BB(6) hunt
Machines / #36

1RB1RF_1LC0RF_1LE0LD_1RC0LB_---1RA_1LB1RA

#36ParkedBouncer + Chaotic Edgebbchallenge ↗Explorer

Where it stands

Parked at the observation

Parked 2026-10-10 with #13 (auto-applied recommendation): same edge lemma (E'); an (E') proof would now pay #13, #34, #35, #36. Observe done.

Read the full status

The mechanism is exact and validated, and it is identical to parked #13 at the cycle level: the same bulk (1101101101)^a, the same 7 junction words, and the same tail orbit D_n (1,952/1,952 shared cycles; probe map equal on 217/217 tested D, 0 disagree). #35 is this table renamed. Edge map clean to 1.5e7 cycles (about 4.9e16 steps), but no non-halting argument exists for the chaotic tail (#13's open edge lemma E'). Confidence med, so proof work waits on Dennis (decision.json).

Observe
Deepen
Informal proof
Rocq
space-time diagram, first 10,000 steps from blank
The first 10,000 steps from a blank tape. Time runs down; white is 1.
About this picture

The standard picture (bbchallenge.org's Default view): each row is the whole tape at one step, time runs downward, and the head's path draws the pattern.

bbchallenge-defaultStart blankRows steps 0 → 10,000, top to bottomWindow cells −200..199, left to rightColors 1 0 or unvisited

Overview

In one sentence

#13's twin: same period-10 bouncer, same 7 junction words, same chaotic tail D_n cycle for cycle. Tail = whole problem.

Status
Parkedobserving · updated 2026-10-10Not blocking · Stuck at the observing stage. Answer Q1–Q3 whenever; a “no” redirects the work.
idle since Oct 10 13:58 · status.json 2026-10-10
Class
Bouncer + Chaotic Edgeours: Bouncer + Chaotic Edge (sheet) · ours adds: twin of #13 at the cycle level (same bulk, junction and tail orbit D_n) · med confidence
On the list
line 552 of BB6_holdouts_663.txt · historical 740 line 615
#36eqv ×1sim 1e15UBRRBA
Your calls on file 3
priority call (answered)
Park this machine with #13 (same edge lemma), or fund one family stage on that lemma for #13, #34, #35 and #36?
parked
Parked 2026-10-10 with #13 (auto-applied recommendation): same edge lemma (E'); an (E') proof would now pay #13, #34, #35, #36.
decision.json · answered 2026-10-10: park
Park this machine with #13 (same edge lemma), or fund one family stage on that lemma for #13, #34, #35 and #36? Decisions →

Mechanism

How it works: 3 notes
  • Bulk (1101101101)^a on a left wall moves 1 cell left per bounce (62a steps); the junction cycles through #13's 7 words; 1 bounce in 7 touches the tail D.
  • D_n equals #13's D_n in all 1,952 shared cycles, and the probe map P is #13's (217/217 tested, 0 disagree). #35 = this table renamed; #34 = #13 mirrored.
  • Halt only via B reading 0 with 00 on its left. Never seen: the edge map is clean to 1.5e7 cycles (about 4.9e16 steps). The tail has no proof.

Figures for Checkpoint A

Fig 1 · One row per bounce, first 1.1M stepsQ1viewer ↗

Rows are bounces. Left: striped bulk; middle: diagonal seam; right: ragged part. Compare their widths, top vs bottom.

One row per bounce, first 1.1M steps
bbchallenge-defaultStart blankRows left recordsWindow cells -399..597, 1 px = 1 cellColors 1 0 or never visited head just moved right/left
Rows: the start, then one row per new leftmost cell = one row per bounce (400 rows, steps 0..1,140,693; rows are not equal time) · Window: cell 0 = the start cell
Fig 2 · Our tail vs #13's tail, per cycleQ2

One row per 7-bounce cycle. Compare left and middle panels row by row; the right panel shows differences in red.

