“Every block of seven letters reshuffled the same way, before or after the table”
The copper's doublets line up in sevens. Experiment 6 tested columnar routes, which read a grid down its columns. This tests the other way a grid of seven can be used: each row of seven letters reshuffled by the same order and read across, before or after a repeating secret word on the table, for all 5,040 orders.
Five of the six doubled lettersthe doublets Places where a carved letter is immediately followed by the same letter. fall in the same column when the copper is written in rows of seven. One simple way to leave such a trace is to write the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. in rows of seven and shuffle the letters inside every row in the same order, then encipher with a repeating secret wordthe secret word The word or string a cipher needs besides the method., or the other way round. There are only 5,040 orders for seven letters, so we try them all with every length of secret word up to 52 and see whether any fits 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. better than scrambled coppers do.
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
Reshuffling every block of seven letters the same way could make the copper's width-7 pattern. Here a reshuffle comes before or after a repeating key on the table.
the check
Reversing each row, places 22 and 29 are carved at 28 and 35 and force M and R in the same slot: that reshuffle with a 7-letter word is dead.
the bound
All 5,040 ways of reshuffling seven letters, both orders, key lengths 1 to 52: lengths up to 24 die; 25 and 26 survive as often as on shuffled coppers, and no fragment spells a word.
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: encipher first, then reverse each row.
Place 22 is the first letter of its row, so after reversing it is carved at place 28, P.
Known E over carved P, on the sculpture's alphabet with Vigenere, needs key letter M, in slot 0 of a 7-letter word.
Place 29 is also first in its row, carved at place 35, O; known T over O needs R, also slot 0.
M and R disagree, so this order with a 7-letter word is dead.
Repeat for all 5,040 orders, both ways round, every word length and 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.
every row of seven reversed, after a 7-letter key; sculpture's alphabet, Vigenere
place carved at carved known key slot
22 28 P E M 0
29 35 O T R 0
23 27 Q A G 1
30 34 S H M 1
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Family 6 re-entered with a stated reason (it had rested after eleven reverts): the width-7 doublet column (0056, real at corrected p 0.005) is a period-7 positional trace, and the one width-7 transposition not yet tested is the block permutation, write in rows of 7 and permute the letters within every row by one fixed order (a 'matrix' in the C1(a) sense; also the one-row-at-a-time reading of C27(a)), which keeps each letter in its row of seven. 0006 covered columnar read-outs only. Orders: (A) periodic tableau then block permutation, (B) block permutation then periodic tableau. All 5,040 permutations of 7 (the last block of 6 keeps the relative order of its letters), periods 1-52, plain and cipher alphabets standard or KRYPTOS, Vigenere or Beaufort (8 classes; the key alphabet drops out of a consistency test, and variant Beaufort equals Vigenere here). A (permutation, class, period) survives if every key slot's forced values agree. Ten shuffled coppers through the same sweep. Clue entries: C1(a), C11a ('interval 7'), C27(a).
6
Prediction
Written before the run; the result decides it.
In both orders, no permutation makes any period up to 26 consistent under any class, except where the shuffled coppers do the same at a similar rate; periods 27-29 stay consistent as always. No period is singled out at multiples of 7. Zero candidates pass.
7
Method
For each order of seven, map each known place to the carved place it lands on.
For each class, compute the key value each known letter forces, and its slot: the plaintext place in order A, the carved place in order B.
For each period up to 52, check that each slot's values agree.
Count consistent cases per period, and the smallest consistent period.
Repeat on ten shuffled coppers.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.