/UncombedCoconut/bbchallenge-nfa-verification Link]
|Python 3
|[[Finite Automata Reduction]] verifier
|-
|Pavel Kropitz @uni
|[https://github.com/univerz/bbc Link]
|Rust
|
|-
|Dan Briggs
|[https://github.com/danbriggs/Turing Link]
|Java
|TM proofs/writing
|-
|Thomas Vigouroux @Vigoux
|[https://git.sr.ht/~vigoux/busybeaver/tree/master Link]
|Lean 4
|An attempt at formalising results regarding Busy Beavers. Contains deciders and their proof of correctness.
|-
|@mxdys
|[https://github.com/ccz181078/TM Link]
|C++
|Enumerating 7x2 TMs
|}

Piecewise Affine Function

2026-02-12T02:47:07Z

Int-y1: /* Formal Definition */

A '''Piecewise Affine Function (PAF)''' is a piecewise-defined function where each case is affine (and the case constraints are polyhedra). Many [[Cryptids]] are modeled by iterated PAFs, for example, [[BMO1]]. Like [[Generalized Collatz Problems]], iterated PAFs are also proven to be [[Turing complete]]. On the [[bbchallenge]] Discord, these were originally called "Linear-Inequality Affine Transformation Automata (LIATA)" before we knew about the existing name in published literature.
== Formal Definition ==
A ''n''-dimension, ''p''-region PAF is a piecewise defined partial function <math>f: \Z^n \to \Z^n</math>:<math display="block">f(\vec{x}) = \begin{cases}
f_0(\vec{x}) & \text{for } \vec{x} \in H_0 \\
f_1(\vec{x}) & \text{for } \vec{x} \in H_1 \\
& \vdots \\
f_{p-1}(\vec{x}) & \text{for } \vec{x} \in H_{p-1}
\end{cases}</math>Where each <math>f_i(\vec{x}) = A_i \vec{x} + \vec{b_i}</math> is an affine function and the <math>H_i \subset \Z^n</math> are non-overlapping "polyhedral regions" (defined below). If <math>\vec{x}</math> is not in any region <math>H_i</math>, we say that it halts on that configuration.

We define a closed, rational ''half-space'' to be a region <math>\{ \vec{x} \in \mathbb{R}^n : \vec{c} \cdot \vec{x} + d \ge 0 \}</math> for some <math>\vec{c} \in \mathbb{Q}^n</math> and <math>d \in \mathbb{Q}</math>. In other words, it is the half of n-dimensional Euclidean space on one side of a hyperplane (a subspace defined by an affine function). And let an open, rational half-space be the same thing but using strict inequality (<math>\{ \vec{x} \in \mathbb{R}^n : \vec{c} \cdot \vec{x} + d > 0 \}</math>).

Finally, define a ''polyhedral regions'' (or simply ''polyhedron'') as the intersection of a finite number of rational half-spaces (any combination of closed and open ones). So, for example, the following are all polyhedral regions:

* <math>a<b</math> represented formally as <math>b-a > 0</math>
* <math>3a \le b < 4a</math> represented formally as <math>(b-3a \ge 0) \and (4a-b > 0)</math>
* <math>a = 2</math> represented formally as <math>(a - 2 \ge 0) \and (-a + 2 \ge 0)</math>

Given a PAF ''f'', we say that it halts in ''k'' steps starting from configuration <math>\vec{x}</math> iff <math>f^k(\vec{x})</math> is undefined.

== Example ==
An example of a PAF are the rules for [[BMO1]]:<math display="block">f(a,b) = \begin{cases}
(a-b, 4b+2) & \text{if } a > b \\
(2a+1, b-a) & \text{if } a < b
\end{cases}</math>where <math>f(n,n)</math> is undefined. BMO1 halts iff there exists k such that <math>f^k(1,2)</math> is undefined (in other words <math>f^{k-1}(1,2) = (n,n)</math> for some n).

This is a 2-dimension, 2-region PAF. The 2 dimensions are the parameters ''a,b'' and the two regions are the half-spaces <math>a<b</math> and <math>a>b</math>. For each case the parameters are transformed via an affine transformation.

== Turing Complete ==
PAF are Turing complete. This has been proven by implementing Generalized Collatz Problems and Minsky machines as iterated PAF problems: Amir M. Ben-Amram proved in 2015 that 2-dimensional PAF are Turing complete by implementing arbitrary GCP<ref>Ben-Amram 2015. Section 2: Undecidability in Two Dimensions</ref> and also that 2-region PAF are Turing complete by implementing arbitrary Minsky machines.<ref>Ben-Amram, Genaim, Masud 2012. Section 5: Loops with Two Linear Pieces</ref>

Independently, similar results were proven on the bbchallenge Discord in 2025 (all by reduction to Minsky machines): @Bard proved that 3-dim PAF are Turing complete: [https://discord.com/channels/960643023006490684/1239205785913790465/1420457986564030641 Discord link], @star proved that 2-dim PAF are Turing complete: [https://discord.com/channels/960643023006490684/1239205785913790465/1421271424588451915 Discord link] and Shawn Ligocki wrote up a proof sketch that 2-region PAF are Turing complete: [https://discord.com/channels/960643023006490684/1239205785913790465/1422772752980639866 Discord link].

In other words, we know that there exists a Universal PAF (which can simulate all TMs ... and thus all other PAFs) with 2-dimensions and some unspecified number of regions and another with 2-regions and an unspecified number of dimensions. Similar to [[Universal Turing Machine|Universal Turing Machines]], we could explore the "pareto frontier" of minimal Universal PAFs by calculating exact values for the number of regions/dimensions in those specific constructions and searching for minimal values here and for more minimal sizes with 3+ dimensions and 3+ regions.

Note: the fact that PAF in general are Turing complete does not prove anything about specific PAF problems. For example, although 2-dim PAF (and 2-region PAF) are known to be Turing complete it is not known if 2-dim, 2-region PAF are. And even if it were known that 2-dim, 2-region PAF are Turing complete, it does not mean that the BMO1 PAF specifically is. And even if the BMO1 PAF specifically was known to be Turing complete, it does not mean that the specific trajectory followed by BMO1 is undecidable. In other words, there is really no hope of using the fact that PAF are Turing complete to prove anything rigorously about a specific example (like BMO1). Instead, the proofs that PAF in general are Turing complete simply provides some intuition for why PAFs in general are probably challenging problems to solve. This is similar to how the proof that Generalized Collatz Problems are Turing complete provides some intuition for why Generalized Collatz Problems in general are probably challenging problems to solve.

== References ==
<references />

* Amir M. Ben-Amram. 2015. [https://drops.dagstuhl.de/storage/00lipics/lipics-vol020-stacs2013/LIPIcs.STACS.2013.514/LIPIcs.STACS.2013.514.pdf Mortality of iterated piecewise affine functions over the integers: Decidability and complexity]. ''Computability''. 2015;4(1):19-56. {{doi|10.3233/COM-150032}}
* Amir M. Ben-Amram, Samir Genaim, and Abu Naser Masud. 2012. On the Termination of Integer Loops. ''ACM Transactions on Programming Languages and Systems'' 34, 4, Article 16 (December 2012), 24 pages. {{doi|10.1145/2400676.2400679}}

[[Category:Functions]]

Fractran

2026-01-02T04:05:20Z

Int-y1: /* Champions */ holdouts update

'''Fractran''' (originally styled FRACTRAN) is an esoteric [[Turing complete]] model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right. There are three extra deciders: [https://discord.com/channels/960643023006490684/1438019511155691521/1449775657554022531 Spanning Vectors Masked,] which should be very effective, but implementing it is in-progress, a version of Spanning Vectors Masked - [https://discord.com/channels/960643023006490684/1438019511155691521/1453217977385091092 Masked Linear Invariant] - which is very powerful, and some holdouts were removed by [[User:Sligocki|Shawn Ligocki]] with [https://lsv.ens-paris-saclay.fr/Software/fast/ FAST] (Fast Acceleration of Symbolic Transition systems), a pre-existing general tool.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|Decider: Jason Yuen (@-d)
([https://github.com/int-y1/BBFractran/tree/main/holdout Enum+initial])
Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438559507579011194 13] and [https://discord.com/channels/960643023006490684/1438019511155691521/1438996636389998773 14 Nov 2025]

Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7] and [https://discord.com/channels/960643023006490684/1438019511155691521/1453213088630444168 24 Dec 2025]
6 Holdouts: Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1452913055053778945 23 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|345 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|22
|<math>> 1.146 \times 10^{62}</math>
|<code>[1/12, 9/10, 14/3, 11/2, 5/7, 3/11]</code>
|<math>\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448912286713384961 11 Dec 2025] and Jason Yuen (@-d)<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1448953682237460480 <nowiki>[1]</nowiki>]</sup>
|5682 holdouts remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1456471391938547722 1 Jan 2026]
|-
|23
|
|
|
|
|Known [[Cryptid|Cryptids]]:

# Frankenstein's Monster
# Antihydra-like Cryptid
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

==== BBf(20) ====
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

==== BBf(21) ====
The BBf(21) champion (running >31M steps):
<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>

This program implements a Collatz-like iteration. Let <math>D(n) = [0, 0, n, 0]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1439779341365022852]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & D(1) \\
D(3k) & \xrightarrow{k} & \text{halt} \\
D(3k+1) & \xrightarrow{21k+7} & C(10k+4) \\
D(3k+2) & \xrightarrow{21k+14} & C(10k+7) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:
<math display="block">\begin{array}{ll}
D(1) & \to D(4) \to D(14) \to D(47) \to D(157) \to D(524) \to D(1747) \to D(5824) \to D(19414) \\
& \to D(64714) \to D(215714) \to D(719047) \to D(2396824) \to D(7989414) \to \text{halt} \\
\end{array}</math>

==== BBf(22) ====
The BBf(22) champion (running <math>> 10^{62}</math> steps):
<math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

This program implements a [[Collatz-like]] unbiased psuedo-random walk. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449118888142049421]</sup>
<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,1) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+1) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+2) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

This random walk iterates 275 times until it halts reaching a maximum y value of 14 at iteration 111:
<math display="block">\begin{array}{ll}
S(0,1) & \to S(1,1) \to S(3,2) \to S(6,2) \to S(11, 2) \to S(19, 1) \to S(33, 2) \to S(56, 2) \to S(94, 1) \\
& \to S(158, 2) \to S(264, 1) \to S(441, 1) \to S(736, 1) \to S(1228, 2) \to S(2048, 3) \\
& \vdots \\
& \to S(4065328691604230522442358, 13) \\
& \to S(6775547819340384204070598, 14) \\
& \to S(11292579698900640340117664, 13) \\
& \vdots \\
& \to S(27930059557111373800280446055462487109112535227834136644, 2) \\
& \to S(46550099261852289667134076759104145181854225379723561074, 1) \\
& \to S(77583498769753816111890127931840241969757042299539268458, 2) \\
& \to S(129305831282923026853150213219733736616261737165898780764, 1) \\
& \to S(215509718804871711421917022032889561027102895276497967941, 1) \\
& \to S(359182864674786185703195036721482601711838158794163279903, 2) \\
& \vdots \\
& \to S(5894430516013404355095519889620117404469367857588232386361874, 2) \\
& \to S(9824050860022340591825866482700195674115613095980387310603124, 1) \\
& \to S(16373418100037234319709777471166992790192688493300645517671874, 0)
\end{array}</math>

== Cryptids ==

=== Frankenstein's Monster ===
"Frankenstein's Monster" is a size 23 [[Cryptid]]. It was created by tweaking a single instruction in the size 22 champion. This tweak switches it from a unbiased random walk to a biased one and thus makes halting probviously impossible. It is called Frankenstein's Monster since it was found by a combination of exhaustive search and hand design.<sup>[https://discord.com/channels/960643023006490684/1438019511155691521/1449138938215141478]</sup>

<code>[1/12, 9/10, 14/3, 121/2, 5/7, 3/11]</code> <math display="block">\begin{bmatrix}
-2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 0 & 2 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math>

It's behavior is extremely similar to the size 22 champion. Let <math>S(x,y) = [0, 0, x, 0, y]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{1} & S(0,2) \\
S(x, 0) & = & \text{halt} \\
S(3k, y+1) & \xrightarrow{14k+4} & S(5k+1, y+2) \\
S(3k+1, y+1) & \xrightarrow{14k+10} & S(5k+3, y+4) \\
S(3k+2, y+1) & \xrightarrow{14k+12} & S(5k+4, y) \\
\end{array}</math>

with the only difference that the y values now change by {+1,+3,-1} depending on the value of x mod 3 (instead of {0,+1,-1} in the original size 22 program). The x values follow the exact same path as in the original size 22 champion, but the y values quickly grow linearly with the number of iterations (as expected by the random model):
0: S(0, 1) @ 1 (0.00s)
100_000: S(10^22_185, 100171) @ 10^22_186 (0.87s)
200_000: S(10^44_370, 200187) @ 10^44_371 (3.42s)
300_000: S(10^66_555, 300759) @ 10^66_556 (7.68s)
400_000: S(10^88_740, 400451) @ 10^88_741 (13.64s)
500_000: S(10^110_925, 500421) @ 10^110_925 (21.28s)
600_000: S(10^133_109, 600351) @ 10^133_110 (30.62s)
700_000: S(10^155_294, 700319) @ 10^155_295 (41.64s)
800_000: S(10^177_479, 799911) @ 10^177_480 (54.30s)
900_000: S(10^199_664, 900259) @ 10^199_665 (68.59s)
1_000_000: S(10^221_849, 1000853) @ 10^221_850 (84.51s)
...
4_000_000: S(10^887_395, 4000201) @ 10^887_396 (1474.02s)
...
27_500_000: S(10^6_100_841, 27512703) @ 10^6_100_842 (87616.45s)

=== Antihydra-like Cryptid ===
This Cryptid is a size 23 [[Cryptid]]. This Cryptid was [https://discord.com/channels/960643023006490684/1438019511155691521/1449293536737361973 constructed by Maksandchael] by tweaking Frankenstein's Monster to make it as similar to [[Antihydra]] as possible. <code>[9/10, 1/6, 1331/2, 14/3, 5/7, 3/11]</code>

<math display="block">\begin{bmatrix}
-1 & 2 & -1 & 0 & 0 \\
-1 & -1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 3 \\
1 & -1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1
\end{bmatrix}</math><syntaxhighlight>
H(a, b) = [0, 0, a-2, 0, b]
Start -> H(2, 3)
H(2a, b) -> H(3a, b+2)
H(2a+1, b+1) -> H(3a+1, b)
H(a,0) -> halt
</syntaxhighlight>

=== Hydra ===
A size 25 program was produced and golfed by hand to simulate [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1449829146040467681 Discord]):

<code>[363/14, 125/2, 22/21, 1/3, 7/11, 14/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & -1 & 2 \\
-1 & 0 & 3 & 0 & 0 \\
1 & -1 & 0 & -1 & 1 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 36 program was produced by hand to simulate [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program was produced by hand to simulate [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><syntaxhighlight>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</syntaxhighlight>

== References ==
<references />

[[Category:Functions]]

TMBR: December 2025

2025-12-27T18:46:24Z

Int-y1: fractran

{{TMBRnav|November 2025|January 2026}}

''This edition of TMBR is in progress and has not yet been released. Please add any notes you think may be relevant (including in the form a of a TODO with a link to any relevant Discord discussion).''

This is the last edition of TMBR this year. 2025 was a very productive year for [[Busy Beaver Challenge|BBChallenge]]: about 60% of the next domain, [[BB(6)]], was solved. Furthermore, new champions were discovered for BB(6), [[BB(7)]] and [[BB(4,3)]]. Many models of computation other than Turing Machines were also explored - most notably [[Fractran]] and [[Instruction-Limited Busy Beaver]]. Some new methods were developed, such as [https://discord.com/channels/960643023006490684/1028746861395316776/1442964185599447152 mxdys's new version of FAR.]

This year, [[TMBR: November 2025#Themed Months|Themed Months]] were introduced - first, for [[BB(3,3)]], then for [[BB(2,5)]] - and the result is the clarification and verification of some of the results and techniques on the [https://discord.com/channels/960643023006490684/1259770474897080380 Discord] and [https://wiki.bbchallenge.org/wiki/BB(3,3) wiki]. See [[TMBR: November 2025#Themed Months]] for more information.

An annotated spreadsheet of [[BB(6)]] holdouts was also shared by [[User:RobinCodes|Robin Rovenszky]], which includes links to Discord discussions, classification of machines and is almost always up-to-date. See [https://docs.google.com/spreadsheets/d/1mMp8bAcTFT91j7azn72liX8NSTwc2E_ozKnOGTfRCfw/edit?gid=1330361301#gid=1330361301 Google Sheets]

== This Year in Beaver Research <small><sub>(TYBR - "Thank You Beaver Researchers!")</sub></small> ==

=== Holdouts Reductions. ===

* [[BB(6)]] - Reduced from '''3571''' to '''1343''' holdouts. Hence, '''2228 machines were solved this year'''. This is a '''63% reduction.'''
* [[BB(2,5)]] - Reduced from '''217''' to '''75,''' a '''65.43% reduction.''' (The number of informal holdouts is '''64''').
* [[BB(7)]] - '''Enumeration was completed''', the number of holdouts was reduced from an initial 85,853,789 to '''20,387,509''' machines, a '''76.25%''' reduction.
* [[BB(4,3)]] - Reduced from 460,916,384 to 9,401,447 holdouts, a '''97.96% reduction.'''
* [[BB(3,4)]] - Reduced from 434,787,751 to 13,334,244 holdouts, a '''96.93% reduction.'''
* [[BB(2,6)]] - '''Enumeration was completed''', the number of holdouts was reduced from an inital 2,278,655,696 to '''870,085''' machines, a '''near 100%''' reduction.
* [[BB(2,7)]] - '''Enumeration started''', 90K of the 1M subtasks have been enumerated ('''9%''').

=== Champions. ===
* [[BB(6)]] - On 16 June 2025, mxdys discovered {{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}, running for 10 ↑↑ 11010000 steps. This was surpassed on 25 June when mxdys discovered {{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}}, a TM which runs for <math>10 \uparrow\uparrow 10 \uparrow\uparrow 10 \uparrow\uparrow 8.10237</math> steps.
* [[BB(2,5)]] - [[1RB3LA4RB0RB2LA 1LB2LA3LA1RA1RZ|The champion]], initially discovered by Daniel Yuan on 24 Jun 2024 was [https://discord.com/channels/960643023006490684/1259770421046411285/1379877629288644722 verified by mxdys] on 4 Jun 2025.
* [[BB(7)]] - Within three days of the start of the enumeration of BB(7), three champions were discovered. The first two were discovered by [[User:Sligocki|Shawn Ligocki]]: {{TM|1RB0RF_1LC0RE_1RD1LB_1LA1LD_0RA0LE_1RG0LB_1RZ1RB|halt}} with a sigma score of about 10 ↑↑ 22 and {{TM|1RB1RA_1RC0LC_0LD1LG_1LF0LE_1RZ1LF_0LA1LD_1RA1LC|halt}} with a sigma score of about 10 ↑↑ 35. This was followed by the discovery of {{TM|1RB0LG_1RC0RF_1LD1RZ_1LF0LE_1RA1LD_1LG1RE_0LB0LB|halt}}, achieving a sigma score of about 10 ↑↑ 46, by Terry Ligocki. On 10 May 2025, Pavel Kropitz discovered {{TM|1RB0RA_1LC1LF_1RD0LB_1RA1LE_1RZ0LC_1RG1LD_0RG0RF|halt}}, a TM which runs for over <math>2 \uparrow^{11} 2 \uparrow^{11} 3</math> steps.
* [[BB(4,3)]] - In Feb 2025, Racheline identified {{TM|0RB1RZ0RB_1RC1LB2LB_1LB2RD1LC_1RA2RC0LD|halt}} as the new BB(4,3) champion with a score of over <math>2 \uparrow\uparrow\uparrow 2^{2^{32}}</math>. In Oct 2025 [[User:Polygon|Polygon]] identified a new [[BB(4,3)]] champion with a score of over <math>10 \uparrow^{4} 4</math> ({{TM|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD|halt}}). These TMs were first proven to halt by Pavel Kropitz in May 2024, but their runtimes were not known at the time.
* [[Beeping Busy Beaver|BBB(3,3)]] - In March 2025 Nick Drozd [https://groups.google.com/g/busy-beaver-discuss/c/EuIXSir4Eps discovered] {{TM|1RB0LB2LA_1LA0RC0LB_2RC2RB0LC}}, which [[quasihalt|quasihalts]] after running for more than 10 ↑↑ 6 steps.

