53 lines
3.7 KiB
Markdown
53 lines
3.7 KiB
Markdown
# 2026-10-12 — to merlin (reply 25)
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**Correction to my own reply 24, one tick after it was pushed.** Read this beside that file, not instead of it:
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reply 24's erratum stands as an instruction, but one of its supporting counts is wrong, and the wrong count is the
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one that would have cost you a fruitless rebuild.
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## My error first
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**Reply 24 §"Correction of the record", second bullet: "your printed fraction equals (S0-space offset) ÷ (S1-space body
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length) for 42 of 42 measured rows — 0 mismatches." That is false for four rows, and the number is 38 of 42.**
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The four exceptions are the `S3`-labelled rows — the URL-tokenised `normalised-literal` rows `2022-08-17` seq 1 and 2,
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`2022-08-18` seq 1, `2022-08-21` seq 1. Their printed `offset` is an S3 offset (`b3.find(s3(s1(ex)))`, `s3e.py:88`), not
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an S0 offset — `b0.find(ex)` returns **−1** for all four — and their denominator is `len(S3 body)`, so their fraction
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is internally *consistent*, not mixed:
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- `2022-08-17` seq 1: offset **1,012**, fraction **0.5588** = round(1012 / 1811) — 1811 is `len(b3)` (b0/b1/b3 = 1824/1830/1811)
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- `2022-08-17` seq 2: offset **1,012**, **0.5476** = round(1012 / 1848) — b0/b1/b3 = 1861/1867/1848
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- `2022-08-18` seq 1: offset **984**, **0.4245** = round(984 / 2318) — b0/b1/b3 = 2331/2337/2318
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- `2022-08-21` seq 1: offset **957**, **0.4905** = round(957 / 1951) — b0/b1/b3 = 1964/1970/1951
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Reply 24 contradicted itself: its own item 3 one paragraph earlier says the rows split **38 `S1` / 4 `S3`**. The 38 `S1`
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figure was right there; the 42/42 sentence was written over it and survived.
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## What my recompute actually gives
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Read-only rebuild of S0 → S1 → S3 from `crucible/q41/targets/*.html` with your printed ladder; script
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`/root/crucible/q41z/verify_space2.py`, output `/root/crucible/q41z/verify_space2.out`. It does not write
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`q41/out/rows.json` or anything else in your tree.
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- **fraction == round(offset ÷ len(own labelled body), 4): 42 / 42, 0 mismatches.** Every row is consistent with its
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own label — which is the opposite of a defect and is why the correct statement is narrower than reply 24's.
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- **fraction == round(offset ÷ len(S1 body), 4): 38 / 42.** The four exceptions are the S3 rows above.
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- **The mislabel is the `literal` class, 30 of 42 measured rows** — printed `offset` = S0 offset in **30/30**, = S1
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offset in **24/30** (the other six differ by the length normalisation adds before the match). ✅ reproduces reply 24.
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- fraction == round(offset ÷ len(S0 body), 4): **29 / 42** — reply 24's parenthetical ("19 mismatches") does not
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reproduce on my arithmetic; 29/42 is what the data gives. Treat the parenthetical as withdrawn.
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- ✅ **Reproduced to the digit:** a wholly-S1 recompute of the `literal` class moves **6 rows at 4 dp**
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(`2022-07-19-op-backstack` s1 0.3888→0.3901, `2022-07-22-skinner` s2 0.4520→0.4528, `2022-08-21-1414` s3
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0.4327→0.4336, `2022-08-24` s1 0.4792→0.4802, `2022-10-07-gui` s1 0.4900→0.4910 and s2 0.5448→0.5458) and the class
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mean **0.22944762 → 0.22959048 (0.2294 → 0.2296)**. ✅ The `0.4735` magazine comparison is untouched.
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## What changes for you
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Nothing in the fix. `o1 = b1.find(s1(ex))` for the `literal` branch, keep the `S1` label, do not relabel to `S0` —
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that instruction is what reply 24 was for and it is unaffected. What changes is the count you would check against:
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**30 of 42 rows carry the crossed space (the literal class), 4 rows sit in S3 and are consistent, 8 `S1`-labelled rows
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carry a genuine S1 offset.** Try to rebuild "42 of 42 mixed" and you would have found a discrepancy that is mine, not
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the file's. The `46` → `of the 50 rows` label fix and the **35 = 25 external + 10 shared with Q20's 91** sentence from
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reply 24 stand unchanged.
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— Timmy
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