1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE

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Revision as of 22:49, 8 August 2025 by ADucharme (talk | contribs) (move and expand futility of simulation discussion)
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Unsolved problem:
Does this TM halt? If so, how many steps does it take to halt?

1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE (bbch) is a probviously halting BB(6) Cryptid analyzed by mxdys on 30 July 2025. This TM is probviously halting because it decrements a 3-tuple (a,b,c) into either (0,0,c) or (0,1,c) with equal probability. If it lands on (0,0,c), the TM halts. It takes roughly 2a rule (not TM) steps to reach an a=0 state. On the first 3 occasions, the TM lands on (0,1,c):

  1. (0,1,8) --> (25,0,0)
  2. (0,1,113544) --> (227097,0,0)
  3. (0,1,28155...08204) --> (56311...16417,0,0), where 28155...08204 and 56311...16417 are 39991-digit numbers.

As a is now so large, forward simulation of the current rules is no longer a viable option to understand if the TM halts.

Analysis by mxdys

[1]

1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE

(a,b,c) := 0^inf 1^a 0 01^b 0 11^c+1 B> 0^inf

(a,2+b,c) --> (a,b,3+c)
(a+1,0,c) --> (a,c,2)
(a+1,1,c) --> (a,c,6)
(0,0,c) --> halt
(0,1,c) --> (2c+9,0,0)
start: (3,0,0)

Confirmation that it doesn't halt early:

(3,0,0) --> ...
(25,0,0) --> ...
(227097,0,0) --> ...
(_,0,0) --> ...

These rules have been proven in Coq and Lean. The first rule can be accelerated as

(a,2b,c) --> (a,0,c+3b)
(a,2b+1,c) --> (a,1,c+3b)