“A repeating key on unknown mixed alphabets (a free Quagmire) after or before a K3-style move”
0080-0094 used the standard or KRYPTOS alphabet, fixed. 0096 found that any key repeating every 3, 7 or 21 carved letters after a move makes the width-21 repeats 50 times more often than chance; on a fixed alphabet those periods are dead. Here the alphabets are unknown: the tableau is built from an unknown mixed alphabet (Quagmire III) or an unknown pair (Quagmire IV), shifted by a repeating key, after or before every move of the type-1 library, and the cribs are tested exactly by linear algebra.
K1the earlier messages The three solved parts of the sculpture. and K2 used a table built from a scrambled alphabet that starts KRYPTOS. Sanborn said his table for K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. was slightly different. Here we let the scrambled alphabet be anything at all, unknown to us, and still ask whether a short repeating secret word on such a table, before or after a grida grid Writing the message into rows of some width and reading it out in another order, column by column for instance, which shuffles the letters without changing them. move, can explain 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.. We do not need to guess the alphabet: arithmetic on the 24 letters shows whether any alphabet could work.
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 alphabets
K1 and K2 used a mixed alphabet on the table. Here the alphabets are left unknown, one (Quagmire III) or a pair (IV), with a key of 3, 7 or 21 letters, before or after each move.
the check
Each known letter gives an equation between letter positions and a key shift. Solved mod 2 and mod 13, the equations show whether two letters are forced into one position; if none are, some alphabet fits.
measured after the run
A pair of alphabets at period 3 after the move fits 16 settings against 1-12 on scrambled coppers; counted as distinct equation sets, 12 against 1-11. Marginal, and every other count sits inside the range.
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.
Pick a move and a key length, say 7, after the move.
Each known letter lands at a carved place j and gives: index of the carved letter = index of the known letter + key slot j mod 7, all unknown.
If E lands twice in the same slot under two different carved letters, those two carved letters would need the same index: impossible.
Longer chains of such equations can force the same thing; solving the equations mod 2 and mod 13 finds every forced pair.
A setting with no forced pair is consistent: some alphabet and key fit.
Repeat for every move, key length 1-26, both orders and both table types, and on scrambled coppers.
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.
no move, key length 7, unknown alphabets: known letters grouped by slot (place mod 7)
slot 0: E->F (22), T->R (29), B->N (64), L->Z (71)
slot 1: A->L (23), H->N (30), E->Y (65), O->F (72)
slot 2: S->R (24), E->G (31), R->P (66), C->P (73)
slot 3: T->V (25), A->K (32), L->V (67), K->K (74)
slot 4: N->Q (26), S->S (33), I->T (68)
slot 5: O->Q (27), T->S (34), N->T (69)
slot 6: R->P (28), C->M (70)
free alphabet per slot (functional and one-to-one): False
one mixed alphabet pair shifted by the key (Quagmire IV): not reached
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Type 2 open item 'width 7 with two free alphabets' and the type-1 gap 'two unknown alphabets at periods 13-26' (directions.md, prior-work database). Sanborn: 'slightly different tableau' (C11a, 0071); K1-K2 use a keyed Quagmire III. 0096: periods dividing 21 after a move reproduce the width-21 repeats (1.5% against 0.03%). Model: index(c_j) = index'(m_j) + k[slot], slot = carved place mod p after the move (order A) or message place mod p before it (order B); index and index' unknown bijections (equal for Quagmire III). Each known letter gives a homogeneous linear equation over Z26 in the unknown letter indices and key values. By the Chinese remainder theorem, two letters are forced to the same index mod 26 exactly when the solution spaces over GF(2) and GF(13) both force it; that, or three letters forced together mod 13, contradicts the bijection. Filter first by the weaker free-alphabet-per-slot test (functional and injective within a slot), periods 1-26; the linear test at periods 3, 7 and 21, the divisors of 21 that 0096 ties to the repeats (the time budget allows no more). Moves as in 0090 plus no move; ten shuffled coppers.
6
Prediction
Written before the run; the result decides it.
Under Quagmire III and IV the consistent counts at periods 3, 7 and 21 lie inside the ten shuffled coppers' range, give or take one period. Zero candidates.
7
Method
Build the route library.
For each route, order, period and text, apply the free-alphabet-per-slot filter.
For survivors at periods 3, 7 and 21, write the 24 linear equations and compute the solution space over GF(2) and GF(13).
Reject if two different letters of one alphabet are forced to the same index in both fields, or three to the same index mod 13.
Count survivors per period, model and order, on K4 and ten shuffled coppers.
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
27,853 of 27,853 routes x 2 orders x periods 1-26 x 2 models. K4 live at periods 3/7/14/21: {'III A': [0, 87, 0, 6286], 'III B': [0, 481, 0, 5964], 'IV A': [16, 1344, 0, 12478], 'IV B': [0, 6733, 0, 17210]}. Periods outside the control range: {'III A': [], 'III B': [], 'IV A': [3], 'IV B': []}. K4 live at periods 3, 7, 21: 50599.
What was searched, as the harness recorded it.
family
periodic key on unknown mixed alphabets (Quagmire III: one alphabet; IV: a pair), type-1 moves plus no move