{
"tm": "1RB1LA_1RC1RE_1RD0RD_1LA1RC_1RF0LE_1RA---",
"rank": 45,
"class_sheet": "Shift-overflow structural counter",
"class_guess": "SO structural counter: binary counter stored between 6-cell-block segments, widening into a finite reservoir",
"confidence": "low",
"summary": "Binary counter over segments of 6-cell blocks; it eats a finite reservoir, and what happens after (~1e13+) is unseen",
"mechanism": [
"Cycle: from the right edge the head zigzags left (rule Z: 20 steps per block, 101110 -> 111110), bounces off the first boundary that does not carry, then runs right (RS: 6 steps per block). The right edge grows 4 cells per cycle. Halting needs `0 11 by step 7.5e9, losing 5 then 2 per new top bit. qsim (block 12): the reservoir is gone by 3.4e12, older prefix gadgets are eaten by 7.7e12, and there is no halt to 8e13. The prefix left after that, 1^4 (01)^11 1 (10)^4, is the open question."
],
"diagrams": [
{
"mode": "records",
"png": "img/records_4e5.png",
"html": "img/viewer.html",
"steps": 400000,
"caption": "Records to 400k: early left counter, then left edge frozen",
"title": "Left/right records, first 400,000 steps",
"why": "Top quarter: the early phase grows on both sides. Below it, compare the frozen left block with the right part.",
"primary": true,
"start": "blank",
"rows": "left/right records: the start, then one row each time the head reaches a new leftmost or rightmost cell (1,162 rows, steps 0..400,000); rows are not equal time",
"window": "cells -59..1102, 1 px = 1 cell",
"legend": "default colours: white = 1, black = 0 or never visited, red/green = head just moved right/left",
"style": "bbchallenge-default"
},
{
"mode": "custom",
"png": "img/cycle_depth_256.png",
"html": null,
"steps": 4879538,
"caption": "One row per cycle: how far left each sweep went",
"title": "How far left each of 272 cycles reaches",
"why": "One row per cycle. Compare where the orange bars start, row to row; their right parts are all alike.",
"primary": true,
"start": "mid-run: step 9406481 (rows are the real run from blank; no rebuilt tape)",
"rows": "one row (2 px tall) per cycle, i.e. per new right record after a leftward excursion of more than 20 cells: 272 cycles, steps 9,406,481..14,286,019",
"window": "cells -60..1539, 1 px = 2 cells (the right part of the tape, up to cell ~5,900, is cut off)",
"legend": "orange = cells the head swept during that cycle, dark grey = tape it did not reach in that cycle, black = never visited",
"style": "custom"
},
{
"mode": "every:1",
"png": "img/zigzag_closeup.png",
"html": null,
"steps": 120,
"caption": "Rule Z up close: 20-step zigzag per 6-cell block",
"title": "The leftward zigzag, step by step",
"why": "120 steps from the real tape. Follow the head colours through one 20-step repeat; compare blocks before and after.",
"primary": true,
"start": "mid-run: step 9848484 (the real tape at that step, passed in with --from-config)",
"rows": "every step (121 rows, steps 9,848,484..9,848,604)",
"window": "cells 4902..4957, 1 px = 0.1 cells (each cell is 10 px wide; drawn as -44..11 relative to the head at the start)",
"legend": "explore colours: white = 1, slate = 0; the head cell is coloured by state: A red, B orange, C blue, D green, E magenta, F cyan",
"style": "bbchallenge-explore"
},
{
"mode": "blaze",
"png": "img/blaze_1e7.png",
"html": "img/viewer.html",
"steps": 10000000,
"caption": "Blaze 1e7: sqrt-like growth to the right, fixed left part",
"title": "Blaze, 10 million steps",
"why": "The overall shape: span grows about like sqrt(t) to the right while the left end stays put.",
"start": "blank",
"rows": "Blaze time blocks of 16,384 steps (averaged), 611 blocks stretched to 500 px, so 1 px row ~ 20,000 steps (steps 0..10,000,000)",
"window": "cells -60..4984, 1 px = 6.3 cells (about: 1,261 engine px of 4 cells each, stretched to 800 px)",
"legend": "white = 0, orange = 1, paler orange = mixed (each pixel averages its cells over its time block); no head drawn",