Our tail vs #13's tail, per cycle
customStart mid-run: step 679,041Rows one row per 7-bounce cycle n = 0..399Window cells 0..423, 1 px = 1 cellColors 1 0 past the end of D same cell cells differ
Start: row 0 = cycle start C_0(58, D_0) of the run from blank; #13's row 0 is its C_0(59, D_0) at step 784,041 · Rows: taken at each cycle start (this machine: steps 679,041..47,105,452) · Window: of D in each of 3 panels (cell 0 = first cell right of the 54-cell junction X0); panels: this machine | #13 | XOR · Colors: blue-grey bar = panel separator; XOR panel
Fig 3 · Right end only, 100 million stepsviewer ↗

Rows are shifted so the right edge stays put. Compare the ragged band's width at top and bottom.

Right end only, 100 million steps
bbchallenge-defaultStart blankRows left recordsWindow cells -1199..0, 1 px = 1 cellColors 1 0 or never visited head just moved right/left
Rows: the start, then one row per new leftmost cell = one row per bounce (4,580 rows, steps 0..100,000,000; rows are not equal time) · Window: counted from the right edge (each row is shifted so its rightmost visited cell is at 0; the bulk continues off-picture to the left)
More figures (2)Fig 4–5: Blaze view, 10 million steps · bbchallenge Default view, first 10,000 steps
Fig 4 · Blaze view, 10 million stepsviewer ↗

Whole run to 1e7 steps. Compare the smooth left edge with the ragged right edge.

Blaze view, 10 million steps
bbchallenge-blazeStart blankRows Blaze time blocks of 16,384 stepsWindow cells -1346..1137, 1 px = 3.1 cellsColors 0 1 mixed
Rows: (averaged), 611 blocks stretched to 500 px, so 1 px row ~ 20,000 steps (steps 0..10,000,000) · Colors: paler orange: each pixel averages its cells over its time block; no head drawn
Fig 5 · bbchallenge Default view, first 10,000 stepsviewer ↗

For orientation: the start of the run as bbchallenge.org shows it.

