1RB1RF_1LC0RF_1LE0LD_1RC0LB_---1RA_1LB1RA

Status: CHECKPOINT_A (confidence med: mechanism exact and validated; the edge has no non-halting argument, so proof work waits) Class guess (agent / Dennis): Bouncer + Chaotic Edge. It is #13's twin at the cycle level: the same period-10 bulk, the same 7 junction words and the same chaotic tail orbit D_n, cycle for cycle / _ Spreadsheet notes / Discord thread: Results row 908 has a blank status, class "Bouncer + Chaotic Edge", and no notes, sources or "done by". The only Discord mention is mxdys's UBRRBA_TODO list (2026-08-24, line 76; Dennis ruled that list fair game). UBRRBA is a halting-only tool (atlas/acronyms.md), so mxdys may suspect a late halt. Byte-exact on data/holdouts/2026-10-07_778.txt line 651. Not SKIPPED. Viewer: https://bbchallenge.org/1RB1RF_1LC0RF_1LE0LD_1RC0LB_---1RA_1LB1RA

Observation

• At 1e9 steps: lo=-14847, hi=6897 (21,745 cells), 15,171 nonzeros, no halt. Both edges grow like sqrt(t): lo ≈ -0.47√t, hi ≈ +0.22√t.
• qsim (block 3, 200k loops) reaches only 1.16M steps with 1 trivial rule: no acceleration. compress track to 1e8 finds 300 distinct skeletons out of 300 (the tail never repeats). rulefit was not run: as for #13, there are no whole-tape hubs to fit.
• Halting: E0 is the only undefined cell, and E is entered only from C0. A D0→C0 entry finds the 1 that D just wrote, so it is safe. Halting therefore needs B reading 0 with 00 on its left (B0, C0, E0): 00 B> 0 or 000 <B halts in 3 steps (40/40). It never happens to 1e9.
• Bulk. At every left record the tape is 0^inf <D 0 (1101101101)^a ...: the same bulk word as #13 (whose wall is 0^inf <A (1101101101)^a).
• Right sweep: 12 steps per block. Left sweep: 50 steps per block. So one bounce costs 62a + β.
• 7-bounce cycle, exact from step 657,139 to at least 1e9 (2,079 cycles):
• C_k(a, D) := 0^inf <D 0 (1101101101)^a X_k D 0^inf, with junction words X_0..X_6 (below) identical to #13's. Here X_1..X_6 include the 3 cells 011 that #13's notes write as part of "011 P(D)".
• Six bounces never touch D. One bounce (the probe, C_0 → C_1) rewrites D to P(D).
• The bulk gains a block at C_5 → C_6. The cut moves +3 cells per cycle.
• Init. The first steady configuration is C_1(57, D_0) at step 657,139. The first cycle start is C_0(58, D_0) at step 679,041 (#13: C_0(59, D_0) at 784,041). D_0 is #13's work/D_cycle0.txt, 261 cells.
• Same tail orbit as #13 (the main finding):
• The real D_n equals #13's D_n for every n = 0..1951 that both 1e9 runs cover.
• Our probe map equals #13's probe map:
  • 120/120 instances on orbit D (random a);
  • 200 random D: 97 give the same P(D), 103 break the junction in both machines, 0 disagree.
• scripts/edgemap.c is #13's edge map ported to this table (D state, extra 0 at the wall). It reproduces our real D at n=1951 and #13's D_edgemap_1951.txt byte for byte. Its 5.5e6-cycle log matches #13's line for line (lenD/min/max at every 5e5 checkpoint).
• Not a relabeling of #13. The best state renaming of #13 (A,B,C,D,E,F → C,F,B,A,D,E) still differs in C0/D0/E1. Those cells are used about 560 times per cycle. The coincidence is at the level of left-record configurations, not transitions.
• #35 is this machine's table, exactly. #35 1RB0LF_1LC0LA_---1RD_1RF1RE_1LF1RD_1LB0RE renamed A,B,C,D,E,F → D,C,E,A,F,B is our table (unique; no mirror), started in #35's state D.

• 35 also follows #13's D_n (892 consecutive cycles to 2e8).

• 34 1RB0LC_1LC0RE_1RA1LD_1LA1LC_1RF0RA_---0LA follows the mirror of D_n (836 cycles to 2e8).

