“A6.2: the climbs' power on Sanborn-style plaintext: X stops, K1-K3 windows and three changed letters planted through F1.5's design 2 (one free alphabet at 23, two at 13) and 0125's running-key search, judged by quadgrams and by the blind judge”
A6.2 (Stage 2, step 4, issue #84). The sampled closes of F1.1 (0139-0142), F1.2 (0143-0146) and F1.3 (0125) measured their power on clean Carter. Here the planted messages are the judge bench's Sanborn-style classes (reports/judge-bench.md): Carter with X for every full stop (xstops), 97-letter windows of K1-K3's own plaintexts with their misspellings and X stops (k123), and clean English with three letters changed (sub3). Four of each go through each frozen search unchanged: 0140's tempering climb at one free alphabet, Beaufort form, length 23; 0144's two-alphabet climb at 13; 0125's running-key reading of the forced key. Each plant is found or not by quadgrams (the verdicts' rule) and by the judge, against the same copies the verdict's power used.
Earlier experiments checked that their searches could find a hidden test message, but the test messages were clean English from a book. Sanborn's real messages are messier: X between sentences, deliberate misspellings, odd phrasing. Here we hide messy messages of that kind, run the same searches on them unchanged, and see how often the search, scored by letter statistics and separately by the judgethe judge A language model that scores how much of a reading is English, from 0 to 100, allowing misspellings and X between sentences. (a language model that reads one text at a time without knowing where it came from), picks the hidden message out from scrambled copiesa shuffled control The same experiment run on the carved letters shuffled into a random order. of itself.
Words in the margin are defined on the glossary page.
1.1
The method at work
The carved letters go in at the top; the reading comes out at the bottom.
the idea
Earlier power tests planted clean book English. Sanborn writes with X stops, misspellings and odd phrasing; here the planted messages are written that way.
the check
The three frozen searches behind F1.1, F1.2 and F1.3's sampled closes, run unchanged on 36 planted texts and each text's own scrambled copies; found by letter statistics, and separately by the judge.
the bound
Letter statistics find 10 of 12 at length 23 and 12 of 12 for 0125, above their bars; 9 of 12 at length 13, one plant under its bar of 4 in 5, so F1.2's close at 13 goes back to open. The judge finds 1 and 0 of 12 in the climbed readings.
2
Experiment lineage
What this experiment was built on, from the lineage page: 14 experiments before it, back to the root of its line. Its number is in bold.
keptreverted, in its family's colour×error or timeoutreverted under today's rulesbuilt on a parent in the same familybuilt on a parent in another family
3
Best candidate against the known letters
This experiment produced no candidate: the search ruled everything out before a full text was built. Its product is the bounded negative under Result and bounds below.
4
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Take a messy test message, say a window of K2: ...THEYUSEDTHEEARTHSMAGNETICFIELDXTHEINFORMATION....
Put EASTNORTHEAST and BERLINCLOCK at their places.
Encipher it by hand the way the item says: one shuffled alphabet, a key of 23 numbers, Beaufort form.
Make six scrambled copies of the ciphertext that the same key length can still fit.
Run the same search on the seven texts and keep each best reading.
The message is found if its reading scores above all six copies', once by letter statistics and once by the judge.
the worksheet: the first steps by hand, on squared paper
5
Worked example, generated
The same steps run by the code on the real letters.
A6.2 plants Sanborn-style messages. A K1-K3 window with the known words set:
message BETWEENSUBTLESHADINGAEASTNORTHEASTFLIGHTLIESTHENUANCEOFIQLUSIONBERLINCLOCKYINVISIBLEHOWSTHATPOSSI
judge sees BETWEENSUBTLESHADINGA#############FLIGHTLIESTHENUANCEOFIQLUSION###########YINVISIBLEHOWSTHATPOSSI
enciphered with one shuffled alphabet and a key of 23 (Beaufort form), climbed by 0140's search
with six scrambled copies; found if its reading beats all six, by quadgrams and by the judge.
