“K3's own matrix run through all the possible shifts, then the tableau”
Route set replaced. Instead of columnar grids, the routes are K3's transposition rule with every shift, plus K3-style double rotations of a grid. Survival test unchanged.
The third messagethe earlier messages The three solved parts of the sculpture. was scrambled by a fixed rule: take every 192nd letter, wrapping around. The artist once said he was surprised nobody had taken that same rule and tried all its settings. So we do: every step size and every starting point for the 97 letters, and also the grid-turning trick the code-breakers described for the third message. For each setting we work out where 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. would have gone and ask whether a short secret word on 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. could then produce the carved lettersthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. there. A setting that allows a short word is the signal.
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 rule
K3 was solved by reading its letters with a fixed start and step. Here every start and step is tried before the table: drawn for start 5, step 7, where the E at place 22 comes from carved place 55.
the forced letters
Each known letter and the carved letter it lands on force one key letter. Two places whose distance is a multiple of the key length share a slot, so their letters must agree.
the histogram
Over 21,025 rules, and double rotations, under 24 ways of using the table, no secret word shorter than 22 letters survives, and no puzzle word fits the forced letters.
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 rule: start at 5, step 7. The first plaintext letter comes from position 5, the second from 12, the third from 19, and so on, wrapping past 97.
Position 22 of the plaintext, the E of EAST, is the 22nd letter read, so it comes from 5 + 7 x 21 = 152, which wraps to 55.
The ciphertext at 55 is therefore the enciphered E; under the sculpture's alphabet and Vigenere the key letter that sends E there is forced.
Do this for all 24 known letters to get 24 forced key letters at 24 positions.
Two positions whose distance is a multiple of p share a key slot at period p; if their forced letters differ, p is dead.
The smallest live p is the shortest secret word the rule allows.
Slide the puzzle's words across the survivors' pattern.
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: K3 rule, start 5, step 7 (plaintext[i] comes from position 5 + 7 i, 0-indexed); convention kry/kry/kry Vigenere
plain pos letter lands at cipher there forced key letter
22 E 31 G Y
23 A 45 E T
24 S 59 W L
25 T 73 P Z
64 B 37 W I
65 E 51 T N
66 R 65 Y R
67 L 79 Z B
consecutive crib letters land 7 apart, so period 7 and period 1 need their forced letters equal.
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Sanborn, 2003, secondhand (clue C2): surprised no one had tried 'recovering the original matrix and running it through all the possible shifts'. K3's matrix is known: Stehle's rule reads plaintext[i] = ciphertext[(191 + 192 i) mod 337] (data/k3.json), and the NSA described it as a double rotation of a grid. Reading (a): K4 is the same matrix with a different shift. For n = 97, prime, every pair (a, b) with b in 1..96 is a valid rule, 9,312 routes, each applied forward and inverse; add the double rotations, write in rows of w1 and read columns bottom-up, then again with w2, for w1, w2 in 2..50. Each route goes before a periodic tableau with separate alphabets, as in 0006, and the cribs force key letters at their new positions; the divisor rule gives the smallest live period. Live periods up to 30 are matched against the vocabulary and K1-K3 windows. Bounds: complete for this route set, 24 conventions, periods 1-97.
6
Prediction
Written before the run; the result decides it.
As in 0006, the smallest live period clusters between 24 and 50. No route allows a period of 15 or less. No forced fragment matches a vocabulary word or K1-K3 window; zero candidates pass.
7
Method
Build the routes: plaintext[i] = inter[(a + b i) mod 97] for a in 0..96, b in 1..96, forward and inverse; double rotations of a grid for widths 2-50 twice.
For each route, send each crib position to its intermediate position.
For each convention, force the key letter from the crib plaintext and the ciphertext at the destination.
Period p is dead if it divides a distance between two destinations with different forced letters; the smallest live p is the route's score.
For live p up to 30, match the forced fragment against the vocabulary and 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.