BusyBeaverWiki - User contributions [en]

Beaver Math Olympiad

2026-04-10T05:25:17Z

Mxdys: /* 8. {{TM|1RB0LD_0RC0RF_0RD0RA_1LE0RD_1LF---_0LA1LA|undecided}} */

'''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?

=== 8. {{TM|1RB0LD_0RC0RF_0RD0RA_1LE0RD_1LF---_0LA1LA|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) = (10, 12)</math> and

<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}
(a_n-\lfloor\frac{b_n}{2}\rfloor-3, 3\lfloor\frac{b_n+1}{2}\rfloor+6) & \text{if }a_n > \lfloor\frac{b_n}{2}\rfloor \\
(3a_n+5, b_n-2a_n) & \text{if }a_n \le \lfloor\frac{b_n}{2}\rfloor
\end{cases}</math>

for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_n = \lfloor\frac{b_n}{2}\rfloor + 1</math>?

== 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

Formalised solution: [https://discord.com/channels/960643023006490684/1259770421046411285/1488737894943166604 Initial announcement], [https://discord.com/channels/960643023006490684/1259770421046411285/1488743526882738276 Lean proof], [https://discord.com/channels/960643023006490684/1259770421046411285/1488781537699696821 LLM-translated Rocq proof], [https://discord.com/channels/960643023006490684/1259770421046411285/1488898995865784442 Proof of closure of existing mid-level rules].

=== 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>

=== 7. {{TM|1RB1RF_1RC0RA_1LD1RC_1LE0LE_0RA0LD_0RB---|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>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)

Link to Discord discussion: https://discord.com/channels/960643023006490684/1421782442213376000/1431483206208852001

== Practice Problems ==
Problems that are not BMO-level, but provide counter-examples to certain [[probvious]] intuition:

* {{TM|1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD}}
* {{TM|1RB0RD_0LC1RA_0RA1LB_1RE1LB_1LF1LB_---1LE}}

[[Category:Individual machines]]

Beaver Math Olympiad

2026-04-10T05:21:07Z

Mxdys: /* 8. {{TM|1RB0LD_0RC0RF_0RD0RA_1LE0RD_1LF---_0LA1LA|undecided}} */

'''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?

=== 8. {{TM|1RB0LD_0RC0RF_0RD0RA_1LE0RD_1LF---_0LA1LA|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) = (10, 12)</math> and

<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}
(a_n-\lfloor\frac{b_n}{2}\rfloor-3, 3\lfloor\frac{b_n+1}{2}\rfloor+6) & \text{if }a_n \ge \lfloor\frac{b_n}{2}\rfloor + 2 \\
(0, 0) & \text{if }a_n = \lfloor\frac{b_n}{2}\rfloor + 1 \\
(3a_n+5, b_n-2a_n) & \text{if }a_n \le \lfloor\frac{b_n}{2}\rfloor
\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 = 0</math>?

== 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

Formalised solution: [https://discord.com/channels/960643023006490684/1259770421046411285/1488737894943166604 Initial announcement], [https://discord.com/channels/960643023006490684/1259770421046411285/1488743526882738276 Lean proof], [https://discord.com/channels/960643023006490684/1259770421046411285/1488781537699696821 LLM-translated Rocq proof], [https://discord.com/channels/960643023006490684/1259770421046411285/1488898995865784442 Proof of closure of existing mid-level rules].

=== 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>

=== 7. {{TM|1RB1RF_1RC0RA_1LD1RC_1LE0LE_0RA0LD_0RB---|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>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)

Link to Discord discussion: https://discord.com/channels/960643023006490684/1421782442213376000/1431483206208852001

== Practice Problems ==
Problems that are not BMO-level, but provide counter-examples to certain [[probvious]] intuition:

* {{TM|1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD}}
* {{TM|1RB0RD_0LC1RA_0RA1LB_1RE1LB_1LF1LB_---1LE}}

[[Category:Individual machines]]

Beaver Math Olympiad

2026-04-10T05:12:00Z

Mxdys: /* Unsolved problems */

'''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?

=== 8. {{TM|1RB0LD_0RC0RF_0RD0RA_1LE0RD_1LF---_0LA1LA|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) = (10, 12)</math> and

<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}
(a_n-\lfloor\frac{b_n}{2}\rfloor-3, 3\lfloor\frac{b_n+1}{2}\rfloor+6) & \text{if }a_n \ge \lfloor\frac{b_n}{2}\rfloor + 3 \\
(2, 3\lfloor\frac{b_n+1}{2}\rfloor+8) & \text{if }a_n = \lfloor\frac{b_n}{2}\rfloor + 2 \\
(0, 0) & \text{if }a_n = \lfloor\frac{b_n}{2}\rfloor + 1 \\
(3a_n+5, b_n-2a_n) & \text{if }a_n \le \lfloor\frac{b_n}{2}\rfloor
\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 = 0</math>?

== 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

Formalised solution: [https://discord.com/channels/960643023006490684/1259770421046411285/1488737894943166604 Initial announcement], [https://discord.com/channels/960643023006490684/1259770421046411285/1488743526882738276 Lean proof], [https://discord.com/channels/960643023006490684/1259770421046411285/1488781537699696821 LLM-translated Rocq proof], [https://discord.com/channels/960643023006490684/1259770421046411285/1488898995865784442 Proof of closure of existing mid-level rules].

=== 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>

=== 7. {{TM|1RB1RF_1RC0RA_1LD1RC_1LE0LE_0RA0LD_0RB---|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>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)

Link to Discord discussion: https://discord.com/channels/960643023006490684/1421782442213376000/1431483206208852001

== Practice Problems ==
Problems that are not BMO-level, but provide counter-examples to certain [[probvious]] intuition:

* {{TM|1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD}}
* {{TM|1RB0RD_0LC1RA_0RA1LB_1RE1LB_1LF1LB_---1LE}}

[[Category:Individual machines]]

1RB1LA 1RC1RE 1LD0RB 1LA0LC 0RF0RD 0RB---

2025-11-29T09:34:51Z

Mxdys: /* Analysis by mxdys */

{{TM|1RB1LA_1RC1RE_1LD0RB_1LA0LC_0RF0RD_0RB---|undecided}} is a [[BB(6)]] holdout.

=Analysis by mxdys=
https://discord.com/channels/960643023006490684/1239205785913790465/1441124403801755730
<pre>
1RB1LA_1RC1RE_1LD0RB_1LA0LC_0RF0RD_0RB---

start: S(18)

S(n) --> S((n+(3^i*6+i+4))/2), n mod 2 = i mod 2, 3^i*2-i-2 <= n <= 3^i*6-i-6
S(n) --> S(3^i*12-1), n mod 2 = (i+1) mod 2, 3^i*2-i <= n <= 3^i*6-i-10
</pre>

These rules are closed if <pre>(3^i*2+i+5)/(2^v2(3^i*2+i+5))>=2i+14 for all i>=50 </pre>
= Further Analysis (Pomme, Autumn Pan, et al.) =
Although it has not been proven in Rocq, it is known that the inequality holds. This has been proven informally by Pomme, and an alternative proof is underway by Autumn Pan.

1RB--- 0RC1RD 1LA0LA 1LE0RB 1LF1LE 1LC0LD

2025-10-03T19:59:25Z

Mxdys:

{{machine|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}}{{Stub}}
{{TM|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}} is an undecided [[BB(6)]] machine. @mxdys referred to it as "chaotic 1dCA in bell".

1D — CA, look like rule 30 or rule 110 but doesn't work in the same way.

When the TM starts on any starting state, it exhibits the same behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}}.
* When the starting state is B, the TNF is not in TNF-1RB format, and can be (mostly) ignored.
* When the starting state is C, the TNF is {{TM|1RB0RB_1LC---_0LA1LD_1RE0LC_1RF1RE_1RA0RD}}.
* When the starting state is D, the TNF is {{TM|1RB0LF_1RC1RB_1RD0RA_1RE0RE_1LF---_0LD1LA}}.
* When the starting state is E, the TNF is {{TM|1RB1RA_1RC0RF_1RD0RD_1LE---_0LC1LF_1RA0LE}}.
* When the starting state is F, the TNF is {{TM|1RB0RE_1RC0RC_1LD---_0LB1LE_1RF0LD_1RA1RF}}.

[[Category:BB(6)]]

1RB--- 0RC1RD 1LA0LA 1LE0RB 1LF1LE 1LC0LD

2025-10-03T19:58:13Z

Mxdys: Undo revision 4205 by Mxdys (talk)

{{machine|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}}{{Stub}}
{{TM|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}} is an undecided [[BB(6)]] machine. @mxdys referred to it as "chaotic 1dCA in bell".

1D — CA, rule 30 or rule 110.

When the TM starts on any starting state, it exhibits the same behaviour:

* When the starting state is A, the [[Tree Normal Form|TNF]] is {{TM|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}}.
* When the starting state is B, the TNF is not in TNF-1RB format, and can be (mostly) ignored.
* When the starting state is C, the TNF is {{TM|1RB0RB_1LC---_0LA1LD_1RE0LC_1RF1RE_1RA0RD}}.
* When the starting state is D, the TNF is {{TM|1RB0LF_1RC1RB_1RD0RA_1RE0RE_1LF---_0LD1LA}}.
* When the starting state is E, the TNF is {{TM|1RB1RA_1RC0RF_1RD0RD_1LE---_0LC1LF_1RA0LE}}.
* When the starting state is F, the TNF is {{TM|1RB0RE_1RC0RC_1LD---_0LB1LE_1RF0LD_1RA1RF}}.

[[Category:BB(6)]]

1RB--- 0RC1RD 1LA0LA 1LE0RB 1LF1LE 1LC0LD

2025-10-03T19:57:30Z

Mxdys:

{{machine|1RB---_0RC1RD_1LA0LA_1LE0RB_1LF1LE_1LC0LD}}

1D — CA, look like rule 30 or rule 110
[[Category:Stub]]

Beaver Math Olympiad

2025-09-28T10:19:09Z

Mxdys: /* 7. 1RB1RF_1RC0RA_1LD1RC_1LE0LE_0RA0LD_0RB--- (bbch) */

'''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>

Beaver Math Olympiad

2025-09-28T09:34:07Z

Mxdys: /* 7. 1RB1RF_1RC0RA_1LD1RC_1LE0LE_0RA0LD_0RB--- (bbch) */

'''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?

== 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>

Beaver Math Olympiad

2025-09-28T09:31:41Z

Mxdys: add BMO7

'''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>k</math> is even?

== 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>

TMBR: August 2025

2025-09-02T14:07:55Z

Mxdys: /* Holdouts */

[[File:Conference poster for DNA31 by Tristan Stérin.png|thumb|396x396px|Conference poster for the [https://dna31.sciencesconf.org/ 31st International Conference on DNA Computing and Molecular Programming], [https://discord.com/channels/960643023006490684/960643023530762341/1409904231468761159 made by Tristan Stérin (cosmo)]]]

[[:Category:This Month in Beaver Research|This Month in Beaver Research]] for August 2025. This month, Tristan Stérin presented a poster (see right) at DNA 31, there were significant holdouts reduction in numerous domains, a fast algorithm for Antihydra (and similar Collatz-like problems) was re-discovered, and multisymbol support was added to the Blaze TM visualizer.

== Cryptids ==
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
* 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.

== Holdouts ==
In August there were significant reductions in [[Holdouts lists]] across many [[BB Domains]]
{| class="wikitable"
|+BB Holdout Reduction by Domain
!Domain
!New Holdout Count
!July Holdout Count
!# TMs Decided
!% Reduction
|-
|[[BB(6)]]
|2,592
|2,728
|136
|5.0%
|-
|[[BB(2,6)]]
|18,054,938
|22,302,296
|4,247,358
|19.0%
|-
|[[BB(7)]]
|59,727,905
|86,129,304
|26,401,399
|30.7%
|}
* [[BB(6)]] holdouts: Reduced by a total of 136 holdouts by 4 people.
** XnoobSpeakable found 9 new halting TMs in the high exponential runtime range (~<math>10^{100000}</math>) by running Enumerate.py out to extremely high parameters. https://discord.com/channels/960643023006490684/1239205785913790465/1401470301467836556
** [https://discord.com/channels/960643023006490684/1239205785913790465/1407580922831831152 Andrew Ducharme found a surprisingly short running halting TM] in the [[BB(6)]] holdouts list with runtime ~<math>10^{78}</math>, to which [https://discord.com/channels/960643023006490684/1239205785913790465/1407659104910053387 Peacemaker replied with another TM] that was almost identical, and soon, simulation showed it to halt in the same number of steps. Later on the 28th, [https://discord.com/channels/960643023006490684/1239205785913790465/1410465964528504872 Ducharme found another one] with surprisingly low runtime: ~<math>10^{11}</math>. In response, [https://discord.com/channels/960643023006490684/1239205785913790465/1410475867200815114 Peacemaker found an almost identical machine], which also halts with similar runtime.
** Peacemaker shared a list of BB(6) holdouts and how many steps are required to use all defined transitions. https://discord.com/channels/960643023006490684/1239205785913790465/1410437756777398344
** @mxdys shared a list of 7 holdouts that he solved using [https://github.com/ccz181078/busycoq/blob/b1e53d74e5053c3379645aaf75fa8c7a72d00547/verify/RWLAcc.v his RWLAcc decider in Rocq] (previously known as Coq). https://discord.com/channels/960643023006490684/1239205785913790465/1408304281039409212 He also shared results featuring 6 holdouts that were solved in "50 Random Holdouts", see https://discord.com/channels/960643023006490684/1400456788955893840/1409115537631613020
** On August 30, @mxdys shared the [https://wiki.bbchallenge.org/wiki/Holdouts_lists#Downloadable_Holdout_Lists new holdouts list], consisting of 2,592 TMs. (The additional 110 TMs not listed here were solved by @mxdys using deciders or individual proofs.)
*After the [[BB(7)#Phase 1|enumeration of BB(7)]] was completed, Andrew Ducharme ran [[BB(7)#Phase 2|several deciders]] on the holdouts list, filtering the original 86,129,304 holdouts down to 59,727,905 in 10 days. https://drive.google.com/drive/u/0/folders/17U0BRpJHTMLtB0poBlOSZhGGp4FkCHIO
*[[BB(3,3)|BB(3,3):]] 9 holdouts were proven non-halting in Rocq (previously known as Coq) by mxdys. [https://wiki.bbchallenge.org/wiki/BB(3,3)#Holdouts 10 holdouts remain, 4 of them solved with moderate rigor.] https://discord.com/channels/960643023006490684/1259770474897080380/1410308974275985428
*[https://discord.com/channels/960643023006490684/1259770421046411285/1411488532500971631 @mxdys proved] 2 [[BB(2,5)]] machines are non-halting in Rocq.
*Andrew Ducharme reduced the [[BB(2,6)]] holdouts from 22,302,296 to 18,054,938. https://discord.com/channels/960643023006490684/1084047886494470185/1412345770241163294

==BB Adjacent==
*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.

