/machines/1RB0RF_1LC0RE_0LC0LD_1RD1LB_0RF1RA_0RB---/notes.md · raw

1RB0RF_1LC0RE_0LC0LD_1RD1LB_0RF1RA_0RB---

Status: KILLED_NONHALT (2026-10-10; see ## Kill at the bottom) Class guess (agent / Dennis): Bell eats counter = SOMDCv6 TM7 + a "+1 pass at each 110 entered" / – Spreadsheet notes / Discord thread: Results row 1035: class "Bell eats counter"; Status, Notes, Sources, Done By all blank; no links, comments or strikethrough. Not on either Discord list in data/discord/. Its twin 1RB0RA_… (A1=0RA, row 1048) is also open. Viewer: https://bbchallenge.org/1RB0RF_1LC0RE_0LC0LD_1RD1LB_0RF1RA_0RB---

Observation

The pictures (img/)

Both pairs start from the same constructed configuration: S(L^4 1L1) = 0^inf 11 D> 110 110 110 110 100 110 100 0^inf. Neither machine passes through it from blank. Each pair shows every step, both panels top-aligned at the same scale. To regenerate: python3 -B scripts/compare_tm7.py. - eat_pair.png: one eating pass each, from the start until the eater reads the 1-digit at cell 12. - Both eaters eat the same 4 L's (k = 4). - TM7 makes an 11-step wiggle at each L: back 1 cell, forward 2, back 3, forward 4. - Ours starts the same wiggle. At A1 (its 4th step), instead of turning back, it runs one +1 pass (blue box) and comes back. - Cut ours' 4 passes (38 + 14 + 20 + 2 = 74 extra steps) and what is left is TM7's run: the same state and head cell at every step, 118 − 74 = 44. - sweep_pair.png: one full sweep each. - After the first stop, a drain pass reaches the L at cell 15 and fires A1 in both machines. TM7 turns back. Ours runs +1 pass 5. In both, the eater then restarts. - With the passes cut, the head paths agree up to TM7 step 70. There TM7 reads a 1 at cell 18 that ours' +1s had made 0. - So ours eats 2 more digits (k = 6 vs 4) and drains 6 passes vs 4. - Ours ends at S(L^9 111) after 421 steps, TM7 at S(L^7 101) after 237. Both end with the wall 1 cell further left. - The drain zigzags contain no A1, so they use the same transitions in both machines. Only the blue boxes are extra. - sweep_ours.html and sweep_tm7.html: zoomable viewers of the same two sweeps. Hovering shows the step, state and cell. - A real example (not drawn): ours visits S(L^8 0L11) at step 809. - Both machines eat L^8 0 L the same way (9 A1s). - Then TM7 meets the 1-digit at cell 30: it stops at step 102 with k = 10. - Ours' +1s had turned the next digits into 0s. It eats 3 more and stops at cell 39 at step 381, with k = 13. - The sweeps take 1,121 vs 641 steps. - Removed: the old eater_ours.png / eater_tm7.png. They showed the same 420-step window from that config, which is a different phase for each machine. Ours was still eating (eating ends at step 381). TM7 had finished eating at step 102 and was mostly draining. So the pair could not isolate the difference.

Candidate rules

The digit form is the same as TM7's in SOMDCv6.v.

Sweep: S(r) -->+ S(eat(r)), with the wall one cell further left.

eat is TM7's RIncs hx 0, plus the two clauses marked NEW. Together they say: every L entered fires exactly one +1, whether the eater enters it or a drain pass does.

eater, k:   0 -> L, k+1        L -> L, k+1, +1 to the rest (NEW)        1 -> L, enter count mode with k
count, k>=1: L -> L, back to eater with k, +1 to the rest (NEW)
             1 -> L, k-1       0 -> 0, (k-1)/2 if k odd       0 -> 1, (k-2)/2 if k even
right end:  k=0 -> 1;  k odd -> 0, (k-1)/2;  k even -> 1, (k-2)/2   (repeat)
+1 = little-endian binary over the 0/1 digits; L digits are skipped; overflow appends 1

Where the +1 comes from: A1 fires exactly when an L is entered, once per L digit of r in every sweep. - TM7's A1 = 0RC turns straight back. - Ours' A1 = 0RF runs one ordinary h-pass over everything to its right, then comes back. It ends exactly where TM7 is one step after its A1: state C, one cell right of the A1 cell. The only change is that the digits to the right are +1. The pass itself: - B> 100 -> 000 B> (carry); - B> 000 -> <C 100; - B> 110 -> 011 B>, then on the way back 011 <C -> <C 110. - Eater enters L: 1 D> 110 X -> 1 111 D> inc(X) in 6j+13 steps (j = digits before the first 000 in X). TM7: -> 1 111 D> X in 11 steps. - Drain pass enters L: B> 1 110 X -> 1 111 D> inc(X) in 6j+12 steps. TM7: -> 1 111 D> X in 10 steps. The eater restarts there.

Tests (all by simulation; run python3 -B check_rules.py):

check result
Sweep, from blank, every left record to 1e9 208/208 (TM7's map alone: 1/208)
Sweep, random r, first half / second half 120/120 and 174/174
L entry, ours vs TM7 (--lentry): eater and drain contexts, every X up to 6 digits plus 300 random X up to 60 digits 1393/1393 in each of the 4 cases
Eater enters L, random X (in low_level) 150/150
Each low-level chain (a = 0..1500) 45/45
Abstract map from S(L1) 9,432 sweeps (~2e14 steps), ending at n = 109,794, without leaving the domain of eat; the smallest pass count seen in count mode is k = 1

Not a cryptid: the rules are counter arithmetic (+1, halving a pass count), with no branching on the parity of huge numbers.

Checkpoint A (Dennis), 2026-10-10

Informal proof

Not started (waits for Checkpoint A). Plan: 1. Copy TM7. 2. Add Inductive Inc, proved as sideRLs tm [h] (toRC r) (toRC r'). 3. Put Inc into RIncs_hx_ld (eater enters L) and RIncs_h1_ld (drain pass enters L). 4. Redo RIncs_nxt and the domain lemma; this is the main risk. 5. Prove nonhalt exactly as TM7 does.

Formalization log

Outcome

Card data pass (2026-10-10)

Kill

Checkpoint B (Dennis), 2026-10-10

Submission, 2026-10-10