1RB3LA1LA1RA1RA 2LB2RA---4RB1LB

1RB3LA1LA1RA1RA_2LB2RA---4RB1LB (bbch) is a BB(2,5) TM analyzed by mxdys on 12 Mar 2025 that appears to be a Cryptid.

Analysis by mxdys and Racheline

1RB3LA1LA1RA1RA_2LB2RA---4RB1LB
start: A(4,4)
A(a,3+a+b) --> A(4+2a,b)
A(a,2+a) --> B(2a+2,3)
A(a,1+a) --> halt
A(b+a,b) --> B(a,2b)

B(3a+1,b) --> A(4,2+5a+b)
B(3a+2,b) --> B(5+5a+b,2)
B(3a+0,b) --> B(3+5a+b,0)

A(a,b) := 0^inf 1 4^a 1^4 A> 1^b 2 0^inf
B(a,b) := 0^inf 1 4^a 1^3+b A> 11 0^inf

Let's rewrite the first rule as:
A(a, b) --> A(2a + 4, b - a - 3) if b >= a + 3
Then, after n applications of rule 1 (assuming b is large enough):
A(a, b) --> A(2^n * a + 2^(n + 2) - 4, b - 2^n * a + a - 2^(n + 2) + n + 4)
Considering that a is always set to 4 when leaving B, this simplifies to:
A(4, b) --> A(2^(n + 3) - 4, b - 2^(n + 3) + n + 8)
Now consider the halting configuration: A(a, a + 1):
For this to be in a halting configuration, the following must be true:
2^(n+3)-4 + 1 = b-2^(n+3)+n+8
This simplifies to:
b = 2^(n+4)-n-11
--> For any positive integer n (and 0): A(4, 2^(n+4)-n-11) --> halt
Considering that all rules either keep the overall score or increase it, and that the transition from B(3a+1,b) moves the score onto b, it is guaranteed that b will always increase between A(4,b) configs.
As a result of this, once a halting value of b is passed, it cannot be reached again. However, as b increases, the distance between these halting values of b increases exponentially, making it less and less likely for the TM to halt.