== Misc ==
* Iijil shared an algorithm for converting an arbitrary n-state m-symbol TM into a 2-state TM with 3(n+1)m symbols. https://gist.github.com/Iijil1/0d611dbf0a9d52984f72cb14e66a4b28
* Carl K updated his TM web-visualizer to support multi-symbol machines. [https://carlkcarlk.github.io/busy_beaver_blaze/v0.2.6/index.html#palette=edit&colors=000000%2Cff0000%2Cffff00%2Cff00ff%2C00ffff&run=true https://carlkcarlk.github.io/busy_beaver_blaze/v0.2.6/index.html] He also extended his series of videos showing TM simulation accompanied by classical music out to some multi-symbol TMs:
** [[Bigfoot]]: https://youtu.be/YvOHWbQNMoY
** [[Surprise in a Box|Brady's Surprise in a Box]]: https://youtu.be/vIG2CvJShRc
** [[1RB3LA4RB0RB2LA 1LB2LA3LA1RA1RZ|BB(2,5) champ:]] https://youtu.be/QpYBzYDdLEY
** Some interesting BB(2,5) machines: https://youtu.be/CSEKxTpXrDE
*@mxdys Introduced "50 Random Holdouts", a thread on the Discord server, where 50 random TMs are selected from the BB(6) holdout list, and everybody focuses on these 50 machines. This month, 6/50 TMs were solved by @mxdys single-handedly.
*The community (especially, Andrew Ducharme) [https://discord.com/channels/960643023006490684/992572017683472514/1408559249067479145 proposed a concept "BB(n,m) month",] where the community mainly focuses on a single domain, i.e. BB(3,3). The motive of this focused month is to make genuine progress in the one selected domain, with the ultimate goal to reduce all holdouts to [[Cryptids]], with all remaining TMs having been proven in Rocq.

==Blog Posts==
* 1 Sep 2025. Katelyn Doucette. [https://katelyndoucette.com/articles/all-about-space-needle All About Space Needle].

==In the News==
* 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].

==Interesting TMs==
A collection of interesting TMs that were mentioned on Discord, mostly because of their space-time diagrams or general behavior.

* {{TM|1RB0LE_1RC0RF_1RD---_0LA1RB_1RB1LE_1LD1RF}}: Wavy Machine
* {{TM|1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE}}: [[Probviously]] halting Cryptid
* <code>[[1RB1LE_1LC0RA_0RF0LD_1LE1LA_1RC0LB_---1RC]]</code> ([https://bbchallenge.org/1RB1LE_1LC0RA_0RF0LD_1LE1LA_1RC0LB_---1RC bbch]): Unsolved BB(6) TM with pseudorandom behaviour

[[Category:This Month in Beaver Research|2025-08]]

Beaver Math Olympiad

2025-08-29T15:47:39Z

Mxdys:

Beaver Math Olympiad

2025-08-29T15:46:43Z

Mxdys: f^0 is necessary for the inductive definition of iteration

Beaver Math Olympiad

2025-08-27T18:17:13Z

Mxdys: /* 6. 1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD (bbch) */

Beaver Math Olympiad

2025-08-27T18:12:39Z

Mxdys: /* 6. 1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD (bbch) */

1RB1RA 1RC1RZ 1LD0RF 1RA0LE 0LD1RC 1RA0RE

2025-06-28T13:59:45Z

Mxdys: /* Analysis by mxdys */

{{machine|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE}}
{{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 25 June 2025.

It's in a family of 4 machines with the halting time and sigma score between 2↑↑2↑↑2↑↑10 and 2↑↑2↑↑2↑↑11:
<pre>
1RB1RA_1RC---_1LD0RF_1RA0LE_0LD1RC_1RA0RE (hereafter referred to as TM1)
1RB---_1LC0RF_1RE0LD_0LC1RB_1RA1RE_1RE0RD (TM2)
1RB0LE_1RC1RB_1RD---_1LA0RF_0LA1RD_1RB0RE (TM3)
1RB0RF_1RC1RB_1RD---_1LE0RA_1RB0LF_0LE1RD (TM4)
</pre>

== Analysis by mxdys ==
<pre>
Inc2:
S1(len0,a0+1,2,a ,b ) -->
S1(len0,a0 ,1,a+b+2,2^b-1)

Inc1:
S1(len0,a0+1,1,a ,b ) -->
S1(len0,a0 ,0,a+b+2,2^b-1)

Inc0:
S1(len0,a0+1,0,a ,b ) -->
S1(len0,a0 ,2,a+b+1,2^b-1)

Rst0:
S1(a0,0,0,a,b) -->
halt

Rst1:
S1(a0,0,1,a,b) -->
S1(a0+a+2,(2^(a0+2)-1)*2^a-1,2,b,2^b-1)

start: S1(3,7,2,6,63)

the rules are used in the following order:
Inc2,Inc1,Inc0, Inc2,Inc1,Inc0, Inc2, Rst1,
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2, Rst1,
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2,Inc1, Rst0.

where
S1(len0,a0,m,a,b) = 0^inf LH LC(len0,a0) d0 10 1^m LC(a,0) <X 0 11100 111^(1+b) 0^inf
d0 = 100
d1 = 111
LC(0,0) = ""
LC(n+1,2x) = LC(n,x) d1
LC(n+1,2x+1) = LC(n,x) d0
for TM2, X=D, LH=111011
for TM3, X=E, LH=11

TM1 is equivalent to TM2 after several steps
TM4 is equivalent to TM3 after several steps
TM1 has the highest halting time among this family
TM1,TM2 have the highest sigma score among this family
</pre>
estimation of time/score:
<pre>
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2
n mod 3 = 1:
S1(len0,n,2,b,2^b-1) -->
S1(len0,0,1,st2(n,b)+floor(n/3)*5+2,t2(n+1,b))

Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2,Inc1, Rst0
n mod 3 = 2:
S1(len0,n,2,b,2^b-1) -->
S1(len0,0,0,st2(n,b)+floor(n/3)*5+4,t2(n+1,b)) -->
halt

Rst1:
S1(len0,0,1,a,b) -->
S1(len0+a+2,2^(len0+a+2)-2^a-1,2,b,2^b-1)

where
t2(0,b) = b, t2(a+1,b) = 2^t2(a,b)-1
st2(a,b) = t2(0,b) + t2(1,b) + ... + t2(a,b)

S1(3,7,2,6,63) -->
S1(3,0,1,st2(7,6)+12,t2(8,6)) -->
S1(≈t2(7,6),≈t2(8,6),2,_,_) -->
S1(≈t2(7,6),0,1,≈2^^t2(8,6),_) -->
S1(≈2^^t2(8,6),≈2^^t2(8,6),2,_,_) -->
S1(≈2^^t2(8,6),0,1,≈2^^2^^t2(8,6),≈2^^2^^t2(8,6)) -->
halt with time/score ≈2^^2^^((2^)^8 6)
2^^^5 < 2^^2^^2^^10 < 2^^2^^((2^)^8 6) < 2^^2^^2^^11 < 2^^^6
</pre>

BB(6)

2025-06-26T10:39:07Z

Mxdys:

The 6-state, 2-symbol Busy Beaver problem, '''BB(6)''', refers to the unsolved 6<sup>th</sup> value of the [[Busy Beaver function]]. With the discovery of the [[Cryptid]] machine [[Antihydra]] in June 2024, we now know that we must solve a [[Collatz-like]] problem in order to solve BB(6) and thus [https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard].

The current BB(6) champion {{TM|1RB1RA_1RC---_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}} was discovered by mxdys in June 2025, proving the lower bound:

<math display="block">S(6) > \Sigma(6) > 2 \uparrow\uparrow\uparrow 5</math>

== Techniques ==
Simulating tetrational machines, such as the former champion {{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE|halt}}, requires [[Accelerated simulator|accelerated simulation]] that can handle Collatz Level 2 [[Inductive rule|inductive rules]]. In other words, it requires a simulator that can prove the rules:

<math display="block">\begin{array}{lcl}
C(4k) & \to & {\operatorname{Halt}}\Big(\frac{3^{k+3} - 11}{2}\Big) \\
C(4k+1) & \to & C\Big(\frac{3^{k+3} - 11}{2}\Big) \\
C(4k+2) & \to & C\Big(\frac{3^{k+3} - 11}{2}\Big) \\
C(4k+3) & \to & C\Big(\frac{3^{k+3} + 1}{2}\Big) \\
\end{array}</math>

and also compute the remainder mod 3 of numbers produced by applying these rules 15 times (which requires some fancy math related to [[wikipedia:Euler's_totient_function|Euler's totient function]]).

== Cryptids ==
Several [[Turing machines]] have been found that are [[Cryptids]], considered so because each of them have a [[Collatz-like]] halting problem, a type of problem that is generally difficult to solve. However, probabilistic arguments have allowed all but one of them to be categorized as [[probviously]] halting or probviously non-halting.

Probviously non-halting Cryptids:

* {{TM|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA}}, [[Antihydra]]
* {{TM|1RB1RC_1LC1LE_1RA1RD_0RF0RE_1LA0LB_---1RA|undecided}}, a variant of [[Hydra]] and Antihydra
* {{TM|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC|undecided}}, similar to Antihydra
* {{TM|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---|undecided}}, similar to Antihydra
* {{TM|1RB0LB_1LC0RE_1LA1LD_0LC---_0RB0RF_1RE1RB|undecided}}, similar to Antihydra

Probviously halting Cryptids:

* {{TM|1RB0RD_0RC1RE_1RD0LA_1LE1LC_1RF0LD_---0RA}}, [[Lucy's Moonlight]]
* {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC|undecided}}, a family of 16 related TMs
* {{TM|1RB1RE_1LC1LD_---1LA_1LB1LE_0RF0RA_1LD1RF}}
* {{TM|1RB0RE_1LC1LD_0RA0LD_1LB0LA_1RF1RA_---1LB}}
* {{TM|1RB0LC_0LC0RF_1RD1LC_0RA1LE_---0LD_1LF1LA}}
* {{TM|1RB0LC_1LC0RD_1LF1LA_1LB1RE_1RB1LE_---0LE}}
Although {{TM|1RB1LE_0LC0LB_1RD1LC_1RD1RA_1RF0LA_---1RE}} behaves similarly to the probviously halting Cryptids, it is estimated to have a 3/5 chance of becoming a [[translated cycler]] and a 2/5 chance of halting.

There are a few machines considered notable for their chaotic behaviour, but which have not been classified as Cryptids due to seemingly lacking a connection to any known open mathematical problems, such as Collatz-like problems.

Potential Cryptids:

* {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}
* {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}}
* {{TM|1RB0RB_1LC1RE_1LF0LD_1RA1LD_1RC1RB_---1LC|undecided}}
* {{TM|1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD}}

== Top Halters ==
Below is a table of the machines with the 10 highest known runtimes.<ref>Shawn Ligocki's list of 6-state, 2-symbol machines with large runtimes ([https://github.com/sligocki/busy-beaver/blob/main/Machines/bb/6x2.txt Link])</ref> Their sigma scores are expressed using an extension of Knuth's up-arrow notation.<ref>Shawn Ligocki. 2022. [https://www.sligocki.com/2022/06/25/ext-up-notation.html "Extending Up-arrow Notation"]</ref>
{| class="wikitable"
|+Top Known BB(6) Halters
!Standard format
!(approximate) Σ
|-
|{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}
|10 ↑↑ 11010000
|-
|{{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE}}
|10 ↑↑ 15.60465
|-
|{{TM|1RB0LF_1RC1RB_1LD0RA_1LB0LE_1RZ0LC_1LA1LF}}
|10 ↑↑ 7.52390
|-
|{{TM|1RB0LF_1RC1RB_1LD0RA_1RF0LE_1RZ0LC_1LA1LF}}
|10 ↑↑ 7.52390
|-
|{{TM|1RB0LF_1RC1RB_1LD0RA_1LF0LE_1RZ0LC_1LA1LF}}
|10 ↑↑ 7.52390
|-
|{{TM|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}
|10 ↑↑ 7.23619
|-
|{{TM|1RB1RA_1LC1LE_1RE0LD_1LC0LF_1RZ0RA_0RA0LB}}
|10 ↑↑ 6.96745
|-
|{{TM|1RB0RF_1LC0RA_1RZ0LD_1LE1LD_1RB1RC_0LD0RE}}
|10 ↑↑ 5.77573
|-
|{{TM|1RB0LA_1LC1LF_0LD0LC_0LE0LB_1RE0RA_1RZ1LD}}
|10 ↑↑ 5.63534
|-
|{{TM|1RB1RE_1LC1LF_1RD0LB_1LE0RC_1RA0LD_1RZ1LC}}
|10 ↑↑ 5.56344
|}
The runtimes are presumed to be about <math>\text{score}^2</math> which is roughly indistinguishable in tetration notation.