=== New Methods. ===

* New FAR using DFA generator by mxdys.<sup>[https://discord.com/channels/960643023006490684/1028746861395316776/1442964185599447152 <nowiki>[1]</nowiki>][https://discord.com/channels/960643023006490684/1239205785913790465/1443990614483013632 <nowiki>[2]</nowiki>]</sup>
* @Bricks shared a method to estimate susceptibility to [[Block Analysis]] and a [https://docs.google.com/spreadsheets/d/1j00LBxxp9W7uz1wZdMIvDCZ56eReuH0IGO9Z8-yybcQ/edit?usp=sharing spreadsheet] of [[BB(6)]], [[BB(3,3)]] and [[BB(2,5)|BB(2,5]]) holdouts quantified by it.<sup>[https://discord.com/channels/960643023006490684/1239205785913790465/1430227817957953638 <nowiki>[3]</nowiki>][https://discord.com/channels/960643023006490684/1239205785913790465/1430651610102632579 <nowiki>[4]</nowiki>]</sup>

TODO: Before July

=== Misc. ===

* A fast algorithm for [[Consistent Collatz]] simulation was re-discovered and popularized. Using it,
** apgoucher simulated [[Antihydra]] to <math>2^{38}</math> iterations. This is actually a result from one year ago, but was rediscovered and added to the wiki. [https://discord.com/channels/960643023006490684/1026577255754903572/1271528180246773883 Source]
** [[User:Sligocki|Shawn Ligocki]] simulated {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC}} out to one additional Collatz reset, demonstrating that (if they halt, which they probviously should) they will have sigma scores <math>> 10^{10^{10^7}}</math>.
** This algorithm has near linear runtime (in the number of iterations simulated), but also linear memory growth since the parameters grow exponentially. This memory limit seems to be the main bottleneck to simulating Antihydra and other Consistent Collatz iterations further. There has been some discussion on more efficient memory usage or a distributed algorithm to support further scaling, but no results are available yet.
* Andrew Wade claims to have proven that BB(432) is [[Independence from ZFC|independent of ZF]]. [https://codeberg.org/ajwade/turing_machine_explorer Source]
* [[Piecewise Affine Function|Piecewise Affine Functions]] (PAF) were explored as a generalization of the [[BMO1]] rules:
** @Bard proved that 3 dimension PAF are [[Turing complete]].<sup>[https://discord.com/channels/960643023006490684/1239205785913790465/1420457986564030641]</sup>
** @star proved that 2 dimension PAF are Turing complete.<sup>[https://discord.com/channels/960643023006490684/1239205785913790465/1421271424588451915][https://discuss.bbchallenge.org/t/bmo1-type-problems-are-turing-complete/305]</sup>
** Shawn Ligocki wrote up a proof sketch that 2-region PAF are Turing complete.<sup>[https://discord.com/channels/960643023006490684/1239205785913790465/1422772752980639866]</sup>
** It was discovered that Amir Ben-Amram had already proven both of these results in 2015 (both the 2-dim and the 2-region results).
** BMO1 is a 2-dim, 2-region PAF so this provides some sense for the difficulty of the problem.
** This introduces a new type of [[Cryptids]] separate from previous [[Collatz-like]] ones.
* @coda [[TMBR: October 2025#Misc|shared a mechanical implementation]] of [[Antihydra]]<sup>[https://discord.com/channels/960643023006490684/1362008236118511758/1425894649280598066]</sup> and @zts439 3d-printed a prototype.<sup>[https://discord.com/channels/960643023006490684/1362008236118511758/1427103960317296826]</sup>
* @vonhust created a fast TM simulator that averages 2 billion steps / s. It uses fixed-block [[Macro Machine|Macro Machines]] with each block bit-packed into integers. It is about 10x faster than direct simulators across most TMs.<sup>[https://discord.com/channels/960643023006490684/1226543091264126976/1438890558499061821]</sup>

TODO: Before July

=== BB Adjacent. ===
* [[Instruction-Limited Busy Beaver]] was introduced and calculated up to BBi(7).
* [[Reversible Turing Machine]] Busy Beaver values were calculated up to BB<sub>rev</sub>(5).
* [[Terminating Turmite]]s (Relative Movement Turing Machines) were introduced.
* John Tromp introduced the <math>BB \lambda _1(n)</math> function for [[Busy Beaver for lambda calculus#Oracle Busy Beaver|Busy Beaver for lambda calculus with an oracle]] and computed it up to <math>BB \lambda _1(22)</math>.
* Instruction-Limited Greedy Busy Beaver gBBi(n) and an [[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants|Instruction-Limited variant]] of the [[Blanking Busy Beaver]] (BLBi(n)) were introduced. gBBi(n) was computed up to n = 13 and BLBi(n) was computed up to n = 7.
* @savask shared the [[Bug Game]] (and fast-growing <math>Bug(H,W)</math> function).
* [[Fractran|Busy Beaver for Fractan]] (BBf) was introduced on 1 Nov by Jason Yuen.<sup>[https://discord.com/channels/960643023006490684/1362008236118511758/1433148101170040924]</sup> Exact values have been proven up to BBf(19) = 370 and exhaustive enumeration has been run up to size 22 (with <math>BBf(22) > 1.146 \times 10^{62}</math> and 9829 holdouts).
* [[Cyclic Tree Busy Beaver]] (CTBB) was introduced by @Jack on 14 Nov.<sup>[https://discord.com/channels/960643023006490684/1438694294042181742]</sup> The exact value is known for CTBB(2) = 5 and lower bounds have been found up to size 7 with CTBB(7) > 4↑↑↑↑(4↑↑↑3).
TODO: Before July

=== In the News. ===
* 6 January 2025. It Boltwise. [https://www.it-boltwise.de/durchbruch-im-busy-beaver-problem-eine-neue-aera-der-mathematik.html Durchbruch im Busy Beaver Problem: Eine neue Ära der Mathematik] (German) (English: Breakthrough in the Busy Beaver problem: A new era of mathematics).
* 24 March 2025. Nick Drozd. [https://nickdrozd.github.io/2025/03/24/bbb-3-3.html BBB(3, 3) > 10 ↑↑ 6].
* 21 Apr 2025. Shawn Ligocki. [https://www.sligocki.com/2025/04/21/lucys-moonlight.html Lucy's Moonlight: The 5% Champion].
* 9-13 June 2025. Terence Tao mentioned bbchallenge in their talk "The Equational Theories Project: advancing collaborative mathematical research at scale" ([https://www.youtube.com/watch?v=T4DE27uk0jw video] / [https://terrytao.wordpress.com/wp-content/uploads/2025/06/math-experiments.pdf slides]) at the [https://www.newton.ac.uk/event/bprw03/ 2025 Big Proof workshop]. The talk is about the [https://teorth.github.io/equational_theories/ Equational Theories Project], a large-scale mathematical collaboration that crowd-sourced a proof in Lean. Tao mentions bbchallenge as the only other example of a large-scale mathematical collaboration to prove a single result that he knows of.
* 28 June 2025. Scott Aaronson. [https://scottaaronson.blog/?p=8972 BusyBeaver(6) is really quite large].
* 1 July 2025. The Quanta Podcast. [https://discord.com/channels/960643023006490684/1285212639399776256/1389643208811745310 How Amateurs Solved a Major Computer Science Puzzle].
* 2 July 2025. Manon Bischoff. Spektrum. [https://www.spektrum.de/news/mathematik-die-sechste-fleissige-biber-zahl-ist-gigantisch/2274249 Wie der sechste Fleißige Biber die Mathematik an ihre Grenzen bringt].
* 3 July 2025. Nick Drozd. [https://nickdrozd.github.io/2025/07/03/busy-beaver-backwards.html Busy Beaver Backwards].
* 7 July 2025. Karmela Padavic-Callaghan. New Scientist. [https://www.newscientist.com/article/2487058-mathematicians-are-chasing-a-number-that-may-reveal-the-edge-of-maths/ Mathematicians are chasing a number that may reveal the edge of maths]. (Paywalled)
* 9 July 2025. David Roberts. [https://thehighergeometer.wordpress.com/2025/07/09/bb547176870-bb6-is-astronomically-larger/ BB(5)=47,176,870: BB(6) is … astronomically larger].
* 11 July 2025. New Scientist podcast [https://www.newscientist.com/podcasts/how-geoengineering-could-save-us-from-climate-disaster-have-we-broken-mathematics-why-exercise-reduces-cancer-risk/ episode 311]. Discusses mxdys's [[BB(6)]] pentation result "We’re brushing up against the edge of mathematics".
* 11 July 2025. Darren Orf. Popular Mechanics. [https://www.popularmechanics.com/science/math/a65357535/busy-beaver-six/ Mathematicians Say There’s a Number So Big, It’s Literally the Edge of Human Knowledge].
* 14 July 2025. Joe Brennan. Dario AS. [https://en.as.com/latest_news/meet-the-busy-beaver-number-a-number-so-huge-that-mathematicians-call-it-the-frontier-of-mathematical-knowledge-n/ Meet the Busy Beaver number, a number so huge that mathematicians call it the frontier of mathematical knowledge]
* 15 July 2025. Nick Drozd. [https://nickdrozd.github.io/2025/07/15/performance-hacks-for-bradys-algorithm.html Performance Hacks for Brady's Algorithm].
* 18 July 2025 https://francis.naukas.com/2025/07/18/espeluznante-nueva-cota-inferior-para-la-funcion-castor-afanoso-bb6/
* 22 Aug 2025. Ben Brubaker. Quanta Magazine. [https://www.quantamagazine.org/busy-beaver-hunters-reach-numbers-that-overwhelm-ordinary-math-20250822/ Busy Beaver Hunters Reach Numbers That Overwhelm Ordinary Math].
* 25-29 Aug 2025. [[User:Cosmo|Tristan Stérin]] presented [[:File:Conference poster for DNA31 by Tristan Stérin.png#file|a poster]] at [https://dna31.sciencesconf.org/ DNA 31].
* 1 Sep 2025. Katelyn Doucette. [https://katelyndoucette.com/articles/all-about-space-needle All About Space Needle].
* 12 Sep 2025. Katelyn Doucette. [https://katelyndoucette.com/articles/bugs-mazes-and-bradys-algorithm Bugs, Mazes, and the Unreasonably Effective Brady's Algorithm].
* 14 Sep 2025. Ben Brubaker. Wired. [https://www.wired.com/story/the-quest-to-find-the-longest-running-simple-computer-program/ The Quest to Find the Longest-Running Simple Computer Program]. (Reprint of Quanta article from last month).
* 17 Sep 2025. Hacker News. [https://news.ycombinator.com/item?id=45273999 Determination of the fifth Busy Beaver value].
* 18 Sep 2025. Tuomas Kangasniemi. Tekniikkatalous. [https://www.tekniikkatalous.fi/uutiset/a/85aafdaf-f506-4ce0-8035-a4dbe15ee4ff Iso matematiikan ongelma ratkesi 63 v jälkeen] (Finnish) (English: A big math problem solved after 63 years).
* 23 Sep 2025. Katelyn Doucette. [https://katelyndoucette.com/articles/building-the-busy-beaver-ladder Building the Busy Beaver Ladder].
* 30 Sep 2025. Nick Drozd. [https://nickdrozd.github.io/2025/09/30/shape-of-a-turing-machine.html The Shape of a Turing Machine].
* 22 Oct 2025. Ben Brubaker. [https://benbrubaker.com/why-busy-beaver-hunters-fear-the-antihydra/ Why Busy Beaver Hunters Fear the Antihydra]. ([https://news.ycombinator.com/item?id=45723359 Hacker News thread])
* 27 Oct 2025. [[User:Cosmo|Tristan Stérin]] gave a talk about [[bbchallenge]] and the [[BB(5)]] proof at Collège de France: [https://www.youtube.com/watch?v=YYrSdaB-6cE Le cinquième nombre Busy Beaver] (in French).<sup>[https://discord.com/channels/960643023006490684/1242208042460647575/1435724346051006516 <nowiki>[1]</nowiki>]</sup>
* 7-9 Nov 2025. Carl Kadie gave a talk on BB during the PyData Seattle 2025 conference: [https://www.youtube.com/watch?v=wSiF1Bm8f3s ''How to make Python programs run very slow (and new Turing Machine results)''].<sup>[https://discord.com/channels/960643023006490684/960643023530762343/1440090541936214017 <nowiki>[2]</nowiki>]</sup>
TODO: Before July

==BB Adjacent==
TODO. [[Register machine|Register machines]], [[General Recursive Function|General Recursive Functions]], [[Fractran]] progress.

== [[General Recursive Function|Holdouts]] ==
* [[BB(6)|BB(6):]]
**There are 14 holdouts left to simulate up to 1e12 steps, and 288 to simulate up to 1e13 steps<sup>[https://discord.com/channels/960643023006490684/1239205785913790465/1447303829400846482 <nowiki>[1]</nowiki>]</sup>. The two lists can be found [https://docs.google.com/spreadsheets/d/1mMp8bAcTFT91j7azn72liX8NSTwc2E_ozKnOGTfRCfw/edit?gid=806905077#gid=806905077 here].
**The possibility of simulating computationally tractable machines which nonetheless has large time and memory requirements [https://discord.com/channels/960643023006490684/1448725136340422717 was discussed]. [[User:RobinCodes/Machines at the Edge#1RB0RE 1LC1LD 0RA0LD 1LB0LA 1RF1RA ---1LB (bbch) CRYPTID|List]]
** mxdys [https://discord.com/channels/960643023006490684/1239205785913790465/1450455364179857410 shared a new holdouts list,] consisting of '''1343''' machines, which means 73 solved TMs. This is a 5.4% reduction. There is one extra machine that is solved formally, but unverified.
** The old spreadsheet was replaced with a newer one, see [https://docs.google.com/spreadsheets/d/1mMp8bAcTFT91j7azn72liX8NSTwc2E_ozKnOGTfRCfw/edit?gid=1330361301#gid=1330361301 Google Sheets]
* [[BB(7)|BB(7):]]
**Further enumeration by Andrew Ducharme has reduced the number of holdouts from 20,405,295 to 20,387,509, a 0.09% reduction.
*[[BB(3,4)|BB(3,4):]]
**[[User:XnoobSpeakable|XnoobSpeakable]] and [[User:WarpedWartWars|Lúkos]] ran stages 8, 9, 10A & 10B of [[BB(3,4)#Phase 2|Phase 2]], reducing the number of holdouts from 15,136,283 TMs to 13,334,244 holdouts. This is a 11.9% reduction.
* [[BB(2,7)|BB(2,7):]]
** Terry Ligocki enumerated 50K more subtasks, increasing the number of holdouts to 274,623,183. A total of 90K subtasks out of the 1 million subtasks (or '''9%''') have been enumerated.

[[Category:This Month in Beaver Research|2025-12]]

Fractran

2025-12-12T07:31:14Z

Int-y1: /* Champions */

'''Fractran''' (originally styled FRACTRAN) is an esoteric Turing complete model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|≥ 746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|29 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz20_29.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|602 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz21_602.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1442928279995809882 25 Nov 2025]
|-
|22
|≥ 7548863488598188537
|<code>[7/30, 27/2, 8/35, 5/3, 7/5]</code>
|<math>\begin{bmatrix}
2 & -1 & -1 & 0 & 0 \\
-1 & -2 & 0 & 0 & 0 \\
-1 & 1 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 & 1 \\
0 & 0 & 1 & -1 & 0 \\
1 & 0 & 0 & 0 & -1
\end{bmatrix}</math>
|Shawn Ligocki (@sligocki) [https://discord.com/channels/960643023006490684/1438019511155691521/1448920959745654836 11 Dec 2025]
|10458 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz22_10458.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1448806255261913199 11 Dec 2025]
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

=== BBf(20) ===
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

== Cryptids ==
No fractran [[Cryptids]] have been found yet via enumeration, but some have been constructed by hand.

=== Hydra ===
A size 29 program that simulates [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1438230006571470919 Discord]):

<code>[507/22, 26/33, 245/2, 5/21, 1/3, 11/13, 22/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & 0 & -1 & 2 \\
1 & -1 & 0 & 0 & -1 & 1 \\
-1 & 0 & 1 & 2 & 0 & 0 \\
0 & -1 & 1 & -1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 0 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 36 program which simulates [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program which simulates [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><syntaxhighlight>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</syntaxhighlight>

== References ==
<references />

[[Category:Functions]]

Fractran

2025-12-12T05:21:51Z

Int-y1: /* Champions */

'''Fractran''' (originally styled FRACTRAN) is an esoteric Turing complete model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|≥ 746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|29 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz20_29.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|602 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz21_602.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1442928279995809882 25 Nov 2025]
|-
|22
|≥ 29,872,960,885
|<code>[27/70, 25/2, 4/21, 7/5, 3/7]</code>
|<math>\begin{bmatrix}
-1 & 3 & -1 & -1 \\
-1 & 0 & 2 & 0 \\
2 & -1 & 0 & -1 \\
0 & 1 & 0 & -1
\end{bmatrix}</math>
|@Peacemaker II [https://discord.com/channels/960643023006490684/1438019511155691521/1448864464203157646 11 Dec 2025]
|10458 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz22_10458.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1448806255261913199 11 Dec 2025]
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

=== BBf(20) ===
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

== Cryptids ==
No fractran [[Cryptids]] have been found yet via enumeration, but some have been constructed by hand.

=== Hydra ===
A size 29 program that simulates [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1438230006571470919 Discord]):

<code>[507/22, 26/33, 245/2, 5/21, 1/3, 11/13, 22/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & 0 & -1 & 2 \\
1 & -1 & 0 & 0 & -1 & 1 \\
-1 & 0 & 1 & 2 & 0 & 0 \\
0 & -1 & 1 & -1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 0 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 36 program which simulates [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program which simulates [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><syntaxhighlight>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</syntaxhighlight>

== References ==
<references />

[[Category:Functions]]

Fractran

2025-12-12T05:20:40Z

Int-y1: /* Champions */

'''Fractran''' (originally styled FRACTRAN) is an esoteric Turing complete model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|≥ 746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|29 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz20_29.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|602 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz21_602.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1442928279995809882 25 Nov 2025]
|-
|22
|≥ 29,872,960,885
|<code>[27/70, 25/2, 4/21, 7/5, 3/7]</code>
|<math>\begin{bmatrix}
-1 & 3 & -1 & -1 \\
-1 & 0 & 2 & 0 \\
2 & -1 & 0 & -1 \\
0 & 1 & 0 & -1
\end{bmatrix}</math>
|@Peacemaker II [https://discord.com/channels/960643023006490684/1438019511155691521/1448864464203157646 11 Dec 2025]
|91123 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz22_91123.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1448214077028569098 10 Dec 2025]
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

=== BBf(20) ===
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

== Cryptids ==
No fractran [[Cryptids]] have been found yet via enumeration, but some have been constructed by hand.

=== Hydra ===
A size 29 program that simulates [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1438230006571470919 Discord]):

<code>[507/22, 26/33, 245/2, 5/21, 1/3, 11/13, 22/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & 0 & -1 & 2 \\
1 & -1 & 0 & 0 & -1 & 1 \\
-1 & 0 & 1 & 2 & 0 & 0 \\
0 & -1 & 1 & -1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 0 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 36 program which simulates [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program which simulates [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><syntaxhighlight>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</syntaxhighlight>

== References ==
<references />

[[Category:Functions]]

Fractran

2025-12-12T04:48:30Z

Int-y1: sz21 has 587+15 holdouts now

'''Fractran''' (originally styled FRACTRAN) is an esoteric Turing complete model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|≥ 746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|29 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz20_29.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|602 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz21_602.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1442928279995809882 25 Nov 2025]
|-
|22
|
----
|
----
|
----
|
----
|91123 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz22_91123.txt]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1448214077028569098 10 Dec 2025]
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

=== BBf(20) ===
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

== Cryptids ==
No fractran [[Cryptids]] have been found yet via enumeration, but some have been constructed by hand.

=== Hydra ===
A size 29 program that simulates [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1438230006571470919 Discord]):

<code>[507/22, 26/33, 245/2, 5/21, 1/3, 11/13, 22/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & 0 & -1 & 2 \\
1 & -1 & 0 & 0 & -1 & 1 \\
-1 & 0 & 1 & 2 & 0 & 0 \\
0 & -1 & 1 & -1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 0 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 36 program which simulates [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program which simulates [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><syntaxhighlight>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</syntaxhighlight>

== References ==
<references />

[[Category:Functions]]

Fractran

2025-12-08T07:03:12Z

Int-y1: add Shawn Ligocki's explanation of BBf(15) and BBf(16)

'''Fractran''' (originally styled FRACTRAN) is an esoteric Turing complete model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN.

Discord user Coda came up with a way to transform any Fractran program into a Turing Machine, see [https://discord.com/channels/960643023006490684/1438019511155691521/1441844795613122560 source].

