Claude Fable 5.1 cracks Urquhart’s Cyphral Distich

Claude Fable 5.1 was asked to pick an unsolved historical cipher and crack it. It chose Sir Thomas Urquhart’s Cyphral Distich and decoded it in 44 minutes, using 176,000 tokens, with no interjections from the person running it. The key turned out to be the pages of Urquhart’s own book. The author, writing on the Vals blog on September 7, says the model “appears to have actually solved it”, and calls the answer “quite embarrassing for humans in hindsight”.

The puzzle is short: 64 numbers in two lines of 32, printed at the end of Logopandecteision, Urquhart’s 1653 book. There is nothing else to go on. Notes and Queries posed it as an open problem in 1899, it came up again in 20th-century cryptography literature, and the historical-cipher researcher Klaus Schmeh put it on his Top 50 list of unsolved encrypted messages. People tried frequency analysis, simple substitution and homophonic substitution, the variant where several numbers can stand for the same letter. None of these produced readable text.

All of those methods assume an external key: a cipher alphabet, or some mapping from numbers to letters, that has to be rebuilt from outside the text. Fable 5.1 tried a few approaches of its own and then made two observations about the book.

The first concerns the 32 Proquiritations, a numbered series of passages that sits immediately before the cryptogram. Urquhart goes out of his way to stress that number: “there can no number like that of two and thirty … be pitched upon.” Each line of the cipher has 32 numbers. The second is in the poem printed with the cipher, which promises that an honest reader will find in it “his own heart’s wishes, and the Author’s minde.” The Proquiritations themselves keep ending on phrases like “is the desire,” “wish” or “hope of.”

Together these give the rule. For the i-th number in a line, go to the i-th Proquiritation, count that many words in, and keep the first letter of the word. It is a book cipher, where each number is an address into a text both sides already hold, so no alphabet is needed at all. Applied to both lines, it gives:

O GOD UPHOLD KING CHARLS THE SECOND AND MAKE HIM THE SUPREME RULER OF THIS LAND

The result checks itself. Each line comes out at exactly 32 letters, and the lines end on “and” and “land”, a rhyming couplet that fits the promised distich. One coordinate depends on treating the Latin “hinc inde” as a single word; counted as two, that position shifts by one. The content fits the author too. Urquhart was a committed royalist: Charles I was executed in 1649, and Urquhart was imprisoned in 1651-1652 for fighting on behalf of Charles II.

The same approach then decoded a cipher eight times longer. Urquhart left the Cyphral Octastich in The Jewel (1652), which has 285 numbers and was also unsolved. The Jewel has exactly 284 numbered pages, and the eight lines plus a closing Decagram hold 285 numbers. Here the k-th number is a word index into page k, so pages take the place of paragraphs. As in the Distich, Urquhart almost always used the first word on the page that started with the letter he needed. In the EEBO-TCP transcription, 231 of 275 readable positions are exact first-occurrence hits. The other 44 are one to three words off, for reasons that can be traced to the transcription: hyphenated words at the top of a page, hyphenated compounds, ”&”, paragraph numbers, and a Greek phrase that was never transcribed.

The plaintext is a royalist prayer in ottava rima, rhymed ABABABCC and dated London, March 1652. It opens “GREAT LORD, MANTAINE THAT REGAL FAMILIE WHEREOF KING CHARLS THE SECOND IS THE HEAD”, ends “LET HIM BE OUR SOLE CESAR, ARTUR, HECTOR, OUR EMPEROUR, KING, MONARCH AND PROTECTOR”, and the Decagram adds “AMEN, SO BE IT.” One word in it, “sweigh”, is the Scots “swey”, meaning sway or controlling power. The Dictionary of the Older Scots Tongue records the exact idiom “to bear the swey” from around 1600, and the word rhymes with “familie” and “authoritie”.

The write-up lists its caveats. Line 4 is enciphered I-R-S-H, which is either a slip for IRISH or a contraction that gives the line exactly ten syllables. Nine letters in line 5 (pages 149-157) come out as C-O-N-E-R-T-H-T-O. Eight of those nine are exact first-occurrence hits, so this really is what the transcribed text yields, but it does not read as English. The model tested page-shift, dropped-letter, misprinted-number and dictionary-lattice hypotheses, and none produced English. Either Urquhart made a mistake there or some numbers were misprinted; the rest of the line, “THIS USURP’D AUTHORITIE”, is certain. From position 159 onward every number points to page k−1, so either one page was used twice or the printed “5.5” in line 5 is a dittography, a number accidentally printed twice. Lines 6-8 and the Decagram decode cleanly once that shift is applied.

Checking those positions is harder than it sounds. A subagent searched Google Books, HathiTrust, the Internet Archive and the National Library of Scotland and found no free page images of a 1652 Jewel. The octastich is printed on a final gathering of leaves that the TCP copy lacks, although the main text and pagination are the same in every copy. Confirming the ±1 positions and the nine letters needs a physical copy or the 1983 Jack & Lyall edition. The run left behind a verification script, verify_octastick.py, which prints all 285 positions with page, word, letter and offset.

The author says they had spent months trying to get models to solve an important unsolved cipher, and no other frontier model they tried produced a verified solve. The prompt was simple: solve an unsolved cipher, look online at Fable’s strongest feats, especially the math problems it has solved, keep in mind that this should be easy by comparison, and think creatively. It came with two constraints. One was to skip ciphers that already have solutions or could support many plausible answers, because with the creators long dead such answers may never be confirmed. The other was to stay away from the very hardest targets, the ones where thousands of people or organizations like the CIA had already put in serious effort. Kryptos K4 was the example; the author doubts Fable 5.1 could solve it in a reasonable amount of time yet.

Fable 5.1 looked over several candidates and dropped the ones that weren’t budging. The author thinks this was what it did well. Once it reached the Distich it spotted the clue almost immediately, and the author does not claim other frontier models would necessarily have failed, since “the clue is actually extremely simple.” What stood out was that the model saw this problem as unusually tractable. The author’s broader point is that problems like this were held back by human attention, by whether anyone would spend hours reading obscure material and testing unpromising ideas, and that this limit “is seemingly disappearing.”

“The answer was simple in hindsight,” the author writes. “It just kept looking until it found it.”