== Holdouts ==
@mxdys's informal [[Holdouts lists|holdouts list]] has 3335 machines up to equivalence as of June 2025.

== References ==
<references />
[[Category:BB Domain]]

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-26T10:36:49Z

Mxdys:

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Former [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 7 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
1RB1LE_1LC0RA_1RB1LD_1LC0LC_1RF0LB_---1RE
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
else: assert 0
</syntaxhighlight>

Tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

Execution process (shared by these 7 TMs):
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095,(W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>

In each iteration of (a,b,x) --> (a',b',x'), a'≈2^a. It halts after 11010064 iterations of (a,b,x) --> (a',b',x'). The omitted numbers in x and b are much smaller than a, and much larger than 66060289.

The halting config (reversed):<pre>
H3 = 0^inf 1 Z> 00110 110110^2
(W1,n,l) = l 110110^n 11 110110^2
(W2,n,l) = l 110110^n 1111 110110^3

(a,b,x) = x 110110^b 11^1+a 0^inf
</pre>

where a is between 10↑↑11010000 and 10↑↑11011000.

The first 6 TMs in this family have exactly the same halting config, while <code>1RB1LE_1LC0RA_1RB1LD_1LC0LC_1RF0LB_---1RE</code> has <code>H3 = 0^inf 1 Z> 11110 110110^2</code>, which results in a sigma score increase of 2.

We can see by looking at the space-time diagram that <code>1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB</code> takes the longest time to halt.

==References==
<references />

Main Page

2025-06-26T10:36:31Z

Mxdys:

The [[Busy Beaver function]] BB (called ''S'' originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to ''m''-symbol Turing machines:<ref>Rado, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x</ref><ref>Brady, Allen H, and the Meaning of Life, 'The Busy Beaver Game and the Meaning of Life', in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.</ref>

{| class="wikitable"
|-
| BB(''n'', ''m'') = Maximum number of steps taken by a halting ''n''-state, ''m''-symbol Turing machine starting from a blank (all 0) tape
|}

The 2-symbol case BB(''n'', 2) is abbreviated as BB(''n''). The busy beaver function is not computable, but a few of its values are known:

{| class="wikitable"
|+ Small busy beaver values<ref>P. Michel, "[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]".</ref>
! !!2-state!!3-state !!4-state!!5-state!!6-state
!7-state
|-
! 2-symbol
| [[BB(2)]] = 6
| [[BB(3)]] = 21
| [[BB(4)]] = 107
| [[BB(5)]] = 47,176,870
| style="background: orange;" | [[BB(6)]] > <math>2 \uparrow \uparrow \uparrow 5</math>
| style="background: #ffe4b2;" | [[BB(7)]] > <math>2 \uparrow^{11} 2 \uparrow^{11} 3</math>
|-
! 3-symbol
| [[BB(2,3)]] = 38
| style="background: orange;" | [[BB(3,3)]] > <math>10^{17}</math>
| style="background: #ffe4b2;" | [[BB(4,3)]] > <math>2 \uparrow\uparrow\uparrow 2^{2^{32}}</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|-
! 4-symbol
| [[BB(2,4)]] = 3,932,964
| style="background: #ffe4b2;" | [[BB(3,4)]] > <math>2 \uparrow^{15} 5</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|-
! 5-symbol
| style="background: orange;" | [[BB(2,5)]] > <math>10\uparrow\uparrow 4</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|-
! 6-symbol
| style="background: #ffe4b2;" | [[BB(2,6)]] > <math>10 \uparrow\uparrow\uparrow 3</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|}

In the above table, <span style="background: orange">cells are highlighted in orange</span> when there are known [[Cryptids]] (mathematically-hard machines) in that class, and <span style="background: #ffe4b2">cells are highlighted in light orange</span> when the existence of a Cryptid is given by using a known one with less states or symbols.

== About bbchallenge ==
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt. Other efforts also include:

* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Coq_(software) Coq])
* Maintaining [[Holdouts lists]] for small busy beaver values
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]
* Finding [[Cryptids]] (mathematically-hard machines)
* Searching for new [[Champions]]
* Building [[Accelerated Simulator]]s to simulate halting machines faster
* Writing papers and giving talks about busy beaver, see [[Papers & Talks]]

In June 2024, bbchallenge achieved a significant milestone by proving in Coq / Rocq that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.<ref>H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html</ref>

== Contribute to this wiki ==
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:

<inputbox>
type=create
width=100
break=no
buttonlabel=Create new article
default=(Article title)
</inputbox>

==Notes==
<references />

1RB1RA 1RC1RZ 1LD0RF 1RA0LE 0LD1RC 1RA0RE

2025-06-26T10:35:03Z

Mxdys: Created page with "{{machine|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE}} {{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}} Current BB(6) Champion. Discovered by mxdys on 25 June 2025. It's in a family of 4 machines with the halting time and sigma score between 2↑↑2↑↑2↑↑9 and 2↑↑2↑↑2↑↑10: <pre> 1RB1RA_1RC---_1LD0RF_1RA0LE_0LD1RC_1RA0RE (hereafter referred to as TM1) 1RB---_1LC0RF_1RE0LD_0LC1RB_1RA1RE_1RE0RD (TM2) 1RB0LE_1RC1RB_1RD---_1LA0RF_0LA1RD_1RB0RE (T..."

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-20T18:13:11Z

Mxdys:

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 7 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
1RB1LE_1LC0RA_1RB1LD_1LC0LC_1RF0LB_---1RE
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
else: assert 0
</syntaxhighlight>

Tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

Execution process (shared by these 7 TMs):
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095,(W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>

In each iteration of (a,b,x) --> (a',b',x'), a'≈2^a. It halts after 11010064 iterations of (a,b,x) --> (a',b',x'). The omitted numbers in x and b are much smaller than a, and much larger than 66060289.

The halting config (reversed):<pre>
H3 = 0^inf 1 Z> 00110 110110^2
(W1,n,l) = l 110110^n 11 110110^2
(W2,n,l) = l 110110^n 1111 110110^3

(a,b,x) = x 110110^b 11^1+a 0^inf
</pre>

where a is between 10↑↑11010000 and 10↑↑11011000.

The first 6 TMs in this family have exactly the same halting config, while <code>1RB1LE_1LC0RA_1RB1LD_1LC0LC_1RF0LB_---1RE</code> has <code>H3 = 0^inf 1 Z> 11110 110110^2</code>, which results in a sigma score increase of 2.

We can see by looking at the space-time diagram that <code>1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB</code> takes the longest time to halt.

==References==
<references />

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-18T05:47:07Z

Mxdys:

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
1RB1LE_1LC0RA_1RB1LD_1LC0LC_1RF0LB_---1RE
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
else: assert 0
</syntaxhighlight>

Tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

Execution process:
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095,(W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>

In each iteration of (a,b,x) --> (a',b',x'), a'≈2^a. It halts after 11010064 iterations of (a,b,x) --> (a',b',x'). The omitted numbers in x and b are much smaller than a, and much larger than 66060289.

==References==
<references />

[[Category:Stub]]

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-17T13:17:09Z

Mxdys: /* Analysis */

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
else: assert 0
</syntaxhighlight>

Tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

Execution process:
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095,(W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>

In each iteration of (a,b,x) --> (a',b',x'), a'≈2^a. It halts after 11010064 iterations of (a,b,x) --> (a',b',x'). The omitted numbers in x and b are much smaller than a, and much larger than 66060289.

==References==
<references />

[[Category:Stub]]

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-17T13:16:38Z

Mxdys: /* Analysis */

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
else: assert 0
</syntaxhighlight>

Tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

Execution process:
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095,(W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>

In each iteration of (a,b,x) --> (a',b',x'), a'≈2^a. It halts after 11010064 iterations of (a,b,x) --> (a',b',x').The omitted numbers in x and b are much smaller than a, and much larger than 66060289.

==References==
<references />

[[Category:Stub]]

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-17T12:44:49Z

Mxdys: fix typo

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
pass
else: assert 0
</syntaxhighlight>

tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

execution process:
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095,(W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>
==References==
<references />

[[Category:Stub]]

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-17T06:17:03Z

Mxdys:

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025.

It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
pass
else: assert 0
</syntaxhighlight>

tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

execution process:
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095 (W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>
==References==
<references />

[[Category:Stub]]

1RB1LC 1LA1RE 0RD0LA 1RZ1LB 1LD0RF 0RD1RB

2025-06-17T06:15:45Z

Mxdys: Created page with "{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}} {{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}} Current BB(6) Champion. Discovered by mxdys on 16 June 2025. The halting t steps. It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000: <pre> 1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB 1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB 1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB 1RB1LC_1LA1RD_1LA0LA_1LD0RE_0R..."

{{machine|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}}
{{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB|halt}}

Current [[BB(6)]] Champion. Discovered by mxdys on 16 June 2025. The halting t steps.

It's in a family of 6 machines with the halting time and sigma score between 10↑↑11010000 and 10↑↑11011000:
<pre>
1RB1LC_1LA1RE_0RD0LA_---1LB_1LD0RF_0RD1RB
1RB1LC_1LA1RE_0RD0LA_---1LB_1LE0RF_0RD1RB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LB
1RB1LC_1LA1RD_1LA0LA_1LD0RE_0RF1RB_---1LC
1RB1LC_1LA1RD_1LA0LA_1LF0RE_0RF1RB_---1LB
1RB1LD_1LC1RE_---1LD_1LA0LA_1LE0RF_0RC1RB
</pre>

== Analysis ==
It's behavior can be described by the python code below:
<syntaxhighlight lang="python" line="1">
def LBS(x):
if x=='H1': return 'H1',1,1
if x=='H2': return 'H4',2,2
if x=='H3': print('Halted'); exit()
if x=='H4': return 'H5',1,1
if x=='H5': return 'H1',1,2
w,k0,x = x
if w=='W1':
if k0==0 and x=='H1': return 'H5',1,2
x0,n,k = LBS(x)
k1=k0+k-2
return ('W1',k1,x0),((n+1)*(1<<k1)-2)*4+5,2
if w=='W2':
if k0==0 and x=='H1': return 'H3',1,3
if k0==1 and x=='H1': return 'H2',1,3
x0,n,k = LBS(x)
k1=k0+k-3
return ('W2',k1,x0),((n+1)*(1<<k1)-2)*8+5,3
assert 0

a,b,x = 57,5,'H1'
while 1:
r,m2 = a%3,a//3
x0,n,k = LBS(x)
if r==2:
m1 = b-3
a,b,x = ((((n+1)*(1<<(m1+k))-2)*4+2)*(1<<(m2+1))+1), m2, ('W2',m1+k,x0)
elif r==1:
m1 = b-2
a,b,x = ((((n+1)*(1<<(m1+k))-2)*2+2)*(1<<(m2+1))+1), m2, ('W1',m1+k,x0)
elif r==0:
m1 = b-1
a,b,x = ((((n+1)*(1<<(m1+k))-1))*(1<<(m2+1))+1), (m2+m1+k+1), x0
pass
else: assert 0
</syntaxhighlight>

tape encoding (reversed):
<pre>
H1 = 0^inf 1^7
H2 = 0^inf 1^5 101111 111111 1111
H3 = 0^inf 1^5 111111 1111
H4 = 0^inf 1^7 10 111111
H5 = 0^inf 1^7 111111
(W1,n,l) = l 111111^n 10 111111^2
(W2,n,l) = l 111111^n 1010 111111^3

(a,b,x) = x 111111^b <F0 11^1+a 0^inf
</pre>

execution process:
<pre>
(57,5,H1) -->
(66060289,25,H1) -->
(_,22020096,(W1,24,H1)) -->
(_,_,(W2,22020095 (W1,23,H1))) --> ...
(_,_,(...(W2,22020074,(W1,2,H1))...)) -->
(_,_,(...(W2,22020073,(W1,1,H1))...)) -->
(_,_,(...(W2,22020072,(W1,0,H1))...)) -->
(_,_,(...(W2,22020071,H5)...)) -->
(_,_,(...(W2,22020070,H1)...)) --> ...
(_,_,(...(W2,4,H1)...)) -->
(_,_,(...(W2,2,H1)...)) -->
(_,_,(...(W2,0,H1)...)) -->
(_,_,(...H3...)) -->
halt
</pre>
==References==
<references />

