Experiment 119

“One alphabet in an unknown order on both sides of the table (Quagmire III), no move: does any order fit the 24 known letters, key lengths 1-26, solved exactly?”

F1.1, first step (exact consistency). 0097 tested one unknown alphabet only at key lengths 3, 7 and 21, and only with a linear relaxation that can refute but not confirm (it rejects when two letters are forced to the same place mod 2 and mod 13; a system can still have no one-to-one solution without any such forced pair). Here every key length 1-26 is solved exactly by a search over the alphabet with propagation, in both table forms (Vigenere-like: place(c) = place(p) + k; Beaufort-like: place(c) = k - place(p)), with no move, on K4 and on 100 scrambled coppers. Key lengths where K4 has no fitting alphabet are closed exactly; the rest go to the English climb (next experiment).

revertedstatus
experiment 97parents
11 mintime, all phases
0candidates
0/24known letters
1

The experiment method

What the method does with the 24 known letters.

K1the earlier messages The three solved parts of the sculpture. and K2 were enciphered on a table whose rows are one scrambled alphabet (it starts KRYPTOS), each row shifted one place, with a repeating key wordthe secret word The word or string a cipher needs besides the method. choosing the row for each letter. Sanborn said his own table was slightly different. Here the scrambled alphabet may be any of the 400 million million million million orders. We do not guess it: for each key length from 1 to 26 a careful search asks whether any order at all, with any key, turns the carved lettersthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. into the 24 letters we know. If none does, that key length is finished for good.

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.

OBKRUOXOGHULBSOLIFBBWEASTNORTHEASTOTWTQSJQSSEKZZWATJKLUDIAWINFBBERLINCLOCKWGDKZXTJCDIGKUHUAUEKCARIN: THE COPPER, ALL 97 LETTERS, THE 24 KNOWN PLACES IN ACCENTtwo known letters encrypt to themselves: S under S at 33, K under K at 74EASTNORTHEASTBERLINCLOCKsame alphabet on both sides, key added: a letter to itself means key 0 in its slot (33 and 74)at key length 26, place 22 shares 74's slot: E would have to encrypt to itself, but the copper has Fkey lengths 1-26, added form (Quagmire III): an alphabet fits?1510152026no alphabet fitssome alphabet fitsthe same, Beaufort form (carved place = key minus known place)151015202326K4: 0 lengths fit; scrambled coppers 0-15, middle 90% 0-10K4: 2 lengths fit (23, 26); coppers 1-11, middle 90% 2-9coppers that fit at 23: 44%, at 26: 85%OUT: NO READING; THIS STEP ONLY DECIDES WHICH KEY LENGTHS ANY ALPHABET CAN FITEASTNORTHEASTBERLINCLOCK
the idea

K1 and K2 used one scrambled alphabet on both sides of the table. Sanborn said his table was slightly different, so here the alphabet may be any of the 26! orders, and every key length 1-26 is decided exactly by algebra, not by guessing alphabets.

the check

S encrypts to S and K to K. With the key added, that forces a zero key in their slots, and at every length 1-26 another known letter shares one of those slots without encrypting to itself: dead at every length.

the bound

Added form: no alphabet fits at any length 1-26. Beaufort form: only 23 and 26 fit, where 7 and 1 checks are left. Scrambled coppers fit at 0-15 and 1-11 lengths: K4 is inside their range. Lengths 23 and 26 go to the English test.

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.

