{ "tm": "1RB0RF_1RC1LB_1LD1RE_1LB0LD_1RF0RC_1RA---", "rank": 409, "class_sheet": "Shift-overflow counter", "class_guess": "Shift-overflow counter (SOC_2_5_v1 TM10 with A1 = 0RF); halting hinges on the bits of a counter value: cryptid-like", "confidence": "med", "summary": "TM10's two binary counters; it can halt only if a counter value x ever has every 1-bit at an odd position", "mechanism": [ "SOC_2_5_v1.v TM10's macro rules (A1 = 0LB there, 0RF here): a left counter LC (digits 11/10) counts down, a right counter RC (digits 10000/00000) counts up, and overflows shift digits between them. Every TM10 lemma except ROv3_0/ROv3_1 compiles for this string.", "From blank (cfgL(3,6,1) at step 35) every cycle is LOv at cfgL(L,0,2x+1), RC1_Ov_0, c >= 0 RC4_Ov_1, RC4_Ov_0. The chain eats x's 1-bits from the bottom and stops at the first one at an even position, which gives an explicit map (L,x) -> (L',x'). All 20,000 cycles computed (L to 172,193) are like this. The history table has one row per cycle (step column = cycle t).", "If every 1-bit of x is at an odd position, the chain empties and the run leaves TM10's cycle into ROv3, where A reads 1. With 0RF that can halt (cfgL(8,0,17) halts at step 10,189,718). Even-position 1-bits grow ~0.36 per cycle (min 7 at cycle 13); chain lengths look random. No invariant is known, so ground rule 6 applies." ], "diagrams": [ { "mode": "default", "png": "img/default_site.png", "html": null, "steps": 10000, "caption": "bbchallenge default view, first 10,000 steps", "title": "bbchallenge default view, 10,000 steps", "why": "The site's view. Left: the slowly growing left counter. Right: the right counter's carry sweeps.", "primary": true, "start": "blank", "rows": "every step: steps 0..9,999 squeezed into 500 px rows as on bbchallenge.org (1 px row = 20 steps)", "window": "cells -200..199, 1 px = 1 cell", "legend": "white = 1, black = 0 or never visited, red/green = head after a right/left move (pale = blended by the squeeze)", "style": "bbchallenge-default" }, { "mode": "custom", "png": "img/xbits_low340.png", "html": null, "steps": 20000, "caption": "Bits of x at each left overflow, cycles 1..240", "title": "Counter value x, one row per cycle", "why": "Each row is x in binary, lowest bit at the right. Look for any row without orange.", "primary": true, "start": "blank", "rows": "one row per left overflow cfgL(L,0,2x+1), cycles t = 1..240 (2 px per row); computed with the cycle map F (scripts/Fmap.py), which matches the TM10 macro rules and the TM itself", "window": "cells 0..339, 1 px = 1/2 cell: here a 'cell' is a bit position of x (2 px per bit), bit 0 at the right edge; for t >= 46 the top bits of x lie further left and are cut off", "legend": "orange = 1-bit at an even position, blue = 1-bit at an odd position, white = 0-bit, grey = above the top bit of x", "style": "custom" }, { "mode": "explore", "png": "img/halt_r3_explore.png", "html": null, "steps": 91, "caption": "Constructed RC3 overflow halts at step 91", "title": "Where A1 = 0RF matters: a halt", "why": "Row 0 is TM10's ROv3 situation. Watch the red A on a 1 head right; F (cyan) later reads 1.", "primary": true, "start": "constructed: LC(43,k) 10 C> RC3(4,0,15), reached from cfgL(8,0,17) (full tape in data/r3_overflow_config.txt)", "rows": "every step, steps 0..91 (6 px per row); the last row is the halt", "window": "cells -89..47, 1 px = 1/6 cell (6 px per cell)", "legend": "slate = 0, white = 1, head cell coloured by state: A red, B orange, C blue, D green, E magenta, F cyan", "style": "bbchallenge-explore" }, { "mode": "records", "png": "img/records_1e10.png", "html": null, "steps": 10000000000, "caption": "Left and right records to 1e10 steps", "title": "Records to 10^10 steps", "why": "Rows are records only. The staircase of right records is the right counter's carry sweep writing 1s.", "primary": false, "start": "blank", "rows": "the start, then one row per new leftmost or rightmost cell (218 rows, steps 0..1e10; 3 px per row); not evenly spaced in time", "window": "cells -6..211, 1 px = 1/3 cell (3 px per cell)", "legend": "white = 1, black = 0 or never visited, red/green = head after a right/left move", "style": "bbchallenge-default" }, { "mode": "explore", "png": "img/explore_400.png", "html": null, "steps": 400, "caption": "First 50 steps from blank; A reads 1 only at step 10", "title": "From blank: the only A1 step", "why": "Row 10: the red A on a white 1. Through 3e10 steps A never reads 1 again.", "primary": false, "start": "blank", "rows": "every step, steps 0..49 (10 px per row; the site's 800x500 Explore view)", "window": "cells -40..39, 1 px = 1/10 cell (10 px per cell)", "legend": "slate = 0, white = 1, head cell coloured by state: A red, B orange, C blue, D green, E magenta, F cyan", "style": "bbchallenge-explore" } ], "history": { "skeleton": "0^inf 