[[Category:Stub]]

Main Page

2025-06-17T05:11:46Z

Mxdys: update BB(6) lower bound

The [[Busy Beaver function]] BB (called ''S'' originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to ''m''-symbol Turing machines:<ref>Rado, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x</ref><ref>Brady, Allen H, and the Meaning of Life, 'The Busy Beaver Game and the Meaning of Life', in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.</ref>

{| class="wikitable"
|-
| BB(''n'', ''m'') = Maximum number of steps taken by a halting ''n''-state, ''m''-symbol Turing machine starting from a blank (all 0) tape
|}

The 2-symbol case BB(''n'', 2) is abbreviated as BB(''n''). The busy beaver function is not computable, but a few of its values are known:

{| class="wikitable"
|+ Small busy beaver values<ref>P. Michel, "[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]".</ref>
! !!2-state!!3-state !!4-state!!5-state!!6-state
!7-state
|-
! 2-symbol
| [[BB(2)]] = 6
| [[BB(3)]] = 21
| [[BB(4)]] = 107
| [[BB(5)]] = 47,176,870
| style="background: orange;" | [[BB(6)]] > <math>10 \uparrow \uparrow 10^7</math>
| style="background: #ffe4b2;" | [[BB(7)]] > <math>2 \uparrow^{11} 2 \uparrow^{11} 3</math>
|-
! 3-symbol
| [[BB(2,3)]] = 38
| style="background: orange;" | [[BB(3,3)]] > <math>10^{17}</math>
| style="background: #ffe4b2;" | [[BB(4,3)]] > <math>2 \uparrow\uparrow\uparrow 2^{2^{32}}</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|-
! 4-symbol
| [[BB(2,4)]] = 3,932,964
| style="background: #ffe4b2;" | [[BB(3,4)]] > <math>2 \uparrow^{15} 5</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|-
! 5-symbol
| style="background: orange;" | [[BB(2,5)]] > <math>10\uparrow\uparrow 4</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|-
! 6-symbol
| style="background: #ffe4b2;" | [[BB(2,6)]] > <math>10 \uparrow\uparrow\uparrow 3</math>
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
| style="background: #ffe4b2;" |
|}

In the above table, <span style="background: orange">cells are highlighted in orange</span> when there are known [[Cryptids]] (mathematically-hard machines) in that class, and <span style="background: #ffe4b2">cells are highlighted in light orange</span> when the existence of a Cryptid is given by using a known one with less states or symbols.

== About bbchallenge ==
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt. Other efforts also include:

* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Coq_(software) Coq])
* Maintaining [[Holdouts lists]] for small busy beaver values
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]
* Finding [[Cryptids]] (mathematically-hard machines)
* Searching for new [[Champions]]
* Building [[Accelerated Simulator]]s to simulate halting machines faster
* Writing papers and giving talks about busy beaver, see [[Papers & Talks]]

In June 2024, bbchallenge achieved a significant milestone by proving in Coq / Rocq that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.<ref>H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html</ref>

== Contribute to this wiki ==
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:

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==Notes==
<references />

Closed Tape Language

2025-05-08T18:35:00Z

Mxdys: /* DFA generator */

Closed tape language (CTL) is a class of methods to prove that a Turing machine does not stop. A Turing machine may go through an infinite number of configurations while running, but we can classify them into a finite number of classes in a suitable way, and prove that each configuration in a class reaches another configuration (instead of halting) in some class after running for one step.

== regular CTL ==
See also: https://www.sligocki.com/2022/06/10/ctl.html

We use two DFAs (one with state sets L and R respectively) and an accepting set A to describe each classes.

The elements in set A have the form (l,r,q,s): L×R×Q×Σ, where Q is the state set of the TM, Σ is the symbol set of the TM, and (l,r,q,s) means the class of configurations in the form of 0^inf x q> s y 0^inf s.t. the DFA L gets state l when the string x is input, the DFA R gets state r when the string y (reversed) is input.

A variant is use (l,r,q,d): L×R×Q×{-1,+1}, and d=-1 means 0^inf x <nowiki><q y 0^inf, d=+1 means 0^inf x q> y 0^inf.</nowiki>

=== DFA generator ===
Assume we have set S, function f:S×Σ→S and s0∈S, we can build a DFA incrementally: the starting state of the DFA is s0, the transtion (s0,0) is defined as s0, all other transitions are undefined at the beginning, and whenever we need to use a transition (a,b): S×Σ, we define it as f(a,b).

The DFA L,R and the accepting set A can be generated using this method.

Initially we have A={(s0,s0,q0,+1)}, where q0∈Q is the initial state of the TM, then we update A until it's closed under running for one step:

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

A state of a DFA is visited iff it appears at least once in A.

If we visit TM(q,s) and it's a halting transition, this method failed.

TM: Q×Σ→Q×Σ×{-1,+1} is the transition table of the turing machine.

This algorithm may not halt when |S| is infinite (or very slow when |S| is large), so we usually add a time limit or step limit to it.

==== RWL_mod ====
<pre>
s0 = []
f([(x,n,m)]+z,x) = g([(x,min(n+1,N),(m+1)%M)]+z)
f(z,x) = g([(x,1,1)]+z), otherwise
g(a+[b]) = a, len(a+[b])>=L
g(a) = a, len(a)<L
where L,N,M are parameters, typical values are N=2, M=1,2,3, 1<=L<=16.
</pre>

==== CPS_LRU ====
<pre>
s0 = ([],[],[])
f((s1,s2, s3),x) = (s1,s2,[x]+s3), len(s3)<N3
f((s1,s2, s3),x) = ([x]+s1,s2,s3), len(s1)<N1
f((s1,s2+[z], s3),x) = (s1',[y]+s2,s3), len(s2+[z])=N2 and y not in s2 and [x]+s1=s1'+[y]
f((s1,s2, s3),x) = (s1',[y]+s2,s3), y not in s2 and [x]+s1=s1'+[y]
f((s1,s2+[y]+s2',s3),x) = (s1',[y]+s2+s2',s3), y not in s2 and [x]+s1=s1'+[y]
when multiple rules match, take the first one.
s1 is a queue, s2 is an LRU cache, s3 is a stack that never pops.
N1,N2,N3 are parameters, typical values are N1=0,1, N3=0,1, N2=0,1,2,3.
</pre>

==== n-gram CPS ====
n-gram CPS is the special case of CPS_LRU where N2=N3=0, N1=n.

=== FAR-direct ===
TODO

See https://github.com/bbchallenge/bbchallenge-deciders/tree/main/decider-finite-automata-reduction

=== machine transform ===
n-gram CPS can be much more powerful if we transform the TM properly. Such a transformation should not cause a halting Turing machine to nonhalt.

==== macro machine ====
Divide the tape into fixed size blocks, each block is a symbol in the transformed machine.

==== local transition history ====
The Turing machine can be modified to write down and update extra data at each step. We use the f defined in CPS_LRU (update the extra data as f(h,(q,s)) when the machine is at state q, read symbol (s,h) where s is the original symbol and h is extra data).

== MITMWFAR ==
TODO

See https://github.com/Iijil1/MITMWFAR
[[Category:Deciders]]
[[Category:Stub]]

Busy Beaver for lambda calculus

2025-04-12T19:22:33Z

Mxdys: update BBλ(46) champion (trivial)

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > (2\uparrow\uparrow)^{15} 33 > 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{15} 33</math> || mxdys and racheline
|-
|41 || <math>\ge 5 \left(3^{3^{85}}\right) + 6 > 10^{10^{40}}</math>
| <code>(\1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{85}})</math>||mxdys
|-
|42 ||<math> \ge BB\lambda(40)+2</math>
|<code>\(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || ||
|-
|43 ||<math> > 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 8</math>
|<code>(\1 1) (\1 (\1 (\\2 (2 1)) 2))</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;(\lambda y.y\;C(2)\;T(n))</math> and <math>k > 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 8</math>||mxdys
|-
|44 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33</math> ||
|-
|45 || <math> \ge BB\lambda(43)+2</math>
| <code>\(\1 1) (\1 (\1 (\\2 (2 1)) 2))</code>
| || ||
|-
|46 || <math> \ge BB\lambda(44)+2</math>
| <code>\(\1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33 - 1} 33</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Busy Beaver for lambda calculus

2025-04-11T17:55:02Z

Mxdys: update BBλ(45) champion (trivial)

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > (2\uparrow\uparrow)^{15} 33 > 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{15} 33</math> || mxdys and racheline
|-
|41 || <math>\ge 5 \left(3^{3^{85}}\right) + 6 > 10^{10^{40}}</math>
| <code>(\1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{85}})</math>||mxdys
|-
|42 ||<math> \ge BB\lambda(40)+2</math>
|<code>\(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || ||
|-
|43 ||<math> > 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 8</math>
|<code>(\1 1) (\1 (\1 (\\2 (2 1)) 2))</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;(\lambda y.y\;C(2)\;T(n))</math> and <math>k > 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 8</math>||mxdys
|-
|44 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33</math> ||
|-
|45 || <math> \ge BB\lambda(43)+2</math>
| <code>\(\1 1) (\1 (\1 (\\2 (2 1)) 2))</code>
| || ||
|-
|46 ||
|
| || ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33 - 1} 33</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Busy Beaver for lambda calculus

2025-04-11T13:46:54Z

Mxdys: update BBλ(43) champion

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > (2\uparrow\uparrow)^{15} 33 > 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{15} 33</math> || mxdys and racheline
|-
|41 || <math>\ge 5 \left(3^{3^{85}}\right) + 6 > 10^{10^{40}}</math>
| <code>(\1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{85}})</math>||mxdys
|-
|42 ||<math> \ge BB\lambda(40)+2</math>
|<code>\(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || ||
|-
|43 ||<math> > 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 8</math>
|<code>(\1 1) (\1 (\1 (\\2 (2 1)) 2))</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;(\lambda y.y\;C(2)\;T(n))</math> and <math>k > 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 8</math>||mxdys
|-
|44 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33</math> ||
|-
|45 || <math> > 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3) > 10 \uparrow\uparrow\uparrow (7.6 \times 10^{12})</math>
| <code>(\1 1 1) (\1 (\\2 (2 (2 1))) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(3)\;T(n)</math> and <math>k > 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3)</math> ||
|-
|46 ||
|
| || ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33 - 1} 33</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Busy Beaver for lambda calculus

2025-04-11T10:54:05Z

Mxdys: update BBλ(42) champion (trivial)

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > (2\uparrow\uparrow)^{15} 33 > 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{15} 33</math> || mxdys and racheline
|-
|41 || <math>\ge 5 \left(3^{3^{85}}\right) + 6 > 10^{10^{40}}</math>
| <code>(\1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{85}})</math>||mxdys
|-
|42 ||<math> \ge BB\lambda(40)+2</math>
|<code>\(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || ||
|-
|43 ||
|
| || ||
|-
|44 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33</math> ||
|-
|45 || <math> > 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3) > 10 \uparrow\uparrow\uparrow (7.6 \times 10^{12})</math>
| <code>(\1 1 1) (\1 (\\2 (2 (2 1))) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(3)\;T(n)</math> and <math>k > 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3)</math> ||
|-
|46 ||
|
| || ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33 - 1} 33</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Busy Beaver for lambda calculus

2025-04-11T10:17:11Z

Mxdys: add champion of size 41

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > (2\uparrow\uparrow)^{15} 33 > 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{15} 33</math> || mxdys and racheline
|-
|41 || <math>\ge 5 \left(3^{3^{85}}\right) + 6 > 10^{10^{40}}</math>
| <code>(\1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{85}})</math>||mxdys
|-
|42 ||
|
| || ||
|-
|43 ||
|
| || ||
|-
|44 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33</math> ||
|-
|45 || <math> > 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3) > 10 \uparrow\uparrow\uparrow (7.6 \times 10^{12})</math>
| <code>(\1 1 1) (\1 (\\2 (2 (2 1))) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(3)\;T(n)</math> and <math>k > 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3)</math> ||
|-
|46 ||
|
| || ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 10 \uparrow\uparrow\uparrow 16</math>
| <code>(\1 1 1 1 1) (\1 (\\2 (2 1)) 1)</code>
| || <math>\lambda x.T(k)</math> where <math>T(0)=x,\;T(n+1)=T(n)\;C(2)\;T(n)</math> and <math>k > (2\uparrow\uparrow)^{(2\uparrow\uparrow)^{(2\uparrow\uparrow)^{15} 33 - 1} 33 - 1} 33</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Busy Beaver for lambda calculus

2025-04-10T20:04:08Z

Mxdys: update BBλ(40) champion

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > 3^{2^{2^{2^{32}+4}}} > 10^{10^{10^{10^{9}}}}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))) 1)</code>
| || <math>\lambda x.T(2^{2^{2^{32}+4}}),\;where\;T(0)=x,\;T(n+1)=T(n)\;(T(n)\;C(2))\;T(n)</math>||mxdys
|-
|41 ||
|
| || ||
|-
|42 || <math>\ge 5 \left(2 \uparrow\uparrow 6\right) + 6 > 10^{10^{10^4}}</math>
| <code>(\1 1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2 \uparrow\uparrow 6)</math> ||
|-
|43 || <math>\ge 5 \left(3 \uparrow\uparrow 5\right) + 6 > 10^{10^{10^{12}}}</math>
| <code>(\1 1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3 \uparrow\uparrow 5)</math> ||
|-
|44 || <math> > (2\uparrow)^{14} 2059 > 10 \uparrow\uparrow 15</math>
| <code>(\1 1 1 (\1 1) 1) (\\2 (2 1))</code>
| || <math>C((\lambda x.x^x)^{16}(2))</math> ||
|-
|45 || <math> > 3 \uparrow\uparrow 28 > 10 \uparrow\uparrow 27</math>
| <code>(\1 1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C((\lambda x.x^x)^{27}(3))</math> ||
|-
|46 || <math> > 256 \uparrow\uparrow 257 > 10 \uparrow\uparrow 257</math>
| <code>(\1 (\1 (\1 1) 1) 1) (\\2 (2 1))</code>
| || <math>C((\lambda x.x^x)^{256}(256))</math> ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > (2\uparrow)^{65534} 2059 > 10 \uparrow\uparrow 65535</math>
| <code>(\1 1 1 1 (\1 1) 1) (\\2 (2 1))</code>
| || <math>C((\lambda x.x^x)^{65536}(2))</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Busy Beaver for lambda calculus

2025-04-10T18:26:12Z

Mxdys: fix typo in BBλ(36)

'''Busy Beaver for lambda calculus''' ('''BBλ''') is a variation of the [[Busy Beaver]] problem for [https://en.wikipedia.org/wiki/Lambda_calculus lambda calculus] invented by John Tromp. BBλ(n) = the maximum normal form size of any closed lambda term of size n. If you are not familiar with lambda calculus and beta-reduction, I recommend starting with that article.

