/machines/1RB1RF_0RC0LE_1RD0RF_1LB---_1LB1LE_1RA0LC/notes.md · raw

W6-7 observe notes: 1RB1RF_0RC0LE_1RD0RF_1LB---_1LB1LE_1RA0LC

740 line 198. Sheet: Unknown, no status, no "done by". No claim in data/discord. Not in upstream busycoq at cdaf693. Claimed 2026-10-14 (coord/claims.tsv, "Wave 6 W6-7"). Observe took about 1 h 55 min.

Table

state read 0 read 1
A 1RB 1RF
B 0RC 0LE
C 1RD 0RF
D 1LB halt
E 1LB 1LE
F 1RA 0LC

Halting pattern (from the table)

Local rules (exact; scripts/validate.py, 200 random instances each with random left/right context words)

LS     L 1^(a+1) 0 1^b <E R   ->  L 1^a <E 0 1^(b+1) R    b+2 steps
WALL   L 0 0 <E R             ->  L 0 0 F> R              3 steps (also L = 0^inf)
RS3    L F> 0111^n R          ->  L 1011^n F> R           6n steps
RSodd  L F> 0 1^(2k+1) R      ->  L 1 01^k 1 F> R         4k+2 steps
RSeven L F> 0 1^(2k) 0 1 R    ->  L 1 01^k 1 <E 0 R       4k+3 steps
HALT   L A> 0 0 0 1 R         ->  halt                    3 steps

(01^k = (01)^k.) Reading in runs of 1s between single 0s: - the left sweep moves every 0 one cell left (it needs every run >= 1; a 00 or the left end turns it); - the right sweep moves every 0 one cell right through odd runs: run 1 merges with the next run, run 3 just shifts, run 5 splits into 1 + 3. So a bounce leaves a pure 0111 field unchanged and toggles a defect 11111 <-> 10111; - an even run (0, 2, 4) makes A read a 0, and the head turns left there: even runs are internal walls.

Observed dynamics

Siblings (canon over all starts)

Template and cryptid check

Next goal (2 h)

Certificate lane first: a bounded CTL62 parameter sweep (CPS_LRU, NG, RWL_mod; native search before any coqc) on the three list starts, because halting needs F> 0000 1 and four-zero runs only appear next to the head. If nothing is found, build the exact run-length model (conveyor + RSeven wall motion + left-end absorption + right emitter) and test whether the wall phases follow a law.

Files

Wave 6, stage 1 (2026-10-14)

Goal: look for a CTL certificate first. If none turns up, build the run-length model and settle the cryptid question.

Result: cryptid candidate, confidence med. No CTL certificate was found. The wall dynamics are now an exact particle picture, and the halting route is known. The machine halts if two exact-timing coincidences occur, each between a wall birth at the left end and a wall arriving far to the right. Nothing in the data rules these coincidences out. Non-halting would need a hidden timing lock, and no regular (CTL-style) language can express one.

CTL search (scripts/ctl/)

Local languages do not close

Particle picture (head is stateless between runs)

The halting route

Halting needs F> 0 0 0 0 1, which in practice means an interior 000 that a wall then reaches.

The tape never has an interior 000 at left-end bounces. None in 773,707 bounces to 2e9 (pairs.c). To 3e10 (5,248,319 bounces, ones.c → work/stage1/ones_3e10.txt) there is still no interior 000, and a run 1 directly before a 00 (1:) occurred 11 times: 3 in the start-up transient and 8 between 1.24e9 and 2.24e10 (bounces 531,221 to 3,891,278). Every one of the 11 was the leftmost wall structure (no wall to its left).

The chain, with each link checked on synthetic tapes (chain_checks.py):

  1. A frozen 1: must exist: a run 1 directly before a 00 wall.
  2. A moving wall that reaches a frozen 1: turns it into an interior 000: 36/36 (forms 2, 4 and :; shifts 0-3; contexts 1:1, 1:3, 5.1:1). Control: the same with a run 3 before the 00 never does, 12/12.
  3. A moving wall that reaches a frozen 000 halts the machine within a few bounces: 105/105. If F reaches the 000 with no wall in front, it just dissolves it: 4/4, no halt in 200 bounces.

What happens in the real run. Each 1: so far was the leftmost structure, so F dissolved it on the next bounce. Example: 2.2:1 -> 3.1:1 -> 7:3 at steps 1,238,230,449 / 1,238,230,967 / 1,238,231,495. That matches chance: with births on about 1.1 % of bounces at those times, 8 episodes give 0.09 expected coincidences. For a 1: to stay frozen, a wall must be born at exactly that bounce. That wall must then reach the 1: before another birth intercepts it. When it makes the 000, another birth must happen at exactly that bounce, followed by another uninterrupted trip.

A real-tape check of half the chain. Start from the tape at step 1,238,230,449, with one change: the first run 3 becomes 1, i.e. a defect arrives one bounce earlier. Then 1:1 stays frozen, the new wall reaches it, and an interior 000 appears at bounce 52. Without the change there is no 000 in 3,000 bounces (chain_checks.py, real_tape_*).

The full chain on the real tape. Same start, but now with three edits, each of which only changes when a defect reaches the left edge:

With these edits the machine halts at bounce 105. With only the first two edits, the natural birth at bounce 86 intercepts W2, and the 000 dissolves later without a halt.

No timing lock is visible. To 1e8, births and annihilations fall in the same bounce 14 times, against 22.7 expected if they were independent. Births 1-5 bounces after an annihilation: 14/20/13/14/24, against about 22 each.

Verdict and estimate

This is a cryptid candidate. Non-halting comes down to whether exact-bounce coincidences between left-end births and distant wall arrivals ever line up. That is pseudo-random timing rather than arithmetic, but the consequence is the same.

A crude estimate, assuming births are independent of everything else:

If the birth rate levels off near 0.5 %, the expected halt is around 1e16 steps. If it keeps falling like t^-0.33, the expected number of halts grows only like a tiny power of t: heuristically it still halts eventually, but astronomically late. Either way, a simulation hunt is unlikely to find the halt. A halt that far out also could not be certified with the runners we have (1e10 or more bounce-level macro steps).

Siblings 248 (start E, mirrored) and 582 (start C) share the table and the route. Censuses to 1e10 (ones.c with start state C or E; work/stage1/ones_start{C,E}_1e10.txt): no interior 000 in either. 1: occurred once for start C and twice for start E (both in the start-up transient), and every one was the leftmost structure.

Files