“F12.6: Stein's breakthrough, probable words: common English and K1-K3 words placed at every position in the free-alphabet cells the known letters leave undecided, decided exactly”
F12.6. Stein broke K2 by guessing THE, then KNOW, and checking each guess against the cipher. Here the guesses are systematic: 60 probable words (Stein's THE and KNOW, the commonest English words, distinctive K1-K3 words) placed at every position of K4 where they do not contradict the known letters, each placement adding its letters as known, then decided exactly for the cells F1.1 and F1.2 left consistent: one free alphabet, Beaufort form, key lengths 23 and 26 (0119's decider), and two free alphabets at 8, 13, 16, 19, 20, 23, 24, 26 (0123's). A cell's consistent placements on K4 are compared with copies of K4 that fit the same cell; F6.1 and F6.2 (with a move) are left out for cost (hundreds of thousands of fitting settings).
Stein's breakthrough was guessing likely words, THE and then KNOW, and seeing whether the cipher allowed them. For the cipher settings the 24 known lettersthe known letters Sanborn has confirmed that positions 22 to 34 of the message read EASTNORTHEAST and positions 64 to 74 read BERLINCLOCK. cannot decide, we try sixty likely words at every position: each guess adds more known letters, and most guesses become impossible. If the carved textthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. allowed many more guesses than scrambled copies do, that setting would deserve a closer look.
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
Where the 24 known letters leave an unknown alphabet undecided, a likely word placed somewhere else adds more letters to check. If K4 is such a code, the right words should fit it more often than they fit scrambled copies.
the check
Sixty words at every place they can go, each placement decided exactly, in the ten settings still open: one alphabet at key lengths 23 and 26, two alphabets at eight lengths.
the bound
K4 fits no more often than its copies: above all ten in one setting, as the copies themselves manage 0 to 3 times. F12.6 closed.
2
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 in section 7.
3
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Take one scrambled alphabet with a 23-letter key, the Beaufort way: the 24 known letters fit.
Guess that THE stands at places 1-3: three more letters are known.
Each new letter adds an equation (carved place + message place = key value of its slot); decide again whether any alphabet and key satisfy them all.
Try THE at every place, then the other words; count the guesses that survive.
Do the same on scrambled copies that also fit the cell.
the worksheet: the first steps by hand, on squared paper
4
Worked example, generated
The same steps run by the code on the real letters.
one scrambled alphabet, Beaufort form, key length 23: THE guessed at places 1-3
place 1: carved O, guessed T, slot 0: place(O) + place(T) = key value of slot 0
place 2: carved B, guessed H, slot 1: place(B) + place(H) = key value of slot 1
place 3: carved K, guessed E, slot 2: place(K) + place(E) = key value of slot 2
with the 24 known letters, 27 equations: decided exactly; repeated for every word and place
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Stein 1999 (knowledge/sources/stein-1999-cracking-the-courtyard-crypto.txt): 'given that the first letter of Part II could very well be a T, I naturally wanted to see what would happen if H were the second letter and E the third'; 'that KNO in the 15th row was an unexpected gift'. C7(a) (K1-K3 words), C19 (X as a stop). Each placement of a w-letter word adds w equations; in a cell with free alphabets most placements are refuted, so the count of consistent placements is a measure the known letters alone could not give. Statistic fixed before the run: per cell, consistent placements with at least one added check, K4 against ten copies drawn to fit that cell; any cell where K4 is above every copy is replicated.
6
Prediction
Written before the run; the result decides it.
Consistent placements on K4 inside the copies' range in every cell. Zero candidates.
7
Method
Words: THE, KNOW, the commonest English words, distinctive K1-K3 words (60).
For each cell, K4 and 10 copies that fit the cell: every word at every position compatible with the known letters; add its letters; decide exactly (node limits, undecided counted).
Count consistent placements with an added check; compare.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
0candidates returned
0passed every gate
0/24known letters, best
–lowest loss
nonechampion before
one free, Beaufort 23: K4 1939 (copies 1413-2072); one free, Beaufort 26: K4 1814 (copies 1212-2223); two free 8: K4 593 (copies 439-651); two free 13: K4 1271 (copies 1045-1372); two free 16: K4 1793 (copies 1489-2089); two free 19: K4 2317 (copies 1722-2443); two free 20: K4 1834 (copies 1782-2255); two free 23: K4 2380 (copies 2086-2486); two free 24: K4 2564 (copies 2097-2545); two free 26: K4 1585 (copies 621-2721)
What was searched, as the harness recorded it.
family
probable-word placements in the undecided free-alphabet cells (F1.1 Beaufort 23, 26; F1.2 two alphabets at 8 lengths), exact