Size is measured in bits using [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus] which is a binary prefix-free encoding for all closed lambda calculus terms.

== Analogy to Turing machines ==
We evaluate terms by applying ''beta-reductions'' until they reach a ''normal form''. As an analogy to [[Turing machines]]:
* ''Lambda terms'' are like TM configurations (tape + state + position).
* Applying ''beta-reduction'' to a term is like taking a TM step.
* A term is in ''normal form'' if no beta-reductions can be applied. This is like saying the term has halted.
* A term may or may not be reducible to a normal form. If it is, this is like saying the term halts.
* Determining whether a term is reducible to a normal form is an undecidable problem equivalent to the halting problem.

Note: That unlike for Turing machines, evaluating lambda terms is non-deterministic. Specifically, there may be multiple beta-reductions possible in a given term. However, if a term can be reduced to a normal form, that normal form is unique. It is not possible to reduce the original term to any different normal form. A term is '''strongly normalizing''' if any choice of beta-reductions will lead to this normal form and '''weakly normalizing''' if there exist divergent reduction paths which never reach the normal form.

== Binary Lambda Encoding ==
A lambda term using [https://en.wikipedia.org/wiki/De_Bruijn_indices De Bruijn indexes] is defined inductively as:
* Variables: For any <math>n \in \mathbb{Z}^+</math>, Var(''n'') is a term. It represents a variable bound by the lambda expression ''n'' above this one (the De Bruijn index). It is typically written simply as <code>n</code>.
* Lambdas: For any term ''T'', Lam(''T'') is a term. It represents a unary function with function body ''T''. It is typically written <math>\lambda T</math> or <code>\T</code>.
* Applications: For any terms ''T, U'', App(''T, U'') is a term. It represents applying function ''T'' to argument ''U''. It is typically written <code>(T U)</code>.

We can think of this as a tree where each variable is a leaf, a lambda is a node with one child and applications are nodes with 2 children. A term is '''closed''' if every variable is bound. In other words, for every Var(''n'') leaf node, there exists ''n'' Lam() nodes above it in the tree of the term.

Encoding (''blc()'') is defined recursively:
<math display="block">\begin{array}{l}
blc(Var(n)) & = & 1^n 0 \\
blc(Lam(T)) & = & 00 \; blc(T) \\
blc(App(T, U)) & = & 01 \; blc(T) \; blc(U) \\
\end{array}</math>

For example, the [https://en.wikipedia.org/wiki/Church_encoding#Church_numerals Church numeral] 2: <math>\lambda f x. (f \; (f \; x))</math> = <code>\\(2 (2 1))</code> = <code>Lam(Lam(App(Var(2), App(Var(2), Var(1))))</code> is encoded as <code>00 00 01 110 01 110 10</code> or simply <code>0000011100111010</code> (spaces are not part of the encoding, only used for demonstration purposes) and thus has size 16 bits.

== Text Encoding conventions ==
For human readability, a text encoding and set of conventions is used in this article. As described earlier we encode a lambda term as:
* Var(''n'') -> <code>n</code>
* Lam(''T'') -> <code>(\T)</code>
* App(''T, U'') -> <code>(T U)</code>

However, parentheses are also dropped in certain cases by convention:
* The outermost parentheses are dropped: <code>Lam(1)</code> -> <code>\1</code> and <code>App(1, 2)</code> -> <code>1 2</code>.
* Parentheses are dropped immediately inside a Lam: <code>Lam(Lam(1))</code> -> <code>\\1</code> and <code>Lam(App(1, 1))</code> -> <code>\1 1</code>.
* Parentheses are dropped in nested Apps using left associativity: <code>App(App(1, 2), 3)</code> -> <code>1 2 3</code>. (Note: parentheses are still required for <code>App(1, App(2, 3))</code> -> <code>1 (2 3)</code>).

This is the convention used in John Tromp's code and so is used here for consistency.

== Champions ==
There are no closed lambda terms of size 0, 1, 2, 3 or 5 and so BBλ(n) is not defined for those values. The smallest closed lambda term is <code>\1</code> which has size 4.

For the rest of n ≤ 20: BBλ(n) = n is trivial and can be achieved via picking any n bit term already in normal form. For example <code>\\...\1</code> and <code>\\...\2</code> with k lambdas has size 2k+2 and 2k+3 respectively (for k ≥ 1 and k ≥ 2 respectively).

{| class="wikitable"
|+
!n
!BBλ(n)
!Champion
!# Beta reductions
!Normal form
!Discovered By
|-
|21 || = 22 || <code>\(\1 1) (1 (\2))</code>
|1|| <code>\(1 (\2)) (1 (\2))</code> || John Tromp & Bertram Felgenhauer
|-
|22 || = 24 || <code>\(\1 1) (1 (\\1))</code>
|1|| <code>\(1 (\\1)) (1 (\\1))</code> ||Shawn Ligocki
|-
| || || <code>\(\1 1 1) (1 1)</code>
|1|| <code>\(1 1) (1 1) (1 1)</code> || John Tromp & Bertram Felgenhauer
|-
|23 || = 26 || <code>\(\1 1) (1 (\\2))</code>
|1|| <code>\(1 (\\2)) (1 (\\2))</code> || John Tromp & Bertram Felgenhauer
|-
|24 || = 30 || <code>\(\1 1 1) (1 (\1))</code>
|1|| <code>\(1 (\1)) (1 (\1)) (1 (\1))</code> || John Tromp & Bertram Felgenhauer
|-
|25 || = 42 || <code>\(\1 1) (\1 (2 1))</code>
|3|| <code>\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|26 || = 52 || <code>(\1 1) (\\2 (1 2))</code>
|3|| <code>\\2 (\\2 (1 2)) (1 (2 (\\2 (1 2))))</code> || John Tromp & Bertram Felgenhauer
|-
|27 || = 44 || <code>\\(\1 1) (\1 (2 1))</code>
|3|| <code>\\1 (\1 (2 1)) (1 (1 (\1 (2 1))))</code> || John Tromp & Bertram Felgenhauer
|-
|28 || = 58 || <code>\(\1 1) (\1 (2 (\2))))</code>
|3|| <code>\1 (\\1 (3 (\2))) (1 (\2 (\\1 (4 (\2)))))</code>|| John Tromp & Bertram Felgenhauer
|-
| 29 || = 223|| <code>\(\1 1) (\1 (1 (2 1)))</code>
|4|| <pre>
\B (B (1 B))
where:
B = (A (A (1 A)))
A = (1 (\1 (1 (2 1))))
</pre>||John Tromp & Bertram Felgenhauer
|-
|30
|= 160
|<code>(\1 1 1) (\\2 (1 2))</code>
|7
|<pre>
\\2 B A (1 (2 B A))
where:
B = (\\2 A (1 (2 A)))
A = (\\2 (1 2))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|
|
|<code>(\1 (1 1)) (\\2 (1 2))</code>
|5
|Same as above
|Shawn Ligocki
|-
|31
|= 267
|<code>(\1 1) (\\2 (2 (1 2)))</code>
|6
|<pre>
\\2 A (2 A (C (2 A)))
where:
C = (2 A (2 A (1 B (2 A))))
B = (\3 A (3 A (1 (3 A))))
A = (\\2 (2 (1 2)))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|32
|= 298
|<code>\(\1 1) (\1 (1 (2 (\2))))</code>
|
|
|John Tromp & Bertram Felgenhauer
|-
|33
|= 1812
|<code>\(\1 1) (\1 (1 (1 (2 1))))</code>
|5
|<pre>
\C (C (C (1 C)))
where:
C = (B (B (B (1 B)))
B = (A (A (A (1 A)))
A = (1 (\1 (1 (1 (2 1)))))
</pre>
|John Tromp & Bertram Felgenhauer
|-
|34 || <math>= \ 5 \left(2^{2^{2^2}}\right) + 6 = 327\,686</math>
| <code>(\1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^2}})</math> || John Tromp & Bertram Felgenhauer
|-
|35 || <math>\ge 5 \left(3^{3^3}\right) + 6 > 3.8 \times 10^{13}</math>
| <code>(\1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^3})</math> || John Tromp & Bertram Felgenhauer
|-
|36 || <math>\ge 5 \left(2^{2^{2^3}}\right) + 6 > 5.7 \times 10^{77}</math>
| <code>(\1 1) (\1 (1 (\\2 (2 1))))</code>
| || <math>C(2^{2^{2^3}})</math>|| John Tromp & Bertram Felgenhauer
|-
|37 ||
|
| || ||
|-
|38 || <math>\ge 5 \left(2^{2^{2^{2^2}}}\right) + 6 > 10^{10^4}</math>
| <code>(\1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2^{2^2}}})</math> || John Tromp & Bertram Felgenhauer
|-
|39 || <math>\ge 5 \left(3^{3^{3^3}}\right) + 6 > 10^{10^{12}}</math>
| <code>(\1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3^{3^{3^3}})</math> || John Tromp & Bertram Felgenhauer
|-
|40 || <math> > 5 \left(2^{2^{2059}}\right) + 6 > 10^{10^{619}}</math>
| <code>(\1 1 (\1 1) 1) (\\2 (2 1))</code>
| || <math>C(2^{2^{2059}})</math> ||
|-
|41 ||
|
| || ||
|-
|42 || <math>\ge 5 \left(2 \uparrow\uparrow 6\right) + 6 > 10^{10^{10^4}}</math>
| <code>(\1 1 1 1 1 1) (\\2 (2 1))</code>
| || <math>C(2 \uparrow\uparrow 6)</math> ||
|-
|43 || <math>\ge 5 \left(3 \uparrow\uparrow 5\right) + 6 > 10^{10^{10^{12}}}</math>
| <code>(\1 1 1 1 1) (\\2 (2 (2 1)))</code>
| || <math>C(3 \uparrow\uparrow 5)</math> ||
|-
|44 || <math> > (2\uparrow)^{14} 2059 > 10 \uparrow\uparrow 15</math>
| <code>(\1 1 1 (\1 1) 1) (\\2 (2 1))</code>
| || <math>C((\lambda x.x^x)^{16}(2))</math> ||
|-
|45 || <math> > 3 \uparrow\uparrow 28 > 10 \uparrow\uparrow 27</math>
| <code>(\1 1 (\1 1) 1) (\\2 (2 (2 1)))</code>
| || <math>C((\lambda x.x^x)^{27}(3))</math> ||
|-
|46 || <math> > 256 \uparrow\uparrow 257 > 10 \uparrow\uparrow 257</math>
| <code>(\1 (\1 (\1 1) 1) 1) (\\2 (2 1))</code>
| || <math>C((\lambda x.x^x)^{256}(256))</math> ||
|-
|47 ||
|
| || ||
|-
|48 || <math> > (2\uparrow)^{65534} 2059 > 10 \uparrow\uparrow 65535</math>
| <code>(\1 1 1 1 (\1 1) 1) (\\2 (2 1))</code>
| || <math>C((\lambda x.x^x)^{65536}(2))</math> ||
|-
|49
|<math>> f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right) > \text{Graham's number}</math>
|<code>(\1 1) (\1 (1 (\\1 2 (\\2 (2 1)))))</code>
|
|<math>C(N) \text{ for } N \approx f_{\omega+1}\left(\frac{2 \uparrow\uparrow 6}{2}\right)</math>
|[https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam Gustavo Melo]
|-
|...
|
|
|
|
|
|-
|1850
|> Loader's number
|<code>too large to show</code>
|
|
|[https://codegolf.stackexchange.com/questions/176966/golf-a-number-bigger-than-loaders-number/274634#274634 John Tromp]
|}

Where <math>C(n)</math> represents the Church numeral ''n'' (<math>\lambda f x. f^n(x)</math>) written as <code>\\2 (2 ... (2 1)...)</code> with ''n'' 2s in this text representation.