bbchallenge Default view, first 10,000 steps
bbchallenge-defaultStart blankRows every step 0..10,000Window cells -200..199, 1 px = 1 cellColors 1 0 or never visited head just moved right/left
Rows: squeezed 20:1 into 500 px as on bbchallenge.org (px row y blends steps 20y+9 and 20y+10)
Rules (9/9 validated)exact tape rewrites the proof will formalize, each re-checked by simulation
rulefrom ⟶ tocasesstepschecked
WALL
0∞ <D 0 110110 r
⟶ 0∞ 11101111 B> r
—12✓30/30
RS right sweep
B> (1101110110)n
⟶ (0110101111)n B>
—12n✓30/30
LS left sweep
(0111101101)n <D
⟶ <D (0110110111)n
—50n✓30/30
B_k bounce, k=1..5
0∞ <D 0 (1101101101)a X_k D 0∞
⟶ 0∞ <D 0 (1101101101)a+[k=5] X_(k+1) D 0∞
any D (random and orbit); X_k = #13's junction words62a+b_k, b = 13, 433, 55, 49, 73✓200/200
B_6 cycle close
0∞ <D 0 (1101101101)a X6 D 0∞
⟶ 0∞ <D 0 (1101101101)a X0 D 0∞
any D62a+13✓40/40
probe B_0
0∞ <D 0 (1101101101)a X0 D 0∞
⟶ 0∞ <D 0 (1101101101)a X1 P(D) 0∞
P = #13's probe map: orbit D 120/120 same; random D 97 same + 103 break in both, 0 disagree62a+c(D)✓120/120
cycle
0∞ <D 0 (1101101101)a X0 D_n 0∞
⟶ 0∞ <D 0 (1101101101)a+1 X0 D_(n+1) 0∞
D_n = #13's D_n; also 2,079/2,079 real cycles to 1e9434a+698+c(D_n)✓60/60
halt
00 B> 0
⟶ halt
also 000 <B; the only way to halt (B0, C0, E0)3✓40/40
init
0∞ A> 0∞
⟶ 0∞ <D 0 (1101101101)57 X1 D_0 0∞
first cycle start C_0(58, D_0) at step 679,041657139✓1/1
Compressed historythe tape as 0∞ <D 0 (1101101101)a X0 D 0∞ (X0 = 110111100110011101100100100111101100100110110111010011, #13's X0) at 20 moments: where the rules are read off
0∞ <D 0 (1101101101)a X0 D 0∞ (X0 = 110111100110011101100100100111101100100110110111010011, #13's X0)
stepa|D|D (first 24 cells)
679,041582611.1·1022
710,472592581.1·1022
739,157602551.11·1022
771,604612521·1021
801,867622491.1·1022
833,752632461.11·1022
864,985642431·1021
992,699682311.1·1022
1,349,053782011.1·1022
2,711,1631083211·1021
5,831,4031583211.11·1022
15,317,0352582911.11·1022
29,137,4753583511.1·1022
69,822,1115583511·1021
127,853,0137583211.1·1022
247,450,4991,0584111·1021
406,142,4291,3582911.11·1022
603,885,7851,6583211·1021
884,868,2322,0093481.1·1022
999,698,4152,1363271.11·1022
alast 2,136
2,136586.8e51e9log xstep 679041: a = 58step 710472: a = 59step 739157: a = 60step 771604: a = 61step 801867: a = 62step 833752: a = 63step 864985: a = 64step 992699: a = 68step 1349053: a = 78step 2711163: a = 108step 5831403: a = 158step 15317035: a = 258step 29137475: a = 358step 69822111: a = 558step 127853013: a = 758step 247450499: a = 1058step 406142429: a = 1358step 603885785: a = 1658step 884868232: a = 2009step 999698415: a = 2136
|D|last 327
4112016.8e51e9log xstep 679041: |D| = 261step 710472: |D| = 258step 739157: |D| = 255step 771604: |D| = 252step 801867: |D| = 249step 833752: |D| = 246step 864985: |D| = 243step 992699: |D| = 231step 1349053: |D| = 201step 2711163: |D| = 321step 5831403: |D| = 321step 15317035: |D| = 291step 29137475: |D| = 351step 69822111: |D| = 351step 127853013: |D| = 321step 247450499: |D| = 411step 406142429: |D| = 291step 603885785: |D| = 321step 884868232: |D| = 348step 999698415: |D| = 327
D (first 24 cells)last 1.11·1022
1.1e221e216.8e51e9log xstep 679041: D (first 24 cells) = 11001001101101100100110step 710472: D (first 24 cells) = 11001001001100100100110step 739157: D (first 24 cells) = 11101101100100110011011step 771604: D (first 24 cells) = 1001001100100110110010step 801867: D (first 24 cells) = 11001001100100100110111step 833752: D (first 24 cells) = 11101101100100111011011step 864985: D (first 24 cells) = 1001001100100110110010step 992699: D (first 24 cells) = 11001001101101110100110step 1349053: D (first 24 cells) = 11001001101101110100110step 2711163: D (first 24 cells) = 1001001100100110110010step 5831403: D (first 24 cells) = 11101101100100111011011step 15317035: D (first 24 cells) = 11101101100100110011011step 29137475: D (first 24 cells) = 11001001001100100100110step 69822111: D (first 24 cells) = 1001001100100110110010step 127853013: D (first 24 cells) = 11001001101101110100110step 247450499: D (first 24 cells) = 1001001100100110110010step 406142429: D (first 24 cells) = 11101101100100110011011step 603885785: D (first 24 cells) = 1001001100100110110010step 884868232: D (first 24 cells) = 11001001100100100110111step 999698415: D (first 24 cells) = 11101101100100111011011
Template comparisonBouncer_v3.v TM1: partly applies · the nearest solved proof, what carries over, and the transitions that differ
01
A1RB1RF
B1LC0RF
C1LE0LD
D1RC0LB
E---1RA
F1LB1RA
file
Bouncer_v3.v:44
module
TM1
machine
1LB1RA_0RA0RC_---1RD_1RE1RB_1LF0LE_0RD0LE (12 transitions apart)
diff
—
note
sideRLs: a fixed crossing protocol into a one-sided right part fits (ours: 14 fixed round trips 2 cells left of the X0|D cut), but TM1's RC i is an explicit word; our D is chaotic and needs a closed language (#13's lemma E'). The A-C neighbours in RWL_mod_v2.v / CPS_LRU_v1.v are decider certificates: same parameters give false here.
Proof strategy & riskshow the formal proof will go · 3 risks to watch

