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 1.5T high level rules:

    1_000_000_000  A(    1_449_166_375,  1,     3_050_820_388)
   10_000_000_000  A(   39_348_725_977,  1,     5_651_355_297)
  100_000_000_000  A(  150_379_323_247,  1,   299_620_772_649)
  500_000_000_000  A(2_123_188_460_901,  1,   126_811_502_474)
1_000_000_000_000  A(2_018_953_178_979,  1, 2_481_046_137_608)
1_500_000_000_000  A(  415_020_273_351,  1, 6_334_978_834_030)

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 $P(\text{never halt})=\prod _t(1-\frac{1}{2S(t)})=0$. In other words, this random model is guaranteed to eventually halt.

This model supports a survival function $V(T)=\prod _{t=0}^T(1-\frac{1}{2(S_0+\mu t)})$ for $\mu \approx 4.5$ the average increase in S per step. With a bit of approximation you can get $V(t)\approx {(1+\frac{\mu t}{S_0})}^{-\frac{1}{2\mu }}$ and solving for t (to find the time when only P = V(t) have survived) you get $t(P)=\frac{S_0}{\mu }(P^{2\mu }-1)$, if you further assume that $S_0\approx N\mu$ (the sim started from S=0, and after N steps was at $N\mu$), then you see that the median halting time (given that you have already simulated for N steps without halt) is 512N.

I (Shawn) simulated starting from all a,c such that S=a+c=2048 (and all

$b\in \{1,2\}$

and there were the empirical survival rates:

S=2048: 4098 configs, 2264 halted within 20_000_000 steps

Survival at multiples of S:
  step 2_048: 3491/4098 (85.2%)
  step 4_096: 3407/4098 (83.1%)
  step 8_192: 3205/4098 (78.2%)
  step 16_384: 3205/4098 (78.2%)
  step 32_768: 2704/4098 (66.0%)
  step 65_536: 2704/4098 (66.0%)
  step 131_072: 2350/4098 (57.3%)
  step 262_144: 2235/4098 (54.5%)
  step 524_288: 2070/4098 (50.5%)
  step 1_048_576: 2070/4098 (50.5%)
  step 2_097_152: 2070/4098 (50.5%)
  step 4_194_304: 1834/4098 (44.8%)
  step 8_388_608: 1834/4098 (44.8%)
  step 16_777_216: 1834/4098 (44.8%)

These seem to roughly match the model which expects that the median halting time would have been 1,048,576.