== See Also ==

* https://oeis.org/A333479
* [https://tromp.github.io/blog/2023/11/24/largest-number The largest number representable in 64 bits]. 24 Nov 2023. John Tromp.
* [https://tromp.github.io/cl/Binary_lambda_calculus.html Binary Lambda Calculus]. John Tromp.
* https://github.com/tromp/AIT/tree/master/BB

Sequences

2025-04-07T23:50:24Z

Mxdys: group by position in arithmetical hierarchy

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 Turing machine halts in exactly a(n) steps on an initially blank tape.
|LB(1)=2, LB(2)=7, LB(3)=22, LB(4)=72, LB(5)=427
|[[oeis:A337805|A337805]]
|}

=== Noncomputable Sequences ===
The following sequences depend on the specific behavior of programs.

TODO: group by position in arithmetical hierarchy
{| class="wikitable"
!Sequence Name
!Symbol
!Description
!Values
![[oeis:|OEIS]] sequence
!Position in Arithmetical Hierarchy
|-
|[[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]]
|Π1
|-
|[[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]]
|Π1
|-
|
|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
|BB_SPACE(1,2)=2, BB_SPACE(2,2)=4, BB_SPACE(3,2)=7, BB_SPACE(4,2)=16
| -
|Π1
|-
|
|
|Number of n-state Turing machines which halt.
|
|[[oeis:A004147|A004147]]
|Π1
|-
|[[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]]
| -
|Π2
|-
|[https://nickdrozd.github.io/2021/02/14/blanking-beavers.html Blanking Beavers]
|
|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)
|
| -
|Π1
|-
|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])
|
|Π1
|-
|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
|
|
|Π1
|-
|Size of the Runtime Spectrum
|
|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]"
|
|Π1
|-
|
|#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
| -
|
|-
|
|#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 number of non-halting programs with n states which reach infinitely many tape cells
|
| -
|Π1
|-
|
|
|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

=== 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"]]

Closed Tape Language

2025-03-11T18:00:26Z

Mxdys: fix RWL_mod definition

Closed tape language (CTL) is a class of methods to prove that a Turing machine does not stop. A Turing machine may go through an infinite number of configurations while running, but we can classify them into a finite number of classes in a suitable way, and prove that each configuration in a class reaches another configuration (instead of halting) in some class after running for one step.

== regular CTL ==
See also: https://www.sligocki.com/2022/06/10/ctl.html

We use two DFAs (one with state sets L and R respectively) and an accepting set A to describe each classes.

The elements in set A have the form (l,r,q,s): L×R×Q×Σ, where Q is the state set of the TM, Σ is the symbol set of the TM, and (l,r,q,s) means the class of configurations in the form of 0^inf x q> s y 0^inf s.t. the DFA L gets state l when the string x is input, the DFA R gets state r when the string y (reversed) is input.

A variant is use (l,r,q,d): L×R×Q×{-1,+1}, and d=-1 means 0^inf x <nowiki><q y 0^inf, d=+1 means 0^inf x q> y 0^inf.</nowiki>

=== DFA generator ===
Assume we have set S, function f:S×Σ→S and s0∈S, we can build a DFA incrementally: the starting state of the DFA is s0, the transtion (s0,0) is defined as s0, all other transitions are undefined at the beginning, and whenever we need to use a transition (a,b): S×Σ, we define it as f(a,b).

The DFA L,R and the accepting set A can be generated using this method.

Initially we have A={(s0,s0,A,+1)}, then we update A until it's closed under running for one step:

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

A state of a DFA is visited iff it appears at least once in A.

If we visit TM(q,s) and it's a halting transition, this method failed.

TM: Q×Σ→Q×Σ×{-1,+1} is the transition table of the turing machine.

This algorithm may not halt when |S| is infinite (or very slow when |S| is large), so we usually add a time limit or step limit to it.

==== RWL_mod ====
<pre>
s0 = []
f([(x,n,m)]+z,x) = g([(x,min(n+1,N),(m+1)%M)]+z)
f(z,x) = g([(x,1,1)]+z), otherwise
g(a+[b]) = a, len(a+[b])>=L
g(a) = a, len(a)<L
where L,N,M are parameters, typical values are N=2, M=1,2,3, 1<=L<=16.
</pre>

==== CPS_LRU ====
<pre>
s0 = ([],[],[])
f((s1,s2, s3),x) = (s1,s2,[x]+s3), len(s3)<N3
f((s1,s2, s3),x) = ([x]+s1,s2,s3), len(s1)<N1
f((s1,s2+[z], s3),x) = (s1',[y]+s2,s3), len(s2+[z])=N2 and y not in s2 and [x]+s1=s1'+[y]
f((s1,s2, s3),x) = (s1',[y]+s2,s3), y not in s2 and [x]+s1=s1'+[y]
f((s1,s2+[y]+s2',s3),x) = (s1',[y]+s2+s2',s3), y not in s2 and [x]+s1=s1'+[y]
when multiple rules match, take the first one.
s1 is a queue, s2 is an LRU cache, s3 is a stack that never pops.
N1,N2,N3 are parameters, typical values are N1=0,1, N3=0,1, N2=0,1,2,3.
</pre>

==== n-gram CPS ====
n-gram CPS is the special case of CPS_LRU where N2=N3=0, N1=n.

=== FAR-direct ===
TODO

See https://github.com/bbchallenge/bbchallenge-deciders/tree/main/decider-finite-automata-reduction

=== machine transform ===
n-gram CPS can be much more powerful if we transform the TM properly. Such a transformation should not cause a halting Turing machine to nonhalt.

==== macro machine ====
Divide the tape into fixed size blocks, each block is a symbol in the transformed machine.

==== local transition history ====
The Turing machine can be modified to write down and update extra data at each step. We use the f defined in CPS_LRU (update the extra data as f(h,(q,s)) when the machine is at state q, read symbol (s,h) where s is the original symbol and h is extra data).

== MITMWFAR ==
TODO

See https://github.com/Iijil1/MITMWFAR
[[Category:Deciders]]
[[Category:Stub]]

Closed Tape Language

2025-03-11T02:02:29Z

Mxdys: fix format

Closed tape language (CTL) is a class of methods to prove that a Turing machine does not stop. A Turing machine may go through an infinite number of configurations while running, but we can classify them into a finite number of classes in a suitable way, and prove that each configuration in a class reaches another configuration (instead of halting) in some class after running for one step.

== regular CTL ==
See also: https://www.sligocki.com/2022/06/10/ctl.html

We use two DFAs (one with state sets L and R respectively) and an accepting set A to describe each classes.

The elements in set A have the form (l,r,q,s): L×R×Q×Σ, where Q is the state set of the TM, Σ is the symbol set of the TM, and (l,r,q,s) means the class of configurations in the form of 0^inf x q> s y 0^inf s.t. the DFA L gets state l when the string x is input, the DFA R gets state r when the string y (reversed) is input.

A variant is use (l,r,q,d): L×R×Q×{-1,+1}, and d=-1 means 0^inf x <nowiki><q y 0^inf, d=+1 means 0^inf x q> y 0^inf.</nowiki>

=== DFA generator ===
Assume we have set S, function f:S×Σ→S and s0∈S, we can build a DFA incrementally: the starting state of the DFA is s0, the transtion (s0,0) is defined as s0, all other transitions are undefined at the beginning, and whenever we need to use a transition (a,b): S×Σ, we define it as f(a,b).

The DFA L,R and the accepting set A can be generated using this method.

Initially we have A={(s0,s0,A,+1)}, then we update A until it's closed under running for one step:

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

A state of a DFA is visited iff it appears at least once in A.

If we visit TM(q,s) and it's a halting transition, this method failed.

TM: Q×Σ→Q×Σ×{-1,+1} is the transition table of the turing machine.

This algorithm may not halt when |S| is infinite (or very slow when |S| is large), so we usually add a time limit or step limit to it.

==== RWL_mod ====
<pre>
s0 = ()
f((z,(x,n,m)),x) = (z,(x,min(n+1,N),(m+1)%M))
f(z,x) = (z,(x,1,1)), otherwise
where N,M are parameters, typical values are N=2, M=1,2,3.
</pre>

==== CPS_LRU ====
<pre>
s0 = ([],[],[])
f((s1,s2, s3),x) = (s1,s2,[x]+s3), len(s3)<N3
f((s1,s2, s3),x) = ([x]+s1,s2,s3), len(s1)<N1
f((s1,s2+[z], s3),x) = (s1',[y]+s2,s3), len(s2+[z])=N2 and y not in s2 and [x]+s1=s1'+[y]
f((s1,s2, s3),x) = (s1',[y]+s2,s3), y not in s2 and [x]+s1=s1'+[y]
f((s1,s2+[y]+s2',s3),x) = (s1',[y]+s2+s2',s3), y not in s2 and [x]+s1=s1'+[y]
when multiple rules match, take the first one.
s1 is a queue, s2 is an LRU cache, s3 is a stack that never pops.
N1,N2,N3 are parameters, typical values are N1=0,1, N3=0,1, N2=0,1,2,3.
</pre>

==== n-gram CPS ====
n-gram CPS is the special case of CPS_LRU where N2=N3=0, N1=n.

=== FAR-direct ===
TODO

See https://github.com/bbchallenge/bbchallenge-deciders/tree/main/decider-finite-automata-reduction

=== machine transform ===
n-gram CPS can be much more powerful if we transform the TM properly. Such a transformation should not cause a halting Turing machine to nonhalt.

==== macro machine ====
Divide the tape into fixed size blocks, each block is a symbol in the transformed machine.

==== local transition history ====
The Turing machine can be modified to write down and update extra data at each step. We use the f defined in CPS_LRU (update the extra data as f(h,(q,s)) when the machine is at state q, read symbol (s,h) where s is the original symbol and h is extra data).

== MITMWFAR ==
TODO

See https://github.com/Iijil1/MITMWFAR
[[Category:Deciders]]
[[Category:Stub]]

Closed Tape Language

2025-03-10T18:07:28Z

Mxdys: add some MitM CTL tricks used in BB(6)

Closed tape language (CTL) is a class of methods to prove that a Turing machine does not stop. A Turing machine may go through an infinite number of configurations while running, but we can classify them into a finite number of classes in a suitable way, and prove that each configuration in a class reaches another configuration (instead of halting) in some class after running for one step.

== regular CTL ==
See also: https://www.sligocki.com/2022/06/10/ctl.html
We use two DFAs (one with state sets L and R respectively) and an accepting set A to describe each classes.

The elements in set A have the form (l,r,q,s): L×R×Q×Σ, where Q is the state set of the TM, Σ is the symbol set of the TM, and (l,r,q,s) means the class of configurations in the form of 0^inf x q> s y 0^inf s.t. the DFA L gets state l when the string x is input, the DFA R gets state r when the string y (reversed) is input.

A variant is use (l,r,q,d): L×R×Q×{-1,+1}, and d=-1 means 0^inf x <nowiki><q y 0^inf, d=+1 means 0^inf x q> y 0^inf.</nowiki>

=== DFA generator ===
Assume we have set S, function f:S×Σ→S and s0∈S, we can build a DFA incrementally: the starting state of the DFA is s0, the transtion (s0,0) is defined as s0, all other transitions are undefined at the beginning, and whenever we need to use a transition (a,b): S×Σ, we define it as f(a,b).

The DFA L,R and the accepting set A can be generated using this method.

Initially we have A={(s0,s0,A,+1)}, then we update A until it's closed under running for one step:

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

If (l,f(r,s),q,+1)∈A, and r is a visited state in R, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',-1), we add (l,f(r,s'),q',-1) to A.

If (f(l,s),r,q,-1)∈A, and l is a visited state in L, and TM(q,s) = (q',s',+1), we add (f(l,s'),r,q',+1) to A.

A state of a DFA is visited iff it appears at least once in A.

If we visit TM(q,s) and it's a halting transition, this method failed.

TM: Q×Σ→Q×Σ×{-1,+1} is the transition table of the turing machine.

This algorithm may not halt when |S| is infinite (or very slow when |S| is large), so we usually add a time limit or step limit to it.

==== RWL_mod ====
<pre>
s0 = ()
f((z,(x,n,m)),x) = (z,(x,min(n+1,N),(m+1)%M))
f(z,x) = (z,(x,1,1)), otherwise
where N,M are parameters, typical values are N=2, M=1,2,3.
</pre>

==== CPS_LRU ====
<pre>
s0 = ([],[],[])
f((s1,s2, s3),x) = (s1,s2,[x]+s3), len(s3)<N3
f((s1,s2, s3),x) = ([x]+s1,s2,s3), len(s1)<N1
f((s1,s2+[z], s3),x) = (s1',[y]+s2,s3), len(s2+[z])=N2 and y not in s2 and [x]+s1=s1'+[y]
f((s1,s2, s3),x) = (s1',[y]+s2,s3), y not in s2 and [x]+s1=s1'+[y]
f((s1,s2+[y]+s2',s3),x) = (s1',[y]+s2+s2',s3), y not in s2 and [x]+s1=s1'+[y]
when multiple rules match, take the first one.
s1 is a queue, s2 is an LRU cache, s3 is a stack that never pops.
N1,N2,N3 are parameters, typical values are N1=0,1, N3=0,1, N2=0,1,2,3.
</pre>

==== n-gram CPS ====
n-gram CPS is the special case of CPS_LRU where N2=N3=0, N1=n.