Everything except the edge is routine (Bouncer_v3/sideRLs shape; init by direct execution, 657,139 steps). A kill needs #13's edge lemma (E'): a closed language for D under P. It is shared by #13, #34, #35 and #36, so one lemma would mean 4 kills. Until then, park with #13.

  • ▲Edge lemma open: #13 refuted SFT_k invariants for k <= 36 and plain n-gram CPS (n <= 26); no regular invariant is known.
  • ▲mxdys listed the family on UBRRBA_TODO (a halting tool): a late break or halt is possible; edge map clean to 1.5e7 cycles (about 4.9e16 steps).
  • ▲Even with (E'), this machine needs its own .v: same architecture, but the wall, the sweeps (62a) and the step constants differ from #13.
Stats & provenance109 steps simulated · 21,745 cells · the status in full · files
simulated
109 steps
tape
21,745 cells · 15,171 nonzeros
growth
both edges ~ sqrt(t): lo = -0.47 sqrt(t), hi = +0.22 sqrt(t) at 1e9
status now
Parked 2026-10-10 with #13 (auto-applied recommendation): same edge lemma (E'); an (E') proof would now pay #13, #34, #35, #36. Observe done. The mechanism is exact and validated, and it is identical to parked #13 at the cycle level: the same bulk (1101101101)^a, the same 7 junction words, and the same tail orbit D_n (1,952/1,952 shared cycles; probe map equal on 217/217 tested D, 0 disagree). #35 is this table renamed. Edge map clean to 1.5e7 cycles (about 4.9e16 steps), but no non-halting argument exists for the chaotic tail (#13's open edge lemma E'). Confidence med, so proof work waits on Dennis (decision.json). (status.json, 2026-10-10)
observer's note
Mechanism exact and validated, and identical to parked #13 at the cycle level; no non-halting argument for the chaotic tail D (shared edge lemma E'). Confidence med: proof work waits on Dennis (decision.json).
files
observe.json · notes.md · status.json

Proof & checks

No proof work yet.

Files

machines/<tm>/: 7 entries · raw notes, status, scripts, images

namesizechanged
notes.md ↓8.4 KB2026-10-10 13:58
status.json ↓1.0 KB2026-10-10 13:58
observe.json ↓12 KB2026-10-10 13:56
decision.json ↓2.8 KB2026-10-10 13:58
img/7 items2026-10-10 13:40
scripts/3 items2026-10-10 13:44
work/22 items2026-10-10 13:47
status.json in full
{
 "tm": "1RB1RF_1LC0RF_1LE0LD_1RC0LB_---1RA_1LB1RA",
 "status": "STUCK",
 "stage": "observe",
 "updated": "2026-10-10",
 "confidence": "med",
 "reason": "Parked 2026-10-10 with #13 (auto-applied recommendation): same edge lemma (E'); an (E') proof would now pay #13, #34, #35, #36. Observe done. The mechanism is exact and validated, and it is identical to parked #13 at the cycle level: the same bulk (1101101101)^a, the same 7 junction words, and the same tail orbit D_n (1,952/1,952 shared cycles; probe map equal on 217/217 tested D, 0 disagree). #35 is this table renamed. Edge map clean to 1.5e7 cycles (about 4.9e16 steps), but no non-halting argument exists for the chaotic tail (#13's open edge lemma E'). Confidence med, so proof work waits on Dennis (decision.json).",
 "priority_call_answered": "Park this machine with #13 (same edge lemma), or fund one family stage on that lemma for #13, #34, #35 and #36?",
 "parked": "Parked 2026-10-10 with #13 (auto-applied recommendation): same edge lemma (E'); an (E') proof would now pay #13, #34, #35, #36."
}