Fable 5.1 Solves the Cyphral Distich, a 370-year-old cipher

Fable 5.1 破解了 370 年前的“密码对句”（Cyphral Distich）

Problem 问题

We gave Claude Fable 5.1 an open task: solve Sir Thomas Urquhart’s Cyphral Distich. It appears to have actually solved it, and the solution is quite embarrassing for humans in hindsight. At the end of Urquhart’s Logopandecteision is a cryptogram consisting of two lines of 32 numbers each, called the Cyphral Distich. A cryptogram is a short message deliberately encoded so it can’t be read without knowing the rule that produced it. Here the entire puzzle input is these 64 numbers, and the goal is to recover the hidden plaintext: 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.

我们给 Claude Fable 5.1 布置了一个开放性任务：破解托马斯·厄克特爵士（Sir Thomas Urquhart）的“密码对句”（Cyphral Distich）。它似乎真的破解了它，而且事后看来，这个解决方案让现代人类感到相当尴尬。在厄克特的《Logopandecteision》一书末尾，有一段由两行各 32 个数字组成的密码，被称为“密码对句”。密码是一种经过刻意编码的短消息，如果不了解其生成规则就无法阅读。这里的谜题输入就是这 64 个数字，目标是恢复隐藏的明文：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。

This cipher has remained seemingly unsolved for centuries. It was posed as an open problem in Notes and Queries in 1899, appeared again in 20th-century cryptography literature, and was later listed by historical-cipher researcher Klaus Schmeh among his Top 50 unsolved encrypted messages. Various people attempted to decipher it, but it seems they were missing one crucial hint. They tried methods like frequency analysis, substitution, and homophonic substitution, and none of these approaches worked. That’s because they missed one easy clue.

这个密码几个世纪以来似乎一直未被破解。它在 1899 年的《Notes and Queries》杂志上被作为一道公开难题提出，后来又出现在 20 世纪的密码学文献中，并被历史密码研究员克劳斯·施梅（Klaus Schmeh）列入其“50 大未解加密信息”名单。许多人曾尝试破译它，但似乎都漏掉了一个关键线索。他们尝试了频率分析、替换法和同音替换法等方法，但这些方法都无效。这是因为他们错过了一个简单的提示。

Solution 解决方案

After 44 minutes, 176k tokens, and zero interjections from me, Fable 5.1 arrived at a solution. It tried a few approaches, but was finally able to solve it with two central realizations. First: the cryptogram is printed immediately after Urquhart’s 32 Proquiritations, and Urquhart even goes out of his way to emphasize that number. I know, surprising. He says: “there can no number like that of two and thirty … be pitched upon” Second: the poem accompanying the cipher promises that an honest reader will find in it “his own heart’s wishes, and the Author’s minde.” The Proquiritations themselves repeatedly conclude with formulations like “is the desire,” “wish,” or “hope of.” If you put these clues together: 32 Proquiritations. 32 numbers in the first cipher line. 32 numbers in the second. “Wishes.”

在 44 分钟、176k token 的处理量以及我零干预的情况下，Fable 5.1 得出了解决方案。它尝试了几种方法，最终通过两个核心发现破解了它。第一：密码紧接在厄克特的 32 条“恳求”（Proquiritations）之后，厄克特甚至特意强调了这个数字。我知道，这很令人惊讶。他说：“没有哪个数字能像三十二这样……被选中”。第二：随密码附带的诗歌承诺，诚实的读者会在其中找到“他自己内心的愿望，以及作者的思想”。这些“恳求”本身反复以“是……的渴望”、“愿望”或“希望”等措辞结尾。如果你把这些线索放在一起：32 条恳求。第一行密码有 32 个数字。第二行有 32 个数字。“愿望”。

Most historical attempts assumed the key was external: a cipher alphabet, or some mapping of numbers to letters or words, that had to be reconstructed from outside the text. But the key was not an external cipher alphabet at all. The key was the book itself. The rule was simple: 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. With this, you get: O GOD UPHOLD KING CHARLS THE SECOND AND MAKE HIM THE SUPREME RULER OF THIS LAND. And the result is extremely self-verifying. Each line contains exactly 32 letters and ends and / land (a rhyming 2 line verse), consistent with the promised distich. It also makes historical sense: Urquhart was a committed Royalist. Hiding a prayer for Charles II in the text is entirely consistent with his politics.

大多数历史上的尝试都假设密钥是外部的：即一个密码字母表，或者某种必须从文本之外重建的数字到字母或单词的映射。但密钥根本不是外部的密码字母表。密钥就是这本书本身。规则很简单：对于密码行中的第 i 个数字，找到第 i 条“恳求”，将该数字用作单词索引，并取该单词的第一个字母。由此，你得到：O GOD UPHOLD KING CHARLS THE SECOND AND MAKE HIM THE SUPREME RULER OF THIS LAND（上帝啊，请支持查理二世国王，让他成为这片土地的最高统治者）。结果具有极强的自我验证性。每一行恰好包含 32 个字母，并以 and / land 结尾（一个押韵的两行诗），与承诺的“对句”一致。这在历史上也说得通：厄克特是一位坚定的保皇党。在文本中隐藏为查理二世的祈祷，完全符合他的政治立场。

Urquhart left a second, much larger cryptogram in the same style — the Cyphral Octastich in The Jewel (1652), 285 numbers instead of 64, and just as unsolved. From this, Fable 5.1 was also able to decipher it: Result: the Cyfral Octastick is solved (all but nine letters).

