1RB0LD 1RC1RA 1LD0RB 1LE1LA 1RF0RC ---1RE: Difference between revisions

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{{machine|1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE}}{{unsolved|Does this TM halt? If so, how many steps does it take to halt?}}
{{machine|1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE}}{{unsolved|Does this TM halt? If so, how many steps does it take to halt?}}
{{TM|1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE}} is a [[probviously]] halting [[BB(6)]] [[Cryptid]] analzyed by mxdys on 30 July 2025.
{{TM|1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE}} 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 <code>(a,b,c)</code>  into either <code>(0,0,c)</code> or <code>(0,1,c)</code> with equal probability. If it lands on <code>(0,0,c)</code>, 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 <code>(0,1,c)</code>:
 
# <code>(0,1,8) --> (25,0,0)</code>
# <code>(0,1,113544) --> (227097,0,0)</code>
# <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.


== Analysis by mxdys ==
== Analysis by mxdys ==
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</pre>
</pre>


These rules have been proven in Coq and [https://github.com/int-y1/proofs/blob/10a7f272a5570d3c8d5dc506881df8b7c934c0c5/BusyLean/Individual/1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE.lean Lean].
These rules have been proven in Coq and [https://github.com/int-y1/proofs/blob/10a7f272a5570d3c8d5dc506881df8b7c934c0c5/BusyLean/Individual/1RB0LD_1RC1RA_1LD0RB_1LE1LA_1RF0RC_---1RE.lean Lean]. The first rule can be accelerated as<pre>
 
The first rule can be accelerated as
<pre>
(a,2b,c) --> (a,0,c+3b)
(a,2b,c) --> (a,0,c+3b)
(a,2b+1,c) --> (a,1,c+3b)
(a,2b+1,c) --> (a,1,c+3b)
</pre>
</pre>


Andrew Ducharme forward simulated the map with this acceleration 10^7 steps and found the TM had not yet halted.
This TM is probviously halting because <code>(0,0,c)</code> and <code>(0,1,c)</code> should occur equally often. On the first 3 attempts, the TM has landed on <code>(0,1,c)</code>:
# <code>(0,1,8) --> (25,0,0)</code>
# <code>(0,1,113544) --> (227097,0,0)</code>
# <code>(0,1,28155...08204) --> (56311...16417,0,0)</code>, where 28155...08204 and 56311...16417 are 39991-digit numbers.


However, forward simulation will not work for the 4th attempt, because c grows too large.


[[Category:Cryptids]]
[[Category:Cryptids]]

Revision as of 22:49, 8 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):

  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)