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs. Holdouts lists by Daniel Yuan: [https://github.com/int-y1/BBFractran/blob/main/holdout/README.md Holdouts lists]

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

=== Relationship to VAS / Petri Nets ===
Using vector representation, fractran programs are a deterministic version of [[wikipedia:Vector_addition_system|Vector Addition Systems (VAS)]] (and, equivalently, [[wikipedia:Petri_net|Petri Nets]]). VAS are identical to fractran programs in vector representation except that the rules are unordered and non-deterministic, they are used to model distributed systems where precise order of rule execution cannot be predicted. Interestingly, many problems about VAS are actually decidable, but their runtimes are extremely slow. Notably, the reachability problem (given states A and B are there a sequence of rules so that <math>A \to^* B</math>) is "Ackermann-complete" meaning that the optimal algorithm has worst-case runtime akin to the famously fast-growing Ackermann function.<ref>Czerwiński, Wojciech; Orlikowski, Łukasz (2021). ''Reachability in Vector Addition Systems is Ackermann-complete''. 2021 IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS). https://arxiv.org/abs/2104.13866.</ref>

== Deciders ==
[[File:Fractran deciders.png|alt=Fractran deciders|thumb|All Fractran deciders summarized and their relations, shared by Daniel Yuan on [https://discord.com/channels/960643023006490684/1438019511155691521/1439001835904958655 14 Nov 2025]]]Many specialized deciders have been invented to prove fractran programs non-halting. See image at right.

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|≥ 746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|29 holdouts<sup>[https://github.com/int-y1/BBFractran/blob/main/holdout/sz20_29.txt <nowiki>[1]</nowiki>]</sup> remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1447069110541484146 7 Dec 2025]
|-
|21
|≥ 31,957,632
|<code>[7/15, 4/3, 27/14, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
0 & -1 & -1 & 1 \\
2 & -1 & 0 & 0 \\
-1 & 3 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & 2 & -1 & 0
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1439759182587891894 16 Nov 2025]
|587 holdouts[https://github.com/int-y1/BBFractran/blob/main/holdout/sz21_587.txt <nowiki>[2]</nowiki>] remain. [https://discord.com/channels/960643023006490684/1438019511155691521/1442928279995809882 25 Nov 2025]
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(15) Family ====
The BBf(15) and BBf(16) champions are members of a family of programs (parameterized by <math>n \ge 1</math>):

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & n
\end{bmatrix}</math>

Let a = 2, b = 3, and c = 5.

The BBf(15) champion (n = 2) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{7} & b^4 \\
b^2 & \xrightarrow{7} & b^5 \\
b^3 & \xrightarrow{5} & b^2 \\
b^4 & \xrightarrow{5} & b^3 \\
b^{n+5} & \xrightarrow{3} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^4 \to b^3 \to b^2 \to b^5 \to b^0 \to \text{halt}</math>

The BBf(16) champion (n = 3) implements this iteration:

<math display="block">\begin{array}{lcl}
b^0 & \xrightarrow{0} & \text{halt} \\
b^1 & \xrightarrow{10} & b^6 \\
b^2 & \xrightarrow{10} & b^7 \\
b^3 & \xrightarrow{8} & b^4 \\
b^4 & \xrightarrow{8} & b^5 \\
b^5 & \xrightarrow{6} & b^2 \\
b^6 & \xrightarrow{6} & b^3 \\
b^{n+7} & \xrightarrow{4} & b^n \\
\end{array}</math>

which follows a permutation-like trajectory: <math>a \xrightarrow{1} b^1 \to b^6 \to b^3 \to b^4 \to b^5 \to b^2 \to b^7 \to b^0 \to \text{halt}</math>

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

=== BBf(20) ===
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
0 & -1 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 & 1 \\
1 & 1 & 0 & -1 & -1 \\
-1 & 0 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 0
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

== Cryptids ==
No fractran [[Cryptids]] have been found yet via enumeration, but some have been constructed by hand.

=== Hydra ===
A size 29 program that simulates [[Hydra]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1438230006571470919 Discord]):

<code>[507/22, 26/33, 245/2, 5/21, 1/3, 11/13, 22/5]</code>
<math display="block">
\begin{bmatrix}
-1 & 1 & 0 & 0 & -1 & 2 \\
1 & -1 & 0 & 0 & -1 & 1 \\
-1 & 0 & 1 & 2 & 0 & 0 \\
0 & -1 & 1 & -1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & -1 & 0 & 1 & 0
\end{bmatrix}
</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 36 program which simulates [[BMO1]] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1440018895212642424 Discord]):

<code>[153/55, 2/11, 26/35, 3/7, 11/17, 7/13, 25/6, 55/2, 14/3]</code>
<math display="block">
\begin{bmatrix}
0 & 2 & -1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & -1 & 0 & 0 \\
1 & 0 & -1 & -1 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 \\
-1 & -1 & 2 & 0 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & -1 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}
</math>

Let <math>A(a,b) = [a, b, 0, 0, 0, 0, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

=== BMO 6 (“Space Needle”) ===
A size 48 program which simulates [https://wiki.bbchallenge.org/wiki/1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD BMO 6] rules ([https://discord.com/channels/960643023006490684/1438019511155691521/1441137371046482071 Discord])

<code>[77/2, 2/99, 17/33, 13/11, 285/119, 17/19, 1375/51, 1/17, 3/5, 243/7, 10/13]</code>

<math>\begin{bmatrix}
-1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & -1 & 3 & 0 & 1 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}</math><syntaxhighlight>A(a, b) = B^a C^b E or B^(a-2) C^b D E

Start: A(7, 1)

A(1, b) --> halt

A(2a, b) --> A(5a+b+2, 1)

A(2a+1, b) --> A(b-1, b+c+3)</syntaxhighlight>

== References ==
<references />

[[Category:Functions]]

Fractran

2025-11-16T02:51:31Z

Int-y1: holdout progress

'''Fractran''' (originally styled FRACTRAN) is an esoteric Turing complete model of computation invented by John Conway in 1987.<ref>Conway, John H. (1987). "FRACTRAN: A Simple Universal Programming Language for Arithmetic". ''Open Problems in Communication and Computation''. Springer-Verlag New York, Inc. pp. 4–26. <nowiki>http://doi.org/10.1007/978-1-4612-4808-8_2</nowiki></ref> In this model a program is simply a finite list of fractions (rational numbers), the program state is an integer. For more details see https://en.wikipedia.org/wiki/FRACTRAN

'''BB_fractran'''(n) or '''BBf'''(n) is the Busy Beaver function for Fractran programs.

== Definition ==
A fractran program is a list of rational numbers <math>[q_0, q_1, ... q_{k-1}]</math>called rules and a fractran state is an integer <math>s \in \mathbb{Z}</math>. The numerator and denominator of any rational number fraction do not share any prime factors (they are in reduced form). We say that a rule <math>q_i</math> applies to state <math>s</math> if <math>s \cdot q_i \in \mathbb{Z}</math>. If no rule applies, we say that the computation has halted otherwise we apply the first applicable rule at each step. In that case we say <math>s \to t</math> and <math>t = s \cdot q_i</math> and <math>i = \min \{ i : s \cdot q_i \in \mathbb{Z} \}</math>. As with [[Turing machines]], we will write <math>s \xrightarrow{N} t</math> if <math>s \to s_1 \to \cdots \to s_{N-1} \to t </math> (s goes to t after N steps) and <math>s \to^* t</math> or <math>s \to^+ t</math> if <math>s \xrightarrow{N} t</math> for some N≥0 or N≥1 (respectively). We say that a program has runtime N (or halts in N steps) starting in state s if <math>s \xrightarrow{N} t</math> and computation halts on t.

Let <math>\Omega(n)</math> be the total number of prime factors of a positive integer n. In other words, <math>\Omega(2^{a_0} 3^{a_1} \cdots p_n^{a_n}) = \sum_{k=0}^n a_n</math>. Then given a rule <math>\frac{a}{b} </math> we say that <math>\text{size} \left( \frac{a}{b} \right) = \Omega(a) + \Omega(b) </math>. And the size of a fractran program <math>[q_0, q_1, ... q_{k-1}]</math> is <math>k + \sum_{i=0}^{k-1} \text{size}(q_i) </math>.

BB_fractran(n) or BBf(n) is the maximum runtime starting in state 2 for all halting fractran programs of size n. It is a non-computable function akin to the [[Busy Beaver Functions]] since Fractran is Turing Complete.

== Vector Representation ==
Fractran programs are not easy to interpret, in fact it may be completely unclear at first that they can perform any computation at all. One of the key insights is to represent all numbers (states and rules) in their prime factorization form. For example, we can use a vector <math>[ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{Z}^n</math> to represent the number <math>2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math>.

Let the vector representation (for a sufficiently large n) for a state <math>a = 2^{a_0} 3^{a_1} \cdots p_{n-1}^{a_{n-1}}</math> be <math>v(a) = [ a_0, a_1, \dots, a_{n-1} ] \in \mathbb{N}^n</math> and the vector representation for a rule <math>\frac{a}{b}</math> be <math>v \left( \frac{a}{b} \right) = v(a) - v(b) \in \mathbb{Z}^n</math> (Note that this is just an extension of the original definition extended to allow negative <math>a_i</math>).

Now, rule q applies to state s iff <math>v(s) + v(q) \in \mathbb{N}^n</math> (all components of the vector are ≥0) and if <math>s \to t</math> then <math>v(t) = v(s) + v(q)</math>. So the fractran multiplication model is completely equivalent to the vector adding model. For presentation, we will represent a fractran program with a matrix where each row is the vector representation for a rule.

For example, the BBf(15) champion (<code>[1/45, 4/5, 3/2, 25/3]</code>) in vector representation would be:

<math display="block">\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>

In this representation, it becomes much easier to reason about fractran programs and describe general rules. It is also very easy to calculate the size of a rule or program in vector representation. It is the sum of absolute values of all elements in the matrix + number of rules (number of rows).

== Champions ==
The table of champions is split into two pieces: the first for small champions (up to BBf(14)) which all share the same relatively simple behavior (sequential programs) is collapsed by default; the second for champions BBf(15) and beyond which have more complex and varied behavior.

All small champions as well as the first few larger ones were discovered and proven maximal by Jason Yuen (@-d) in their initial enumeration on [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025].

<div class="toccolours mw-collapsible mw-collapsed">'''Small Champions'''<div class="mw-collapsible-content">
{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
|-
| 2 || 1 || <code>[1/2]</code>
|<math>\begin{bmatrix}
-1
\end{bmatrix}</math>
|-
| 3 || 1 || <code>[3/2]</code>
|<math>\begin{bmatrix}
-1 & 1
\end{bmatrix}</math>
|-
| 4 || 1 || <code>[9/2]</code>
|<math>\begin{bmatrix}
-1 & 2
\end{bmatrix}</math>
|-
| 5 || 2 || <code>[3/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 1 \\
0 & -1
\end{bmatrix}</math>
|-
| 6 || 3 || <code>[9/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix}</math>
|-
| 7 || 4 || <code>[27/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 3 \\
0 & -1
\end{bmatrix}</math>
|-
| 8 || 5 || <code>[81/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 4 \\
0 & -1
\end{bmatrix}</math>
|-
| 9 || 6 || <code>[243/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 5 \\
0 & -1
\end{bmatrix}</math>
|-
| 10 || 7 || <code>[729/2, 1/3]</code>
|<math>\begin{bmatrix}
-1 & 6 \\
0 & -1
\end{bmatrix}</math>
|-
| 11 || 10 || <code>[27/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 3 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 12 || 13 || <code>[81/2, 25/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 2 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 13 || 17 || <code>[81/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 4 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|-
| 14 || 21 || <code>[243/2, 125/3, 1/5]</code>
|<math>\begin{bmatrix}
-1 & 5 & 0 \\
0 & -1 & 3 \\
0 & 0 & -1
\end{bmatrix}</math>
|}
</div></div>

{| class="wikitable"
|+
!n
!BBf(n)
!Example Champion
!Vector Representation
!Champion Found
!Holdouts Proven
|-
| 15 || 28 || <code>[1/45, 4/5, 3/2, 25/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 2
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 16 || 53 || <code>[1/45, 4/5, 3/2, 125/3]</code>
|<math>\begin{bmatrix}
0 & -2 & -1 \\
2 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 3
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434033599094587595 1 Nov 2025]
|-
| 17 || 107 || <code>[5/6, 49/2, 3/5, 40/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
3 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1434313398799175710 1 Nov 2025]
|Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1362008236118511758/1434771877376557086 3 Nov 2025]
|-
| 18 || 211 || <code>[5/6, 49/2, 3/5, 80/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
4 & 0 & 1 & -1
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1435313806493614131 4 Nov 2025]
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1362008236118511758/1436661215911870584 8 Nov 2025]
|-
| 19 || 370 || <code>[5/6, 49/2, 3/5, 160/7]</code>
|<math>\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & 2 \\
0 & 1 & -1 & 0 \\
5 & 0 & 1 & -1
\end{bmatrix}</math>
|@creeperman7002 [https://discord.com/channels/960643023006490684/1362008236118511758/1435763150489387090 5 Nov 2025]
|Decider: Daniel Yuan (@dyuan01) [https://discord.com/channels/960643023006490684/1438019511155691521/1438558242388312165 13 Nov 2025]
3 Holdouts: Racheline & Shawn Ligocki
|-
|20
|≥ 746
|<code>[7/15, 22/3, 6/77, 5/2, 9/5]</code>
|<math>\begin{bmatrix}
& -1 & -1 & +1 & \\
+1 & -1 & & & +1 \\
+1 & +1 & & -1 & -1 \\
-1 & & +1 & & \\
& +2 & -1 & &
\end{bmatrix}</math>
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438480761169776733 13 Nov 2025]
|34 holdouts remain
|-
|21
|≥ 55,213
|<code>[7/15, 22/3, 3/77, 175/2, 9/5]</code>
|
|Jason Yuen (@-d) [https://discord.com/channels/960643023006490684/1438019511155691521/1438833384779546634 14 Nov 2025]
|A holdouts list has not been created
|-
|22
|≥ 1,362,233
|<code>[121/6, 35/3, 15/77, 4/5, 9/2]</code>
|
|@creeperman7002 [https://discord.com/channels/960643023006490684/1438019511155691521/1438706712293933138 13 Nov 2025]
|
|-
|23
|≥ 3,441,433
|<code>[11/6, 35/3, 15/77, 16/5, 9/2]</code>
|
|@creeperman7002 [https://discord.com/channels/960643023006490684/1438019511155691521/1438706712293933138 13 Nov 2025]
|
|}

=== Behavior of Champions ===

==== Sequential programs ====
All champions up to BBf(14) have very simple behavior. They are all of the form: <math>\left[ \frac{3^{a_1}}{2}, \frac{5^{a_2}}{3}, ... \frac{p_n^{a_k}}{p_{k-1}}, \frac{1}{p_k} \right]</math> or in vector representation (limited to k=4):

<math display="block">\begin{bmatrix}
-1 & a_1 & 0 & 0 & 0 \\
0 & -1 & a_2 & 0 & 0 \\
0 & 0 & -1 & a_3 & 0 \\
0 & 0 & 0 & -1 & a_4 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}</math>

These champions repeatedly apply the rules in sequence, never going back to a previous rule. They apply the first rule until they've exhausted all 2s, then the second rule until they've exhausted all 3s, etc. They have a runtime of <math>1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots = \sum_{i=0}^k \prod_{j=1}^i a_j</math> and size <math>2k+2 + \sum_{i=1}^k a_i</math>. This grows linearly for k=1 (BBf(5) to BBf(10)) and quadratically for k=2 (BBf(11) to BBf(14)). Letting k grow with the size, the maximum runtime grows exponentially in the program size.

==== BBf(17) Family ====
The BBf(17) to BBf(19) champions are members of a family of programs (parameterized by <math>m,n \ge 0</math>)

<math display="block">\begin{bmatrix}
-1 & -1 & 1 & 0 \\
-1 & 0 & 0 & n \\
0 & 1 & -1 & 0 \\
m & 0 & 1 & -1
\end{bmatrix}</math>

which have size <math>m+n+12</math>

This family obeys the following rules:

# <math>[1, 0, 0, 0] \xrightarrow{1} [0, 0, 0, n]</math>
# if d≥1 and b≤m:<math display="block">[0, b, 0, d] \xrightarrow{m+b+2} [0, b+1, 0, d - 1 + n(m-b)]</math>
# if d≥1 and b≥m:<math display="block">[0, b, 0, d] \xrightarrow{2m+2} [0, b+1, 0, d - 1]</math>
#if d=0: [0,b,0,d] has halted

and furthermore these rules are applied in order since b is always increasing (and d is eventually decreasing). Combining these together we get runtime:<math display="block">1 + n(m+1)(m(m+1)+2) - \frac{m(m+1)}{2}</math>

The optimal choices for n,m for various program sizes are:
{| class="wikitable"
|+
!Size
!n
!m
!Runtime
|-
|16
|1
|3
|51
|-
|17
|2
|3
|107
|-
|18
|2
|4
|211
|-
|19
|2
|5
|370
|-
|20
|2
|6
|596
|-
|21
|3
|6
|904
|}

=== BBf(20) ===
The BBf(20) champion (running 746 steps):

<math display="block">\begin{bmatrix}
& -1 & -1 & +1 & \\
+1 & -1 & & & +1 \\
+1 & +1 & & -1 & -1 \\
-1 & & +1 & & \\
& +2 & -1 & &
\end{bmatrix}</math>

This program implements a [[Collatz-like]] iteration. Let <math>C(n) = [0, 0, n, 2, 0]</math>, then:

<math display="block">\begin{array}{lcl}
[1,0,0,0,0] & \xrightarrow{49} & C(2) \\
C(3k) & \xrightarrow{3k} & \text{halt} \\
C(3k+1) & \xrightarrow{11k+22} & C(4k+3) \\
C(3k+2) & \xrightarrow{11k+22} & C(4k+4) \\
\end{array}</math>

which follows the reasonably "lucky" trajectory:

<math display="block">C(2) \to C(4) \to C(7) \to C(11) \to C(16) \to C(23) \to C(32) \to C(44) \to C(60) \to \text{halt}</math>

== Cryptids ==
No fractran [[Cryptids]] have been found yet via enumeration, but some have been constructed by hand.

