Dorabella Decipher

manuscript → machine-readableEach glyph becomes arcs + direction: 2se is a two-arc symbol facing south-east. Three readings ship with the tool:

primary — the raw first-pass transcription.
resolved — nine glyphs were flagged as ambiguous; this picks, for each, the variant that best fits a 1:1 substitution. Validated against a published reading at 82/87 with zero letter collisions, p < 5×10⁻⁵. The default.
corrected — adds five further re-reads inferred from that published solution. Circular: do not use it to test that solution.

The legend shows all 24 glyphs as currently drawn. If the shapes look wrong against your manuscript, change the glyph convention — not the transcription.

put the glyphs in encipherment orderA pure reordering — no symbol changes, nothing added or lost. It matters only because the order Elgar enciphered in need not be the order the glyphs sit on the page. The output below is the same 87 symbols in their new sequence; with “as written” it is the identity and nothing moves.

the heart of the ideaIf the key was turning as Elgar wrote, every symbol was encoded with the key in a different position. This turns each one back by however many steps the key had taken by that point — the exact inverse of the encipher side.

Read the χ². A 45° turn preserves the ring, so a per-symbol rotation forces the eight direction counts toward 87/8 ≈ 10.9 whatever the key or fill. Dorabella sits at 26.6 on 7 degrees of freedom — Monte Carlo over 40 000 simulated rotating ciphertexts puts that at p = 0.0003. A regular rotation is ruled out, and no key guess rescues it.

The irregular schedules escape that argument: advancing on a repeated symbol, an arc-count change or a mirrored pair is driven by the text, so the counts are never forced flat. Those are the live options.

geometry → lettersLook each unwound cell up in the key grid and write its letter. The only phase that touches the alphabet; everything before it is pure geometry. A · means that symbol has no letter assigned. The glyph row underneath shows each original manuscript glyph with the letter it produced.

is the output language?English is scored with a blended 3+4-gram model. Other languages get bigram models — the method Hauer et al. (HistoCrypt 2021) used for language identification, which ranked Latin first and English third across 380 languages.

Why calibrate. A raw score means nothing alone. The tool enciphers real English passages with random substitutions and attacks them with this same solver at the same effort, then reports where your text falls in that distribution. Comparing a solved score against unsolved English would be unfair — the solver never quite reaches the true text.

German is a full solving target, not a ranking one: a 3+4-gram model over 2.5 M letters of 19th-century prose (Goethe, Fontane, Mann, Kafka, Nietzsche), with controls drawn from a held-out 15% the model never saw.

Caveat: the remaining language corpora are ~1.7 KB of hand-written prose each. Good enough to rank candidates, not to conclude one.

sweep the space, solve each candidateA substitution solver absorbs any fixed relabelling of the 24 symbols — so a different grid fill, a once-off rotation, or “orientation is the letter and arcs are a modifier” are all already covered and need no test. Only reading order and scheduled key motion change what the solver sees. That is why this space is small enough to sweep at all.

Dedupe. Many different settings produce the identical unwound stream. Each is hashed and solved once — a measured 50–56% saving on the larger modes. The ≡ column counts how many settings collapsed into that row.
Two stages. Stage A solves every distinct stream cheaply; stage B re-solves only the top N at full effort. A cheap solve is a noisy but adequate filter, so you pay full price only where it can matter.

Watch the live view: the dial fills in with the best key found so far, and the spark line plots every candidate as it is solved, with a dashed marker at the worst real-English score. A genuine solution would sit near the middle of the English band, not scraping the bottom.

what makes this different from phase 6 Phase 6 sweeps once, at a fixed effort, in one language, and forgets everything when it stops. This one is built to be left running.

The space is addressed by index, not built as a list. Combination number 431,905 is decoded arithmetically, so a sweep of any size costs no memory and “resume where I stopped” is exact rather than approximate.
Survey, then refine forever. Every distinct symbol stream gets one cheap solve per language. After that the engine keeps giving candidates more effort, carrying the elite pool forward, so the answer improves the longer you leave it. It chooses by successive halving: bring the whole field to the same number of rounds, cut the bottom half, double the budget, refill the free places from the leaderboard, repeat. Every comparison is between candidates that got equal effort.
This began as a bandit rule and it fixated badly — 267 of 319 rounds on one candidate. A bandit assumes its value estimate is an unbiased mean, but “best score so far” only rises with more rounds, so the exploit term rewards whatever has already been pulled most, purely for having been pulled most. Equal budgets remove the problem instead of tuning around it.
Two languages at once. Raw scores from an English and a German model are not comparable, so nothing ever compares them. Each candidate is placed in standard deviations against its own language's controls, and those z-scores compete.

And the correction, which is the point. Test sixty thousand candidates and the best of them looks impressive by chance alone. So the headline is not the best score. It is the best score measured against this stream shuffled — same symbols, same frequencies, no linguistic order — solved at the same effort, corrected for how many candidates were tried.