“Both fingerprints at once: every row of seven reshuffled, and the self-feeding autokey at every distance”
Experiment 70 joined the block-of-seven reshuffle to a repeating key, and experiment 63 joined column routes to the self-feeding autokey. The two fingerprints suggest a seven-wide stage and a feedback stage at distance 21, so this joins the block-of-seven reshuffle, all 5,040 orders, to the self-feeding autokey at every distance, in both orders.
The copper carries a mark tied to seven and a mark tied to twenty-one. A seven-letter reshuffle of every row could leave the first, and the autokeyan 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. that feeds on carved letters 21 places back could leave the second. Here the two are joined, in either order, for every one of the 5,040 ways of reshuffling seven letters and every feeding distance, and 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. give an exact yes or no each time. Scrambled coppers through the same test show how many yeses luck gives.
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
The reshuffle of every block of seven could make the width-7 column; ciphertext autokey at distance 21 makes the width-21 repeats. Together they might make both.
the check
Reversing each row and undoing the autokey at distance 21, places 22 and 23 leave 20 and 8; a constant shift needs them equal.
the bound
All 5,040 reshuffles, both orders, every distance, all 32 ways of using the table: survivors at three checks match the shuffled coppers, and distance 21 never passes three.
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 the order that reverses every row of seven, and order A: autokey first, then reverse each row.
Undo it: reverse each row of seven of the copper, giving XOURKBO SBLUHGO WBBFILO PQQVRLF ...
Lag 21, sculpture's alphabet from K = 0, Vigenere. Place 22: the unreversed letter P (3) minus known E (11) is 18; the letter 21 back is X (24); residual 18 - 24 = 20.
Place 23: Q (20) minus A (7) is 13; 21 back O (5); residual 8.
20 and 8 disagree: dead.
Repeat for every order, lag, class and both ways round.
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.
every row of seven reversed after the autokey; undone: XOURKBOSBLUHGOWBBFILOPQQVRLFOSSKGNR...
lag 21, sculpture's alphabet, Vigenere
place known letter 21 back residual
22 E P X 20
23 A Q O 8
24 S Q U 19
25 T V R 17
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Family 1 (ciphertext feedback) with a block transposition, chosen to explain both fingerprints in one construction: the block-of-7 permutation leaves a period-7 positional trace (the doublet column, 0056), and ciphertext autokey at lag 21 is the only generator that made the width-21 repeats (0062). Orders: (A) ciphertext autokey, then every block of 7 permuted by one fixed order; (B) the plaintext's blocks permuted, then ciphertext autokey. Autokey with a constant shift, T[j] = s*A(c[j-L]) + k0, L = 1-96; 5,040 orders (the last block of 6 keeps the order restricted to its offsets); plain, cipher and key alphabets standard or KRYPTOS, Vigenere or Beaufort, sign +1 or -1 (32 classes): 31 million cases, each an exact check of the residuals at the known places past the lag. Ten shuffled coppers through the same sweep. Clue entries: C1(a), C11a, C27(a).
6
Prediction
Written before the run; the result decides it.
No (order, distance, class) case survives with three or more redundant checks beyond the shuffled-copper range, in either order; lag 21 has no case with more than two checks. Zero candidates pass.
7
Method
For each order of seven, compute where each known place lands.
Order A: undo the reshuffle on the copper and run the autokey test on the result. Order B: run the autokey test on the copper with the known letters at their landing places.
For each lag and class, check the residuals agree; tally survivors by redundant checks.
Repeat on ten shuffled coppers.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.