"style": "bbchallenge-blaze"
},
{
"mode": "every:400",
"png": "img/every400_16cycles.png",
"html": null,
"steps": 265000,
"caption": "16 cycles, every 400 steps, whole tape (sampled columns)",
"title": "16 cycles across the whole tape",
"why": "The slanted lines are the zigzag fronts moving left; their lower ends show how deep each cycle went.",
"start": "mid-run: step 9817765 (the real tape at that step, passed in with --from-config)",
"rows": "every 400 steps (663 rows, steps 9,817,765..10,082,765), about 40 rows per cycle",
"window": "cells -59..4998, 1 px = 4 cells (every 4th column sampled, so the 6-cell blocks alias into stripes; drawn as -4994..63 relative to the head at the start)",
"legend": "default colours: white = 1, black = 0 or never visited, red/green = head just moved right/left",
"style": "bbchallenge-default"
}
],
"history": {
"skeleton": "0^inf 1^4 (01)^11 1 (10)^4 1^7 0 1^6 (10)^3 1^6 (10)^4 1^6 (10)^3 1^6 (10)^7 1^6 (10)^m [counter gadgets + block segments] C> 0^inf",
"vars": [
"m (reservoir pairs)",
"right edge cell",
"source"
],
"rows": [
{
"step": 37397,
"values": [
66,
246,
"direct"
]
},
{
"step": 45549,
"values": [
58,
342,
"direct"
]
},
{
"step": 47893,
"values": [
53,
358,
"direct"
]
},
{
"step": 53183,
"values": [
51,
390,
"direct"
]
},
{
"step": 66489,
"values": [
46,
454,
"direct"
]
},
{
"step": 103687,
"values": [
44,
582,
"direct"
]
},
{
"step": 220745,
"values": [
39,
838,
"direct"
]
},
{
"step": 625191,
"values": [
37,
1350,
"direct"
]
},
{
"step": 2115721,
"values": [
32,
2374,
"direct"
]
},
{
"step": 7823015,
"values": [
30,
4422,
"direct"
]
},
{
"step": 30142857,
"values": [
25,
8518,
"direct"
]
},
{
"step": 118403239,
"values": [
23,
16710,
"direct"
]
},
{
"step": 469407113,
"values": [
18,
33094,
"direct"
]
},
{
"step": 1869346983,
"values": [
16,
65862,
"direct"
]
},
{
"step": 7460955529,
"values": [
11,
131398,
"direct"
]
},
{
"step": 3387627872013,
"values": [
0,
"~2.8M (span)",
"qsim: reservoir gone; next gadget (10)^7 down to (10)^2"
]
},
{
"step": 7686039316189,
"values": [
0,
"~4.2M (span)",
"qsim: prefix eaten down to 1^4 (01)^11 1 (10)^4"
]
}
]
},
"rules": [
{
"name": "Z zigzag left (L-pass)",
"from": "(101110)^n = 0, any context (the head never leaves the shown cells)",
"validated": {
"instances": 90,
"passed": 90
},
"source": "manual"
},
{
"name": "RS run right (R-pass)",
"from": "D> (111110)^n",
"to": "(101011)^n D>",
"steps": "6n",
"cases": "any n >= 0, any context",
"validated": {
"instances": 90,
"passed": 90
},
"source": "manual"
},
{
"name": "H the only way to halt",
"from": "0 1",
"from": "X G S0 (110101)^n 01 C> 0^inf",
"to": "X G S1 (110101)^n 01 C> 0^inf",
"steps": "26n+108",
"cases": "G = 1^7 (01)^3 110 1^9 (01)^2 110 1^7 0; S0 = 1^8 (10)^4 1^9 (01)^6 1 (10)^3 1^5; S1 = 1^8 (10)^4 1^8 (10)^4 1^7 (01)^7; X = any left context (not reached)",
"validated": {
"instances": 100,
"passed": 100
},
"source": "manual"
},
{
"name": "RC1 right end, n+1",
"from": "X G S1 (110101)^n 01 C> 0^inf",
"to": "X G S2 (110101)^(n+1) 01 C> 0^inf",
"steps": "26n+44",
"cases": "S2 = 1^8 (10)^4 1^8 (10)^4 1^6 (10)^4 1^5",
"validated": {
"instances": 100,
"passed": 100
},
"source": "manual"
},
{
"name": "RC3 right end, back to S0, n+1",
"from": "X G S3 (110101)^n 01 C> 0^inf",
"to": "X G S0 (110101)^(n+1) 01 C> 0^inf",
"steps": "26n+44",
"cases": "S3 = 1^8 (10)^4 1^8 (10)^6 1^3 (01)^6",
"validated": {
"instances": 100,
"passed": 100
},
"source": "manual"
},
{
"name": "RC2 right end carries left",
"from": "X G S2 (110101)^n 01 C> 0^inf",
"to": "X' G S3 (110101)^n 01 C> 0^inf",
"steps": null,
"cases": "X' = X after one deeper pass (bit flips at the boundaries to the left); seen on the orbit, not yet an exact rule",
"validated": {
"instances": 0,
"passed": 0
},
"source": "manual"
},
{
"name": "D binary-ruler law for cycle depths",
"from": "cycle k (right record after a leftward excursion > 20 cells)",
