“A matrix transposition first, the keyed tableau last: which routes let the 24 cribs force a short periodic key”
New root. Family 6 opens after five reverts in family 1. A columnar transposition stage runs before the periodic tableau, so the cribs constrain the key through the route.
revertedstatus
a new rootparents
4 mintime, all phases
0candidates
0/24known letters
1
The experiment method
What the method does with the 24 known letters.
The artist said he first scrambled the order of the letters by writing them 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 reading it another way, and only then used the big tablethe table The right-hand half of the sculpture, 26 rows of the alphabet each shifted one place further, starting from the word KRYPTOS. on the sculpture. If that is true, the 24 letters we know are no longer where we think: the grid moved them. For each way of making a grid, we work out where the known lettersthe known letters Sanborn has confirmed that positions 22 to 34 of the message read EASTNORTHEAST and positions 64 to 74 read BERLINCLOCK. would have gone, and ask whether a short secret word could then turn them into the carved lettersthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. at those spots. Most grids make the known letters contradict each other, which rules them out. A grid that lets a short secret word do the job would be a real find, and we then check whether any word from the puzzle's own worldthe puzzle's words A list of 62 words tied to the sculpture: the earlier secret words, place names, the names of the people involved, the words in the Morse code, and the themes Sanborn has named. fits.
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 route
Write the 97 letters in rows of 7 and read the columns down. Every known letter is carried to a new place, and the carved letter there is its enciphered form. Drawn for one route; the E at place 22 lands at place 4.
the pair
With the table and a repeating secret word, each landed pair forces one key letter. Thirteen neighbours end up 14 apart, so at a period of 28 they share key slots in pairs, and the forced letters must agree.
the histogram
The experiment ran 2,905 routes under all 24 ways of using the table. No route lets a secret word shorter than 23 letters explain the known letters, and no puzzle word fits the forced letters of any route.
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 route: write the text in rows of 7 and read down the columns, left column first.
Position 22, the E of EAST, is row 4 column 1 (counting from 1). Column 1 holds rows 1 to 14, so E lands at intermediate position 4.
Position 64, the B of BERLIN, is row 10 column 1, so it lands at position 10.
The ciphertext at 4 is R and at 10 is H. Under the sculpture's alphabet with Vigenere, the key letter that sends E to R and the one that sends B to H are forced.
Do this for all 24 known letters; you get 24 forced key letters at 24 positions.
A period of 6 would make positions 4 and 10 share a key letter; if their forced letters differ, period 6 is dead, and so is any period dividing 6.
Cross out every dead period; the smallest survivor is the shortest secret word this route allows.
Write the survivors' forced letters into a pattern with blanks and slide the puzzle's words across it.
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.
route: columnar width 7, read columns left to right; convention kry/kry/kry Vigenere
plain pos letter lands at cipher there forced key letter
22 E 4 R J
23 A 18 F O
24 S 32 K Q
25 T 46 K V
64 B 10 H S
65 E 24 R J
66 R 38 T P
67 L 52 J Z
positions 4 and 10 share a key slot at period 6, 3, 2 and 1; if their forced letters differ, those periods die.
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Sanborn, CNN 2005 (knowledge/sources, clue C1 reading a): matrix code first, double and triple matrixes sometimes, 'then went back to this table to do the ultimate'. Family A in knowledge/06-hypotheses.md: K4 = tableau(T(P)) where T is a columnar transposition and the tableau stage is periodic with separate alphabets. The cribs sit in P at 22-34 and 64-74. Under T they land at intermediate positions j = T(pos); the ciphertext at j then forces the key letter at slot j mod period under each of the 24 conventions. For each route T we find the smallest period whose forced letters never conflict, by the divisor rule: period p is dead if p divides the distance between two intermediate positions that force different key letters. Routes searched: plain columnar at every width 2 to 96, read columns left to right or right to left, applied forward or inverse; keyed columnar from every vocabulary word, forward or inverse; and every double of two plain forward columnars at widths 2 to 50 (C1 'double matrixes', C2 'all the possible shifts'). Survivors with a short period get their forced fragment matched against the vocabulary and K1-K3 windows; matches are decrypted and submitted. Bounds: complete for the route set above under 24 conventions and periods 1-97.
6
Prediction
Written before the run; the result decides it.
A dry run on 40 routes calibrated this. The smallest live period will cluster between 24 and 50 for almost every route and convention. Any route with a live period of 15 or less is rarer than one in a thousand and gets a follow-up. No forced fragment matches a vocabulary word or K1-K3 window, so zero candidates pass.
7
Method
Build the route set: plain columnar widths 2-96 in two column directions, forward and inverse; keyed columnar from each vocabulary word, forward and inverse; doubles of plain forward columnars at widths 2-50.
For each route, map each crib position to its intermediate position.
For each convention, force the key letter at each intermediate position from the crib plaintext and the ciphertext there.
Mark period p dead if it divides a distance between two positions with different forced letters; the smallest live period is the route's score.
For live periods up to 30, match the forced fragment against the vocabulary and the K1-K3 windows; decrypt matches and submit.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.