=== Hydra ===
A size 29 program that simulates [[Hydra]] rules: <code>[507/22, 26/33, 245/2, 5/21, 1/3, 11/13, 22/5]</code> and in vector form (here the 0s have been omitted to avoid clutter):

<math display="block">\begin{bmatrix}
-1 & +1 & & & -1 & +2 \\
+1 & -1 & & & -1 & +1 \\
-1 & & +1 & +2 & & \\
& -1 & +1 & -1 & & \\
& -1 & & & & \\
& & & & +1 & -1 \\
+1 & & -1 & & +1 &
\end{bmatrix}</math>

The intended interpretation is that if we let <math>S(h,w) = [1, 0, 0, w, h-3, 0]
</math> then it follows the following rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & = & S(3, 0) \\
S(2k, 0) & \to^* & \text{halt} \\
S(2k, w+1) & \to^* & S(3k, w) \\
S(2k+1, w) & \to^* & S(3k+1, w+2) \\
\end{array}</math>

=== BMO1 ===
A size 49 program which simulates [[BMO1]] rules:

<math display="block">\begin{bmatrix}
-1 & & & +1 & & & & & & +1 \\
& -1 & +1 & & & & & -1 & -1 & +2 \\
& -1 & & +1 & & & & -1 & & +1 \\
& -1 & & & +1 & & & +1 & -1 & \\
& & & -1 & & +1 & & & +2 & -1 \\
& +1 & & -1 & & & & +1 & & \\
& & & & -1 & & +1 & +1 & & -1 \\
& +1 & & & -1 & & & & +1 & \\
& +1 & -1 & & & & & & & \\
& & & +1 & & -1 & & & & \\
& & & & +1 & & -1 & & & \\
\end{bmatrix}</math>

Let <math>A(a,b) = [0, 1, 0, 0, 0, 0, 0, a, b, 0]</math>, then it follows the rules:

<math display="block">\begin{array}{lcl}
[1,0,\dots] & \to^* & A(1, 2) \\
A(a, b) & \to^* & A(a-b, 4b+2) & \text{if } a > b \\
A(a, b) & \to^* & A(2a+1, b-a) & \text{if } a < b \\
A(a, b) & \to^* & \text{Halt} & \text{if } a = b \\
\end{array}</math>

== References ==
<references />

[[Category:Functions]]

Hydra function

2025-10-08T08:48:19Z

Int-y1: math format

[[category:Functions]]
[[File:Hydra Spiral.png|thumb|185px|A spiral-like figure that gives the first few terms of the Hydra sequences with initial values 2, 5, 8, 11, 14, and 17.]]
The '''Hydra function''' is a [[Collatz-like]] function defined as:
<math display="block">\textstyle H(n)\equiv n+\big\lfloor\frac{1}{2}n\big\rfloor=\Big\lfloor\frac{3}{2}n\Big\rfloor=\begin{cases}
\frac{3n}{2}&\text{if }n\equiv0\pmod{2},\\
\frac{3n-1}{2}&\text{if }n\equiv1\pmod{2}.
\end{cases}</math>
It is named as such because of its connection to the unsolved halting problems for the [[Cryptids]] [[Hydra]] and [[Antihydra]]. Due to its simplicity, simulations for both of these [[Turing machines]] utilize this function instead of what can initially be proven.
== Relationship to Hydra and Antihydra problems==
Using the Hydra function, we can obtain simplified rules for Hydra and Antihydra:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|Let <math>C_H(a,b):=0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)}\;3^b\;2\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{20}&C_H(3,0),\\
C_H(2a,0)&\xrightarrow{54a^2-48a-2}&0^\infty\;3^{9a-8}\;1\;\textrm{A}\textrm{>}\;2\;0^\infty,\\
C_H(2a,b+1)&\xrightarrow{54a^2-39a-5}&C_H(3a,b),\\
C_H(2a+1,b)&\xrightarrow{4b+54a^2-3a+4}&C_H(3a+1,b+2).\\\hline
\end{array}</math>
|Let <math>A_H(a,b):=0^\infty\;1^a\;0\;1^{b-4}\;\textrm{E}\textrm{>}\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{11}&A_H(0,8),\\
A_H(a,2b)& \xrightarrow{2a+3b^2-1}& A_H(a+2,3b),\\
A_H(0,2b+1)&\xrightarrow{3b^2-3b-7}& 0^\infty\;\textrm{<}\textrm{F}\;110\;1^{3b-6}\;0^\infty,\\
A_H(a+1,2b+1)&\xrightarrow{3b^2-7}& A_H(a,3b+1).\\\hline
\end{array}</math>
|}
<div class="toccolours mw-collapsible mw-collapsed">'''Proof'''<div class="mw-collapsible-content">
Recall the high-level rules for Hydra and Antihydra:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|Let <math>C(a,b):=0^\infty\;\textrm{<}\textrm{A}\;2\;0^a\;3^b\;2\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{20}&C(3,0),\\
C(2a,0)&\xrightarrow{6a^2+20a+4}&0^\infty\;3^{3a+1}\;1\;\textrm{A}\textrm{>}\;2\;0^\infty,\\
C(2a,b+1)&\xrightarrow{6a^2+23a+10}&C(3a+3,b),\\
C(2a+1,b)&\xrightarrow{4b+6a^2+23a+26}&C(3a+3,b+2).\\\hline
\end{array}</math>
|Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E}\textrm{>}\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{11}&A(0,4),\\
A(a,2b)& \xrightarrow{2a+3b^2+12b+11}& A(a+2,3b+2),\\
A(0,2b+1)&\xrightarrow{3b^2+9b-1}& 0^\infty\;\textrm{<}\textrm{F}\;110\;1^{3b}\;0^\infty,\\
A(a+1,2b+1)&\xrightarrow{3b^2+12b+5}& A(a,3b+3).\\\hline
\end{array}</math>
|}
Already, both machines appear to be very similar. They have one parameter that increases exponentially with growth factor <math display="inline">\frac{3}{2}</math> and another that takes a pseudo-random walk. Below, the exponentially increasing variables are described by integer sequences:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">a_0=3,a_{n+1}=\begin{cases}\frac{3a_n+6}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{3a_n+3}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|<math display="block">a_0=4,a_{n+1}=\begin{cases}\frac{3a_n+4}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{3a_n+3}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|}
This will make demonstrating the transformation easier. Now we will define a new integer sequence based on the old one and discover the recursive rules for that sequence. This new sequence is <math display="inline">b_n=\frac{1}{3}a_n+2</math> and <math>b_n=a_n+4</math> for Hydra and Antihydra respectively. We start by using <math>b_{n+1}</math> instead and substituting <math>a_{n+1}</math> for its recursive formula. By doing so, we get:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">b_0=3,b_{n+1}=\begin{cases}\frac{a_n+6}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{a_n+5}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|<math display="block">b_0=8,b_{n+1}=\begin{cases}\frac{3a_n+12}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{3a_n+11}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|}
After that, we can substitute <math>a_n</math> for its solution in terms of <math>b_n</math>. What results is the following:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">b_0=3,b_{n+1}=\begin{cases}\frac{3(b_n-2)+6}{2}&\text{if }3(b_n-2)\equiv0\pmod{2}\\\frac{3(b_n-2)+5}{2}&\text{if }3(b_n-2)\equiv1\pmod{2}\end{cases}</math>
|<math display="block">b_0=8,b_{n+1}=\begin{cases}\frac{3(b_n-4)+12}{2}&\text{if }b_n-4\equiv0\pmod{2}\\\frac{3(b_n-4)+11}{2}&\text{if }b_n-4\equiv1\pmod{2}\end{cases}</math>
|}
The <math>\text{if}</math> statements amount to checking if <math>b_n</math> is even or odd. After simplifying, we are done:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">b_0=3,b_{n+1}=\begin{cases}\frac{3b_n}{2}&\text{if }b_n\equiv0\pmod{2}\\\frac{3b_n-1}{2}&\text{if }b_n\equiv1\pmod{2}\end{cases}</math>
|<math display="block">b_0=8,b_{n+1}=\begin{cases}\frac{3b_n}{2}&\text{if }b_n\equiv0\pmod{2}\\\frac{3b_n-1}{2}&\text{if }b_n\equiv1\pmod{2}\end{cases}</math>
|}
Now that we have demonstrated a strong similarity in the behaviour of both Turing machines, we can return to using the high-level rules. Doing that while considering the step counts yields the final result.
</div></div>
Under these rules, the halting problem for Hydra is about whether repeatedly applying the function <math>H(n)</math>, starting with <math>n=3</math>, will eventually generate more even terms than twice the number of odd terms. Similarly, Antihydra halts if and only if repeatedly applying <math>H(n)</math>, starting with <math>n=8</math>, will eventually generate more odd terms than twice the number of even terms.

=== Coding the Hydra and Antihydra problems using the Hydra function ===
Paired with the corresponding even/odd criterion as loop halting condition (implemented as a counter variable) and initial Hydra function value, the Hydra function definition can be used to write computer programs that simulate the abstracted behavior of the Hydra and Antihydra Turing machines. The following Python program is a Hydra simulator:
<syntaxhighlight lang="python" line="1">
# 'a' and 'b' fulfill the same purpose as in the Hydra rules:
# The current hydra function value
a = 3
# The even/odd condition counter
b = 0
# As long as Hydra has not halted, 'b' remains greater than -1.
while b != -1:
# If 'a' is even, decrement 'b', otherwise increase 'b' by 2.
if a % 2 == 0:
b -= 1
else:
b += 2
# This performs one step of the Hydra function H(a) = a + floor(a/2).
# Note that integer division by 2 is equivalent to one bit shift to the right (a >> 1)
a += a//2
</syntaxhighlight>
Replacing <code>a = 3</code> with <code>a = 8</code> and swapping <code>b -= 1</code> with <code>b += 2</code> turns this program into an Antihydra simulator.

Determining whether these programs halt or not (and if so, after how many loop iterations) would resolve these open problems.
==Properties==
The Hydra function can be rewritten as follows:
<math display="block">\begin{array}{l}
H(2n)&=&3n,\\
H(2n+1)&=&3n+1.
\end{array}</math>
Now assume that for some positive integer <math>s</math> and every odd integer <math>t</math>, <math>H^s(2^st)=3^st</math> and <math>H^s(2^st+1)=3^st+1</math>, where <math>H^i(n)</math> is function iteration. Notice that we can write <math>2^{s+1}t=2\cdot2^st</math> and <math>2^{s+1}t+1=2\cdot2^st+1</math>, so if we apply <math>H</math> to these numbers, we get <math>H(2\cdot2^st)=3\cdot 2^st</math> and <math>H(2\cdot2^st+1)=3\cdot2^st+1</math>. Now, if we apply <math>H</math> to these numbers <math>s</math> times, we get <math>H^{s+1}\big(2^{s+1}t\big)=H^s(2^s\cdot3t)=3^{s+1}t</math> and <math>H^{s+1}\big(2^{s+1}t+1\big)=H^s(2^s\cdot3t+1)=3^{s+1}t+1</math>. Therefore, by mathematical induction we have proved the following formulas:
<math display="block">\begin{array}{l}
H^s(2^st)&=&3^st,\\
H^s(2^st+1)&=&3^st+1.
\end{array}</math>
This optimization can be directly applied to the high-level rules for Hydra and Antihydra, producing this result:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|Let <math>C_H(a,b):=0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)}\;3^b\;2\;0^\infty</math>:
<math display="block">\begin{array}{|lll|}\hline
C_H(2^st,b+s)&\xrightarrow{f_1(s,t)}&C_H(3^st,b),\\
C_H(2^st+1,b)&\xrightarrow{f_2(s,t,b)}&C_H(3^st+1,b+2s),\\\hline
\end{array}</math>
where <math display="inline">f_1(s,t)=\frac{3t(3^s-2^s)(18(3^s+2^s)t-65)}{5}-5s</math> and <math display="inline">f_2(s,t,b)=(b+s)4s+\frac{3t(3^s-2^s)(18(3^s+2^s)t-5)}{5}</math>.
|Let <math>A_H(a,b):=0^\infty\;1^a\;0\;1^{b-4}\;\textrm{E}\textrm{>}\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
A_H(a,2^st)& \xrightarrow{f_3(s,t,a)}& A_H(a+2s,3^st),\\
A_H(a+s,2^st+1)&\xrightarrow{f_4(s,t)}& A_H(a,3^st+1),\\\hline
\end{array}</math>
where <math display="inline">f_3(s,t,a)=(2a-3+2s)s+\frac{3t^2(9^s-4^s)}{5}</math> and <math display="inline">f_4(s,t)=\frac{3t^2(9^s-4^s)}{5}-7s</math>.
|}
== Visualizations ==
The four images below depict the first 1000 values of four Hydra sequences with different initial values. Each row of pixels shows a number in binary on the right and its parity on the left (blue for even, red for odd):
<gallery mode="packed" heights="250">
File:HydraFunction-StartingValue2.png|Starting value 2. There are 492 even numbers and 508 odd numbers.
File:HydraFunction-StartingValue5.png|Starting value 5. There are 497 even numbers and 503 odd numbers.
File:Antihydra increasing value.png|Starting value 8. There are 499 even numbers and 501 odd numbers.
File:HydraFunction-StartingValue11.png|Starting value 11. There are 481 even numbers and 519 odd numbers.
</gallery>

Beaver Math Olympiad

2025-10-08T08:46:54Z

Int-y1: math format

'''Beaver Mathematical Olympiad''' (BMO) is an attempt to re-formulate the halting problem for some particular Turing machines as a mathematical problem in a style suitable for a hypothetical math olympiad.

The purpose of the BMO is twofold. First, statements where non-essential details (related to tape encoding, number of steps, etc.) are discarded are more suitable to be shared with mathematicians who perhaps are able to help. Second, it's a way to jokingly highlight how a hard question could appear deceptively simple.

== Unsolved problems ==

=== 1. {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}} ===

Let <math>(a_n)_{n \ge 1}</math> and <math>(b_n)_{n \ge 1}</math> be two sequences such that <math>(a_1, b_1) = (1, 2)</math> and

<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}
(a_n-b_n, 4b_n+2) & \text{if }a_n \ge b_n \\
(2a_n+1, b_n-a_n) & \text{if }a_n < b_n
\end{cases}</math>

for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_i = b_i</math>?

The first 10 values of <math>(a_n, b_n)</math> are <math>(1, 2), (3, 1), (2, 6), (5, 4), (1, 18), (3, 17), (7, 14), (15, 7), (8, 30), (17, 22)</math>.

=== 2. [[Hydra]] and [[Antihydra]] ===

Let <math>(a_n)_{n \ge 0}</math> be a sequence such that <math>a_{n+1} = a_n+\left\lfloor\frac{a_n}{2}\right\rfloor</math> for all non-negative integers <math>n</math>.

# If <math>a_0=3</math>, does there exist a non-negative integer <math>k</math> such that the list of numbers <math>a_0, a_1, a_2, \dots, a_k</math> have more than twice as many even numbers as odd numbers? ([[Hydra]])
# If <math>a_0=8</math>, does there exist a non-negative integer <math>k</math> such that the list of numbers <math>a_0, a_1, a_2, \dots, a_k</math> have more than twice as many odd numbers as even numbers? ([[Antihydra]])

=== 5. {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}} ===

Let <math>(a_n)_{n \ge 1}</math> and <math>(b_n)_{n \ge 1}</math> be two sequences such that <math>(a_1, b_1) = (0, 5)</math> and

<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}
(a_n+1, b_n-f(a_n)) & \text{if } b_n \ge f(a_n) \\
(a_n, 3b_n+a_n+5) & \text{if } b_n < f(a_n)
\end{cases}</math>

where <math>f(x)=10\cdot 2^x-1</math> for all non-negative integers <math>x</math>.

Does there exist a positive integer <math>i</math> such that <math>b_i = f(a_i)-1</math>?

=== 6. {{TM|1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD|undecided}} ===
Let <math>f(b) = b + k + 3a</math>, where <math>k</math> and <math>a</math> are non-negative integers satisfying <math>b = (2a+1)\cdot 2^k</math>.

Now consider the iterated application of the function <math>f^{n+1}(b) = f(f^n(b)))</math>, <math>f^0(b)=b</math>. Does there exist a non-negative integer <math>n</math> such that <math>f^n(6)</math> equals a power of 2?

=== 7. {{TM|1RB1RF_1RC0RA_1LD1RC_1LE0LE_0RA0LD_0RB---|undecided}} ===
Let <math>v_2(n)</math> be the largest integer <math>k</math> such that <math>2^k</math> divides <math>n</math>.

Let <math>f(n) = n+1+(v_2(n+1) \bmod 2)</math>.

Now consider the iterated application of the function <math>f^{n+1}(b) = f(f^n(b)))</math>, <math>f^0(b)=b</math>.

Let <math>(a_n)_{n \ge 0}</math> be a sequence such that <math>a_0=1</math> and <math>a_{n+1} = f^{n+2}\left(\left\lfloor\frac{a_n}{2}\right\rfloor\right)</math> for all non-negative integers <math>n</math>.

Does there exist a non-negative integer <math>k</math> such that <math>a_k</math> is even?

(for simplicity, this question is slightly stronger than the halting problem of this TM)

== Solved problems ==

=== 3. {{TM|1RB0RB3LA4LA2RA_2LB3RA---3RA4RB|non-halt}} and {{TM|1RB1RB3LA4LA2RA_2LB3RA---3RA4RB|non-halt}} ===

Let <math>v_2(n)</math> be the largest integer <math>k</math> such that <math>2^k</math> divides <math>n</math>. Let <math>(a_n)_{n \ge 0}</math> be a sequence such that

<math display="block">a_n = \begin{cases}
2 & \text{if } n=0 \\
a_{n-1}+2^{v_2(a_{n-1})+2}-1 & \text{if } n \ge 1
\end{cases}</math>

for all non-negative integers <math>n</math>. Is there an integer <math>n</math> such that <math>a_n=4^k</math> for some positive integer <math>k</math>?

Link to Discord discussion: https://discord.com/channels/960643023006490684/1084047886494470185/1252634913220591728

=== 4. {{TM|1RB3RB---1LB0LA_2LA4RA3LA4RB1LB|non-halt}} ===

Bonnie the beaver was bored, so she tried to construct a sequence of integers <math>\{a_n\}_{n \ge 0}</math>. She first defined <math>a_0=2</math>, then defined <math>a_{n+1}</math> depending on <math>a_n</math> and <math>n</math> using the following rules:

* If <math>a_n \equiv 0\text{ (mod 3)}</math>, then <math>a_{n+1}=\frac{a_n}{3}+2^n+1</math>.
* If <math>a_n \equiv 2\text{ (mod 3)}</math>, then <math>a_{n+1}=\frac{a_n-2}{3}+2^n-1</math>.

With these two rules alone, Bonnie calculates the first few terms in the sequence: <math>2, 0, 3, 6, 11, 18, 39, 78, 155, 306, \dots</math>. At this point, Bonnie plans to continue writing terms until a term becomes <math>1\text{ (mod 3)}</math>. If Bonnie sticks to her plan, will she ever finish?

<div class="toccolours mw-collapsible mw-collapsed">'''Solution'''<div class="mw-collapsible-content">
How to guess the closed-form solution: Firstly, notice that <math>a_n \approx \frac{3}{5} \times 2^n</math>. Secondly, calculate the error term <math>a_n - \frac{3}{5} \times 2^n</math>. The error term appears to have a period of 4. This leads to the following guess:

<math display="block">a_n=\frac{3}{5}\begin{cases}
2^n+\frac{7}{3} &\text{if } n\equiv 0 \pmod{4}\\
2^n-2 &\text{if } n\equiv 1 \pmod{4}\\
2^n+1 &\text{if } n\equiv 2 \pmod{4}\\
2^n+2 &\text{if } n\equiv 3 \pmod{4}
\end{cases}</math>

This closed-form solution can be proven correct by induction. Unfortunately, the induction may require a lot of tedious calculations.

For all <math>k</math>, we have <math>a_{4k} \equiv 2\text{ (mod 3)}</math> and <math>a_{4k+1} \equiv a_{4k+2} \equiv a_{4k+3} \equiv 0\text{ (mod 3)}</math>. Therefore, Bonnie will never finish.
</div></div>

Directed head notation

2025-10-08T07:34:30Z

Int-y1: math format

'''Directed head notation''' is a notation for specifying a [[Turing machine]] configuration using tape compression and a TM head which "points" either to the left or right. Directed head notation may be used for a complete tape configuration
<math display="block">0^\infty \; 1 \; \textrm{<}\textrm{B} \; 0^3 \; 13^{10} \; 2 \; 0^\infty</math>
or for a partial configuration
<math display="block">101 \; \textrm{A}\textrm{>} \; 1^n</math>

== Tape Compression ==
This notation supports run-length encoding for tape compression using "exponents". Given a "word" (sequence of tape symbols) <math>w</math> and a count <math>n</math>, <math>w^n</math> represents <math>n</math> repetitions of <math>w</math> concatenated. So for example, <math>2^4 = 2222</math> and <math>10^3 = 101010</math>.

It is important to note here that the words "2" and "10" in these examples are sequences of tape symbols and not integers and that the counts "4" and "3" are integer repetition counts, not representing integer exponentiation. It is perhaps unfortunate that these compressed blocks look identical to integer mathematical expressions. Generally, it should be obvious from the context that this is a tape configuration and not a math expression.

Segments of the tape may also be specified without exponents which represent unrepeated (or 1 repeat) of that segment. The notation <math>0^\infty</math> is used to represent the infinite sequence of blank symbols at either end of a configuration. In general, there will be many different notations for identical tape segments. For example:
<math display="block">0^\infty \; 121212 = 0^\infty \; 12 \; 12 \; 12 = 0^\infty \; 12^3 = 0^\infty \; 01 \; 21^2 \; 2</math>

== Directed Head ==
Traditionally, the Turing machine head is considered to be located "on" a tape cell, for example some common traditional notation for a TM configuration include:
<math display="block">\dots 0110 \underset{A}{\underline{2}} 01110 \dots</math>
or
<math display="block">\dots 0110 (A2) 01110 \dots</math>
which indicate that the TM is in state A and currently positioned on the tape cell containing the 2. In these notations there are 3 groupings of symbols: those to the left (<math>\dots 0110</math>), those to the right (<math>01110 \dots</math>) and the single current symbol (2).

In directed head notation, we instead conceptualize the TM head as in the process of moving from one tape cell to another. So the above configuration could be represented by either
<math display="block">\dots 0110 \; \textrm{A}\textrm{>} \; 2 01110 \dots</math>
or
<math display="block">\dots 0110 2 \; \textrm{<}\textrm{A} \; 01110 \dots</math>
depending on which direction it moved immediately previous to this config. In many contexts, these may be considered equivalent configurations (since the forward behavior is identical for either). Here the notation "A>" means that the TM is in state A moving toward the right and "<A" means that it is moving to the left. The "current symbol" is the symbol "small end" of these "directed head" arrows.

== Partial Configurations ==
A directed head configuration includes a directed head and zero or more tape segments on each side. If it includes <math>0^\infty</math> at both ends, then it is a complete configuration (specifying precisely the entire tape), otherwise it is a partial configuration (specifying only a limited context around the TM head). For example, the partial configuration
<math display="block">101\;\textrm{C}\textrm{>}\;1^3</math>
matches any of these complete configurations
<math display="block">\begin{array}{rcl}
0^\infty \; 101 & \textrm{C}\textrm{>} & 1^3 \; 0^\infty \\
0^\infty \; 11101 & \textrm{C}\textrm{>} & 1^8 \; 0^\infty \\
0^\infty \; 01^3 & \textrm{C}\textrm{>} & 11^2 \; 0^\infty
\end{array}</math>
Partial configurations may have no segments specified even on the side the TM head is facing. For example
<math display="block">\textrm{<}\textrm{C} \; 101</math>
is a valid partial configuration. In these cases, we cannot tell what the "current symbol" is until we know the complete configuration. These configurations are common as the result of configuration transitions as described below.

