{ "tm": "1RB0RD_1RC1RF_1LD0LE_1RA0LC_1RE1LC_0RA---", "rank": null, "class_sheet": "Skelet 1-like Class 2", "class_guess": "Skelet 1-like Class 2: kill #3's table renamed by DABCEF (no mirror), started in #3's state B", "confidence": "high", "summary": "Kill #3's table renamed (A->D, B->A, C->B, D->C), started in #3's B: same 9 rules, its own orbit, 3-pair base.", "mechanism": [ "Every transition is #3's under rho = DABCEF (no mirror); the twin's blank run is #3's run from 0^inf B> 0^inf (lockstep 2e6 steps).", "So #3's binary down-counter on (10)^p (01)^q pairs, its 9 macro rules and INV+(e) apply verbatim, with head state D.", "Different orbit, not coupled to #3: base K(2;(47,32,0),(40,33,1),(13,34,1);-) at step 38,732 has 3 pairs (L*=3), not 4." ], "diagrams": [ { "mode": "every:1000000", "png": "img/every1e6_1e9.png", "html": null, "steps": 1000000000, "caption": "This machine: snapshot every 1e6 steps to 1e9", "title": "This machine, every 1M steps to 1e9", "why": "Bands are pair blocks. Compare with Fig 2: same window, same sampling, same steps.", "primary": true, "start": "blank", "rows": "every 1,000,000 steps (1,001 rows, steps 0..1,000,000,000): big jumps between rows", "window": "cells -81,510..409, 1 px = 32 cells (only every 32nd cell is drawn; all drawn cells are even)", "legend": "white = 1, black = 0 on the drawn cells, red/green = head. Sampling every 32nd cell turns the 1010... tape into solid bands, so each colour edge is a pair boundary: vertical edge = digit block, slanted edge = (10)|(01) split; thin horizontal streaks = rows caught mid-sweep", "style": "bbchallenge-default" }, { "mode": "every:1000000", "png": "img/parent3_every1e6_1e9.png", "html": null, "steps": 1000000000, "caption": "Parent kill #3, drawn the same way", "title": "Parent kill #3, same view", "why": "Kill #3 from its own blank start, drawn exactly like Fig 1. Compare band shapes and the left edge.", "primary": true, "start": "blank", "rows": "every 1,000,000 steps (1,001 rows, steps 0..1,000,000,000) of kill #3 1RB0LD_1RC0RA_1RD1RF_1LA0LE_1RE1LD_0RB---: big jumps between rows", "window": "cells -81,510..409, 1 px = 32 cells (only every 32nd cell is drawn; all drawn cells are even)", "legend": "white = 1, black = 0 on the drawn cells, red/green = head. Sampling every 32nd cell turns the 1010... tape into solid bands, so each colour edge is a pair boundary: vertical edge = digit block, slanted edge = (10)|(01) split; thin horizontal streaks = rows caught mid-sweep", "style": "bbchallenge-default" }, { "mode": "records:L", "png": "img/init_recL_4e4.png", "html": null, "steps": 40000, "caption": "Left records to step 38,792: the run up to the base", "title": "Left records up to the base", "why": "One row per new leftmost cell, so roughly 8 rows per decrement. The base is the last rows (step 38,732).", "primary": true, "start": "blank", "rows": "the start, then one row per new leftmost cell (375 rows, steps 0..38,792; not evenly spaced in time), 1 row = 2 px", "window": "cells -374..51, 1 px = 1 cell (only every 2nd cell is drawn, 2 px wide; all drawn cells are even)", "legend": "white = 1, black = 0 on the drawn cells; green = head (always at the new left edge). On the 1010... tape each band is a (10) or (01) run, so colour edges are pair boundaries", "style": "bbchallenge-default" }, { "mode": "default", "png": "img/default_site.png", "html": "img/viewer.html", "steps": 10000, "caption": "bbchallenge default view, first 10k steps", "title": "bbchallenge default view, first 10,000 steps", "why": "The site's view: the tape grows to the left; the right end barely