=== FAR-direct ===
TODO

See https://github.com/bbchallenge/bbchallenge-deciders/tree/main/decider-finite-automata-reduction

=== machine transform ===
n-gram CPS can be much more powerful if we transform the TM properly. Such a transformation should not cause a halting Turing machine to nonhalt.

==== macro machine ====
Divide the tape into fixed size blocks, each block is a symbol in the transformed machine.

==== local transition history ====
The Turing machine can be modified to write down and update extra data at each step. We use the f defined in CPS_LRU (update the extra data as f(h,(q,s)) when the machine is at state q, read symbol (s,h) where s is the original symbol and h is extra data).

== MITMWFAR ==
TODO

See https://github.com/Iijil1/MITMWFAR
[[Category:Deciders]]
[[Category:Stub]]

1RB0LB 1LC0RE 1LA1LD 0LC--- 0RB0RF 1RE1RB

2024-11-25T10:41:57Z

Mxdys: add classic cryptid #5

{{machine|1RB0LB_1LC0RE_1LA1LD_0LC---_0RB0RF_1RE1RB}}
This is a [[BB(6)]] [[Cryptid]] found by @racheline and shared on Discord on 25 Nov 2024.

== Rules ==
<pre>
start: (4,124)
(a+1,3n+0) --> (a+7,4n+4)
(a+1,3n+1) --> (a+0,4n+7)
(a+1,3n+2) --> (a+5,4n+4)
(0,3(27n+21)+1) --> halt

(a,n) := 0^inf 10 01^a 0^n <A 11011 0^inf
</pre>
For simplicity only one potential halting rule is shown. In fact, it also has the potential to become a translated cycler or bouncer when it reaches (0,n).

BB(6)

2024-11-25T10:36:14Z

Mxdys: add classic cryptid #5

The 6-state, 2-symbol Busy Beaver problem '''BB(6)''' is unsolved. With the discovery of [[Antihydra]] in 2024, we now know that we must solve a [[Collatz-like]] problem in order to solve BB(6).

The current BB(6) champion {{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE|halt}} was discovered by Pavel Kropitz in 2022 proving the lower bound:<ref>Shawn Ligocki. 2022. "BB(6, 2) > 10↑↑15". https://www.sligocki.com/2022/06/21/bb-6-2-t15.html</ref>

<math display="block">S(6) > \Sigma(6) > 10 \uparrow\uparrow 15</math>

== Techniques ==
In order to simulate the current BB(6) champion requires [[Accelerated simulator|accelerated simulation]] that can handle Collatz Level 2 [[Inductive rule|inductive rules]]. In other words, it requires a simulator that can prove the rules:

<math display="block">\begin{array}{l}
C(4k) & \to & Halt(\frac{3^{k+3} - 11}{2}) \\
C(4k+1) & \to & C(\frac{3^{k+3} - 11}{2}) \\
C(4k+2) & \to & C(\frac{3^{k+3} - 11}{2}) \\
C(4k+3) & \to & C(\frac{3^{k+3} + 1}{2}) \\
\end{array}</math>

and also compute the remainder mod 3 of numbers produced by applying these rules 15 times (which requires some fancy math related to [[wikipedia:Euler's_totient_function|Euler's_totient_function]]).

== Cryptids ==
Known BB(6) Cryptids:

* [[Antihydra]]
* {{TM|1RB1RC_1LC1LE_1RA1RD_0RF0RE_1LA0LB_---1RA|undecided}} a variant of [[Hydra]] and Antihydra
* {{TM|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC|undecided}} similar to Antihydra
* {{TM|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---|undecided}} similar to Antihydra
* {{TM|1RB0LB_1LC0RE_1LA1LD_0LC---_0RB0RF_1RE1RB|undecided}} similar to Antihydra
* {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC|undecided}} (and 15 related TMs) a family of [[probviously]] halting cryptids

Potential Cryptids:

* {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}
* {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}}
* {{TM|1RB0RB_1LC1RE_1LF0LD_1RA1LD_1RC1RB_---1LC|undecided}}

== Top Halters ==
The current top 10 BB(6) halters (known by [[User:Sligocki|Shawn Ligocki]]) are<pre>
1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE Halt ~10↑↑15.60465
1RB0LA_1LC1LF_0LD0LC_0LE0LB_1RE0RA_1RZ1LD Halt ~10↑↑5.63534
1RB1RE_1LC1LF_1RD0LB_1LE0RC_1RA0LD_1RZ1LC Halt ~10↑↑5.56344
1RB0LE_0RC1RA_0LD1RF_1RE0RB_1LA0LC_0RD1RZ Halt ~10↑↑5.12468
1RB0RF_1LC1LB_0RE0LD_0LC0LB_0RA1RE_0RD1RZ Halt ~10↑↑5.03230
1RB1LA_1LC0RF_1LD1LC_1LE0RE_0RB0LC_1RZ1RA Halt ~10↑↑4.91072
1RB0LE_1LC1RA_1RE0LD_1LC1LF_1LA0RC_1RZ1LC Halt ~10↑↑3.33186
1RB1RF_1LC1RE_0LD1LB_1LA0RA_0RA0RB_1RZ0RD Halt ~10↑↑3.31128
1RB0LF_1LC0RA_1RD0LB_1LE1RC_1RZ1LA_1LA1LE Halt ~10↑↑3.18855
1RB1RZ_0LC0LD_1LD1LC_1RE1LB_1RF1RD_0LD0RA Halt ~10^646456993.24591
</pre>The numbers listed are sigma scores. Runtimes are not available, but are presumed to be about <math>score^2</math> which is roughly indistinguishable in tetration notation. Fractional tetration notation is described in https://www.sligocki.com/2022/06/25/ext-up-notation.html. For a longer list of halting TMs see https://github.com/sligocki/busy-beaver/blob/main/Machines/bb/6x2.txt. For historical perspective see Pascal Michel's [https://bbchallenge.org/~pascal.michel/ha#tm62 '''Historical survey of Busy Beavers'''].

== Holdouts ==
@mxdys's informal [[Holdouts lists|holdouts list]] is down to 5877 TMs as of 4 Aug 2024.

== References ==
<references />

Code repositories

2024-11-24T11:05:17Z

Mxdys: add https://github.com/ccz181078/busycoq/tree/BB6/verify

{| class="wikitable"
|+
!Creator
!Link
!Notes
|-
|bbchallenge
|[https://github.com/bbchallenge/bbchallenge-deciders Link]
|Official bbchallenge deciders
|-
|@mxdys
|[https://github.com/ccz181078/Coq-BB5 Link]
|Coq proof of BB(5) = 47,176,870. See [[Coq-BB5]].
|-
|@mei
|[https://github.com/meithecatte/busycoq Link]
|A hybrid Coq/Rust/OCaml repo implementing very fast deciders plus verified proofs
|-
|Georgi Georgiev (Skelet)
|[https://skelet.ludost.net/bb/ Link]
|See [[bbfind]] and [[Skelet's 43 holdouts]]. Wrapped via [https://gist.github.com/m1el/d514a353cccde531c298b725043404af bbfind-stdin by @wizord]
|-
|@mxdys
|[https://github.com/ccz181078/busycoq/tree/BB6/verify Link]
|Coq proof of some deciders and some solved machines in BB(6)
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/6cdad07e15f6acf992b79dc2baf0492c Link]
|Accelerated TM simulators in Python utilizing memoization (like HashLife)
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/7f5e10169abbb50d1537165c6e71733b Link]
|Forward segment TM decider by @Mateon1 (variant of [[Halting Segment]])
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/b63eabc371ac35e2a14a9c5ce37413bc Link]
|[[Closed Position Set (CPS)]] TM decider implementation
|-
|Mateusz Naściszewski @Mateon1
|[https://gist.github.com/mateon1/c801565e499be605cea1283a5984b4c3 Link]
|[[Closed Tape Language (CTL)]] SAT Solver; '''TODO dependencies still Discord posts'''
|-
|@savask
|[https://github.com/savask/turing Link]
|Collection of deciders
|-
|@savask
|[https://gist.github.com/savask/1c43a0e5cdd81229f236dcf2b0611c3f Link]
|[[Closed Position Set (CPS)]], reverse-engineered from Skelet's [[bbfind]] program
|-
|@savask
|[https://gist.github.com/savask/888aa5e058559c972413790c29d7ad72 Link]
|Decider for [[Bouncers]]
|-
|@savask
|[https://gist.github.com/savask/c7546bb6384984b2fb3cb90fc7925697 Link]
|Reproduction of @mxdys' [[RepWL_ES]] decider (repeated block decider)
|-
|@djmati1111
|[https://github.com/colette-b/bbchallenge Link]
|The first SAT-based [[Finite Automata Reduction (FAR)]] decider
|-
|Frans Faase
|[https://github.com/FransFaase/SymbolicTM Link]
|An early [[Closed Tape Language (CTL)]] verifier
|-
|@Iijil
|[https://github.com/Iijil1/Bouncers Link]
|Decider for [[Bouncers]]
|-
|@Iijil
|[https://github.com/Iijil1/Bruteforce-CTL Link]
|Bruteforce [[Closed Tape Language (CTL)]] decider
|-
|@Iijil
|[https://github.com/Iijil1/MITMWFAR Link]
|[[Meet-in-the-Middle Weighted Finite Automata Reduction (MITMWFAR)]] decider
|-
|@Iijil
|[https://gist.github.com/Iijil1/d325da33be74a86ac2399c161a57166a Link]
|4-symbol to 2-symbol TM compiler
|-
|Jason Yuen @-d
|[https://github.com/int-y1/proofs/tree/master/BusyLean Link]
|Some progress toward a FAR verifier checked by Lean
|-
|Matthew House @LegionMammal976
|[https://github.com/LegionMammal978/bigfoot-sim Link]
|An accelerated [[Bigfoot]] simulator
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bbchallenge-dafny-deciders Link]
|Formally verified deciders using Dafny
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/busy-beaver-dafny-regex-verifier Link]
|Formally verified [[Closed Tape Language (CTL)]] verifier
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bbchallenge-regexy-decider Link]
|[[Closed Tape Language (CTL)]] decider
|-
|Nathan Fenner @nathanf
|[https://github.com/Nathan-Fenner/bb-simple-n-gram-cps Link]
|[[Closed Position Set (CPS)]] 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]
|
|-
|Tony Guilfoyle
|[https://github.com/TonyGuil/bbchallenge Link]
|
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/uncombedcoconut/bbchallenge Link]
|
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/UncombedCoconut/bbchallenge-deciders/tree/FARther/decider-finite-automata-reduction Link]
|[[Finite Automata Reducion (FAR)]] decider
|-
|Justin Blanchard @UncombedCoconut
|[https://github.com/UncombedCoconut/bbchallenge-nfa-verification Link]
|[[Finite Automata Reducion (FAR)]] verifier
|-
|Pavel Kropitz @uni
|[https://github.com/univerz/bbc Link]
|
|-
|Dan Briggs
|[https://github.com/danbriggs/Turing Link]
|TM proofs/writing
|}

Inductive Proof System

2024-11-16T08:14:37Z

Mxdys: unification in inductive decider

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>} \; 1^n \to 1^n \; \textrm{C>}</math>
#* Proof by induction on n
#** Base case: []<math display="block">\textrm{C>} \; 1^0 = 1^0 \; \textrm{C>}</math>
#** Inductive case: [C1, IH]<math display="block">\textrm{C>} \; 1^{n+1} \xrightarrow{C1} 1 \; \textrm{C>} \; 1^n \xrightarrow{IH} 1^{n+1} \; \textrm{C>}</math>
# Shift (L0) Rule: <math>E(n) : 1^n \; \textrm{<E} \to \textrm{<E} \; 1^n</math>
#* Proof by induction on n
#** Base case: []
#** Inductive case: [E1, IH]
# L1 Rule: <math>A(n) : \textrm{A>} \; 1^n \; 0^\infty \to 1^{2n} \; \textrm{A>} \; 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>} \; 1^{n+1} \; 0^\infty
& \xrightarrow{A1} &
0 \; \textrm{C>} \; 1^n \; 0^\infty \\
& \xrightarrow{C(n)} &
0 \; 1^n \; \textrm{C>} \; 0^\infty \\
& \xrightarrow{C0} &
0 \; 1^{n+1} \; \textrm{D>} \; 0^\infty \\
& \xrightarrow{D0} &
0 \; 1^{n+1} \; \textrm{<E} \; 0^\infty \\
& \xrightarrow{E(n+1)} &
0 \; \textrm{<E} \; 1^{n+1} \; 0^\infty \\
& \xrightarrow{E0} &
1 \; \textrm{D>} \; 1^{n+1} \; 0^\infty \\
& \xrightarrow{D1} &
1^2 \; \textrm{A>} \; 1^n \; 0^\infty \\
& \xrightarrow{IH} &
1^{2(n+1)} \; \textrm{A>} \; 0^\infty \\
\end{array}</math>
# L2 Rule: <math>0^\infty \; 1^a \; \textrm{A>} \; 0^\infty \to 0^\infty \; 1^{2a+6} \; \textrm{A>} \; 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>} \; 0^\infty
& \xrightarrow{A0, B0, C1, C0, D0} &
0^\infty \; 1^{a+2} \; \textrm{<E} \; 0^\infty \\
& \xrightarrow{E(a+2)} &
0^\infty \; \textrm{<E} \; 1^{a+2} \; 0^\infty \\
& \xrightarrow{E0,D1} &
0^\infty \; 1^2 \; \textrm{A>} \; 1^{a+1} \; 0^\infty \\
& \xrightarrow{A(a+1)} &
0^\infty \; 1^{2a+4} \; \textrm{A>} \; 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>} \; 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]]