== Configuration Transitions ==
One of the most common use cases for directed head notation is for specifying configuration transition rules. For example, the rule
<math display="block">0^\infty \; 011 \; \textrm{<}\textrm{C} \; 0^\infty \; \xrightarrow{3} \; 0^\infty \; \textrm{<}\textrm{C} \; 101 \; 0^\infty</math>
indicates that if the TM starts in config <math>0^\infty \; 011 \; \textrm{<}\textrm{C} \; 0^\infty</math>, 3 steps later it will be in configuration <math>0^\infty \; \textrm{<}\textrm{C} \; 101 \; 0^\infty</math>. It is common to define these rules for partial configurations. For example,
<math display="block">011 \; \textrm{<}\textrm{C} \; \xrightarrow{3} \; \textrm{<}\textrm{C} \; 101</math>
which means that for any complete configuration <math>w \; 011 \; \textrm{<}\textrm{C} \; u</math> (where <math>w, u</math> are any valid left and right half-tapes), 3 steps later it will be in configuration <math>w \; \textrm{<}\textrm{C} \; 101 \; u</math>.

Transition rules become especially useful once you use induction to prove them over arbitrary repetition counts. For example, the previous transition rule can prove this [[Shift rule|shift rule]] for all <math>n \ge 0</math>:
<math display="block">011^n \; \textrm{<}\textrm{C} \; \xrightarrow{3n} \; \textrm{<}\textrm{C} \; 101^n</math>
[[Category:Analysis Techniques]]

Collatz-like

2025-10-08T07:28:42Z

Int-y1: math format 2

A '''Collatz-like function''' is a partial function defined piecewise depending on the remainder of an input modulo some number. The canonical example is the original Collatz function:<math display="block">\begin{array}{l}
c(2k) & = & k \\
c(2k+1) & = & 3k+2
\end{array}</math>

A '''Collatz-like problem''' is a question about the behavior of iterating a Collatz-like function. Collatz-like problems are famously difficult.

Many [[Busy Beaver Champions]] have '''Collatz-like behavior''', meaning that their behavior can be concisely described via the iterated values of a Collatz-like function.

== Examples ==

=== 5-state busy beaver winner ===
Consider the [[5-state busy beaver winner]] and the generalized configuration:

<math display="block">M(n) = 0^\infty \; \textrm{<}\text{A} \; 1^n \; 0^\infty</math>Pascal Michel showed that:

<math display="block">\begin{array}{lcl}
0^\infty \; \textrm{<}\textrm{A} \; 0^\infty & = & M(0) \\
M(3k) & \xrightarrow{5k^2+19k+15} & M(5k+6) \\
M(3k+1) & \xrightarrow{5k^2+25k+27} & M(5k+9) \\
M(3k+2) & \xrightarrow{6k+12} & 0^\infty\;1\;\textrm{Z}\textrm{>}\;01\;{(001)}^{k+1}\;1\;0^\infty
\end{array}</math>

Starting on a blank tape <math>M(0)</math>, these rules iterate 15 times before reaching the halt config.<ref>[https://bbchallenge.org/~pascal.michel/beh#tm52a Pascal Michel's Analysis of the BB(5, 2) Champion]</ref>

=== Hydra ===
Consider [[Hydra]] and the generalized configuration:

<math display="block">C(a, b) = 0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)} \; 3^b \; 2 \; 0^\infty</math>
Daniel Yuan showed that:<math display="block">\begin{array}{l}
0^\infty \; \textrm{A}\textrm{>} \; 0^\infty & & \xrightarrow{20} & C(3, 0) \\
C(2n, & 0) & \to & \text{Halt}(9n-6) \\
C(2n, & b+1) & \to & C(3n,b) \\
C(2n+1, & b) & \to & C(3n+1,b+2)
\end{array}</math>

Where <math>\text{Halt}(n)</math> is a halting configuration with <math>n</math> non-zero symbols on the tape.

Starting from <math>C(3, 0)</math>, this simulates a pseudo-random walk along the <math>b</math> parameter, increasing it by 2 every time <math>a</math> is odd, decreasing by 1 every time it's even. Deciding whether or not Hydra halts requires determining whether through the process of applying the Collatz-like function
<math display="block">\begin{array}{l}
H(2n) & = & 3n &\text{(even transition)} \\
H(2n+1) & = & 3n+1&\text{(odd transition)}
\end{array}</math>
to 3 recursively, there eventually comes a point where the amount of even transitions applied is more than twice the amount of odd transitions applied.<ref>Shawn Ligocki. [https://www.sligocki.com/2024/05/10/bb-2-5-is-hard.html BB(2, 5) is Hard (Hydra)]. 10 May 2024.</ref> The first few transitions are displayed below:
<math display="block">\begin{array}{l}
\vphantom{\frac{\frac{0}{.}}{.}}3 \xrightarrow{O} 4 \xrightarrow{E} 6\xrightarrow{E} 9 \xrightarrow{O} 13 \xrightarrow{O} 19 \xrightarrow{O} 28 \xrightarrow{E} 42 \xrightarrow{E} 63\xrightarrow{O} \cdots
\end{array}
</math>
=== Exponential Collatz ===
Consider the machine {{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE|halt}}, discovered by Pavel Kropitz in May 2022, and the general configuration:<math display="block">K(n):=0^\infty\;1\;0^n\;11\;0^5\;\textrm{C}\textrm{>}\;0^\infty</math>Shawn Ligocki showed that:
<math display="block">\begin{array}{l}
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{45}&K(5) \\
K(4k) & \to & \operatorname{Halt}\Bigl(\frac{3^{k+3} - 11}{2}\Bigr) \\
K(4k+1) & \to & K\Bigl(\frac{3^{k+3} - 11}{2}\Bigr) \\
K(4k+2) & \to & K\Bigl(\frac{3^{k+3} - 11}{2}\Bigr) \\
K(4k+3) & \to & K\Bigl(\frac{3^{k+3} + 1}{2}\Bigr)
\end{array}</math>

Demonstrating Collatz-like behavior with exponential piecewise component functions.

Starting from config <math>K(5)</math>, these rules iterate 15 times before reaching the halt config leaving over <math>10 \uparrow\uparrow 15</math> non-zero symbols on the tape.<ref>Shawn Ligocki. [https://www.sligocki.com/2022/06/21/bb-6-2-t15.html BB(6, 2) > 10↑↑15]. 21 Jun 2022.</ref>
== References ==
<references />
[[Category:Zoology]]

Collatz-like

2025-10-08T07:27:32Z

Int-y1: math format

A '''Collatz-like function''' is a partial function defined piecewise depending on the remainder of an input modulo some number. The canonical example is the original Collatz function:<math display="block">\begin{array}{l}
c(2k) & = & k \\
c(2k+1) & = & 3k+2
\end{array}</math>

A '''Collatz-like problem''' is a question about the behavior of iterating a Collatz-like function. Collatz-like problems are famously difficult.

Many [[Busy Beaver Champions]] have '''Collatz-like behavior''', meaning that their behavior can be concisely described via the iterated values of a Collatz-like function.

== Examples ==

=== 5-state busy beaver winner ===
Consider the [[5-state busy beaver winner]] and the generalized configuration:

<math display="block">M(n) = 0^\infty \; \textrm{<}\text{A} \; 1^n \; 0^\infty</math>Pascal Michel showed that:

<math display="block">\begin{array}{lcl}
0^\infty \; \textrm{<}\textrm{A} \; 0^\infty & = & M(0) \\
M(3k) & \xrightarrow{5k^2+19k+15} & M(5k+6) \\
M(3k+1) & \xrightarrow{5k^2+25k+27} & M(5k+9) \\
M(3k+2) & \xrightarrow{6k+12} & 0^\infty\;1\;\textrm{Z}\textrm{>}\;01\;{(001)}^{k+1}\;1\;0^\infty
\end{array}</math>

Starting on a blank tape <math>M(0)</math>, these rules iterate 15 times before reaching the halt config.<ref>[https://bbchallenge.org/~pascal.michel/beh#tm52a Pascal Michel's Analysis of the BB(5, 2) Champion]</ref>

=== Hydra ===
Consider [[Hydra]] and the generalized configuration:

<math display="block">C(a, b) = 0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)} \; 3^b \; 2 \; 0^\infty</math>
Daniel Yuan showed that:<math display="block">\begin{array}{l}
\\
0^\infty \; \textrm{A}\textrm{>} \; 0^\infty & & \xrightarrow{20} & C(3, 0) \\
C(2n, & 0) & \to & \text{Halt}(9n-6) \\
C(2n, & b+1) & \to & C(3n,b) \\
C(2n+1, & b) & \to & C(3n+1,b+2)
\end{array}</math>

Where <math>\text{Halt}(n)</math> is a halting configuration with <math>n</math> non-zero symbols on the tape.

Starting from <math>C(3, 0)</math>, this simulates a pseudo-random walk along the <math>b</math> parameter, increasing it by 2 every time <math>a</math> is odd, decreasing by 1 every time it's even. Deciding whether or not Hydra halts requires determining whether through the process of applying the Collatz-like function
<math display="block">\begin{array}{l}
H(2n) & = & 3n &\text{(even transition)} \\
H(2n+1) & = & 3n+1&\text{(odd transition)}
\end{array}</math>
to 3 recursively, there eventually comes a point where the amount of even transitions applied is more than twice the amount of odd transitions applied.<ref>Shawn Ligocki. [https://www.sligocki.com/2024/05/10/bb-2-5-is-hard.html BB(2, 5) is Hard (Hydra)]. 10 May 2024.</ref> The first few transitions are displayed below:
<math display="block">\begin{array}{l}
\vphantom{\frac{\frac{0}{.}}{.}}3 \xrightarrow{O} 4 \xrightarrow{E} 6\xrightarrow{E} 9 \xrightarrow{O} 13 \xrightarrow{O} 19 \xrightarrow{O} 28 \xrightarrow{E} 42 \xrightarrow{E} 63\xrightarrow{O} \cdots
\end{array}
</math>
=== Exponential Collatz ===
Consider the machine {{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE|halt}}, discovered by Pavel Kropitz in May 2022, and the general configuration:<math display="block">K(n):=0^\infty\;1\;0^n\;11\;0^5\;\textrm{C}\textrm{>}\;0^\infty</math>Shawn Ligocki showed that:
<math display="block">\begin{array}{l}
\\
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{45}&K(5) \\
K(4k) & \to & \operatorname{Halt}\Bigl(\frac{3^{k+3} - 11}{2}\Bigr) \\
K(4k+1) & \to & K\Bigl(\frac{3^{k+3} - 11}{2}\Bigr) \\
K(4k+2) & \to & K\Bigl(\frac{3^{k+3} - 11}{2}\Bigr) \\
K(4k+3) & \to & K\Bigl(\frac{3^{k+3} + 1}{2}\Bigr)
\end{array}</math>

Demonstrating Collatz-like behavior with exponential piecewise component functions.

Starting from config <math>K(5)</math>, these rules iterate 15 times before reaching the halt config leaving over <math>10 \uparrow\uparrow 15</math> non-zero symbols on the tape.<ref>Shawn Ligocki. [https://www.sligocki.com/2022/06/21/bb-6-2-t15.html BB(6, 2) > 10↑↑15]. 21 Jun 2022.</ref>
== References ==
<references />
[[Category:Zoology]]

Inductive Proof System

2025-10-08T07:25:35Z

Int-y1: standardize <pre>

An '''Inductive Proof System''' is an [[Accelerated Simulator]] and [[Decider]] which operates by automatically detecting and proving [[Transition rule|transition rules]] using Mathematical Induction.

== Inductive Rule ==
An inductive rule is a general transition rule (a start and end configuration, generalized with variables for repetition counts) along with a proof. The proof generally has two pieces: the base case and the inductive case. Each is a list of steps where each step is (A) a specific TM transition, (B) an application of the inductive hypothesis, (C) an application of a previously defined rule.

=== Rule Levels ===
We can assign levels to any Inductive Rule. A Level 0 (L0) rule does not depend on any previously defined inductive rules. A Level 1 (L1) rule depends only upon previously proven L0 rules, etc. All L0 Rules (which only invoke the inductive hypothesis once) are [[Shift rule|Shift rules]].

== Example Proofs ==

=== Notation ===
In this article we will use the following notation for an Inductive Rule and its proof:

<math>P(n)</math>

* Proof by induction on n:
** Base case: ''List of steps to prove the base case <math>P(0)</math>. Often this is the empty list since <math>P(0)</math> is trivially true (in zero steps).''
** Inductive case: ''List of steps to prove <math>P(n+1)</math> of which some can be "IH" the inductive hypothesis that <math>P(n)</math>.''

=== Bouncer ===
Consider the [[bouncer]] {{TM|1RB0RC_0LC---_1RD1RC_0LE1RA_1RD1LE|non-halt}}. We can prove the following Inductive Rules:

# Shift (L0) Rule: <math>C(n) : \textrm{C}\textrm{>} \; 1^n \to 1^n \; \textrm{C}\textrm{>}</math>
#* Proof by induction on n
#** Base case: []<math display="block">\textrm{C}\textrm{>} \; 1^0 \to 1^0 \; \textrm{C}\textrm{>}</math>
#** Inductive case: [C1, IH]<math display="block">\textrm{C}\textrm{>} \; 1^{n+1} \xrightarrow{C1} 1 \; \textrm{C}\textrm{>} \; 1^n \xrightarrow{IH} 1^{n+1} \; \textrm{C}\textrm{>}</math>
# Shift (L0) Rule: <math>E(n) : 1^n \; \textrm{<}\textrm{E} \to \textrm{<}\textrm{E} \; 1^n</math>
#* Proof by induction on n
#** Base case: []
#** Inductive case: [E1, IH]
# L1 Rule: <math>A(n) : \textrm{A}\textrm{>} \; 1^n \; 0^\infty \to 1^{2n} \; \textrm{A}\textrm{>} \; 0^\infty</math>
#* Proof by induction on n
#** Base case: []
#** Inductive case: [A1, C(n), C0, D0, E(n+1), E0, D1, IH]<math display="block">\begin{array}{l}
\\
\textrm{A}\textrm{>} \; 1^{n+1} \; 0^\infty
& \xrightarrow{A1} &
0 \; \textrm{C}\textrm{>} \; 1^n \; 0^\infty \\
& \xrightarrow{C(n)} &
0 \; 1^n \; \textrm{C}\textrm{>} \; 0^\infty \\
& \xrightarrow{C0} &
0 \; 1^{n+1} \; \textrm{D}\textrm{>} \; 0^\infty \\
& \xrightarrow{D0} &
0 \; 1^{n+1} \; \textrm{<}\textrm{E} \; 0^\infty \\
& \xrightarrow{E(n+1)} &
0 \; \textrm{<}\textrm{E} \; 1^{n+1} \; 0^\infty \\
& \xrightarrow{E0} &
1 \; \textrm{D}\textrm{>} \; 1^{n+1} \; 0^\infty \\
& \xrightarrow{D1} &
1^2 \; \textrm{A}\textrm{>} \; 1^n \; 0^\infty \\
& \xrightarrow{IH} &
1^{2(n+1)} \; \textrm{A}\textrm{>} \; 0^\infty \\
\end{array}</math>
# L2 Rule: <math>0^\infty \; 1^a \; \textrm{A}\textrm{>} \; 0^\infty \to 0^\infty \; 1^{2a+6} \; \textrm{A}\textrm{>} \; 0^\infty</math>
#* Proof: [A0, B0, C1, C0, D0, E(a+2), E0, D1, A(a+1)]<math display="block">\begin{array}{l}
\\
0^\infty \; 1^a \; \textrm{A}\textrm{>} \; 0^\infty
& \xrightarrow{A0, B0, C1, C0, D0} &
0^\infty \; 1^{a+2} \; \textrm{<}\textrm{E} \; 0^\infty \\
& \xrightarrow{E(a+2)} &
0^\infty \; \textrm{<}\textrm{E} \; 1^{a+2} \; 0^\infty \\
& \xrightarrow{E0,D1} &
0^\infty \; 1^2 \; \textrm{A}\textrm{>} \; 1^{a+1} \; 0^\infty \\
& \xrightarrow{A(a+1)} &
0^\infty \; 1^{2a+4} \; \textrm{A}\textrm{>} \; 0^\infty \\
\end{array}</math>

We can see now that the last rule can be applied repeatedly forever, so if it is ever applied once, the TM will never halt. In fact, in this case the start config is equal to <math>0^\infty \; 1^0 \; \textrm{A}\textrm{>} \; 0^\infty</math> and thus this TM will never halt.

== Tape Compression ==
We need some automated way to compress the tape into a tuple of integers, and decompress the tape when needed.

=== Macro Machine ===
We can divide the tape into blocks of fixed size, and merge adjacent identical blocks into one and record the number of repetitions.

Here is an example of how a half-tape is compressed using macro machine:
<pre>
111110110110101010110110 0^inf
= 111 110 110 110 101 010 110 110 000^inf (use block size 3)
= 111 110^3 101 010 110^2 000^inf
</pre>

=== Nested Repeater ===
We can compress the tape by looking for repeaters:
<pre>
101110111 0^inf
= 1 0 1 1 1 0 1 1 1 0^inf
= 1 0 1 1 1 0 1 1^2 0^inf (use a a = a^2 for a = 1)
= 1 0 1 1 1 0 1^3 0^inf (use a a^n = a^(n+1) for a = 1)
= 1 0 1 1^2 0 1^3 0^inf (use a a = a^2 for a = 1)
= 1 0 1^3 0 1^3 0^inf (use a a^n = a^(n+1) for a = 1)
= 1 (0 1^3)^2 0^inf (use a a = a^2 for a = 0 1^3)
</pre>
Decompression is simple: we only decompress the tape when the head points to a repeater:
<pre>
l X> a^(n+1) r = l X> a a^n r
l X> a^0 r = l X> r
</pre>

=== Fixed Length Repeater ===
<pre>
1101111101111110 0^inf = 1101 (11)^2 0 (11)^3 0 0^inf (length = 2)
</pre>

=== Others ===
For counters and other complex pattern, we can design some specific methods to represent them.

A typical binary counter can be represented by:
<pre>
definition of C:
C(d0,d1,dh,1) = dh 0^inf
C(d0,d1,dh,2n+0) = d0 C(d0,d1,n), n>0
C(d0,d1,dh,2n+1) = d1 C(d0,d1,n), n>0
increment rule: l qR QR> C(d0,d1,dh,n) → l <QL qL C(d0,d1,dh,n+1)
where d0,d1,dh,qL,qR are tape segments, l is half-tape, QL,QR are TM states, n is positive integer.
</pre>
For [[Bell eats counter|bell eats counters]], when the head points to C, (i.e. l X> C(_,_,_,_)), if X=QL we can decompress tape in l to see whether it's prefix matches qR, if it matches we can apply the increment rule of the counter, otherwise we decompress C by expanding the definition of C (but not recursively unless the head points to C again) so that we can run the TM further.

For [[Sync bouncer counter|sync bouncer counters]], the counter may overflow, so we can use a different representation of the binary counter:
<pre>
C'(0,0,0) = dh 0^inf
C'(2a,2b+1,0) = d0 C'(a,b,0)
C'(2a+1,2b,0) = d1 C'(a,b,0)
C'(a,b,n) is well formed iff a+b+1=2^n
increment: l qR QR> C'(a+1,b,n) → l <QL qL C'(a,b+1,n)
overflow: l qR' QR'> C'(0,b,n) → l <QL' qL' C'(2b+1,0,n+1)
where a,b,n are natural numbers, l is half-tape, QL,QR are states, and d0,d1,dh,qL,qR are tape segments.
</pre>
We can also support arithmetic sequence:
<pre>
0 1^2 0 1^3 0 1^4 0 1^5 = ((0 1^(2+1*i)) for i in range(4))
</pre>
And some simple structural counters that can be described by a single integer:
<pre>
B(1) = dh 0^inf
B(2n+0) = w^n d0 B(n)
B(2n+1) = w^n d1 B(n)
l qR QR> B(n) → l <QL qL B(n+1)
</pre>

== Find and Prove Rules ==
We need some automated way to find rules and prove/use them.

=== Find New Rule by Generalizing Known Rules ===
We can prove special rules, and generalize them by interpolation (and maybe also need to generalize their proof).

