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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}
A '''Collatz-like function''' is a function iterated repeatedly on its own output (a recurrence) defined piecewise by the remainder of the current value modulo some number. The characteristic of most Collatz-like functions is that each branch combines multiplication by a small prime with division by 2, which causes the iterates to grow and shrink in a pattern that is practically unpredictable. This pseudorandom behavior is empirically well-documented but remains largely mysterious mathematically.
  c(2k)  & = &  k \\
The canonical example is the original Collatz function:
  c(2k+1) & = & 3k+2 \\
<math display="block">
\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.
\begin{array}{l}
 
c(2k)  & = &  k \\
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.
c(2k+1) & = & 3k+2
\end{array}
</math>
where division by 2 on even inputs and multiplication by 3 (plus a correction) on odd inputs produces trajectories that appear random despite being entirely deterministic.
A '''Collatz-like problem''' is a question about the long-run behavior of such an iteration — for instance, whether all starting values eventually reach a fixed point or cycle. Collatz-like problems are famously difficult; even the simplest instances are open.
Many [[Busy Beaver Champions]] exhibit '''Collatz-like behavior''', meaning their execution trace can be concisely described by the iterates of a Collatz-like function. This is not a coincidence. The [[Busy Beaver]] project systematically filters out machines with predictable, non-chaotic behavior, so the machines that survive and dominate the competition are precisely those whose behavior is hardest to predict. Because the ×3, ÷2 process appears to be the simplest mechanism capable of generating pseudorandom dynamics, it shows up throughout the Busy Beaver landscape.
[[Generalized Collatz Function|Generalized Collatz Functions]] (GCFs) are a well-studied formal class of Collatz-like functions on one variable in which every branch is an affine map. The halting problem for GCFs is known to be [[Turing complete]]. [[Consistent Collatz]] functions are a further restriction of GCFs in which all branches share the same linear coefficient.


== Examples ==
== Examples ==
=== 5-state busy beaver winner ===
Consider the [[5-state busy beaver winner]] and the generalized configuration:


=== BB(5, 2) Champion ===
<math display="block">M(n) = 0^\infty \; \textrm{<}\text{A} \; 1^n \; 0^\infty</math>Pascal Michel showed that:
Consider the [[BB(5, 2)]] Champion (<code>1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA</code>) and the generalize configuration:
 
<math display="block">M(n) = 0^\infty \; \textrm{<A} \; 1^n \; 0^\infty</math>Pascal Michel showed that:


<math display="block">\begin{array}{lcl}
<math display="block">\begin{array}{lcl}
   0^\infty \; \textrm{<A} \; 0^\infty & = & M(0) \\
   0^\infty \; \textrm{<}\textrm{A} \; 0^\infty & = & M(0) \\
   M(3k)  & \xrightarrow{5 k^2 + 19 k + 15} & M(5k+6) \\
   M(3k)  & \xrightarrow{5k^2+19k+15} & M(5k+6) \\
   M(3k+1) & \xrightarrow{5 k^2 + 25 k + 27} & M(5k+9) \\
   M(3k+1) & \xrightarrow{5k^2+25k+27} & M(5k+9) \\
   M(3k+2) & \xrightarrow{6k +12} & 0^\infty \;\; 1 \;\; \textrm{H>} \;\; 01 \;\; {(001)}^{k+1} \;\; 1 \;\; 0^\infty \\
   M(3k+2) & \xrightarrow{6k+12} & 0^\infty\;1\;\textrm{Z}\textrm{>}\;01\;{(001)}^{k+1}\;1\;0^\infty
\end{array}</math>
\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>
Starting on a blank tape <math>C(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 ===
=== Hydra ===
Consider Hydra (a [[Cryptids|Cryptid]]) <code>1RB3RB---3LA1RA_2LA3RA4LB0LB0LA</code> and the generalized configuration:
Consider [[Hydra]] and the generalized configuration:


<math display="block">C(a, b) = 0^\infty \; \textrm{ <B } \; 0^{3(a-2)} \; 3^b \; 2 \; 0^\infty</math> Daniel Yuan showed that:<math display="block">\begin{array}{l}
<math display="block">C(a, b) = 0^\infty\;\textrm{<}\textrm{A}\;2\;0^{3(a-2)} \; 3^b \; 2 \; 0^\infty</math>
   0^\infty \; \textrm{A>} \; 0^\infty & & \xrightarrow{19} & C(3, 0) \\
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,  & 0)  & \to & \text{Halt}(9n-6) \\
   C(2n,  & b+1) & \to & C(3n,   & b) \\
   C(2n,  & b+1) & \to & C(3n,b) \\
   C(2n+1, & b)  & \to & C(3n+1, & b+2) \\
   C(2n+1, & b)  & \to & C(3n+1,b+2)
\end{array}</math>Where <math>\textrm{Halt}(n)</math> is a halting configuration with <math>n</math> non-zero symbols on the tape.
\end{array}</math>


Starting from config <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 being able to prove a detailed question about the trajectory of the Collatz-like function<math display="block">\begin{array}{l}
Where <math>\text{Halt}(n)</math> is a halting configuration with <math>n</math> non-zero symbols on the tape.
  h(2n)  & = & 3n  \\
  h(2n+1) & = & 3n+1 \\
\end{array}</math>starting from 3:


<math display="block">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 \cdots</math>Specifically, will it ever reach a point where the cumulative number of <code>E</code> (even transitions) applied is greater than twice the number of <code>O</code> (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>
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>


