The first two messages used a repeating secret wordthe secret word The word or string a cipher needs besides the method. and the 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. We ask: if the fourth messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. used the same trick with a longer secret word, how long could that word be? 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. tell us which lengths are impossible, because two letters that would use the same letter of the secret word must agree. For the lengths that survive, the known letters also tell us some letters of the secret word, so we 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 those letters.
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 slot row
A repeating key of length 28: place p uses slot p mod 28. Each known letter and the carved letter under it force the key letter in its slot. Drawn for one way of using the table.
the blanks
At period 28 the 24 known letters fall in 24 different slots, so nothing clashes and the period survives; 4 slots stay unknown.
the verdict strip
Periods 1 to 26 and 30 to 52 die under every one of the 24 ways of using the table. 27, 28, 29 and 53 upwards survive somewhere, and no word from the puzzle fits their 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.
Choose a period, say 28, and a convention: plaintext on the sculpture's alphabet, ciphertext on the standard one, key on the sculpture's, Vigenere.
Position 22 is plaintext E, ciphertext F; the key letter that does that is forced. Position 50 would use the same key slot, but 50 is not a crib, so no conflict.
Positions 26 and 54: 54 is not a crib either.
Positions 64 and 36: neither pair conflicts.
Collect the forced letters into a 28-slot pattern with blanks.
Slide every word and phrase from the puzzle's vocabulary across the pattern.
A fit gives the whole key; decrypt and check the cribs.
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.
forced key letters, period 28, plain kry / cipher std / key kry, Vigenere
pos plain cipher slot(pos-1 mod 28) forced key letter
22 E F 21 Q
23 A L 22 T
24 S R 23 E
25 T V 24 L
26 N Q 25 W
64 B N 7 O
65 E Y 8 G
66 R P 9 H
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
K4 is a periodic tableau cipher like K1 and K2 but with a period the cribs allow and possibly different alphabets for plaintext, ciphertext and key (Quagmire I, II, III, IV under the ACA names). kryptos.cribsweep showed that for single-alphabet additive systems only periods 27, 28, 29 and 53 to 97 survive; nobody has published the same sweep for separate alphabets (knowledge/05-attacks.md, open). For each of the 24 conventions and each period 1 to 97 we derive the key letter forced at every crib position and test whether positions congruent modulo the period agree. For each surviving (convention, period) we record the forced key fragment, then score it two ways: does any vocabulary word or K1 to K3 phrase of that length contain the fragment at the forced slots, and if so submit the decryption. Clue register C5(a), C8(a). Bounds: complete for periodic single-key tableau systems under these 24 conventions and all periods.
6
Prediction
Written before the run; the result decides it.
Periods 1 to 26 and 30 to 52 are excluded under every convention, as in the single-alphabet sweep. Surviving periods force key fragments that match no vocabulary word. Zero candidates pass.
7
Method
For convention (pa, ca, ka, m) and period P: forced[slot] = key_letter_for(p, c) for each crib; a conflict at any slot excludes (convention, P). For survivors, build the fragment pattern of length P with forced letters and wildcards, and match it against every word and phrase in data/vocabulary.txt plus the K1 to K3 plaintexts as sliding windows of length P. Each match yields a full key; decrypt and submit.
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
0 forced-fragment matches submitted as candidates.
What was searched, as the harness recorded it.
family
periodic tableau, separate alphabets (Quagmire I-IV)