"to": "cycles that reach level j+1 are exactly every 2nd cycle that reaches level j",
"steps": null,
"cases": "levels found as nested local minima of each cycle's leftmost cell; checked on the real orbit, steps 5e4..1e10 (13 levels); one spacing of 3 at steps 41,783..47,893, the end of the early phase",
"validated": {
"instances": 37885,
"passed": 37885
},
"source": "manual"
},
{
"name": "M reservoir use per new top bit",
"from": "1^6 (10)^m [top gadget]",
"to": "1^6 (10)^(m-5), then 1^6 (10)^(m-7)",
"steps": null,
"cases": "the 11 consecutive changes 58 -> 53 -> 51 -> ... -> 16 -> 11 on the real orbit (direct, to 1e10); 66 -> 58 at the start does not fit",
"validated": {
"instances": 11,
"passed": 11
},
"source": "manual"
}
],
"template": {
"file": "verify/SOC_2_4_v4.v:655",
"module": "TM204",
"machine": "1RB1LA_1LC1RE_1RD0LC_1LA1RC_1RF0RB_1RA---",
"diff": "B0 1RC->1LC, C1 0RD->0LC, E1 0LE->0RB",
"applies": "no",
"note": "Nearest proved machine (3 cells away): a plain BinaryCounter_v2 proof with different words. The class templates SBCv6a/SBCv10/SOBLv1-3 are 8-11 cells away. Longitudinal.v, whose segRLs lemmas SOBLv3 uses (k passes become k*2^len), fits the pass-halving segment structure best."
},
"strategy": "First pin down the epoch map: exact rules for the boundary bit flips and the top gadget, then a fast abstract simulator to watch 2-3 overflows.\nOnly if the overflows look self-similar: a Longitudinal.v (segRLs) proof with a binary-counter invariant plus an epoch induction.\ninit covers steps 0..~5e4 (the early left counter).",
"risks": [
"Halting is decided by the overflow, which has not been seen. After the reservoir and prefix gadgets are gone (~1e13), the counter must reach blank tape on the left and seed a new reservoir, as it did once at step ~3.6e4. Maybe ~1e16 steps, past what qsim reaches.",
"The inner boundary gadgets (4 states each, in two block phases) and the top gadget are only read off the orbit; RC2, the deep right-end step, has no exact rule.",
"A proof needs an invariant over an unbounded list of segments plus an epoch-level argument; no template within 2 cells."
],
"checkpoint_questions": [
{
"text": "Fig 2: do the left ends of the orange bars form a binary ruler (every 2nd bar reaches further, every 4th further still, and so on), with no row out of pattern?",
"figs": [
1
],
"kind": "check"
},
{
"text": "Fig 1 and Fig 3: once the left edge stops moving (about a quarter of the way down Fig 1), is every later leftward sweep the 20-step zigzag of Fig 3, followed by a straight run right?",
"figs": [
0,
2
],
"kind": "check"
},
{
"text": "Priority call: the counter eats a finite reservoir (gone by ~3.4e12), then older prefix gadgets (by 7.7e12). Halting depends on the overflow after that, maybe ~1e16 steps out. Should we spend more budget (exact boundary-gadget rules + an abstract simulator to watch overflows), or park this machine?",
"figs": [
1
],
"kind": "decision"
}
],
"discord_check": "Has anyone run 1RB1LA_1RC1RE_1RD0RD_1LA1RC_1RF0LE_1RA--- (sheet: Shift-overflow structural counter) past ~8e13 steps, or analysed its long-run overflow? We see a binary counter stored between segments of 6-cell blocks. It widens by eating a finite reservoir (10)^m, and then the older prefix gadgets. Quick_Sim (block 12) shows no halt to 8e13, but the next overflow into blank tape looks much further out. (AI-assisted analysis.)",
"stats": {
"simulated_steps": 10000000000,
"cells": 152154,
"nonzeros": 107017,
"growth": "span ~ t^0.5 (48,282 cells at 1e9; 152,154 at 1e10); left edge fixed at -59 since step 35,809; qsim (block 12, steps exact) has no halt to 8e13"
},
"status": "CHECKPOINT_A",
"status_reason": "Mechanism found within the current epoch (binary counter over block segments; 6 exact rules validated plus the binary-ruler law at 37,885/37,885). Confidence is low because the long-run overflow (after ~1e13) decides halting and has not been observed. No sign of a cryptid so far."
}