1RB0LD 1RC1RA 1LD0RB 1LE1LA 1RF0RC ---1RE: Difference between revisions
Jump to navigation
Jump to search
(move and expand futility of simulation discussion) |
(add approximation of 4th occasion) |
||
| Line 6: | Line 6: | ||
# <code>(0,1,28155...08204) --> (56311...16417,0,0)</code>, where 28155...08204 and 56311...16417 are 39991-digit numbers. | # <code>(0,1,28155...08204) --> (56311...16417,0,0)</code>, 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. | As a is now so large, the next 3-tuple is <math>(0, ?, \approx 10^{10^{39990}})</math>, and forward simulation of the current rules is no longer a viable option to understand if the TM halts. | ||
== Analysis by mxdys == | == Analysis by mxdys == | ||
Latest revision as of 07:21, 9 August 2025
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):
(0,1,8) --> (25,0,0)(0,1,113544) --> (227097,0,0)(0,1,28155...08204) --> (56311...16417,0,0), where 28155...08204 and 56311...16417 are 39991-digit numbers.
As a is now so large, the next 3-tuple is , and forward simulation of the current rules is no longer a viable option to understand if the TM halts.
Analysis by mxdys
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)