  1. Choose a key length, say 7, and write the 24 known places with their slot (place mod 7).
  2. For each known place write: place-in-alphabet of the carved letter = place of the known letter + key value of the slot.
  3. In slot 3 the known letters R (66) and C (73) both sit under carved P, so R and C would need the same place: no alphabet works, length 7 is dead.
  4. When no such clash shows at once, set the first letter's place to 0 and try values for the busiest slot's key; each choice fixes further places through the equations.
  5. Back up whenever two letters would need the same place or one letter two places.
  6. If a full set of places and key values satisfies all 24 equations, the length is alive; if every branch backs up, it is dead.
  7. Repeat with the Beaufort form (carved place = key value minus known place), for every length 1-26, then on scrambled coppers.
the worksheet: the first steps by hand, on squared paper
WHY THE ADDED FORM DIES AT EVERY KEY LENGTH: A SLOT HOLDING S/S OR K/K NEEDS KEY 0lengthself-lettershares slotknown/carvedkey 0 needs133 (S under S)22E under FE = F? no233 (S under S)23A under LA = L? no333 (S under S)24S under RS = R? no433 (S under S)25T under VT = V? no533 (S under S)23A under LA = L? no633 (S under S)27O under QO = Q? no733 (S under S)26N under QN = Q? no833 (S under S)25T under VT = V? no933 (S under S)24S under RS = R? no1033 (S under S)23A under LA = L? no1133 (S under S)22E under FE = F? no1233 (S under S)69N under TN = T? no1333 (S under S)72O under FO = F? nolengthself-lettershares slotknown/carvedkey 0 needs1474 (K under K)32A under KA = K? no1574 (K under K)29T under RT = R? no1633 (S under S)65E under YE = Y? no1733 (S under S)67L under VL = V? no1833 (S under S)69N under TN = T? no1933 (S under S)71L under ZL = Z? no2033 (S under S)73C under PC = P? no2174 (K under K)32A under KA = K? no2274 (K under K)30H under NH = N? no2374 (K under K)28R under PR = P? no2474 (K under K)26N under QN = Q? no2574 (K under K)24S under RS = R? no2674 (K under K)22E under FE = F? noEvery length 1-26 puts a second known letter in the slot of 33 or 74; with key 0 it would have to be its own carvedletter, and none is. The Beaufort form escapes this (2 x place = key) and fits only at 23 and 26.
4

Worked example, generated

The same steps run by the code on the real letters.

key length 7: known letters 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)
  form V: no alphabet fits
  form B: no alphabet fits

key length 12: known letters by slot (place mod 12)
  slot 0: T->V (25), C->P (73)
  slot 1: N->Q (26), K->K (74)
  slot 2: O->Q (27)
  slot 3: R->P (28), B->N (64)
  slot 4: T->R (29), E->Y (65)
  slot 5: H->N (30), R->P (66)
  slot 6: E->G (31), L->V (67)
  slot 7: A->K (32), I->T (68)
  slot 8: S->S (33), N->T (69)
  slot 9: E->F (22), T->S (34), C->M (70)
  slot 10: A->L (23), L->Z (71)
  slot 11: S->R (24), O->F (72)
  form V: no alphabet fits
  form B: no alphabet fits
5

Hypothesis, for a cryptanalyst

Family, key space, the constraint under test, and the clues it reads.

C27(c) (Sanborn, NYT 20 Nov 2010: his K1/K2 tableau was 'slightly different from the one on Kryptos'), C11a, C1(b) (the table alone, no move). K1 and K2 are Quagmire III on the KRYPTOS alphabet (keys PALIMPSEST, ABSCISSA). Covered so far: standard and KRYPTOS alphabets (0005, crib sweep), keyword alphabets from every dictionary word (0109), alphabets one swap or move from KRYPTOS (0071), any alphabet at lengths 3, 7, 21 by relaxation (0097). Model: one unknown bijection A: letters -> Z26 used for plain and cipher sides; key k of length L (free values, the key alphabet drops out); form V: A(c_i) = A(p_i) + k[i mod L]; form B: A(c_i) = k[i mod L] - A(p_i) (variant Beaufort is form V with k negated). The 24 known letters give 24 linear equations over Z26 = Z2 x Z13. Exact decision: solve mod 13 and mod 2; a one-to-one alphabet needs every residue class mod 13 to hold at most two letters, each such pair differing mod 2. Refute at once when two letters coincide mod 13 and mod 2 in every solution, or three mod 13 (0097's relaxation); otherwise enumerate the mod-13 letter space up to translation and scaling (rank <= 5), or search it depth first with pruning on class size and mod-2 solvability (larger ranks), with a node limit that reports 'undecided' rather than guessing. Every fit yields a witness alphabet and key re-checked against all 24 equations; cross-checked on 40 coppers against an independent propagation search (no fit it found was called dead). Planted texts at every length must come out consistent. Controls: 100 shuffles of K4 (seed 119), same test.

