1RB---_0RC0RD_1LD1RB_0LE0LC_1RA0LF_1LD1LE

1RB---_0RC0RD_1LD1RB_0LE0LC_1RA0LF_1LD1LE (bbch) appears to be a chaotic probviously halting BB(6) TM, but with no estimate for halting time. It is still under analysis as of 24 Apr 2026.

Analysis by Shawn Ligocki

https://discord.com/channels/960643023006490684/1239205785913790465/1497001816741646407

Low level rules:

1RB---_0RC0RD_1LD1RB_0LE0LC_1RA0LF_1LD1LE

Let A(a,b,c) = 0^inf 1^a 10^b C> 1^c 0^inf

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

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

A(0,b,c+5) --> A(2b+4,2,c)
A(0,b,4) --> Halt
A(0,b,3) --> Halt

Start: A(0,1,0)

High level rules:

if a ≥ floor(c/3):
  A(a, b, 3k)    -->  A(4k+2b+1, 1, a-k+1)    X -> X+4
  A(a, b, 3k+1)  -->  A(4k+2b+3, 1, a-k+1)    X -> X+5
  A(a, b, 3k+2)  -->  A(4k+2b+5, 1, a-k+1)    X -> X+6
if a < floor(c/3):
  A(k, b, 3k+3)  -->  Halt
  A(k, b, 3k+4)  -->  Halt
  A(k, b, 3k+5+c)  -->  A(4k+2b+4, 2, c)      X -> X+3

with X=a+2b+c

where

$X=a+2b+c$

is a sort of norm on these configs that we can see grows only linearly.

Simulated out to 100B high level rules:

  1_000_000_000  A(  1_449_166_375,  1,   3_050_820_388)
100_000_000_000  A(150_379_323_247,  1, 299_620_772_649)

Probabilistic Model

Distribution of r values over the first 5B steps (1000 buckets).

Map governing update of r stat in the limit. It is a Skewed or Asymmetric Tent Map.

Consider $r=\frac{a}{a+c}$. This value seems empirically to be be extremely uniform on the range [0,1]. And for large values of a,c there is a clear reason why: The update function is a Skewed Tent Map, a map which is known to have long-run time average distribution completely uniform.

If we assume that this sequence has (pseudo-)random uniform distribution, then we can calculate an exact probability of halting. Let $S=a+c$ and consider at each step that the model has a fixed value of S and b, but the value of a (and consequently also c) are chosen uniformly at random. If S = 4k+3, there is exactly 1/S choices that will lead to halt: A(k, b, 3k+3). The same is true for S = 4k+4. But for S = 4k+{0,1}, all choices must reset (hit a non-halt rule). Assuming that S is equally likely to have each remainder, we can say that there is a $\frac{1}{2S}$ chance of halting at each high level rule. We can see above that X grows linearly with each rule application, the same is true for S since $b\in \{1,2\}$. Thus $\sum _t\frac{1}{2S(t)}\to \mathrm{\infty }$ and thus $P(\text{not halt})=\prod _t(1-\frac{1}{2S(t)})=0$. In other words, this random model is guaranteed to eventually halt.