“Change, turn, change again: a repeating key on the table, a K3-style grid turn, then a second repeating key, solved exactly”
0008 and the public searches (keithadler, nitsuah, Zekios) put one repeating key on one side of a K3-style turn. Here there is a key on both sides: the message is changed on the table with one repeating word, the letters are moved by a grid turn (or K3's every-n-th rule, or a column shuffle), and the moved text is changed again with a second repeating word. With both alphabets the same, the 24 known letters give 24 linear equations in the two keys, which are solved exactly for every route and every pair of key lengths 2-14.
The first two messages were made by shifting letters along the sculpture's tablethe table The right-hand half of the sculpture, 26 rows of the alphabet each shifted one place further, starting from the word KRYPTOS. with a repeating secret wordthe secret word The word or string a cipher needs besides the method.; the third was made by writing the letters in 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. and turning it. Sanborn said he used 'double and triple matrixes' and 'then went back to this table'. Here we try the sandwich: shift with one secret word, turn the grid, shift again with a second secret word. That sounds like too many unknowns, but every known letter of the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. ties one letter of the first word to one letter of the second, and those ties must agree with each other. We check every grid turn and every pair of word lengths from 2 to 14, exactly, and do the same on ten scrambled copies of the copper to see how often agreement happens by luck.
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 sandwich
The message is shifted along the table with one repeating word, scrambled by K3's grid turn, then shifted again with a second word: Sanborn's 'double matrixes', then 'back to this table'.
the check
Each known letter fixes the sum of one letter of each word. With words of 2 and 3 letters the 24 known letters leave 20 independent checks; one disagreement kills the setting.
the bound
26,158 different scramblings, word lengths 2 to 14 each, standard or KRYPTOS table: no setting passes five checks, and the counts at three and four match ten scrambled coppers.
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.
Write the message in rows of 7 and turn the grid a quarter clockwise: read each column from the bottom up, left column first. Counting from 0, place 22 (the E of EAST) is letter 21, in row 3, column 0, and comes out as letter 10 of the turned text (place 11).
Take a first key of length 2 and a second of length
Letter 21 uses slot 21 mod 2 = 1 of the first key; its new position 10 uses slot 10 mod 3 = 1 of the second. 3. On the sculpture's alphabet, the copper letter at place 11, U, minus E gives the sum of those two key letters: U - E = 10.
Every known letter gives such a sum for one slot of each key: 24 sums for 2 + 3 slots.
Five slots need four sums to fix them (one is a free shift between the keys); the other twenty sums are checks.
Go through the sums and see whether they agree; one disagreement kills this turn and these key lengths.
Repeat for every turn, rule and pair of lengths, and on ten 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.
one quarter turn clockwise at width 7; first key length 2, second key length 3; sculpture's alphabet, s = +1
place known new place copper there first slot second slot sum of the two key letters
22 E 11 U 1 1 U-E = 10
23 A 25 V 0 0 V-A = 15
24 S 39 Q 1 2 Q-S = 14
25 T 53 K 0 1 K-T = 22
26 N 67 V 1 0 V-N = 3
27 O 81 T 0 2 T-O = 25
24 sums over 2 + 3 slots: 20 independent checks
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Type 1 of directions.md, gap a3 of the prior-work database (reports/Kryptos K4 prior attempts database.md: 'rotation sandwiched between two different tableau keys... Zekios's stacked-key search used columnar 2-7 and width-21 routes, not rotations'). Clues C1(a) and C1(c) read together (02-sanborn-said.md, CNN 2005: 'double matrixes and triple matrixes... and then went back to this table'), C2(a) (the K3 matrix and its 'shifts'), C5(b) ('key word/s', two words) and C20(a) (LAYER TWO). Model: X[i] = P[i] (+) a[i mod p1] on alphabet A; Y = route(X); C[j] = Y[j] (+) b[j mod p2] on the same alphabet A, each stage Vigenere, Beaufort or variant. Composed, C[pi(i)] = s P[i] + a'[i mod p1] + b'[pi(i) mod p2] (mod 26, indices on A) with s = -1 when exactly one stage is Beaufort, so the key alphabets drop out and four cases remain: A standard or KRYPTOS, s = +1 or -1. Each known letter is one edge between key slot a_u and key slot b_w carrying the value C - sP; the system is consistent iff every cycle of that bipartite graph sums to zero, checked by a spanning tree; the number of independent checks is edges minus (slots touched minus components). Routes (deduplicated on where the 24 known places land): K3's rule pi(i) = a + b i mod 97, all 9,312; one quarter turn, clockwise or anticlockwise, at widths 2-96, each forward and inverse; two clockwise quarter turns at every width pair 2-96, forward and inverse; plain columnar at widths 2-96 in both reading directions, forward and inverse; keyed columnar from every vocabulary word, forward and inverse. Key lengths p1, p2 = 2-14. Exact; ten shuffled coppers through the identical search give the chance rate at each number of checks. Bound left open: the two stages on different alphabets (not linear), key lengths above 14, nulls or padding (gap a1).
6
Prediction
Written before the run; the result decides it.
At every number of checks K4's count of consistent cases lies inside the range of the ten shuffled coppers. No case on K4 survives seven or more independent checks. Zero candidates.
7
Method
Build the routes and keep one per distinct landing pattern of the 24 known places.
For each route, each alphabet and each sign, compute the edge values C at the landing place minus s times the known letter.
For each pair of key lengths, join slot i mod p1 of the first key to slot (landing place) mod p2 of the second key for every known letter, grow a spanning tree, and test every other edge against the tree.
Count consistent cases by number of checks.
Repeat on ten shuffled coppers.
Print any K4 case with five or more checks, with its key slots and the letters they fix.
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
26,158 distinct routes from 28,246 specs, 4,389,273 route x key-length structures with at least one check, x4 cases. K4 consistent cases with at least c checks, c = 1..8: [12568, 800, 34, 4, 0, 0, 0, 0]; controls max: [12726, 815, 51, 6, 0, 0, 0, 0]; controls min: [12393, 707, 37, 0, 0, 0, 0, 0]. Levels outside the control range: [3]. K4 cases with 5+ checks: 0.
What was searched, as the harness recorded it.
family
substitute, move, substitute: two repeating keys on one alphabet around a route, solved exactly
route specs
28,246
distinct routes
26,158
routes done
26,158
route kinds
K3 rule a + b i mod 97 (9,312); one quarter turn cw/ccw widths 2-96 fwd/inv; two cw quarter turns widths 2-96 x 2-96 fwd/inv; plain columnar widths 2-96 both directions fwd/inv; keyed columnar from 62 vocabulary words fwd/inv