6

Prediction

Written before the run; the result decides it.

Key lengths 1-6 have no fitting alphabet on K4 (too many repeats of a slot); from about 8 upward most lengths fit in both forms, because a free alphabet has 25 free places and the known letters give fewer checks than that. K4's number of fitting lengths lies inside the middle 90% of the 100 scrambled coppers in each form. If K4 fits at a length where fewer than 5% of coppers fit, that is outside the coppers' range and stops the queue. Zero candidates (this step returns none).

7

Method

  1. For each key length L from 1 to 26 and each form, write the 24 equations (slot i mod L, known letter, carved letter).
  2. Decide exactly whether a one-to-one alphabet and a key satisfy them, by linear algebra mod 13 and mod 2 with an exhaustive search of the mod-13 solutions; re-check every witness.
  3. Do the same on 100 scrambled coppers.
  4. Report, per length and form, K4 fit or not, the share of coppers that fit, and the number of checks (24 minus the slots that hold a known letter).
  5. Validate on planted texts and against 0097's relaxation (every exact fit must also pass the relaxation).
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

K4 dead (no alphabet fits) at lengths V [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26], B [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25]. Fitting lengths 3-26: K4 V 0, B 2; coppers 5th-95th V (0, 10), B (2, 9). Unusual fits: {'V': [], 'B': []}.

What was searched, as the harness recorded it.

family
one unknown alphabet on both sides (Quagmire III, any of 26! orders), periodic key, no move
key lengths
1, 26
forms
V
A(c) = A(p) + k
B
A(c) = k - A(p)
exhaustive
exact decision per length and form (search with propagation, witness re-checked)
checks by length
1
23
2
22
3
21
4
20
5
19
6
18
7
17
8
16
9
15
10
14
11
13
12
12
13
11
14
11
15
9
16
8
17
7
18
7
19
9
20
11
21
11
22
9
23
7
24
5
25
3
26
1
k4 fits
V
none
B
23, 26
k4 dead
V
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26
B
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25
copper share fitting by length
V
0, 0, 0, 0, 0, 0, 0, 0.02, 0, 0.06, 0.13, 0.18, 0.21, 0.21, 0.01, 0.39, 0.02, 0.39, 0.01, 0.2, 0.21, 0.32, 0.44, 0.62, 0.69, 0.87
B
0, 0, 0, 0, 0, 0, 0, 0, 0, 0.04, 0.15, 0.15, 0.28, 0.2, 0.01, 0.39, 0.01, 0.45, 0.02, 0.26, 0.2, 0.35, 0.44, 0.58, 0.62, 0.85
k4 fitting lengths 3 26
V
0
B
2
coppers fitting lengths 3 26 sorted
V
0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 14, 15
B
1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 11, 11
coppers 5th 95th percentile
V
0, 10
B
2, 9
k4 fits where under 5pct of coppers fit
V
none
B
none
k4 witnesses
B23
places
A
12
B
25
C
0
E
13
F
1
G
5
H
19
I
3
K
6
L
14
M
17
N
15
O
16
P
18
Q
2
R
20
S
23
T
11
V
7
Y
21
Z
4
key
17, 18, 17, 18, 12, 5, 8, 18, 18, 20, 8, None, None, None, None, None, None, 14, 8, 12, 21, 14, 0
B26
places
A
14
B
1
C
15
E
2
F
16
G
3
H
17
I
4
K
9
L
18
M
5
N
6
O
19
P
7
Q
20
R
21
S
8
T
22
V
10
Y
23
Z
24
key
13, 2, 17, 23, 5, 23, 16, 4, None, None, None, 7, 25, 2, 2, 0, 2, 20, 16, 9, 22, 18, 6, 3, 6, 0
relaxation violations
0
k4 undecided
V
none
B
none
controls undecided total
V
0
B
0
texts done
101
plants
3 per length and form, all recovered as consistent
controls
100 shuffled coppers (seed 119), identical test

How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.

Rendered from the record on 2026-10-18 12:26 UTC; updated after every experiment.
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