“The previous carved letters as a key, plus a repeating secret word on top”
Experiment 65 found that a cipher feeding on the previous carved letter, combined with a repeating secret word of 7, is the only recipe that lines doublets up in sevens. Experiment 2 killed the plain ciphertext autokey; this adds the repeating secret word on top, at every distance back and every length of word, and tests it exactly against the known letters.
An autokey cipheran autokey cipher A cipher that uses the message itself, or the scrambled text itself, as the secret word, each letter enciphered with the letter a fixed distance before it. can feed on the carved lettersthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. themselves: each letter is enciphered with the carved letter a fixed distance backthe lag In an autokey cipher, how far back the key letter sits from the letter being enciphered.. Experiment 2 showed that alone cannot produce the known lettersthe known letters Sanborn has confirmed that positions 22 to 34 of the message read EASTNORTHEAST and positions 64 to 74 read BERLINCLOCK.. Here a repeating secret wordthe secret word The word or string a cipher needs besides the method. is added on top, of any length, at any distance back. Because we can see every carved letter, the feeding part can be undone exactly, which leaves an ordinary repeating secret word that the 24 known letters either fit or do not. Scrambled coppers run through the same test show how many fits luck alone produces.
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 cipher
Each carved letter is the message letter plus a repeating secret word plus the previous carved letter, so the copper keys itself and a short word is added on top.
the check
At distance 1 with a 7-letter word, places 22, 29, 64 and 71 share the word's first letter but force T, Q, P, J: dead.
the bound
Every distance and word length from 1 to 96, all 32 ways of using the table: no setting passes three checks, the same as ten shuffled coppers. Distance 1 with 7, and distance 21 at any length, are dead.
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 lag 1 and a 7-letter secret word, with every letter counted along KRYPTOSABCDEFGHIJLMNQUVWXZ from K = 0.
Place 22: known E (11), carved F (12), previous carved W (23): the secret word's letter is 12 - 11 - 23 = -22, which is 4, T, in slot 0.
Place 29, also slot 0: known T (4), carved R (1), previous carved P (3): 1 - 4 - 3 = -6, which is 20, Q.
T and Q disagree, so this lag and word length are dead under this way of using the table.
Places 64 and 71 (slot 0 too) would force P and J.
Repeat for every lag, every word length, and every way of using the table.
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.
lag 1, period 7, sculpture's alphabet throughout, Vigenere, s = +1
place known carved previous key letter slot
22 E F W T 0
29 T R P Q 0
64 B N B P 0
71 L Z M J 0
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Family 1 re-entered with a stated reason: 0065 and 0062 point at ciphertext feedback (lag 1 makes the width-7 doublet column with a period-7 key, lag 21 makes the width-21 repeats), and 0002 tested ciphertext autokey only without an additional key. Construction: two-component key, ciphertext autokey at lag L plus a periodic key of period P: T[i] = sA(c[i-L]) + k[i mod P], where T is the total key index (C - P for Vigenere, C + P for Beaufort; variant Beaufort is s = -1), A reads the earlier carved letter in the key alphabet. Because the carved text is fully known, the autokey part can be subtracted: d[i] = T[i] - sA(c[i-L]) is a pure periodic key, so the 24 cribs give exact equations k[i mod P] = d[i] for every crib with i >= L. Sweep: L = 1..96, P = 1..96, plain, cipher and key alphabets standard or KRYPTOS, Vigenere or Beaufort, s = +1 or -1 (32 classes, 294,912 cases). A case survives if every slot's forced values agree and at least one check exists; survivors are tallied by the number of redundant checks and compared with ten shuffled coppers run through the same sweep. Forced key fragments of survivors are matched against the vocabulary and K1-K3 windows. Clue entries: C11a ('auto-key', 'interval 7'), C5(b) ('key word/s'), C6(c).
6
Prediction
Written before the run; the result decides it.
No case with a period of 26 or less survives with three or more redundant checks under any class, including lag 1 with period 7 (the doublet mechanism) and lag 21 at every period (the repeat mechanism). Survivor counts at each number of checks match the ten shuffled coppers within their range, and no forced fragment matches a word. Zero candidates pass.
7
Method
For every lag, compute d[i] = total key minus the earlier carved letter's value at each known place past the lag.
For every period, group the known places by slot and check that each slot's values agree.
Tally consistent cases by the number of redundant checks.
Repeat on ten shuffled coppers.
Match the forced fragments of survivors against words.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.