moves.", "start": "blank", "rows": "10,000 steps squeezed into 500 px rows (1 px row = 20 steps; the site blends steps 20y+9 and 20y+10)", "window": "cells -200..199, 1 px = 1 cell", "legend": "white = 1, black = 0 or never visited, red/green = head after a right/left move", "style": "bbchallenge-default" }, { "mode": "custom rows: one per decrement, every cell, annotated", "png": "img/zoom_cycles_annotated.png", "html": "img/zoom_cycles_annotated.html", "steps": 21790, "caption": "Every cell around two re-pairs, tinted by role", "title": "Every cell across two re-pairs, annotated", "why": "A slice of Fig 3, one row per decrement, every cell drawn. Follow purple blocks and blue/orange borders down.", "primary": true, "start": "blank", "rows": "one per counter decrement: each time D reaches a new left edge with the tape reading (10)^h 11001... (21 rows, steps 3,520..21,790, all of them rows of Fig 3); not evenly spaced in time; a yellow band marks each re-pair", "window": "cells -152..47, 1 px = 1/3 cell (1 cell = 3 px wide, 1 row = 14 px tall); from step 8,945 on, the (10)^h run at the left end continues past the left edge", "legend": "white = 1, black = 0, blue = (10)-run of a pair, orange = (01)-run of a pair, purple = digit block, grey = (10)^h run at the left end, gold = marker 11001, teal = tail, green = head, black dashes = old (10)|(01) border carried down across a re-pair band, purple dots = middle of an old digit block carried down the same way. Both lines run only across the yellow bands, never over a cell. The tints are translucent over the raw cells and come from parsing each row; every row parses completely. The inset magnifies the red box; the viewer can hide the tints", "style": "custom" } ], "history": { "skeleton": "0^inf =2; touch (p,q,d) = (p+1,q-1,1-d). #3's rule with 0 digit: fuse with the next pair", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "DecMergeTail", "from": "K(h; Z ++ [(p,1,1)]; (10)^t 1)", "to": "K(h+4; touch Z; (10)^(p+t+3) 1)", "steps": null, "cases": "the last pair runs out into the T1 tail", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "DecMergeEmpty", "from": "K(h; Z ++ [(p,1,1)]; -)", "to": "K(h+4; touch Z; (10)^(p+2) 1)", "steps": null, "cases": "the last pair runs out, no tail (never on the orbit)", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "End", "from": "K(h; Z ++ [(p,q+2,0)]; -)", "to": "K(h+4; touch Z ++ [(p+1,q+1,0)]; (10)^1 1)", "steps": null, "cases": "V = 0, no tail: adds tail T1(1)", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "EndLast", "from": "K(h; Z ++ [(p,1,0)]; -)", "to": "K(h+4; touch Z; (10)^(p+4) 1)", "steps": null, "cases": "V = 0, last q = 1: last pair becomes the tail", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "Ov", "from": "K(h; (p1+2,q1,0) :: Y; (10)^t 1)", "to": "K(2; (h+5,p1+1,0) :: ovp q1 Y t; -)", "steps": null, "cases": "V = 0 with tail: re-pairing (q_j, p_{j+1}), one more pair, h back to 2", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "X1", "from": "K(h; (p1+2,q1,0) :: Y ++ [(pa+1,1,0); (pb,1,0)]; -)", "to": "K(2; (h+5,p1+1,0) :: ovpX q1 Y (pa+1); (101)^2 (01)^(pb+3))", "steps": null, "cases": "V = 0, last two q = 1: re-pairing with tail T2", "validated": { "instances": 45, "passed": 45 }, "source": "manual" }, { "name": "X2", "from": "K(h; Z ++ [(p,q+2,0)]; (101)^2 (01)^(k+1))", "to": "K(h+4; touch Z ++ [(p+1,q+1,0); (3,k+1,1)]; -)", "steps": null, "cases": "V = 0 with tail T2: End-like, appends pair (3,k+1,1)", "validated": { "instances": 45, "passed": 45 }, "source": "manual" } ], "template": { "file": null, "module": null, "machine": "1RB0LD_1RC0RA_1RD1RF_1LA0LE_1RE1LD_0RB---", "diff": "none