Experiment 98
0017, 0004, 0021 and 0078 slid running keys along K4 on the standard and KRYPTOS alphabets only. Here the table's alphabet is unknown, any of the 26! orders (Quagmire III, as K1 and K2 use, with the alphabet left free): for each key text and offset the 24 known letters give linear equations in the unknown letter positions, solved exactly mod 2 and mod 13.
What the method does with the 24 known letters.
Experiment 17 used the sculpture's own texts as a long key, but only with the two alphabets we know. Sanborn said his table was slightly different, so here the alphabet along the top of 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. can be any order of the 26 letters. We do not need to guess it: for each key text and starting point, simple arithmetic on 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. shows whether any alphabet at all could make them fit.
Words in the margin are defined on the glossary page.
The carved letters go in at the top; the reading comes out at the bottom.
Experiment 17 used the sculpture's texts as a running key, but only on the two known alphabets. Sanborn called his table slightly different, so here its alphabet may be any order of the 26 letters.
Each known letter gives one equation: the carved letter's position equals the message letter's plus the key letter's. Solved mod 2 and mod 13, they show whether two letters are forced into one position.
80,676 settings: no alphabet fits K4 for any of them, while the scrambled coppers manage 0 to 57. K4 sits at the bottom of the range.
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.
Redo the method on a few letters with pencil and paper.
The same steps run by the code on the real letters.
key text: the Morse as carved, offset 0, Vigenere on an unknown alphabet X
place known carved key equation
22 E F E X(F) = X(E) + X(E)
23 A L T X(L) = X(A) + X(T)
24 S R A X(R) = X(S) + X(A)
25 T V L X(V) = X(T) + X(L)
64 B N Y X(N) = X(B) + X(Y)
65 E Y O X(Y) = X(E) + X(O)
some alphabet fits all 24: False
Family, key space, the constraint under test, and the clues it reads.
Type 3 and type 5 of directions.md together: keys concealed at the sculpture (C6(a), Scheidt) on a table Sanborn called 'slightly different' from the carved one (C11a); 0071 tried only alphabets one swap or move away. Model, Quagmire III with unknown alphabet X: X(c_i) = X(p_i) + s X(k_{o+i}) mod 26, s = +1 (Vigenere) or -1 (variant), and X(c_i) = X(k_{o+i}) - X(p_i) (Beaufort); each known letter gives one homogeneous equation in the 26 unknowns X(letter). X must be a bijection: two letters forced to the same value mod 2 and mod 13, or three forced together mod 13, is a contradiction (as in 0097). Key texts as in 0090: the carved tableau by rows, rows with the index column, and columns; the Morse as carved and in reverse order; the four ciphertexts; the K1-K3 plaintexts; Sanborn's 1989 letter, the 1991 dedication texts and the 2025 open letter; each forward and backward at every cyclic offset. Ten shuffled coppers.
Written before the run; the result decides it.
A random equation set like these is consistent with some alphabet only rarely; K4's consistent (text, offset, rule) count lies inside the ten shuffled coppers' range, and no consistent setting decrypts to English. Zero candidates.
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
80,676 (text, offset, rule) settings. Consistent with some alphabet: K4 0, controls [57, 0, 0, 40, 0, 0, 30, 1, 6, 43].
What was searched, as the harness recorded it.
How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.