0^inf", "to": "cfgL(3,6,1) = LC(3,6) RC(n)", "steps": null, "cases": "k+1 < 2^L; LC(L,k) = 0^inf 1 + L digits of 2^L-1-k (11 = 1, 10 = 0), RC(n) = n LSB first (10000 = 1, 00000 = 0)", "validated": { "instances": 54, "passed": 54 }, "source": "manual" }, { "name": "RC_Inc", "from": "LC(L,k) 10 C> RC(n)", "to": "LC(L,k) RC1(i,2^(i+1)-2,x)", "steps": null, "cases": "n != 2^L (n = 2^L: corner case, 10/10, never reached from blank since n is odd)", "validated": { "instances": 83, "passed": 83 }, "source": "manual" }, { "name": "RCj_Inc + LC_Inc", "from": "LC(L,k+1) 10 C> RCj(r,n+1,m)", "to": "LC(L,k) 10 C> RCj(r,n,m)", "steps": null, "cases": "j = 1, 3, 4 (RCj = TM10's BinDec2 words with tails 10000, 0010000, 00010000)", "validated": { "instances": 181, "passed": 181 }, "source": "manual" }, { "name": "RC1_Ov_0", "from": "LC(L,k) 10 C> RC1(2r,0,m)", "to": "LC(L+1+5r,(2k+1)2^(5r)-1) 10 C> RC4(i,2^(i+1)-2,m') if m = (2m'+1)2^i, else C> RC(1)", "steps": null, "cases": "m > 0 (47/47) and m = 0 (29/29); odd lenR: RC1_Ov_1/_1_0 77/77 (only after an exhaustion)", "validated": { "instances": 76, "passed": 76 }, "source": "manual" }, { "name": "RC4_Ov_0", "from": "LC(L,k) 10 C> RC4(2r,0,m)", "to": "LC(L+2+5r,(2k+1)2^(5r+1)-1) 10 C> RC(2m+1)", "steps": null, "cases": "ends every cycle", "validated": { "instances": 70, "passed": 70 }, "source": "manual" }, { "name": "RC4_Ov_1", "from": "LC(L,k) 10 C> RC4(2r+1,0,m)", "to": "LC(L+5+5r,(2k+1)2^(5r+4)-1) 10 C> RC4(i,2^(i+1)-1,m') if m = (2m'+1)2^i; m = 0: C> RC(0) (RC4_Ov_1_0 = exhaustion)", "steps": null, "cases": "m > 0 (46/46), m = 0 (33/33)", "validated": { "instances": 79, "passed": 79 }, "source": "manual" }, { "name": "cycle map F", "from": "cfgL(L,0,2x+1)", "to": "cfgL(L',0,2x'+1), x' = (2k_c+1)2^(5q) + (x >> s)", "steps": null, "cases": "x has a 1-bit at an even position; the chain stops there (q = half its gap, s = bits consumed, k_c from the chain). Checked on the real TM at the next left record: chain length 0 (28), 1 (19), 2 (3), L' <= 21; F = macro-rule simulator on 783 orbit cycles", "validated": { "instances": 50, "passed": 50 }, "source": "manual" }, { "name": "RC3 overflow, A1 = 0RF", "from": "LC(L,k) 10 C> RC3(2r,0,2m+1)", "to": "halt", "steps": "30r+30 (r = 0: 30)", "cases": "TM10's ROv3_0 needs A1 = 0LB. Here A reads the 1 of the 1 00 10000 tail and moves right; with m odd, F later reads a 1. Other RC3 cases: 13 halts, 56 no TM10 result in 2e5 steps", "validated": { "instances": 16, "passed": 16 }, "source": "manual" } ], "template": { "file": "SOC_2_5_v1.v:5186", "module": "TM10", "machine": "1RB0LB_1RC1LB_1LD1RE_1LB0LD_1RF0RC_1RA---", "diff": "A1: 0LB -> 0RF", "applies": "partly", "note": "Every TM10 lemma except ROv3_0/ROv3_1 compiles with this string (the scout's Iter.v, which admits those two). They are false here, not just unproved: TM10's invariant P contains halting configurations of this machine (cfgL(8,0,17) halts at step 10,189,718). A proof needs P plus an orbit-specific digit condition (the chain never empties), and that condition is the open part. No other upstream file proves this table from any start." }, "strategy": "No non-halting proof is known. Rule 6: formalize the reduction (Rocq: 'no cycle empties the chain -> ~halts tm c0', from TM10's lemmas and F), then park unless a digit invariant turns up.", "risks": [ "Non-halting depends on the binary digits of an exponentially growing orbit (all 1-bits at odd positions must never happen): cryptid-like.", "The scout's Iter.v 'nonhalt' admits ROv3_0/ROv3_1, which are false for this machine; it is not evidence.", "After an exhaustion the dynamics are unmodelled; halting is shown only from constructed starts, not from blank." ], "checkpoint_questions": [ { "text": "Counter value x: every row should contain orange (a 1-bit at an even position). A row with no orange is what halting needs. Does every row show orange?", "figs": [ 1 ], "kind": "check" }, { "text": "Halt run: from TM10's ROv3 situation, A reads a 1 (row 3) and heads right instead of left; F reads a 1 at step 91. Do you see the A1 = 0RF difference doing this?", "figs": [ 2 ], "kind": "check" }, { "text": "Priority call: halting hinges on the digits of x (cryptid-like, ground rule 6). Should we park it now, or fund one 2 h stage to formalize the reduction in Rocq and search for a digit invariant?", "figs": [], "kind": "decision" } ], "discord_check": null, "stats": { "simulated_steps": 30000000011, "cells": 228, "nonzeros": 70, "growth": "exponential (two binary counters); A reads 1 only at step 10 through 3e10 steps; abstract orbit checked for 20,000 cycles (L to 172,193)" }, "status": "CRYPTID_CANDIDATE", "status_reason": "Mechanism understood (TM10's macro rules, validated); non-halting needs the orbit of an explicit digit map never to put every 1-bit of x at an odd position. Heuristically never; no invariant known." }