Vals AI handed Claude Fable 5.1 a two-line cryptogram of 64 numbers printed at the end of Sir Thomas Urquhart's Logopandecteision (1653) and asked it to solve it. It took 44 minutes, 176k tokens, and zero interjections from the operator. The writeup is dated 2026-08-31 and describes the Cyphral Distich — posed as an open problem in Notes and Queries in 1899 and later listed by historical-cipher researcher Klaus Schmeh among his Top 50 unsolved encrypted messages.
Details like this are worth reading for the shape of the task, not the history.
The Key Was In The Book, Not In An Alphabet
The entire input is these 64 numbers:
5.3.27.38.32.14.21.8.66.8.70.39.5.9.12.18.2.3.56.5.1.7.3.2.13.19.3.25.9.3.16.6.
25.15.13.6.11.20.5.1.2.12.1.20.20.49.20.20.35.33.4.6.8.35.5.33.5.5.18.10.3.11.32.42.
Earlier attempts went at it with frequency analysis, substitution and homophonic substitution. None worked, because the key was never an external cipher alphabet. It was the book itself. Two clues — that the cryptogram sits right after Urquhart's 32 Proquiritations (a number he calls out explicitly: "there can no number like that of two and thirty … be pitched upon"), and that the accompanying poem promises the honest reader "his own heart's wishes, and the Author's minde" — collapse into one rule:
For the i-th number in a cipher line, go to the i-th Proquiritation, use that number as a word index, and take the first letter of that word.
Out comes a rhyming distich, 32 letters per line:
O GOD UPHOLD KING CHARLS THE SECOND AND MAKE HIM THE SUPREME RULER OF THIS LAND
The self-verification is the interesting part. Both lines land on exactly 32 letters and end and / land; the result is consistent with Urquhart's Royalism. Historical-cipher work usually can't be checked this cheaply.
The Second Cipher: 285 Numbers, Nine Letters Missing
Urquhart left a larger cryptogram in the same style — the Cyphral Octastich in The Jewel (1652), 285 numbers instead of 64. Fable 5.1 decoded that too, with the rule shifted from paragraphs to pages: The Jewel has exactly 284 numbered pages, and the k-th number (across eight lines plus the Decagram) indexes a word on page k. It came out as an ottava rima royalist prayer, ABABABCC, written in London in March 1652 — "GREAT LORD, MANTAINE THAT REGAL FAMILIE / WHEREOF KING CHARLS THE SECOND IS THE HEAD" — closing with "AMEN, SO BE IT."
That solve is not complete, and the writeup says so plainly:
| Caveat | Detail |
|---|---|
| Line 4 | Enciphered I-R-S-H (pages 127–130); a slip for IRISH, or a deliberate contraction |
| Line 5, letters 4–12 | Pages 149–157 yield C-O-N-E-R-T-H-T-O — eight of nine are exact first-occurrence hits, and it isn't readable |
| From position 159 | Every number keys to page k−1 (a page used twice, or a dittography in the printed "5.5") |
| Page images | No free page images of the 1652 Jewel exist; confirming the ±1 positions needs a physical copy or the 1983 Jack & Lyall edition |
Of 275 readable positions, 231 are exact first-occurrence hits in the EEBO-TCP text; the other 44 are 1–3 words off for identifiable transcription reasons: hyphenated words at page tops, hyphenated compounds, "&", paragraph numbers, an untranscribed Greek phrase.
What I'd Take From This, As An Engineer
The valuable claim here isn't that a model did cryptanalysis. Vals says the opposite — this was not an extraordinary feat of cryptanalysis, the answer was simple in hindsight. Their stated point is that the model kept looking, and knew when a problem wasn't budging: it surveyed candidate problems, rejected ones that were unverifiable or already solved, and steered away from Kryptos K4 as "a little too hard and convoluted for current models." Across months of attempts, the author says no other frontier model produced a verified solve.
The reproducible part for me is the harness shape, not the puzzle: one goal, a handful of artifacts to look at, two constraints (prefer quickly verifiable problems, avoid ones with thousands of human-hours already sunk in), no human in the loop. That's how I'd scope long research-style agent runs: cheap-to-check targets, an explicit stop criterion, and a written record of what didn't work — here, verify_octastick.py, firstocc.py/lattice.py, plus an updated SOLUTION.md and FAMOUS-CIPHERS-STATUS.md.
The catch is verification cost, and this story shows it at full size. A model can produce a plausible plaintext; confirming it required DOST/SND dictionary entries for Scots swey ("to bear the swey"), the ESTC record R203867, and a physical copy of a 1652 book nobody has imaged. Budget for the check, not just the solve.
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