under rho = DABCEF: all 12 entries equal after renaming states (no mirror); the start state differs (#3's B)", "applies": "yes", "note": "Template = our kill #3's .v (machines/1RB0LD_1RC0RA_1RD1RF_1LA0LE_1RE1LD_0RB---/, sha256 cfa603a3..., check_kill PASS): rename {{A}}->{{D}}, {{B}}->{{A}}, {{D}}->{{C}}; new 38,732-step esx init and 3-pair base; J, Jstep, J_base script verbatim. Same recipe as #25." }, "strategy": "Delta proof (proof.md): Lemma R (renaming) transports #3's rules, INV+(e) and the period theorem; only init and base change.\nFormal: #3's .v with rho applied to the state names, init = one esx of 38,732 steps to the 3-pair base, J_base by computation.\nThen tools/check_kill.sh.", "risks": [ "The base has 3 pairs (L* = 3, INV+(e)'s lower edge); the theorem covers it and J holds there, but it is the one structural difference from #3/#25/#24.", "A 38,732-step esx init; #3's 78,710-step one took about 30 s, so this should be fine.", "Discord #bb6 not searched for this string yet; search before posting." ], "checkpoint_questions": [ { "text": "Fig 1 and Fig 2 are this machine and kill #3, drawn the same way. Do both show the same kind of bands (vertical digit edges, slanted splits), with this machine's left edge farther out?", "figs": [ 0, 1 ], "kind": "check", "answer": "yes", "date": "2026-10-10" }, { "text": "zoom_cycles_annotated: each row is the tape at the start of one count-down step. A pair = a run of 10s (blue) then a run of 01s (orange); its digit block = the next 4 cells (purple), 0001 = 1, 0010 = 0. Re-pairing (the yellow bands, above rows 6,965 and 17,738) = the tape is regrouped into new pairs and the counter (right margin) restarts, at 6, then at 2; it happens at the counter's second 0. At the first 0 the step is End instead: it raises the counter (0 → 1 at row 5,100; 0 → 3 at row 12,389; 0 → 1 at row 20,870) and changes the right end: '+ tail' adds a 3-cell teal tail after the last pair, 'pair 3 → tail' absorbs pair 3 into the tail. (a) Between the bands: each purple block stays in the same 4 columns; only the 1 inside it moves, 0010 ↔ 0001 (pair 3 becomes the teal tail at row 12,389, pair 2 joins it at row 14,453). The blue→orange borders jump 2 cells right: pair 1's on every row, pair 2's on every other row (the magnified box at the top shows one such step). (b) At each band: the black dashed lines (old borders) end on the new purple blocks, and the purple dotted lines (middles of the old purple blocks) end 1 cell left of the new borders (pair 1's starts next to the old gold marker), so the two kinds of column swap roles. The added pair (2 → 3 pairs, then 1 → 2) has its orange run over the old teal tail and its purple block (the rightmost) just past the tail's end, so no line leads to it. Is (a) what you see? Is (b)? (In init_recL_4e4 the edges kink at each swap.)", "figs": [ 4 ], "kind": "check" }, { "text": "OK to submit with the 3-pair base after re-pairing 6, rather than the 4-pair E-start after re-pairing 8 (step 104,510)?", "figs": [], "kind": "decision", "answer": "settled", "note": "The Rocq proof uses the 3-pair base and passed check_kill; nothing left to decide.", "date": "2026-10-10" } ], "discord_check": null, "stats": { "simulated_steps": 1000000000, "cells": 79454, "nonzeros": 39718, "growth": "left edge ~ t^0.5 (cells -79,241..212 at 1e9; #3: 67,343 cells, 33,665 nonzeros)" }, "status": "CHECKPOINT_A", "status_reason": "Card for the delta; the live status is in status.json (CHECKPOINT_B: delta proof.md validated by check_proof.py)", "sweep_id": "F2" }