Inductive Proof System

2024-11-15T21:47:10Z

Mxdys: Find and Prove Rules in Inductive Decider

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>} \; 1^n \to 1^n \; \textrm{C>}</math>
#* Proof by induction on n
#** Base case: []<math display="block">\textrm{C>} \; 1^0 = 1^0 \; \textrm{C>}</math>
#** Inductive case: [C1, IH]<math display="block">\textrm{C>} \; 1^{n+1} \xrightarrow{C1} 1 \; \textrm{C>} \; 1^n \xrightarrow{IH} 1^{n+1} \; \textrm{C>}</math>
# Shift (L0) Rule: <math>E(n) : 1^n \; \textrm{<E} \to \textrm{<E} \; 1^n</math>
#* Proof by induction on n
#** Base case: []
#** Inductive case: [E1, IH]
# L1 Rule: <math>A(n) : \textrm{A>} \; 1^n \; 0^\infty \to 1^{2n} \; \textrm{A>} \; 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>} \; 1^{n+1} \; 0^\infty
& \xrightarrow{A1} &
0 \; \textrm{C>} \; 1^n \; 0^\infty \\
& \xrightarrow{C(n)} &
0 \; 1^n \; \textrm{C>} \; 0^\infty \\
& \xrightarrow{C0} &
0 \; 1^{n+1} \; \textrm{D>} \; 0^\infty \\
& \xrightarrow{D0} &
0 \; 1^{n+1} \; \textrm{<E} \; 0^\infty \\
& \xrightarrow{E(n+1)} &
0 \; \textrm{<E} \; 1^{n+1} \; 0^\infty \\
& \xrightarrow{E0} &
1 \; \textrm{D>} \; 1^{n+1} \; 0^\infty \\
& \xrightarrow{D1} &
1^2 \; \textrm{A>} \; 1^n \; 0^\infty \\
& \xrightarrow{IH} &
1^{2(n+1)} \; \textrm{A>} \; 0^\infty \\
\end{array}</math>
# L2 Rule: <math>0^\infty \; 1^a \; \textrm{A>} \; 0^\infty \to 0^\infty \; 1^{2a+6} \; \textrm{A>} \; 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>} \; 0^\infty
& \xrightarrow{A0, B0, C1, C0, D0} &
0^\infty \; 1^{a+2} \; \textrm{<E} \; 0^\infty \\
& \xrightarrow{E(a+2)} &
0^\infty \; \textrm{<E} \; 1^{a+2} \; 0^\infty \\
& \xrightarrow{E0,D1} &
0^\infty \; 1^2 \; \textrm{A>} \; 1^{a+1} \; 0^\infty \\
& \xrightarrow{A(a+1)} &
0^\infty \; 1^{2a+4} \; \textrm{A>} \; 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>} \; 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>

=== 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]]

Inductive Proof System

2024-11-15T20:04:23Z

Mxdys: tape compression methods for inductive decider

1RB0LD 1RC1RF 1LA0RA 0LA0LE 1LD1LA 0RB---

2024-10-15T00:26:59Z

Mxdys: Created page with "{{machine|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---}} This is a 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> For simplicity only one potential halting rule is shown."

{{machine|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---}}
This is a [[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>
For simplicity only one potential halting rule is shown.

BB(6)

2024-10-15T00:20:54Z

Mxdys: add 4th non-halt cryptid in bb6

The 6-state, 2-symbol Busy Beaver problem '''BB(6)''' is unsolved. With the discovery of [[Antihydra]] in 2024, we now know that we must solve a [[Collatz-like]] problem in order to solve BB(6).

The current BB(6) champion {{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE|halt}} was discovered by Pavel Kropitz in 2022 proving the lower bound:<ref>Shawn Ligocki. 2022. "BB(6, 2) > 10↑↑15". https://www.sligocki.com/2022/06/21/bb-6-2-t15.html</ref>

<math display="block">S(6) > \Sigma(6) > 10 \uparrow\uparrow 15</math>

== Techniques ==
In order to simulate the current BB(6) champion requires [[Accelerated simulator|accelerated simulation]] that can handle Collatz Level 2 [[Inductive rule|inductive rules]]. In other words, it requires a simulator that can prove the rules:

<math display="block">\begin{array}{l}
C(4k) & \to & Halt(\frac{3^{k+3} - 11}{2}) \\
C(4k+1) & \to & C(\frac{3^{k+3} - 11}{2}) \\
C(4k+2) & \to & C(\frac{3^{k+3} - 11}{2}) \\
C(4k+3) & \to & C(\frac{3^{k+3} + 1}{2}) \\
\end{array}</math>

and also compute the remainder mod 3 of numbers produced by applying these rules 15 times (which requires some fancy math related to [[wikipedia:Euler's_totient_function|Euler's_totient_function]]).

== Cryptids ==
Known BB(6) Cryptids:

* [[Antihydra]]
* {{TM|1RB1RC_1LC1LE_1RA1RD_0RF0RE_1LA0LB_---1RA|undecided}} a variant of [[Hydra]] and Antihydra
* {{TM|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC|undecided}}
* {{TM|1RB0LD_1RC1RF_1LA0RA_0LA0LE_1LD1LA_0RB---|undecided}}
* {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC|undecided}} (and 15 related TMs) a family of [[probviously]] halting cryptids

Potential Cryptids:

* {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}
* {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}}
* {{TM|1RB0RB_1LC1RE_1LF0LD_1RA1LD_1RC1RB_---1LC|undecided}}

== Top Halters ==
The current top 10 BB(6) halters (known by [[User:Sligocki|Shawn Ligocki]]) are<pre>
1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE Halt ~10↑↑15.60465
1RB0LA_1LC1LF_0LD0LC_0LE0LB_1RE0RA_1RZ1LD Halt ~10↑↑5.63534
1RB1RE_1LC1LF_1RD0LB_1LE0RC_1RA0LD_1RZ1LC Halt ~10↑↑5.56344
1RB0LE_0RC1RA_0LD1RF_1RE0RB_1LA0LC_0RD1RZ Halt ~10↑↑5.12468
1RB0RF_1LC1LB_0RE0LD_0LC0LB_0RA1RE_0RD1RZ Halt ~10↑↑5.03230
1RB1LA_1LC0RF_1LD1LC_1LE0RE_0RB0LC_1RZ1RA Halt ~10↑↑4.91072
1RB0LE_1LC1RA_1RE0LD_1LC1LF_1LA0RC_1RZ1LC Halt ~10↑↑3.33186
1RB1RF_1LC1RE_0LD1LB_1LA0RA_0RA0RB_1RZ0RD Halt ~10↑↑3.31128
1RB0LF_1LC0RA_1RD0LB_1LE1RC_1RZ1LA_1LA1LE Halt ~10↑↑3.18855
1RB1RZ_0LC0LD_1LD1LC_1RE1LB_1RF1RD_0LD0RA Halt ~10^646456993.24591
</pre>The numbers listed are sigma scores. Runtimes are not available, but are presumed to be about <math>score^2</math> which is roughly indistinguishable in tetration notation. Fractional tetration notation is described in https://www.sligocki.com/2022/06/25/ext-up-notation.html. For a longer list of halting TMs see https://github.com/sligocki/busy-beaver/blob/main/Machines/bb/6x2.txt. For historical perspective see Pascal Michel's [https://bbchallenge.org/~pascal.michel/ha#tm62 '''Historical survey of Busy Beavers'''].

== Holdouts ==
@mxdys's informal [[Holdouts lists|holdouts list]] is down to 5877 TMs as of 4 Aug 2024.

== References ==
<references />

1RB1LD 1RC1RE 0LA1LB 0LD1LC 1RF0RA ---0RC

2024-10-14T19:56:52Z

Mxdys: Created page with "{{machine|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC}} This is a BB(6) Cryptid found by @mxdys and shared on Discord on 20 Aug 2024. == Analysis by @mxdys == <pre> 1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC start: (0,0) (3x+0,y) --> (4x+4,y+1) (3x+1,y+1) --> (4x+5,y+2) (3x+2,y) --> (4x+8,max(0,y-1)) (3x+1,0) --> halt (x,y) := 0^inf 110 <B 11011 01^x 011^y 0^inf </pre>"

{{machine|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC}}
This is a [[BB(6)]] [[Cryptid]] found by @mxdys and shared on Discord on 20 Aug 2024.

== Analysis by @mxdys ==
<pre>
1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC

start: (0,0)

(3x+0,y) --> (4x+4,y+1)
(3x+1,y+1) --> (4x+5,y+2)
(3x+2,y) --> (4x+8,max(0,y-1))

(3x+1,0) --> halt

(x,y) := 0^inf 110 <B 11011 01^x 011^y 0^inf
</pre>

BB(6)

2024-10-14T19:53:41Z

Mxdys: add 3rd non-halt cryptid in bb6

The 6-state, 2-symbol Busy Beaver problem '''BB(6)''' is unsolved. With the discovery of [[Antihydra]] in 2024, we now know that we must solve a [[Collatz-like]] problem in order to solve BB(6).

The current BB(6) champion {{TM|1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE|halt}} was discovered by Pavel Kropitz in 2022 proving the lower bound:<ref>Shawn Ligocki. 2022. "BB(6, 2) > 10↑↑15". https://www.sligocki.com/2022/06/21/bb-6-2-t15.html</ref>

<math display="block">S(6) > \Sigma(6) > 10 \uparrow\uparrow 15</math>

== Techniques ==
In order to simulate the current BB(6) champion requires [[Accelerated simulator|accelerated simulation]] that can handle Collatz Level 2 [[Inductive rule|inductive rules]]. In other words, it requires a simulator that can prove the rules:

<math display="block">\begin{array}{l}
C(4k) & \to & Halt(\frac{3^{k+3} - 11}{2}) \\
C(4k+1) & \to & C(\frac{3^{k+3} - 11}{2}) \\
C(4k+2) & \to & C(\frac{3^{k+3} - 11}{2}) \\
C(4k+3) & \to & C(\frac{3^{k+3} + 1}{2}) \\
\end{array}</math>

and also compute the remainder mod 3 of numbers produced by applying these rules 15 times (which requires some fancy math related to [[wikipedia:Euler's_totient_function|Euler's_totient_function]]).

== Cryptids ==
Known BB(6) Cryptids:

* [[Antihydra]]
* {{TM|1RB1RC_1LC1LE_1RA1RD_0RF0RE_1LA0LB_---1RA|undecided}} a variant of [[Hydra]] and Antihydra
* {{TM|1RB1LD_1RC1RE_0LA1LB_0LD1LC_1RF0RA_---0RC|undecided}}
* {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC|undecided}} (and 15 related TMs) a family of [[probviously]] halting cryptids

Potential Cryptids:

* {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}}
* {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}}
* {{TM|1RB0RB_1LC1RE_1LF0LD_1RA1LD_1RC1RB_---1LC|undecided}}

== Top Halters ==
The current top 10 BB(6) halters (known by [[User:Sligocki|Shawn Ligocki]]) are<pre>
1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE Halt ~10↑↑15.60465
1RB0LA_1LC1LF_0LD0LC_0LE0LB_1RE0RA_1RZ1LD Halt ~10↑↑5.63534
1RB1RE_1LC1LF_1RD0LB_1LE0RC_1RA0LD_1RZ1LC Halt ~10↑↑5.56344
1RB0LE_0RC1RA_0LD1RF_1RE0RB_1LA0LC_0RD1RZ Halt ~10↑↑5.12468
1RB0RF_1LC1LB_0RE0LD_0LC0LB_0RA1RE_0RD1RZ Halt ~10↑↑5.03230
1RB1LA_1LC0RF_1LD1LC_1LE0RE_0RB0LC_1RZ1RA Halt ~10↑↑4.91072
1RB0LE_1LC1RA_1RE0LD_1LC1LF_1LA0RC_1RZ1LC Halt ~10↑↑3.33186
1RB1RF_1LC1RE_0LD1LB_1LA0RA_0RA0RB_1RZ0RD Halt ~10↑↑3.31128
1RB0LF_1LC0RA_1RD0LB_1LE1RC_1RZ1LA_1LA1LE Halt ~10↑↑3.18855
1RB1RZ_0LC0LD_1LD1LC_1RE1LB_1RF1RD_0LD0RA Halt ~10^646456993.24591
</pre>The numbers listed are sigma scores. Runtimes are not available, but are presumed to be about <math>score^2</math> which is roughly indistinguishable in tetration notation. Fractional tetration notation is described in https://www.sligocki.com/2022/06/25/ext-up-notation.html. For a longer list of halting TMs see https://github.com/sligocki/busy-beaver/blob/main/Machines/bb/6x2.txt. For historical perspective see Pascal Michel's [https://bbchallenge.org/~pascal.michel/ha#tm62 '''Historical survey of Busy Beavers'''].

== Holdouts ==
@mxdys's informal [[Holdouts lists|holdouts list]] is down to 5877 TMs as of 4 Aug 2024.

== References ==
<references />