=== Tetration Machine ===
== Why is Collatz randomness mysterious? ==
Consider the current [[BB(6, 2)]] Champion (discovered by Pavel Kropitz in May 2022) <code>1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE</code> and consider the general configuration:<math display="block">K(a, b) = 0^\infty \; 1 \; 0^n \; 11 \; 0^5 \; \textrm{C>} \; 0^\infty</math>Shawn Ligocki showed that:<math display="block">\begin{array}{l}
The (above displayed) Collatz function produces numbers whose even/odd property (parity) passes even NIST randomness tests.<ref>Ren, Wei. "Collatz computation sequence for sufficiently large integers is a pseudo-random bit sequence." Mathematics Open 5 (2026): 2550023.</ref> Consequently, pseudo random number generators (PRNG) have been proposed using this or similar functions.<ref>Xu, David, and Dan E. Tamir. "Pseudo-random number generators based on the Collatz conjecture." International Journal of Information Technology 11.3 (2019): 453-459. https://scholar.google.com/scholar?cluster=3536660581322604740&hl=en&as_sdt=0,5</ref> However, the mathematical proof that such systems have this property in general is remarkably elusive.
  0^\infty \; \textrm{A>} \; 0^\infty & \xrightarrow{45} & K(5) \\
  K(4k)   & \to & \text{Halt}(9n-6) \\
  K(4n+1) & \to & K(\frac{3^{k+3} - 11}{2}) \\
  K(4n+2) & \to & K(\frac{3^{k+3} - 11}{2}) \\
  K(4n+3) & \to & K(\frac{3^{k+3} +  1}{2}) \\
\end{array}</math>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 ==
<references />
<references />
[[Category:Zoology]]

Latest revision as of 16:00, 3 April 2026

A Collatz-like function is a function iterated repeatedly on its own output (a recurrence) defined piecewise by the remainder of the current value modulo some number. The characteristic of most Collatz-like functions is that each branch combines multiplication by a small prime with division by 2, which causes the iterates to grow and shrink in a pattern that is practically unpredictable. This pseudorandom behavior is empirically well-documented but remains largely mysterious mathematically. The canonical example is the original Collatz function: c(2k)=kc(2k+1)=3k+2 where division by 2 on even inputs and multiplication by 3 (plus a correction) on odd inputs produces trajectories that appear random despite being entirely deterministic. A Collatz-like problem is a question about the long-run behavior of such an iteration — for instance, whether all starting values eventually reach a fixed point or cycle. Collatz-like problems are famously difficult; even the simplest instances are open. Many Busy Beaver Champions exhibit Collatz-like behavior, meaning their execution trace can be concisely described by the iterates of a Collatz-like function. This is not a coincidence. The Busy Beaver project systematically filters out machines with predictable, non-chaotic behavior, so the machines that survive and dominate the competition are precisely those whose behavior is hardest to predict. Because the ×3, ÷2 process appears to be the simplest mechanism capable of generating pseudorandom dynamics, it shows up throughout the Busy Beaver landscape. Generalized Collatz Functions (GCFs) are a well-studied formal class of Collatz-like functions on one variable in which every branch is an affine map. The halting problem for GCFs is known to be Turing complete. Consistent Collatz functions are a further restriction of GCFs in which all branches share the same linear coefficient.

Examples

5-state busy beaver winner

Consider the 5-state busy beaver winner and the generalized configuration:

M(n)=0<A1n0Pascal Michel showed that:

0<A0=M(0)M(3k)5k2+19k+15M(5k+6)M(3k+1)5k2+25k+27M(5k+9)M(3k+2)6k+1201Z>01(001)k+110

Starting on a blank tape M(0), these rules iterate 15 times before reaching the halt config.[1]

Hydra

Consider Hydra and the generalized configuration:

C(a,b)=0<A203(a2)3b20 Daniel Yuan showed that:0A>020C(3,0)C(2n,0)Halt(9n6)C(2n,b+1)C(3n,b)C(2n+1,b)C(3n+1,b+2)

Where Halt(n) is a halting configuration with n non-zero symbols on the tape.

Starting from C(3,0), this simulates a pseudo-random walk along the b parameter, increasing it by 2 every time a 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 H(2n)=3n(even transition)H(2n+1)=3n+1(odd transition) 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.[2] The first few transitions are displayed below: 0..3O4E6E9O13O19O28E42E63O

Exponential Collatz

Consider the machine 1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE (bbch), discovered by Pavel Kropitz in May 2022, and the general configuration:K(n):=010n1105C>0Shawn Ligocki showed that: 0A>045K(5)K(4k)Halt(3k+3112)K(4k+1)K(3k+3112)K(4k+2)K(3k+3112)K(4k+3)K(3k+3+12)

Demonstrating Collatz-like behavior with exponential piecewise component functions.

Starting from config K(5), these rules iterate 15 times before reaching the halt config leaving over 1015 non-zero symbols on the tape.[3]

Why is Collatz randomness mysterious?

The (above displayed) Collatz function produces numbers whose even/odd property (parity) passes even NIST randomness tests.[4] Consequently, pseudo random number generators (PRNG) have been proposed using this or similar functions.[5] However, the mathematical proof that such systems have this property in general is remarkably elusive.

References

  1. Pascal Michel's Analysis of the BB(5, 2) Champion
  2. Shawn Ligocki. BB(2, 5) is Hard (Hydra). 10 May 2024.
  3. Shawn Ligocki. BB(6, 2) > 10↑↑15. 21 Jun 2022.
  4. Ren, Wei. "Collatz computation sequence for sufficiently large integers is a pseudo-random bit sequence." Mathematics Open 5 (2026): 2550023.
  5. Xu, David, and Dan E. Tamir. "Pseudo-random number generators based on the Collatz conjecture." International Journal of Information Technology 11.3 (2019): 453-459. https://scholar.google.com/scholar?cluster=3536660581322604740&hl=en&as_sdt=0,5