• R6 1RB1LD_1RC0LA_1LA0RE_1LB1LA_1RF0RB_---0LB does not match D_n in this test (not investigated).
• So #13, #34, #35 and #36 share one edge system.
• Probe interface. Two cells left of the X0|D cut the crossing sequence is fixed: 14 round trips, the same in all 2,079 cycles (listed under Template check). So the left part acts on D only through a fixed protocol, as in #13.
• Tail D. |D| is 198–438 on the real orbit to 1e9 and 144–567 over the edge-map run (no drift). The first 24 cells take 8 values. Deeper cells are chaotic. Edge map to 1.5e7 cycles, which is 48,825,447,516,092,311 ≈ 4.9e16 real steps by the step law: no halt and no junction break. D does not repeat within either run segment: 0–5.5e6 is hashed (work/edgemap_5.5e6.txt) and 5.5e6–1.5e7 is hashed separately (work/edgemap_5.5e6_to_1.5e7.txt).
• Step law (exact; 20/20 checks at n = 100..2000 step 100, plus n = 1951):
• step(C_0(58+n, D_n)) = 679,041 + E(n) + 434·(53n + n(n−1)/2), where E(n) = edge-map steps at a=5.
• So 5.5e6 cycles ≈ 6.6e15 real steps, and 1.5e7 cycles ≈ 4.88e16.

Junction words (paper order):

X0 110111100110011101100100100111101100100110110111010011
X1 1001001001100100100100110110110011101100100110110111010011
X2 11011001001100100100100110110110011101100100110110111010011
X3 101110100110110010011101100100100111101100100110110111010011
X4 1101001101100100110011101100100100111101100100110110111010011
X5 11011011001001100110011101100100100111101100100110110111010011
X6 10011100110011101100100100111101100100110110111010011

Candidate rules

All validated by direct simulation from constructed configurations at random a and D (work/validate.py → work/validate.json, work/validate_shift.py → work/validate_shift.json; tools/sim.py for the shift rules, an independent bytearray stepper for the bounces).

Template check: - verify/Bouncer_v3.v TM1 proves S' i -->+ S'(1+i) with sideRLs: a fixed head-crossing protocol hRL^^128 into a right part RC i that is an explicit word in i, plus a wall on the left. Our probe has a fixed protocol too, so the architecture fits. Two cells left of the X0|D cut the head makes 14 round trips, (in, out) = (B,C)(F,B)(A,D)(F,B)(F,C)(F,B)(F,B)(F,B)(F,B)(A,D)(A,D)(A,D)(B,C)(B,C), the same in all 2,079 real cycles. One cell left of the cut there are 2 sequences. #13 has 16 round trips, one cell left of its cut. Our right part D is chaotic, though, so no explicit RC i exists. We need a closed language for D instead. This is #13's lemma (E′). Applies: partly. - The solved neighbours share A–C (1RB1RF_1LC0RF_1LE0LD_…): RWL_mod_v2.v nonhalt30756/34063 and CPS_LRU_v1.v nonhalt12009. They are decider certificates. With their exact parameters, the extracted CTL62 decider (machines/<#7>/scripts/ctl_extract/ctlx) returns true on the neighbours (0.01 s) and false on ours (RWL_mod 1001 30000 30000 4 320 2 1 2 0; CPS_LRU 1001 3000 3000 4 3200 1 2 1 0). Does not apply.

Cryptid check (ground rule 6): no unbounded pseudo-random arithmetic. a grows by exactly 1 per cycle with no branching. The only open part is a bounded-length chaotic word D under a fixed map P. A bounded D would make the orbit eventually periodic, but the state space is about 2^564. Not a cryptid in the Collatz sense, so not CRYPTID_CANDIDATE. The edge lemma is open, as for #13.

Checkpoint A (Dennis)

(pending; decision.json filed: park with #13 vs. a family stage on the shared edge lemma)

Informal proof

Not started. It would be #13's proof.md with our wall, sweeps and step constants. It stays conditional on (E′).

Formalization log

-

Outcome

-

Decision (2026-10-10, auto-applied)

Question: Park this machine with #13 (same edge lemma), or fund one family stage on that lemma for #13, #34, #35 and #36? Answer: park (Park with #13: STUCK at observe, keep the card and rules as the reduction). auto-applied: Claude's recommendation, under Dennis's standing instruction 'do your recommendations and continuously orchestrate' (2026-10-10, while he was at lunch). Override any time. Parked 2026-10-10 with #13 (auto-applied recommendation): same edge lemma (E'); an (E') proof would now pay #13, #34, #35, #36.