Example:
<pre>
0^inf A> 0^inf → 0^inf A> 0 0^inf
0^inf A> 0 0^inf → 0^inf 1 A> 0^inf
0^inf 1 A> 0^inf → 0^inf 1 A> 0 0^inf
0^inf 1 A> 0 0^inf → 0^inf 1 1 A> 0^inf
0^inf 1 1 A> 0^inf → 0^inf 1^2 A> 0^inf

0^inf 1^2 A> 0^inf → 0^inf 1^2 A> 0 0^inf
0^inf 1^2 A> 0 0^inf → 0^inf 1^2 1 A> 0^inf
0^inf 1^2 1 A> 0^inf → 0^inf 1^3 A> 0^inf

0^inf 1^3 A> 0^inf → 0^inf 1^3 A> 0 0^inf
0^inf 1^3 A> 0 0^inf → 0^inf 1^3 1 A> 0^inf
0^inf 1^3 1 A> 0^inf → 0^inf 1^4 A> 0^inf

interpolation:
0^inf 1^(2+i) A> 0^inf → 0^inf 1^(2+i) A> 0 0^inf
0^inf 1^(2+i) A> 0 0^inf → 0^inf 1^(2+i) 1 A> 0^inf
0^inf 1^(2+i) 1 A> 0^inf → 0^inf 1^(3+i) A> 0^inf

merge:
0^inf 1^(2+i) A> 0^inf → 0^inf 1^(3+i) A> 0^inf

induction:
0^inf 1^2 A> 0^inf → 0^inf 1^(2+n) A> 0^inf
</pre>

=== Find New Rule by Specializing Known Rules ===
We can maintain a sequence of used rules, and keep the rules as general as possible, then specialize the rules when needed.

Example:
<pre>
l A> 0^inf → l A> 0 0^inf
l A> 0 r → l 1 A> r
l A> 0^inf → l A> 0 0^inf
l A> 0 r → l 1 A> r
l 1 1 A> r → l 1^2 A> r

l A> 0^inf → l A> 0 0^inf
l A> 0 r → l 1 A> r
l 1^n 1 A> r → l 1^(n+1) A> r

l A> 0^inf → l A> 0 0^inf
l A> 0 r → l 1 A> r
l 1^n 1 A> r → l 1^(n+1) A> r

to merge
l1 A> 0^inf → l1 A> 0 0^inf
l2 A> 0 r2 → l2 1 A> r2
we need to solve equation:
l1 A> 0 0^inf = l2 A> 0 r2
then we have l1=l2, 0^inf=r2 and substitute them:
l2 A> 0^inf → l2 A> 0 0^inf
l2 A> 0 0^inf → l2 1 A> 0^inf
and then these two rules are ready to be merged as
l2 A> 0^inf → l2 1 A> 0^inf
then we can merge
l2 A> 0^inf → l2 1 A> 0^inf
l 1^n 1 A> r → l 1^(n+1) A> r
in a similar way into
l 1^n A> 0^inf → l 1^(n+1) A> 0^inf

induction:
l 1^n A> 0^inf → l 1^(n+m) A> 0^inf
</pre>
A method for solving equations (similar to unification in type theory):
<pre>
First we use ((X -> A→B) /\ (Y -> C→D)) -> (B=C -> X -> Y -> A→D) to get a rule A→D that has premises B=C,X,Y.
Then we simplify the rule (B=C -> X -> Y -> A→D) by solving equantions in the premise:
If (a,b)=(c,d) is in the premise list, we can replace it with a=c, b=d, where (x,y) can be tape segment concatenation (x y), tape segment repeation (x^y) or something similar.
If a=b is in the premise list, a is variable, we substitute all occurence of a by b in the rule.
If a*x+b=c*y+d is in the premise list, x,y are natural number variables, a,b,c,d are natural number constants, we can divide a,b,c,d by gcd(a,b,c,d) and then substitute x,y by a'*x'+b',c'*x'+d' based on some number theory methods.
For an expression of natural numbers, we can simplify it as a1*x1+a2*x2+...+an*xn+b, where a1,a2,...,an,b are natural number constants.
We can remove a=a from the premise list.

If the premise list is not empty after using these methods repeatly, we just leave it as part of the rule.
</pre>

=== Recursive Record-Breaking Analysis ===
We can run a Turing machine on a tape segment until it leaves the tape segment.

In this process, every time the Turing machine visits a new position, we recursively consider when the Turing machine will leave the positions it has already visited (i.e. visit the next new position or leave the current tape segment).

We can use memoize search in this process for acceleration, and regard the process of leaving a tape segment as a rule. Then the rule sequence (or rule dependency graph) may be generalized to new induction rules.
[[Category:Deciders]]

Hydra function

2025-10-08T07:22:07Z

Int-y1: math format

[[category:Functions]]
[[File:Hydra Spiral.png|thumb|185px|A spiral-like figure that gives the first few terms of the Hydra sequences with initial values 2, 5, 8, 11, 14, and 17.]]
The '''Hydra function''' is a [[Collatz-like]] function defined as:
<math display="block">\textstyle H(n)\equiv n+\big\lfloor\frac{1}{2}n\big\rfloor=\Big\lfloor\frac{3}{2}n\Big\rfloor=\begin{cases}
\frac{3n}{2}&\text{if }n\equiv0\pmod{2},\\
\frac{3n-1}{2}&\text{if }n\equiv1\pmod{2}.
\end{cases}</math>
It is named as such because of its connection to the unsolved halting problems for the [[Cryptids]] [[Hydra]] and [[Antihydra]]. Due to its simplicity, simulations for both of these [[Turing machines]] utilize this function instead of what can initially be proven.
== Relationship to Hydra and Antihydra problems==
Using the Hydra function, we can obtain simplified rules for Hydra and Antihydra:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|Let <math>C_H(a,b):=0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)}\;3^b\;2\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{20}&C_H(3,0),\\
C_H(2a,0)&\xrightarrow{54a^2-48a-2}&0^\infty\;3^{9a-8}\;1\;\textrm{A}\textrm{>}\;2\;0^\infty,\\
C_H(2a,b+1)&\xrightarrow{54a^2-39a-5}&C_H(3a,b),\\
C_H(2a+1,b)&\xrightarrow{4b+54a^2-3a+4}&C_H(3a+1,b+2).\\\hline
\end{array}</math>
|Let <math>A_H(a,b):=0^\infty\;1^a\;0\;1^{b-4}\;\textrm{E}\textrm{>}\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{11}&A_H(0,8),\\
A_H(a,2b)& \xrightarrow{2a+3b^2-1}& A_H(a+2,3b),\\
A_H(0,2b+1)&\xrightarrow{3b^2-3b-7}& 0^\infty\;\textrm{<}\textrm{F}\;110\;1^{3b-6}\;0^\infty,\\
A_H(a+1,2b+1)&\xrightarrow{3b^2-7}& A_H(a,3b+1).\\\hline
\end{array}</math>
|}
<div class="toccolours mw-collapsible mw-collapsed">'''Proof'''<div class="mw-collapsible-content">
Recall the high-level rules for Hydra and Antihydra:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|Let <math>C(a,b):=0^\infty\;\textrm{<}\textrm{A}\;2\;0^a\;3^b\;2\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{20}&C(3,0),\\
C(2a,0)&\xrightarrow{6a^2+20a+4}&0^\infty\;3^{3a+1}\;1\;\textrm{A}\textrm{>}\;2\;0^\infty,\\
C(2a,b+1)&\xrightarrow{6a^2+23a+10}&C(3a+3,b),\\
C(2a+1,b)&\xrightarrow{4b+6a^2+23a+26}&C(3a+3,b+2).\\\hline
\end{array}</math>
|Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E}\textrm{>}\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
0^\infty\;\textrm{A}\textrm{>}\;0^\infty&\xrightarrow{11}&A(0,4),\\
A(a,2b)& \xrightarrow{2a+3b^2+12b+11}& A(a+2,3b+2),\\
A(0,2b+1)&\xrightarrow{3b^2+9b-1}& 0^\infty\;\textrm{<}\textrm{F}\;110\;1^{3b}\;0^\infty,\\
A(a+1,2b+1)&\xrightarrow{3b^2+12b+5}& A(a,3b+3).\\\hline
\end{array}</math>
|}
Already, both machines appear to be very similar. They have one parameter that increases exponentially with growth factor <math display="inline">\frac{3}{2}</math> and another that takes a pseudo-random walk. Below, the exponentially increasing variables are described by integer sequences:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">a_0=3,a_{n+1}=\begin{cases}\frac{3a_n+6}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{3a_n+3}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|<math display="block">a_0=4,a_{n+1}=\begin{cases}\frac{3a_n+4}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{3a_n+3}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|}
This will make demonstrating the transformation easier. Now we will define a new integer sequence based on the old one and discover the recursive rules for that sequence. This new sequence is <math display="inline">b_n=\frac{1}{3}a_n+2</math> and <math>b_n=a_n+4</math> for Hydra and Antihydra respectively. We start by using <math>b_{n+1}</math> instead and substituting <math>a_{n+1}</math> for its recursive formula. By doing so, we get:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">b_0=3,b_{n+1}=\begin{cases}\frac{a_n+6}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{a_n+5}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|<math display="block">b_0=8,b_{n+1}=\begin{cases}\frac{3a_n+12}{2}&\text{if }a_n\equiv0\pmod{2}\\\frac{3a_n+11}{2}&\text{if }a_n\equiv1\pmod{2}\end{cases}</math>
|}
After that, we can substitute <math>a_n</math> for its solution in terms of <math>b_n</math>. What results is the following:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">b_0=3,b_{n+1}=\begin{cases}\frac{3(b_n-2)+6}{2}&\text{if }3(b_n-2)\equiv0\pmod{2}\\\frac{3(b_n-2)+5}{2}&\text{if }3(b_n-2)\equiv1\pmod{2}\end{cases}</math>
|<math display="block">b_0=8,b_{n+1}=\begin{cases}\frac{3(b_n-4)+12}{2}&\text{if }b_n-4\equiv0\pmod{2}\\\frac{3(b_n-4)+11}{2}&\text{if }b_n-4\equiv1\pmod{2}\end{cases}</math>
|}
The <math>\text{if}</math> statements amount to checking if <math>b_n</math> is even or odd. After simplifying, we are done:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|<math display="block">b_0=3,b_{n+1}=\begin{cases}\frac{3b_n}{2}&\text{if }b_n\equiv0\pmod{2}\\\frac{3b_n-1}{2}&\text{if }b_n\equiv1\pmod{2}\end{cases}</math>
|<math display="block">b_0=8,b_{n+1}=\begin{cases}\frac{3b_n}{2}&\text{if }b_n\equiv0\pmod{2}\\\frac{3b_n-1}{2}&\text{if }b_n\equiv1\pmod{2}\end{cases}</math>
|}
Now that we have demonstrated a strong similarity in the behaviour of both Turing machines, we can return to using the high-level rules. Doing that while considering the step counts yields the final result.
</div></div>
Under these rules, the halting problem for Hydra is about whether repeatedly applying the function <math>H(n)</math>, starting with <math>n=3</math>, will eventually generate more even terms than twice the number of odd terms. Similarly, Antihydra halts if and only if repeatedly applying <math>H(n)</math>, starting with <math>n=8</math>, will eventually generate more odd terms than twice the number of even terms.

=== Coding the Hydra and Antihydra problems using the Hydra function ===
Paired with the corresponding even/odd criterion as loop halting condition (implemented as a counter variable) and initial Hydra function value, the Hydra function definition can be used to write computer programs that simulate the abstracted behavior of the Hydra and Antihydra Turing machines. The following Python program is a Hydra simulator:
<syntaxhighlight lang="python" line="1">
# 'a' and 'b' fulfill the same purpose as in the Hydra rules:
# The current hydra function value
a = 3
# The even/odd condition counter
b = 0
# As long as Hydra has not halted, 'b' remains greater than -1.
while b != -1:
# If 'a' is even, decrement 'b', otherwise increase 'b' by 2.
if a % 2 == 0:
b -= 1
else:
b += 2
# This performs one step of the Hydra function H(a) = a + floor(a/2).
# Note that integer division by 2 is equivalent to one bit shift to the right (a >> 1)
a += a//2
</syntaxhighlight>
Replacing <code>a = 3</code> with <code>a = 8</code> and swapping <code>b -= 1</code> with <code>b += 2</code> turns this program into an Antihydra simulator.

Determining whether these programs halt or not (and if so, after how many loop iterations) would resolve these open problems.
==Properties==
The Hydra function can be rewritten as follows:
<math display="block">\begin{array}{l}
H(2n)&=&3n,\\
H(2n+1)&=&3n+1.\\
\end{array}</math>
Now assume that for some positive integer <math>s</math> and every odd integer <math>t</math>, <math>H^s(2^st)=3^st</math> and <math>H^s(2^st+1)=3^st+1</math>, where <math>H^i(n)</math> is function iteration. Notice that we can write <math>2^{s+1}t=2\cdot2^st</math> and <math>2^{s+1}t+1=2\cdot2^st+1</math>, so if we apply <math>H</math> to these numbers, we get <math>H(2\cdot2^st)=3\cdot 2^st</math> and <math>H(2\cdot2^st+1)=3\cdot2^st+1</math>. Now, if we apply <math>H</math> to these numbers <math>s</math> times, we get <math>H^{s+1}\big(2^{s+1}t\big)=H^s(2^s\cdot3t)=3^{s+1}t</math> and <math>H^{s+1}\big(2^{s+1}t+1\big)=H^s(2^s\cdot3t+1)=3^{s+1}t+1</math>. Therefore, by mathematical induction we have proved the following formulas:
<math display="block">\begin{array}{l}
H^s(2^st)&=&3^st,\\
H^s(2^st+1)&=&3^st+1.\\
\end{array}</math>
This optimization can be directly applied to the high-level rules for Hydra and Antihydra, producing this result:
{|class="wikitable"
! Hydra !! Antihydra
|-align="center"
|Let <math>C_H(a,b):=0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)}\;3^b\;2\;0^\infty</math>:
<math display="block">\begin{array}{|lll|}\hline
C_H(2^st,b+s)&\xrightarrow{f_1(s,t)}&C_H(3^st,b),\\
C_H(2^st+1,b)&\xrightarrow{f_2(s,t,b)}&C_H(3^st+1,b+2s),\\\hline
\end{array}</math>
where <math display="inline">f_1(s,t)=\frac{3t(3^s-2^s)(18(3^s+2^s)t-65)}{5}-5s</math> and <math display="inline">f_2(s,t,b)=(b+s)4s+\frac{3t(3^s-2^s)(18(3^s+2^s)t-5)}{5}</math>.
|Let <math>A_H(a,b):=0^\infty\;1^a\;0\;1^{b-4}\;\textrm{E}\textrm{>}\;0^\infty</math>:<math display="block">\begin{array}{|lll|}\hline
A_H(a,2^st)& \xrightarrow{f_3(s,t,a)}& A_H(a+2s,3^st),\\
A_H(a+s,2^st+1)&\xrightarrow{f_4(s,t)}& A_H(a,3^st+1),\\\hline
\end{array}</math>
where <math display="inline">f_3(s,t,a)=(2a-3+2s)s+\frac{3t^2(9^s-4^s)}{5}</math> and <math display="inline">f_4(s,t)=\frac{3t^2(9^s-4^s)}{5}-7s</math>.
|}
== Visualizations ==
The four images below depict the first 1000 values of four Hydra sequences with different initial values. Each row of pixels shows a number in binary on the right and its parity on the left (blue for even, red for odd):
<gallery mode="packed" heights="250">
File:HydraFunction-StartingValue2.png|Starting value 2. There are 492 even numbers and 508 odd numbers.
File:HydraFunction-StartingValue5.png|Starting value 5. There are 497 even numbers and 503 odd numbers.
File:Antihydra increasing value.png|Starting value 8. There are 499 even numbers and 501 odd numbers.
File:HydraFunction-StartingValue11.png|Starting value 11. There are 481 even numbers and 519 odd numbers.
</gallery>

1RB1RE 1LC0RA 0RD1LB ---1RC 1LF1RE 0LB0LE

2025-09-30T05:53:32Z

Int-y1: move info

{{machine|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}{{unsolved|Does this TM run forever?}}
{{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}, called '''[[Beaver Math Olympiad|BMO]] #1''', is a [[holdout]] [[BB(6)]] TM discovered by @-d on 25 Jun 2024 which is modeled by

<pre>
A(a, b) -> A(a-b, 4b+2) if a > b
A(a, b) -> A(2a+1, b-a) if a < b
A(a, b) -> Halt if a = b

Start A(1, 2)
</pre>

for <code>A(a, b) = 0^inf 10^{a-1} 0 1^b E> 0^inf</code>. The configuration is always valid because <math>a \ge 1</math> is always maintained. Forward simulations of this model by @-d and Andrew Ducharme do not halt after 100,000,000 iterations of the map <code>A(a,b)</code>. It is the inspiration for the [[Beaver Math Olympiad]].

== Relation to Other Math Problems ==
This TM is a possible [[Cryptid]] since it seems hard to predict whether we could ever end up in <code>A(n, n)</code>. However, it is not (yet) classified as a Cryptid because there are no arguments for why this TM is hard (e.g., by comparing it to an open/unsolved/hard mathematical problem).

The map <code>A(., .)</code> is related to the study of discontinuous [https://en.wikipedia.org/wiki/Dynamical_system dynamical systems]. When +2 and +1 are ignored, the value of <math>\ln(a/b)</math> follows this chaotic map:

<math display="block">F(x) = \begin{cases}
\ln\left(\frac{e^x-1}{4}\right) & \text{if }x > 0 \\
\ln\left(\frac{2}{e^{-x}-1}\right) & \text{if }x < 0
\end{cases}</math>

Another potential BB(6) cryptid {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}} similarly iterates upon a tuple (x,y) with one function if x < y, another function if x > y, and halts if x = y.

This Turing Machine was the inspiration for the description of Affine Inequality Machines. It was proven that there is no general method to prove an arbitrary Affine Inequality Machine halts, [https://discord.com/channels/960643023006490684/1239205785913790465/1420457986564030641 initially] for 3 dimensions or higher, and [https://discord.com/channels/960643023006490684/1239205785913790465/1420490366980849785 shortly after] for the 2 dimensional case.

== A Backward Reasoning Approach ==
An alternative approach is to work backwards and study the points which A(a,b) maps to the line a = b. For example, the points (2,7), (3,25), (4,71), and (5,181) all eventually map to a = b. (2,7) is immediately sent to a = b because, under the a < b rule, <math display="block">A(2, 7) = (4 + 1, 7 - 2) = (5,5)</math>(5,181) requires more steps:

<math display="block">(5,181) \rightarrow (11, 176) \rightarrow (23, 165) \rightarrow (47, 142) \rightarrow (95, 95)</math>How are these points found? It's faster not to consider individual points, but as Racheline identified, the set of lines which are mapped to a = b. Introducing an index <math>n</math>, we can rewrite A(a,b) as<math display="block">(x_{n+1}, y_{n+1}) = \begin{cases} (x_n - y_n, 4y_n + 2), & \text{if } x_n \geq y_n \\ (2x_n + 1, y_n - x_n), & \text{if } x_n < y_n \end{cases}</math> If <math>x_{n+1} = y_{n+1}</math>{{machine|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE}}, then the lines with slopes <math>y_{n+1} = m x_{n+1} + b</math> map to that line. The two possible slopes are, for <math>x_n > y_n</math>and <math>x_n < y_n</math> respectively,

<math>4y_n + 2 = m (x_n - y_n) + b \iff y_n = \frac{m}{m + 4} x_n + \frac{b - 2}{m + 4}

</math> and<math>y_n - x_n = m (2 x_n + 1) + b \iff y_n = (2m + 1) x_n + (b + m).</math>

Indexing m and b in (slope, y-intercept) space gives

<math>(m_{k-1}, b_{k-1}) = \begin{cases}\left( \frac{m_k}{m_k + 4}, \frac{b_k - 2}{m_k + 4} \right), &x_{k-1} > y_{k-1} \\ (2m_k + 1, b_k + m_k), & x_{k-1} < y_{k-1} \end{cases}</math>.

Iteratively applying this (m,b) map finds an infinite set of points which A sends to a = b. The starting point is said line, with (m, b) = (1,0). Then the two lines of points which map to points on the line (m,b) = (1,0) under a single application A are have slopes and y-intercepts of <math>\left(\frac{1}{5}, -\frac{2}{5} \right) = \left(\frac{1}{1 + 4}, \frac{0-2}{1+4} \right) </math> and (3,1) = (2 + 1, 0 + 1). The four lines of points which map to (1,0) under two applications of A have slopes and y-intercepts <math>\left(\frac{1}{21}, -\frac{4}{7} \right), \left(\frac{7}{5}, -\frac{1}{5} \right), \left(\frac{3}{7}, -\frac{1}{7} \right)</math> and (7,4).
[[File:Slope intercept space after 5 iterations.png|thumb|Each point in the plotted (slope, intercept) space represents a line of points in (x,y) space that are eventually mapped to y = x by iterated application of the operation A. These points require five or fewer iterations to do so.]]
[[File:(m,b) space after 9 iterations with halting condition line.png|thumb|Each plotted point in (slope, intercept) space represents a line of points in (x,y) space which is mapped to y = x by nine or fewer iterated applications of A. If any point is on the black line, the TM in question halts.]]
To apply this to the problem at hand, we want to know if any of these generated lines contain the point (1,2), our TM-derived initial condition. The set of all lines (in position space, or (x,y) space) containing (1,2) are described by the line in (m,b) space 2 = m + b. If any of the generated points in (m,b) space sit on 2 = m + b, then the point (x,y) = (1,2) is eventually mapped by A to a point where x = y. In the second image, of (m,b) points after nine iterations, we see there are three possible locations for this to occur.
[[File:MbSpace iterations10 smallmb wline.png|center|thumb|575x575px|Each plotted point in (slope, intercept) space represents a line of points in (x,y) space which is mapped to y = x by ten or fewer iterated applications of A, with m > 2 and b > 0. If any point is on the black line, the TM in question halts. There are three possible locations. Two do not hit the line (in finite time). The third's status is unknown.]]