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
A6 (knowledge/12-assumptions.md), with C19 (Scheidt 2015, 'what's a period... special characters don't work': X and Q as stops) and C17/C10 (Sanborn's deliberate misspellings, 'rife with mistakes on purpose', 2005). F1.5's design 2 (parallel tempering in the set of keys that keep the 24 known letters; 10 runs x 16 replicas x 625 rounds x 1000 moves) decided F1.1's Beaufort length 23 at 6 of 8 clean plants above their own 6 fitting shuffles (0139) and F1.2's length 13 at 5 of 5 above 5 kept-place shuffles (0143); 0125's running-key test (208,062 dictionary tables x 18 readings, the 24-letter forced key scored as English) found 11 of 20 Carter running keys above 50 shuffles. Those powers were measured on Carter's prose, which a quadgram model scores well. Sanborn's own plaintexts are not that: K2 has X stops and UNDERGRUUND, K1 IQLUSION, K3 DESPARATLY. The test: the same frozen searches on 12 plants per arm in the three bench classes; per plant, found when its score is strictly above every own copy's (6 fitting shuffles of the plant's ciphertext at 23, 6 kept-place shuffles at 13, 0125's 50 shuffles of K4 for the running key), by quadgram and, separately, by the judge in one blind call per arm. Null per plant: 1/7 (0125: 1/51). Pre-registered rule: an arm whose quadgram power falls below the verdict's stated power (23: 6/8; 13: 5/5, bar 4/5; 0125: 11/20) weakens the close it carried; the close stands only if the judge reaches that power here and A6.1 (0221) kept K4 inside its copies under the judge at that cell. Otherwise the item goes back to open in its family table and waits for Alexis.
7
Prediction
Written before the run; the result decides it.
Quadgram power falls on the messy classes: at 23 about 6 of 12 (the climb's quadgram target fights X stops and K1-K3's misspellings less than it fights noise, so k123 and xstops lose most); at 13 about 9 of 12; 0125 about 5 of 12 (a 24-letter window with an X or a changed letter loses quadgrams). The judge does at least as well as quadgrams on every arm and reaches the stated power at 13 but not at 23, where the climbed readings are only a third right. So F1.1's length 23 is weakened, F1.2's length 13 stands, and F1.3 is weakened for quadgrams but rescued by the judge if 0221's rejudge held. Zero candidates.
8
Method
Plaintexts: per arm 4 xstops (tools/judge_bench.py's xstops_plant), 4 k123 windows (the bench's 12, split 4 per arm), 4 sub3 (kryptos.plants.english_plant with 3 free letters changed, the bench's damage); the 24 known letters set.
Arm 23: Beaufort form, random alphabet and key of 23 (0120's make_plants rule); 6 own copies, shuffles of the plant's ciphertext that 0120's exact decider fits at 23 (0140's rule); every text through 0140's frozen _job.
Arm 13: two random alphabets and key of 13 (0144's plant4 rule); witness from 0123's decider; 6 own copies keeping the 24 known places and shuffling the other 73; every text through 0144's frozen _any.
Arm 0125: the class text is the running key, under English plaintext (english_plant), on a random table, rule and key alphabet from 0125's sets; through 0125's frozen _job; its 50 copies regenerated with 0125's seed and checked against its record.
Found by quadgram: the climb's score strictly above every own copy's. Found by the judge: one call per arm on the plants' readings and their copies', the plant strictly above every own copy; None counts as not found.
9
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
F1.3, running key on 208,062 dictionary tables, 0125's frozen search, the class text as the key
null per plant
23
1/7
13
1/7
0125
1/51
runs done
min
10
max
10
decryption checked
yes
judge texts sent
230
judge calls estimated
230
expected total time
about 31,000 core-seconds (84 texts at about 110 s in arm 23, 84 at about 235 s in arm 13, 62 at about 35 s for 0125; from 0140, 0144, 0125's runs), so 65-90 min on 8 cores