View source for 1RB1RA 1RC0LC 0LD1LG 1LF0LE 1RZ1LF 0LA1LD 1RA1LC

{{machine|1RB1RA_1RC0LC_0LD1LG_1LF0LE_1RZ1LF_0LA1LD_1RA1LC}}
{{TM|1RB1RA_1RC0LC_0LD1LG_1LF0LE_1RZ1LF_0LA1LD_1RA1LC}} is a halting tetrational [[BB(7)]] TM that runs for over 10↑↑35 steps found by Shawn Ligocki on 8 May 2025 based on @mxdys's enumeration system https://github.com/ccz181078/TM.

Analysis by Shawn Ligocki:
<pre>
1RB1RA_1RC0LC_0LD1LG_1LF0LE_1RZ1LF_0LA1LD_1RA1LC

B(a,b,c,d) = 0^inf 1^a B> 1^b 01^c 011^d 0^inf
D(a) = B(a,0,0,0) = 0^inf 1^a B> 0^inf

D(3k) -> Halt(2k+1)

D(9k+1)  -->  D(32 2^k - 27)
D(9k+4)  -->  D(32 2^k - 23)
D(9k+7)  -->  D(32 2^k - 18)

D(9k+2)  -->  D(34 2^k - 27)
D(9k+5)  -->  D(34 2^k - 23)
D(9k+8)  -->  D(34 2^k - 18)

Start: D(1)
</pre>

Basic transitions used to build this:
<pre>
0 1^2n+1 B> 1   -->  1^2n+3 B>
0 1^2n+1 B> 0 1^m 0  -->  1^2n+m+4 B>   for m >= 1

0^4 1^2n B> 1  -->  1 B> 010 1^2n 0
0^5 1^2n B> 1  -->  1^5 B> 1^2n 0

0   1^3k   B> 00  -->  1 Z> 011^k 00
0^2 1^3k+1 B> 00  -->  1 B> 01   011^k 00
0^3 1^3k+2 B> 00  -->  1 B> 0111 011^k 00

$ 1^a B> 011^3  -->  $ 1^2a+23 B>  (for a odd)
</pre>

== Model ==
At first it might appear that there is a 1/3 chance of halting each iteration, but in fact, that is not quite right since only they 9k+4 and 9k+5 rules can possibly lead to a halt. Based on manual simulation it appears that the chance of halting each turn reaches a steady state of about 1/7.85. Legion suggested this model: Consider all remainders mod 18 and all the remainders mod 18 that they can lead to:<pre>
0 -> halt
1 -> 5,11,17
2 -> 1,7,13
3 -> halt
4 -> 3,9,15
5 -> 5,11,17
6 -> halt
7 -> 2,8,14
8 -> 4,10,16
9 -> halt
10 -> 1,7,13
11 -> 5,11,17
12 -> halt
13 -> 5,11,17
14 -> 3,9,15
15 -> halt
16 -> 4,10,16
17 -> 2,8,14
</pre>(all weights equal) with stationary distribution <code>(0, 1/18*(3-sqrt(5)), 1/18*(-1+sqrt(5)), 1/18*(3-sqrt(5)), 1/18*(3-sqrt(5)), 1/9*(-1+sqrt(5)), 0, 1/18*(3-sqrt(5)), 1/18*(-1+sqrt(5)), 1/18*(3-sqrt(5)), 1/18*(3-sqrt(5)), 1/9*(-1+sqrt(5)), 0, 1/18*(3-sqrt(5)), 1/18*(-1+sqrt(5)), 1/18*(3-sqrt(5)), 1/18*(3-sqrt(5)), 1/9*(-1+sqrt(5)))</code>, halting probability <code>1/6*(3-sqrt(5)) ≈ 0.12732</code>, and expected lifetime <code>3/2*(3+sqrt(5)) ≈ 7.8541</code>.


This model can be simplified to the Markov Chain:<pre>
A->A,A,B
B->A,B,H
</pre>Where here <code>A</code> represents any value = 1,5 (mod 6), <code>B</code> any value = 2,4 (mod 6) and <code>H</code> any value = 3 (mod 6) (which will lead to halt next turn).