=== Location 1: Approaching (m = 1, b= 1) from below ===
There's two ways to see how (1,1) is not actually an intersection point. First, from Racheline, the elementary one: if m > b, m will always be greater than b, so a point where m = b is impossible.

Second, from Andrew, if one wants to generate a point near m = 1, you should create the largest possible value M to swamp out the additive +4 in the denominator of M/(M+4). M can be maximized with n iterations of the (m,b) map by only using the x < y operation. Then apply the x > y operation once for a near-unitary slope. The pair <math>(m_n, b_n)</math> formed by applying the x < y operation n times to (1,0) is <math>(2^{n+1}-1, 2^{n+1} -n-2)</math>. Apply the x > y operation to get

<math>\left( \frac{2^{n+1} - 1}{2^{n+1} + 3}, \frac{2^{n+1} -n-4}{2^{n+1}+3} \right)</math>.

Taking the limit as n goes to infinity, we get (1,1), so this TM does not reach the line 2 = m + b here in finite time.

=== Location 2: Approaching (m = 13/7, b = 4/21) from above ===
The rightmost possible intersection is the least likely: its approach towards 2 = m + b slows considerably with increasing iterations (because its limit will end up being to above this line). Empirically examining these points, they appear to be near (1.85, 0.19). After 6 through 9 iterations, the closest of the generated (m,b) points to (1.85, 0.19) are (285/151, 37/151), (1117/599, 123/599), (4445/2391, 155/797), and (17757/9559, 1831/9559). Such points are formed by n applications of the (m,b) map when x > y, then two applications of the (m,b) map when x < y, then a single additional application of the x > y and x < y (m,b) maps, in that order. A more concise notation would be <math>(>^n, <^2, >, <)</math>.

A closed form equation in n for <math>(>^n, <^2, >, <)</math> starting from (1,0) is <math>\left( \frac{13 * 4^{n+1} + 23}{7* 4^{n+1} + 5}, \frac{4^{n+2} + 6n + 65}{21 * 4^{n+1} + 15} \right)</math>. Its limit as <math>n \rightarrow \infty</math> is <math>\left( \frac{13}{7}, \frac{4}{21} \right)</math>, which is safely above 2 = m + b. This strand of points will never cause a halting condition.

Racheline showed the same by applying the bounds m > 0 and <math>b > - \frac{2}{3}</math> on the sequence <math>(<^2, >, <)</math>:

<math>\begin{align} m_0 > 0, b_0 > - \frac{2}{3} &\implies m_1 > 1, b_1 > -\frac{2}{3} \\
&\implies m_2 > 3, b_2 > \frac{1}{3} \\
&\implies m_3 > \frac{3}{7}, b_3 > -\frac{5}{21} \\
&\implies m_4 > \frac{13}{7}, b_4 > \frac{4}{21}
\end{align}</math>

Since the combination of operations is empirically determined, this isn't strictly a proof, but the moderate distance of these points from 2 = m + b means the burden of proof isn't that high.

=== Location 3: Jumping above and below b = 2 - m around m = 1.76...infinitely? ===
Notes from my (Andrew) analysis, from when I tried to repeat looking at the points closest to the not-quite-intersection of the sparse set and line b = 2 - m. I went 25 iterations of the slope map before realizing it was likely a fractal and zooming in would give me more of the same:
<u>Slopes of points (m,b) closest to the line m = 2 - b from above and below</u>

13/7 for n = 5 is above and to the right, 119/69 for n = 6 is below and to the left, 223/125 for n = 7 is above and to the right, 669/401 for n = 8 is below left.
Things start to get more complicated at n = 9, as the doubling of dots becomes a little overwhelming.
So many dots at n = 13 you have to zoom in
n = 9 above: 223/125, n = 9 below: 1145/651
n = 10 above: 2037/1151, n = 10 below: 1145/651 (still)
n = 11 above: 6617/3755, n = 11 below: 1145/651 (still)
n = 12 above: 6617/3755 (still), n = 12 below: 24937/14171
n = 13 above: 6617/3755 (still), n = 13 below: 38171/21681
n = 14 above: 6617/3755 (still), n = 14 below: 64639/36701
n = 15 above: 117575/66741, n = 15 below: 64639/36701 (still)
n = 16 above: 117575/66741 (still), n = 16 below: 376131/213545
n = 17 above: 611281/347027, n = 17 below: 376131/213545 (still).
b for the new point here is 2 / 347027 from being on the line!!
At n = 17, have we crossed the Rubicon? This super close point is just above the line, so is anything new now above it?
No! yowza.
n = 18 above: 611281/347027 (still), n = 18 below: 611281/347027.
n = 19 above: 611281/347027 (still), n = 19 below: 2115805/1201207
n = 20 above: 611281/347027 (still), n = 20 below: 3338367/1895261
n = 21 above:611281/347027 (still) , n = 21 below: 10673739/6059585
There's a new point next to our hero which is just to the right and just above. Above trend hitting limit point?
NO!!
n = 22 above: 208917121/118603171, n = 22 below: 20454235/11612017 I might be seeing a new trend line forming
n = 23 above: 834868865/473958819, n = 23 below: 40015227/22716881
n = 24 above: 3338675841/1895381411, n = 24 below: 40015227/22716881 (still)
n = 25 above: 79137211/44926609, n = 25 below: 239198119/135794133

I think I'm zooming in on a fractal structure and will never see the end of this.
[[File:Screenshot of (m,b) set as we zoom in on possible intersection.png|center|thumb|636x636px|To demonstrate the fractal, the generated (m,b) set after n iterations for n = 9, n = 15, n = 20, and n = 25. Note the change in scale on the horizontal and vertical axes. I'm roughly specifying another decimal place in the axes bounds as I look at smaller and smaller region. Largely a copy of post @aducharme [https://discord.com/channels/960643023006490684/1026577255754903572/1267352007811797012 July 27th, 2024 10:24 PM PT]]]
Unless we actually hit the halting condition, we can generate an infinite number of blue points arbitrarily close to the yellow line from both above and below.

It does appear that points within roughly the range 1.724 < m < 1.8556 are generated from finding all possible combinations of operations > and < after n applications, then applying <math>(<,<,<,>,<,>,<) = (<^4, (>,<)^2) = (<^4, >,<,>,<)</math>. This is equivalent to <math display="block">(m_f,b_f) = \left( \frac{104m_n + 119}{56 m_n + 69}, \frac{23m_n + b_n + 12}{56m_n + 69} \right).</math>

== Study of Simplified Model ==
@savask [https://discord.com/channels/960643023006490684/1026577255754903572/1255778174986485780 06/26/2024 11:54 PM]
Maybe we could tackle a simplified version of this problem? For example, let's remove the constant terms:

(a, b) -> (a-b, 4*b) if a > b
(a, b) -> (2*a, b-a) if b > a
(a, b) -> Halt if a = b

And let's start from config (3,17), since it appears in the run of the original machine.
The average increase in <math>\ln a</math> is

<math>\frac{-\frac{\pi^2}{12} + 3 \frac{(\ln 2)^2 }{2} + \text{Li}_2(1/4)}{\ln 3} \approx 0.15097694</math>

when modeled from the PDF

<math>f(r) = \frac{2}{\ln 3 (r+1)(r+3)}</math>

in <math>r = b/a</math>, as found by [https://discord.com/channels/960643023006490684/1026577255754903572/1256315391995809842 @LegionMammal978] and Math.SE user "Sil".<ref>https://math.stackexchange.com/questions/4939082/solving-the-functional-equation-fx-frac4x42f-frac-xx42f2x1/4939136#4939136</ref>

Note that the simplified iteration trivially cannot halt as written: after 2 steps, it reaches <math>(12, 8)</math> and becomes trapped within the families <math>(a, b) \equiv (0, 4)</math> and <math>(a, b) \equiv (4, 0) \pmod{8}</math>.

== Development of Model from TM Tape ==
@-d [https://discord.com/channels/960643023006490684/1026577255754903572/1255047256688824390 25 Jun 2024 2:29 AM ET]:
<pre>
Let f(a,b) represent (10)^a D> 1^b. By inspection, the machine visits f(0,3) f(2,2) f(1,7) f(4,5) f(0,19) f(2,18) f(6,15) f(14,8).

I got these rules for f(a,b) and verified them up to 0<=a<=100 and 1<=b<=100.
f(a,b) -> f(a-b+1,4b-1) if b < a+2
f(a,a+2) -> halt
f(a,b) -> f(2a+2,b-a-1) if b > a+2

I ran these rules from f(0,3) and gave up after reaching a,b > 2^1000000. It looks like halting becomes less and less likely. Is there a way to show this runs forever (or miraculously halts)?
</pre>

@Shawn Ligocki — [https://discord.com/channels/960643023006490684/1026577255754903572/1255371132694167591 25 Jun 2024 at 11:56 PM ET]:
<pre>
I can confirm roughly the same rules (I chose a slightly different standard config):
1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE

E(a, b, c) = 0^inf 10^a 0 10^b 1^c E> 0^inf

Base Rules:
E(a+1, b, c+2) -> E(a, b+2, c+1)
E(a+1, b, 1) -> E(a, 0, 2b+6)
E(0, b, c+2) -> E(b+2, 0, c+1)
E(0, b, 1) -> Halt(b+4)

High-level Rules:
E(a, 0, c+1) --> E(a-c-1, 0, 4c+6) if a > c
E(a, 0, c+1) --> E(2a+2, 0, c-a) if a < c
E(a, 0, c+1) --> Halt(2c+4) if a = c

Connecting to your notation @-d : f(a, b+2) -> E(a, 0, b+1)
</pre>

@Shawn Ligocki — 26 Jun 2024 at 12:25 AM
<pre>
Hm, this is interesting. I decided to track "spread" akin to what @savask and I have been looking at for the similar 3x3 TM we're looking at in ⁠bbxy . Here I define spread = abs(a - c) in config E1(a, c) = E(a, 0, c+1). Here's the results I have so far:
1 E1(2, 0) spread: [2, 2] (0.06s)
100_001 E1(10^6_557, 10^6_555) spread: [1, 10^6_557] (0.44s)
200_001 E1(10^13_124, 10^13_125) spread: [10^6_557, 10^13_125] (1.27s)
300_001 E1(10^19_670, 10^19_669) spread: [10^13_123, 10^19_670] (2.52s)
400_001 E1(10^26_235, 10^26_237) spread: [10^19_668, 10^26_237] (4.23s)
500_001 E1(10^32_799, 10^32_799) spread: [10^26_235, 10^32_799] (6.39s)
600_001 E1(10^39_350, 10^39_352) spread: [10^32_799, 10^39_352] (8.98s)
700_001 E1(10^45_930, 10^45_930) spread: [10^39_350, 10^45_930] (12.06s)
800_001 E1(10^52_486, 10^52_486) spread: [10^45_929, 10^52_486] (15.54s)
900_001 E1(10^59_046, 10^59_047) spread: [10^52_485, 10^59_047] (19.57s)
1_000_001 E1(10^65_613, 10^65_612) spread: [10^59_045, 10^65_613] (24.05s)
1_100_001 E1(10^72_154, 10^72_153) spread: [10^65_609, 10^72_154] (28.92s)
1_200_001 E1(10^78_728, 10^78_729) spread: [10^72_153, 10^78_729] (34.21s)
1_300_001 E1(10^85_288, 10^85_289) spread: [10^78_729, 10^85_289] (40.03s)
1_400_001 E1(10^91_830, 10^91_829) spread: [10^85_288, 10^91_829] (46.40s)
1_500_001 E1(10^98_394, 10^98_393) spread: [10^91_829, 10^98_394] (53.11s)
1_600_001 E1(10^104_961, 10^104_959) spread: [10^98_393, 10^104_961] (60.18s)
1_700_001 E1(10^111_523, 10^111_523) spread: [10^104_958, 10^111_523] (67.75s)
1_800_001 E1(10^118_088, 10^118_088) spread: [10^111_523, 10^118_088] (75.77s)
1_900_001 E1(10^124_631, 10^124_630) spread: [10^118_088, 10^124_631] (84.39s)
2_000_001 E1(10^131_185, 10^131_185) spread: [10^124_630, 10^131_185] (93.41s)
...
3_000_001 E1(10^196_733, 10^196_733) spread: [10^190_163, 10^196_733] (217.87s)

Here spread: [X, Y] means that the spread values (since last print) had min X, max Y
So notice that the spread is basically uniformly increasing. In fact these [X, Y] intervals hardly even overlap!
That makes me think there's something about this that makes small spread impossible (and thus halt impossible)
Or that it's like Bigfoot and it ... could go back to zero, but the chances are infinitesimal!
</pre>

savask — 26 Jun 2024 at 1:02 AM
<pre>
Maybe we can simplify a little bit more. Set (a, c) = E(a, 0, c+1) then the rules are
(a, c) -> (a-c-1, 4c+5) if a > c
(a, c) -> (2a+2, c-a-1) if a < c
(a, c) -> Halt if a = c

I like that this way they are more symmetrical with a-c-1 vs c-a-1
</pre>

Shawn Ligocki — 26 Jun 2024 at 2:33 AM
<pre>
FWIW, there is a value x ~= 1.76 such that as c -> inf: (xc, c) -> (xyc, yc) via path "rll" (right->0 [ie a > c], left->0 [ie a < c], left->0). Which would repeat forever if it were exact, but it is an unstable cycle so any slight deviation and it gets further and further away, so that doesn't seem like a reasonable proof direction :/
</pre>dyuan01 — 26 Jun 2024 at 2:32 PM
If we let A(a, c) = (a-1, c-1), we can get these rules:
```
A(a, c) -> A(a-c, 4c+2) if a > c
A(a, c) -> A(2a+1, c-a) if a < c
A(a, c) -> Halt if a = c
```
And we start with (1, 2)

I'm not sure if it even helps, but at least the rules look slightly more "human-manageable"

[[Category:BB(6)]]

1RB1RA 1LC1LE 1RE0LD 1LC0LF 1RZ0RA 0RA0LB

2025-09-28T07:38:08Z

Int-y1: start on different state

{{Machine|1RB1RA_1LC1LE_1RE0LD_1LC0LF_---0RA_0RA0LB}}
{{TM|1RB1RA_1LC1LE_1RE0LD_1LC0LF_1RZ0RA_0RA0LB|halt}} is a tetrational halting [[BB(6)]] TM first discovered and analyzed by @poppuncher. It was shared on Discord on 5 Jun 2025 ([https://discord.com/channels/960643023006490684/1380384286942822561 Discord Link]). It is a translated counter that halts with a final sigma score of roughly <math>10 \uparrow \uparrow 6.96745</math>.

As of Sep 21 2025, this TM does not have a Rocq proof.

== Analysis by @poppuncher ==
Most of the initial analysis was done on its sister machine {{TM|1RB0RF_1LC0RA_1RZ0LD_1LE1LD_1RB1RC_0LD0RE|halt}} which has identical behavior but with a different starting state. The sister machine halts with a smaller, though still enormous, sigma score of roughly <math>10 \uparrow \uparrow 5.77573</math>. The following explanation describes this sister machine, but aside from the starting conditions and states, it applies to the larger machine as well.

<pre>
basically, there is a unary counter on one side and a weird binary counter on the other side.

the weird binary counter behaves as if counting the number of bit flips required to increment a binary number (let's call it A) of a given length n, until it reaches 2^n. the number of flips are written to the unary counter, which then resets, and the process starts again with a new binary string.

For example, let n = 6 and A = 48, which is 110000 in binary, then it will count up to 1000000. there is a formula for the number of bit flips required, which is 2^(n+1) - 2*A - 1 + pop_count(A), where pop_count is the number of '1's in the binary representation of A. In this example
it evaluates to 2^7-2*48-1+2 = 33.

Note that leading zeroes are allowed, so if n = 9 and A = 48, then
we are counting 000110000 until it reaches 1000000000, and the count will be 2^10-2*48-1+2 = 417.

now for the halting part, i can prove that the following rules:
A> 0 1^(5n+0) -> (10 110)^n <D 01
A> 0 1^(5n+1) -> (10 110)^n 10 <D 01
A> 0 1^(5n+2) -> (10 110)^n 110 <D 01
A> 0 1^(5n+3) -> (10 110)^n 10 10 10 <D 01
A> 0 1^(5n+4) -> Halt(3n+3)

the binary counter encodes the binary string such that "10" on the tape corresponds to 0, and "110" corresponds to 1, (also the most significant bit is on the right, so we need to flip it)

for example at t = 159, the tape is 0^inf 10 10 10 10 110 110 <D 01 0^inf, which corresponds to 110000 (this is also the example earlier, with n = 6 and A = 48).
with this in mind, i got the following rules:
A(n) is the numerical value of the binary string, L(n) is the length of the string, P(n) is the count of '1's in the string, and R(n) is the amount of increments made on the unary counter, counting the number of bit flips before it resets
initial conditions: A(1) = 0, L(1) = 2
P(n) := pop_count(A(n))
R(n) := 2^(1 + L(n)) - 2*A(n) - 1 + P(n)
k(n) := R(n)//5 (division without remainder)
r(n) := R(n) % 5, (so R(n) = 5*k(n) + r(n))

then we have P(1) = 0, R(1) = 7;

the rules:
if R(n) mod 5 == 4 then halt
otherwise
let a(r) = {0, 1, 1, 3} (for 0 <= r < 4)
then
L(n+1) = L(n) + 1 + 2 * k(n) + a(r(n))
A(n+1) = 2 ^ (2 + L(n)) * ( 2^(2*k(n)) - 1 ) / 3 + (if r(n) = 2 then 2 ^ ( L(n)+1 + 2*k(n) ) otherwise 0)
P(n+1) = pop_count(A(n+1)) = k(n) + (if r(n) = 2 then 1 otherwise 0)
</pre>

== Analysis by [[User:isokate|Katelyn Doucette]]==
The machine follows these rules:

<pre>
Let s = 2^{b+2} - 2 + 2^{b+2}((4^k - 1)/3) + k

START = A(1, 3)

A(5k + 0, b) -> A(s + 1, b + 2k + 1)

A(5k + 1, b) -> A(s + 1 + 2^{(b+2k+1)+1}, b + 2k + 2)

A(5k + 2, b) -> A(s + 2, b + 2k + 2)

A(5k + 3, b) -> A(s + 1 + 2^{b+2k+2}(7), b + 2k + 4)

A(5k + 4, b) -> Halt(b + 3k + 4)
</pre>

And runs through seven of these resets before reaching <math>5k + 4</math> and halting.

These configurations are identifiable as the transition of this form:
<code>1^(5k+r) 01 <B (01)^b</code>

== Evaluation==
Both @poppuncher and [[User:isokate|Katelyn Doucette]]'s rules were evaluated on both machines and line up with one another providing high confidence that the analyses and final sigma scores are correct.

@poppuncher's rules were evaluated by [[User:sligocki|Shawn Ligocki]].

[[Category:BB(6)]]

1RB0RF 1LC0RA ---0LD 1LE1LD 1RB1RC 0LD0RE

2025-09-28T07:36:20Z

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1RB0RF 1LC0RA 1RZ0LD 1LE1LD 1RB1RC 0LD0RE

2025-09-28T07:35:17Z

Int-y1: Redirected page to 1RB1RA 1LC1LE 1RE0LD 1LC0LF 1RZ0RA 0RA0LB

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1RB1RA 1LC1LE 1RE0LD 1LC0LF ---0RA 0RA0LB

2025-09-28T07:07:51Z

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1RB--- 0RC0RA 1RD1RC 1LE1LF 0LB0LD 0LE0RC

2025-09-28T07:02:53Z

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1RB1RE 0RC0RA 0LD0LF 1LA1LD 0RB0LD 1LC---

2025-09-28T07:02:37Z

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1RB1RA 1LC1LE 0LD0LB 0RA0RF 0LC0RA 1RD---

2025-09-28T07:01:41Z

Int-y1: similar tm's

{{machine|1RB1RA_1LC1LE_0LD0LB_0RA0RF_0LC0RA_1RD---}}
{{TM|1RB1RA_1LC1LE_0LD0LB_0RA0RF_0LC0RA_1RD---}} is a [[BB(6)]] Turing machine, posted by mxdys on 26 July 2024.<ref>https://discord.com/channels/960643023006490684/1239205785913790465/1266428071817379862</ref> Racheline originally suggested that it "maybe pentational BB(6) cryptid".<ref>https://discord.com/channels/960643023006490684/1239205785913790465/1267546343816171565</ref> But after further analysis, it does not appear to have any clear path to halting, so it does not yet qualify as a [[Cryptid]].