厄克特以同样的风格留下了第二个更大的密码——《The Jewel》（1652 年）中的“密码八行诗”（Cyphral Octastich），包含 285 个数字而非 64 个，同样一直未被破解。Fable 5.1 也成功破译了它：结果：密码八行诗已破解（除 9 个字母外全部正确）。

Rule 规则

The Jewel (1652) has exactly 284 numbered pages, and the octastick + decagram contain 285 numbers. The k-th number (counting straight through the eight lines and the Decagram) is a word index into page k of the book; take the word’s first letter. Same idea as the Distich (number i → Proquiritation i), with pages instead of paragraphs — and, as in the Distich, Urquhart almost always picked the first word on the page starting with the letter he needed.

《The Jewel》（1652 年）恰好有 284 个编号页，而八行诗加十行诗（decagram）共包含 285 个数字。第 k 个数字（贯穿八行诗和十行诗连续计数）是书中第 k 页的单词索引；取该单词的第一个字母。与“对句”的思路相同（数字 i → 第 i 条恳求），只是用页码代替了段落——而且和“对句”一样，厄克特几乎总是选择页面上第一个以他所需字母开头的单词。

Plaintext 明文

(ottava rima, ABABABCC — a royalist prayer written in London, March 1652): GREAT LORD, MANTAINE THAT REGAL FAMILIE WHEREOF KING CHARLS THE SECOND IS THE HEAD, AND GRANT THAT HE MAY BEARE THE SUPREME SWEIGH WHERE ENGLISH, SCOTS AND IR[I]SH ARE BORNE AND BRED, AND [·········] THIS USURP’D AUTHORITIE REIGNE IN HIS ROYAL PREDECESSORS STEAD; LET HIM BE OUR SOLE CESAR, ARTUR, HECTOR, OUR EMPEROUR, KING, MONARCH AND PROTECTOR. AMEN, SO BE IT. (the Decagram)

（八行诗体，ABABABCC——1652 年 3 月写于伦敦的保皇党祈祷文）：伟大的主，维护那以查理二世国王为首的皇室家族，并赐予他至高无上的统治权，让英格兰人、苏格兰人和爱尔兰人在此出生和成长，并……取代这篡夺的权威，在他皇室先辈的位置上统治；让他成为我们唯一的凯撒、亚瑟、赫克托耳，我们的皇帝、国王、君主和保护者。阿门，愿它如此。（十行诗部分）

Sweigh is Scots swey “sway, controlling power” — DOST records the exact idiom “to bear the swey” (c. 1600), and it rhymes with familie/authoritie.

Sweigh 是苏格兰语 swey，意为“统治、控制力”——《苏格兰舌头词典》（DOST）记录了确切的习语“to bear the swey”（约 1600 年），它与 familie/authoritie 押韵。

Caveats, stated plainly: 明确说明的注意事项：

Line 4 is enciphered I‑R‑S‑H (pages 127–130): a slip for IRISH, or a deliberate contraction — “Irsh” gives the line exactly ten syllables.
Line 5, letters 4–12 (pages 149–157) come out C‑O‑N‑E‑R‑T‑H‑T‑O — eight of the nine are exact first-occurrence hits, so this is genuinely what the TCP text yields, and it isn’t readable. I tried page-shift, dropped-letter, misprinted-number and dictionary-lattice hypotheses; none gives English. Either Urquhart slipped here or numbers of this line were misprinted. The rest of the line (THIS USURP’D AUTHORITIE) is certain.
From position 159 onward every number keys to page k−1: one page was used twice (or the printed “5.5” in line 5 is a dittography). Lines 6–8 and the decagram decode cleanly with that shift.
第 4 行加密为 I-R-S-H（第 127-130 页）：可能是 IRISH 的笔误，或者是故意的缩写——“Irsh”使该行恰好有十个音节。
第 5 行，第 4-12 个字母（第 149-157 页）解出为 C-O-N-E-R-T-H-T-O——其中 9 个字母中有 8 个是精确的首词匹配，所以这确实是 TCP 文本产生的结果，且不可读。我尝试了页码偏移、漏字、数字印刷错误和字典格假设；没有一个能得出英语。要么是厄克特在这里写错了，要么是这一行的数字印错了。该行的其余部分（THIS USURP’D AUTHORITIE）是确定的。
从第 159 位开始，每个数字都指向第 k-1 页：有一页被使用了两次（或者第 5 行印刷的“5.5”是重复抄写）。第 6-8 行和十行诗在进行这种偏移后可以清晰地解码。