When the TM starts on these states, it exhibits similar behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB1RA_1LC1LE_0LD0LB_0RA0RF_0LC0RA_1RD---}}.
* When the starting state is B, the TNF is {{TM|1RB1RE_0RC0RA_0LD0LF_1LA1LD_0RB0LD_1LC---}}.
* When the starting state is F, the TNF is {{TM|1RB---_0RC0RA_1RD1RC_1LE1LF_0LB0LD_0LE0RC}}.

==Analysis by Racheline==
Posted by Racheline on 2 August 2024:<ref>https://discord.com/channels/960643023006490684/1267928440506421283/1268996655122878474</ref>

A(a,b+3,2c) -> A(a,b,3c)
A(a,b+1,2c+1) -> A(a,b,3c+1)
A(a+2,0,2c) -> A(a,15c-17,2m+1)
A(a+2,1,2c) -> A(a,15c-7,2)
A(a+8,2,4c) -> A(a,45c+15m-10,2)
A(a+7,2,4c+2) -> B(a,45c+15m+8,1,2)
A(a+7,0,4c+1) -> B(a,45c+15m-7,1,2)
A(a+8,0,4c+3) -> A(a,45c+15m+20,2)

B(a,b,c+3,2d) -> B(a,b,c,3d)
B(a,b,c+1,2d+1) -> B(a,b,c,3d+1)
B(a,b+3,0,2d) -> B(a,b,15d-10,2)
B(a,b+1,1,2d) -> B(a,b,15d-11,2)
B(a+3,2b,2,2d) -> A(a,3b+15d-3,2)
B(a+2,2b+1,2,2d) -> B(a,3b+15d-6,1,2)
B(a+2,2b,0,2d+1) -> B(a,3b+15d-1,1,2)
B(a+3,2b+1,0,2d+1) -> A(a,3b+15d+5,2)

A(a,b,c) = (100)^a (01)^b 0 (01)^(5c-9) 00101101 A> 0^inf
B(a,b,c,d) = (100)^a (01)^b 0 (01)^c 0 (01)^(5d-9) 00101101 A> 0^inf

oh and m = 78201748517382892561132455470720498421856223797368814812889407 which i think is (HydraMap^352(2)-1)/2

== References ==

[[Category:BB(6)]]

1RB0LE 1RC1RB 1RD0RA 0RE--- 1LF1LA 1LA1LF

2025-09-28T06:23:03Z

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1RB1RA 1LC0RF 1LD1LC 1LE0LB 0LF--- 1RA1RB

2025-09-28T06:22:44Z

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1RB1RC 1RC1RB 1LD0RA 1LE1LD 1LF0LC 0LA---

2025-09-28T06:22:32Z

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1RB0RE 0RC--- 1LD1LE 1LE1LD 1RF0LC 1RA1RF

2025-09-28T06:19:16Z

Int-y1:

{{machine|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}{{Stub}}
{{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}} is a holdout [[BB(6)]] TM.

When the TM starts on these states, it exhibits similar behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}.
* When the starting state is C, the TNF is {{TM|1RB1RC_1RC1RB_1LD0RA_1LE1LD_1LF0LC_0LA---}}. This TM is a holdout.
* When the starting state is D, the TNF is {{TM|1RB1RA_1LC0RF_1LD1LC_1LE0LB_0LF---_1RA1RB}}. This TM is a holdout.
* When the starting state is E, the TNF is {{TM|1RB0LE_1RC1RB_1RD0RA_0RE---_1LF1LA_1LA1LF}}. This TM is a holdout.
* When the starting state is F, the TNF is {{TM|1RB1RA_1RC0RF_0RD---_1LE1LF_1LF1LE_1RA0LD}}. This TM halts after 17825053 steps.

== Analysis by mxdys ==

Chaotic 1D - CA in bell

[https://discord.com/channels/960643023006490684/1239205785913790465/1260916010685300746 mxdys — 11 Jul 2024 at 7:10 AM ET]

<pre>
1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF (chaotic 1dCA in bell)

(...0 a1 a2 ... an > b1 b2 ... bm 0...) := (0^inf 0 1^a1 0 1^a2 ... 0 1^an E> 1^b1 0 1^b2 0 ... 1^bm 0 0^inf)
(...0 a1 a2 ... an < b1 b2 ... bm 0...) := (0^inf 0 1^a1 0 1^a2 ... 0 1^an <E 1^b1 0 1^b2 0 ... 1^bm 0 0^inf)

start from:
...0 1 < 2 0...

local rewrite rules:
a > 0 b c+1
----------- (R)
a+b+2 0 > c

a > b+2
--------- (RL)
a+1 < b+1

a > 1 0...
---------- (RL'')
a+1 < 0...

a 2b+1 <
--------- (L)
< a+2 1^b

2a <
------- (LR)
> 0 1^a

a > 0 b 0...
-------------- (RL')
a+b+3 < 2 0...
</pre>

The machine is likely halting if these rules are not closed.

[[Category:BB(6)]]

1RB0RE 0RC--- 1LD1LE 1LE1LD 1RF0LC 1RA1RF

2025-09-28T06:18:50Z

Int-y1: F halts

{{machine|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}{{Stub}}
{{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}} is a holdout [[BB(6)]] TM.

When the TM starts on these states, it exhibits similar behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}.
* When the starting state is C, the TNF is {{TM|1RB1RC_1RC1RB_1LD0RA_1LE1LD_1LF0LC_0LA---}}. This TM is a holdout.
* When the starting state is D, the TNF is {{TM|1RB1RA_1LC0RF_1LD1LC_1LE0LB_0LF---_1RA1RB}}. This TM is a holdout.
* When the starting state is E, the TNF is {{TM|1RB0LE_1RC1RB_1RD0RA_0RE---_1LF1LA_1LA1LF}}. This TM is a holdout.
* When the starting state is F, the TNF is {{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}. This TM halts after 17825053 steps.

== Analysis by mxdys ==

Chaotic 1D - CA in bell

[https://discord.com/channels/960643023006490684/1239205785913790465/1260916010685300746 mxdys — 11 Jul 2024 at 7:10 AM ET]

<pre>
1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF (chaotic 1dCA in bell)

(...0 a1 a2 ... an > b1 b2 ... bm 0...) := (0^inf 0 1^a1 0 1^a2 ... 0 1^an E> 1^b1 0 1^b2 0 ... 1^bm 0 0^inf)
(...0 a1 a2 ... an < b1 b2 ... bm 0...) := (0^inf 0 1^a1 0 1^a2 ... 0 1^an <E 1^b1 0 1^b2 0 ... 1^bm 0 0^inf)

start from:
...0 1 < 2 0...

local rewrite rules:
a > 0 b c+1
----------- (R)
a+b+2 0 > c

a > b+2
--------- (RL)
a+1 < b+1

a > 1 0...
---------- (RL'')
a+1 < 0...

a 2b+1 <
--------- (L)
< a+2 1^b

2a <
------- (LR)
> 0 1^a

a > 0 b 0...
-------------- (RL')
a+b+3 < 2 0...
</pre>

The machine is likely halting if these rules are not closed.

[[Category:BB(6)]]

1RB0RE 0RC--- 1LD1LE 1LE1LD 1RF0LC 1RA1RF

2025-09-28T06:06:56Z

Int-y1: add similar tm's

{{machine|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}{{Stub}}
{{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}} is a holdout [[BB(6)]] TM.

When the TM starts on these states, it exhibits similar behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF}}.
* When the starting state is C, the TNF is {{TM|1RB1RC_1RC1RB_1LD0RA_1LE1LD_1LF0LC_0LA---}}.
* When the starting state is D, the TNF is {{TM|1RB1RA_1LC0RF_1LD1LC_1LE0LB_0LF---_1RA1RB}}.
* When the starting state is E, the TNF is {{TM|1RB0LE_1RC1RB_1RD0RA_0RE---_1LF1LA_1LA1LF}}.

== Analysis by mxdys ==

Chaotic 1D - CA in bell

[https://discord.com/channels/960643023006490684/1239205785913790465/1260916010685300746 mxdys — 11 Jul 2024 at 7:10 AM ET]

<pre>
1RB0RE_0RC---_1LD1LE_1LE1LD_1RF0LC_1RA1RF (chaotic 1dCA in bell)

(...0 a1 a2 ... an > b1 b2 ... bm 0...) := (0^inf 0 1^a1 0 1^a2 ... 0 1^an E> 1^b1 0 1^b2 0 ... 1^bm 0 0^inf)
(...0 a1 a2 ... an < b1 b2 ... bm 0...) := (0^inf 0 1^a1 0 1^a2 ... 0 1^an <E 1^b1 0 1^b2 0 ... 1^bm 0 0^inf)

start from:
...0 1 < 2 0...

local rewrite rules:
a > 0 b c+1
----------- (R)
a+b+2 0 > c

a > b+2
--------- (RL)
a+1 < b+1

a > 1 0...
---------- (RL'')
a+1 < 0...

a 2b+1 <
--------- (L)
< a+2 1^b

2a <
------- (LR)
> 0 1^a

a > 0 b 0...
-------------- (RL')
a+b+3 < 2 0...
</pre>

The machine is likely halting if these rules are not closed.

[[Category:BB(6)]]

1RB1RA 0RC1RF 1LD--- 0LE1LE 1RA1LD 1LD0LF

2025-09-28T06:00:46Z

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1RB0LD 1LC0RA 1RA1LB 1LA1LE 1RF0LC ---0RE

2025-09-28T05:59:40Z

Int-y1:

{{machine|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE}}
{{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE}} is a potential [[BB(6)]] [[Cryptid]] found by @mxdys and shared on Discord on 7 Aug 2024. Andrew Ducharme forward simulated its representative map 33,342,087,612,867 steps, at which point both entries in the tuples (x,y) were on the order 10^(2.00705*10^6). The TM had not yet halted.

When the TM starts on these states, it exhibits similar behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE}}.
* When the starting state is C, the TNF is {{TM|1RB1LC_1RC0LD_1LA0RB_1LB1LE_1RF0LA_---0RE}}.
* When the starting state is D, the TNF is {{TM|1RB1RE_1LC0RA_1RD0LB_1LB1RC_1LF0RD_---0LE}}.

== Analysis by @mxdys ==
<pre>
1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE

start from (2,3)
(a,b+a+2) --> (2a+3,b)
(a+b+1,b) --> (2,a+4b+5)
(a,a) --> halt not used?
(a,a+1) --> (2,2a+4) only used once?

(a,b) := 0^inf 01^a 110 A> 1^(2b+1) 0^inf

example:
(2,3)-->
(2,8)-->(7,4)-->
(2,23)-->(7,19)-->(17,10)-->
(2,51)-->(7,47)-->(17,38)-->(37,19)-->
(2,98)-->(7,94)-->(17,85)-->(37,66)-->(77,27)-->
(2,162)-->(7,158)-->(17,149)-->(37,130)-->(77,91)-->(157,12)-->
(2,197)-->(7,193)-->(17,184)-->(37,165)-->(77,126)-->(157,47)-->
(2,302)-->(7,298)-->(17,289)-->(37,270)-->(77,231)-->(157,152)-->
(2,617)-->(7,613)-->(17,604)-->(37,585)-->(77,546)-->(157,467)-->(317,308)-->
...
</pre>

<pre>
start from <0,4>
b >= a(i)+2: <i,b> --> <i+1,b-a(i)-2>
b <= a(i)-1: <i,b> --> <i,3b+i+7>
b = a(i)+0: <i,b> --> halt unused?
b = a(i)+1: <i,b> --> <i+1,i+5> unused?

<i,b> = (a(i),b)
a(i) = 10*2^i-3
</pre>
These rules have been proved in [https://github.com/ccz181078/busycoq/blob/BB6/verify/1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE.v Rocq].

== Analysis by @dyuan01 ==
If you apply the b=a(i)+1 rule into the b >= a(i)+2 rule, you get (i+1, -1). Applying b <= a(i)-1 gives us (i+1, i+5), which is the result you have shown.

So we can just remove the last rule and add its condition to the condition of the first rule:
<pre>
start from <0,4>
b >= a(i)+1: <i,b> --> <i+1,b-a(i)-2>
b <= a(i)-1: <i,b> --> <i,3b+i+7>
b = a(i)+0: <i,b> --> halt unused?

<i,b> = (a(i),b)
a(i) = 10*2^i-3
</pre>
Since b can equal -1, I want to shift the second term up by 1:
<pre>
start from <0,5>
b >= a(i)+2: <i,b> --> <i+1,b-a(i)-2>
b <= a(i)+0: <i,b> --> <i,3b+i+5>
b = a(i)+1: <i,b> --> halt unused?

<i,b> = (a(i),b+1)
a(i) = 10*2^i-3
</pre>

I also want to redefine a(i) to mean 10*2^i-1, so we get
<pre>
start from <0,5>
b >= a(i)+0: <i,b> --> <i+1,b-a(i)>
b <= a(i)-2: <i,b> --> <i,3b+i+5>
b = a(i)-1: <i,b> --> halt unused?

<i,b> = (a(i)-2,b+1)
a(i) = 10*2^i-1
</pre>

[[Category:BB(6)]]

1RB1RE 1LC0RA 1RD0LB 1LB1RC 1LF0RD ---0LE

2025-09-28T05:57:32Z

Int-y1: Redirected page to 1RB0LD 1LC0RA 1RA1LB 1LA1LE 1RF0LC ---0RE

#REDIRECT [[1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE]]

1RB1LC 1RC0LD 1LA0RB 1LB1LE 1RF0LA ---0RE

2025-09-28T05:57:19Z

Int-y1: Redirected page to 1RB0LD 1LC0RA 1RA1LB 1LA1LE 1RF0LC ---0RE

#REDIRECT [[1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE]]

1RB0LD 1LC0RA 1RA1LB 1LA1LE 1RF0LC ---0RE

2025-09-28T05:57:00Z

Int-y1: similar tm's

{{machine|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE}}
{{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE}} is a potential [[BB(6)]] [[Cryptid]] found by @mxdys and shared on Discord on 7 Aug 2024. Andrew Ducharme forward simulated its representative map 33,342,087,612,867 steps, at which point both entries in the tuples (x,y) were on the order 10^(2.00705*10^6). The TM had not yet halted.

When the TM starts on these states, it exhibits the same behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE}}.
* When the starting state is C, the TNF is {{TM|1RB1LC_1RC0LD_1LA0RB_1LB1LE_1RF0LA_---0RE}}.
* When the starting state is D, the TNF is {{TM|1RB1RE_1LC0RA_1RD0LB_1LB1RC_1LF0RD_---0LE}}.

== Analysis by @mxdys ==
<pre>
1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE

start from (2,3)
(a,b+a+2) --> (2a+3,b)
(a+b+1,b) --> (2,a+4b+5)
(a,a) --> halt not used?
(a,a+1) --> (2,2a+4) only used once?

(a,b) := 0^inf 01^a 110 A> 1^(2b+1) 0^inf

example:
(2,3)-->
(2,8)-->(7,4)-->
(2,23)-->(7,19)-->(17,10)-->
(2,51)-->(7,47)-->(17,38)-->(37,19)-->
(2,98)-->(7,94)-->(17,85)-->(37,66)-->(77,27)-->
(2,162)-->(7,158)-->(17,149)-->(37,130)-->(77,91)-->(157,12)-->
(2,197)-->(7,193)-->(17,184)-->(37,165)-->(77,126)-->(157,47)-->
(2,302)-->(7,298)-->(17,289)-->(37,270)-->(77,231)-->(157,152)-->
(2,617)-->(7,613)-->(17,604)-->(37,585)-->(77,546)-->(157,467)-->(317,308)-->
...
</pre>

<pre>
start from <0,4>
b >= a(i)+2: <i,b> --> <i+1,b-a(i)-2>
b <= a(i)-1: <i,b> --> <i,3b+i+7>
b = a(i)+0: <i,b> --> halt unused?
b = a(i)+1: <i,b> --> <i+1,i+5> unused?

<i,b> = (a(i),b)
a(i) = 10*2^i-3
</pre>
These rules have been proved in [https://github.com/ccz181078/busycoq/blob/BB6/verify/1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE.v Rocq].

== Analysis by @dyuan01 ==
If you apply the b=a(i)+1 rule into the b >= a(i)+2 rule, you get (i+1, -1). Applying b <= a(i)-1 gives us (i+1, i+5), which is the result you have shown.

So we can just remove the last rule and add its condition to the condition of the first rule:
<pre>
start from <0,4>
b >= a(i)+1: <i,b> --> <i+1,b-a(i)-2>
b <= a(i)-1: <i,b> --> <i,3b+i+7>
b = a(i)+0: <i,b> --> halt unused?

<i,b> = (a(i),b)
a(i) = 10*2^i-3
</pre>
Since b can equal -1, I want to shift the second term up by 1:
<pre>
start from <0,5>
b >= a(i)+2: <i,b> --> <i+1,b-a(i)-2>
b <= a(i)+0: <i,b> --> <i,3b+i+5>
b = a(i)+1: <i,b> --> halt unused?

<i,b> = (a(i),b+1)
a(i) = 10*2^i-3
</pre>

I also want to redefine a(i) to mean 10*2^i-1, so we get
<pre>
start from <0,5>
b >= a(i)+0: <i,b> --> <i+1,b-a(i)>
b <= a(i)-2: <i,b> --> <i,3b+i+5>
b = a(i)-1: <i,b> --> halt unused?

<i,b> = (a(i)-2,b+1)
a(i) = 10*2^i-1
</pre>

[[Category:BB(6)]]

1RB0RC 1RC0LC 0RD1LE 1LE1RF 0LB--- 0RA1LA

2025-09-28T04:47:37Z

Int-y1: not holdout

{{machine|1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA}}
{{TM|1RB0RC_1RC0LC_0RD1LE_1LE1RF_0LB---_0RA1LA}} is a non-halting [[BB(6)]] TM proven infinite in Rocq ([https://github.com/ccz181078/busycoq/blob/0a61549704e6305f752a4ad1f0faa5cf41dc0e61/verify/SBCv3.v#L8-L201 link]) by @mxdys on 6 Dec 2024.

Analysis by @mxdys:
<pre>
f(n0, ()) := n0
f(n0, (n,ls)) := 2*(f(n0, ls) - n)

start: (2, (), 1)
(k, ls, n) --> (k+1, (n,ls), 2*(f(2^k, ls)-(n+2))), f(2^k, ls)-(n+2) >= 0
</pre>

Follow up by [[User:sligocki]]:
<pre>
Let (k, ls, n) --> (k+1, [n, ls], n')

Then:
n'+2 = 2 (f(2^k, ls) - n) - 2
f(2^{k+1}, [n, ls]) = 2 (f(2^{k+1}, ls) - n)

So:
f(2^{k+1}, [n, ls]) - (n'+2) = 2 (f(2^{k+1}, ls) - f(2^k, ls)) + 2
</pre>
Which indicates that the condition <math>f(2^k, ls)-(n+2) \ge 0</math> should always hold. But this is not detectible nor provable by any existing deciders.

[[Category:BB(6)]]

1RB0LD 1RC1RF 1LA0RA 0LA0LE 1LD1LA 0RB---

2025-09-28T04:43:06Z

Int-y1: similar tm

{{machine|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---}}{{unsolved|Does this TM run forever?}}
{{TM|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---}} is a [[probviously]] non-halting [[BB(6)]] [[Cryptid]] found by @dyuan01 and shared on Discord on 04 Sep 2024.

== Rules ==
<pre>
start: (1,0)
(3b+0,c+1) --> (4b+5,c)
(3b+1,c) --> (4b+2,c+4)
(3b+2,c+1) --> (4b+6,c+7)
(3(3(3(3b+2)+1)+1)+0,0) --> halt

(b,c) := 0^inf 110 A> 0^b 10^c 01 0^inf
</pre>
[[File:B 100k 1RB0LD 1RC1RF 1LA0RA 0LA0LE 1LD1LA 0RB---.png|thumb|Plot of the second entry in (b,c) over the first 100k steps of this TM. The TM halts or enters a new phase if x hits 0. This is extremely unlikely.]]
For simplicity only one potential halting rule is shown.

== Similar TMs ==

If the starting state is set to C and put into [[Tree Normal Form|TNF]], the TM becomes {{TM|1RB0LB_1LC0RE_1LA1LD_0LC---_0RB0RF_1RE1RB}}.

[[Category:BB(6)]][[Category:Cryptids]]