“Two key words in two passes, on tables rebuilt from the puzzle's words”
Experiment 14's exact linear test, with the plain and cipher alphabets drawn from the keyword mixings of every vocabulary word instead of the standard and KRYPTOS orders. Word lengths with a + b up to 24, where the 24 cribs leave at least one redundant check.
Experiment 14 asked whether two short secret words, used one after the other 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 explain 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.; none could. The table could have been built from a different word, which changes every sum, so we rebuild it from each word the puzzle knows and ask again. Because two words of total length up to 24 leave spare known letters to check against, every pair of lengths gets a yes or no.
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 slots
Two secret words used one after the other on the table add up: each place uses one slot of each word, and each known letter forces their total.
the tables
Drawn for the PALIMPSEST alphabet on the message and the ABSCISSA alphabet on the copper; the experiment tried all 63 x 63 pairs built from the puzzle's words.
the bound
666,792 cases: 770 pairs of lengths survive, none with a total below 22, which is what chance leaves; no survivor is made of puzzle words.
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.
Build the alphabet from PALIMPSEST for the message and from ABSCISSA for the carved letters.
Lengths 7 and 8, Vigenere both passes: position 22 needs a total key index that sends E, index 6 in the PALIMPSEST alphabet, to F, index 11 in the ABSCISSA alphabet: 5.
Position 22 uses slot 0 of the 7-letter word and slot 5 of the 8-letter word, so x0 + y5 = 5.
Position 29 uses slot 0 and slot 4: x0 + y4 equals its own total.
Subtracting fixes y5 - y4; walk all 24 positions and every loop must close.
A contradiction kills this pair of lengths for this pair of alphabets.
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.
plain PALIMSETBCDFGHJKNOQRUVWXYZ
cipher ABSCIDEFGHJKLMNOPQRTUVWXYZ
lengths 7 and 8, Vigenere: totals and slots
pos plain carved total x slot y slot
22 E F 1 0 5
23 A L 11 1 6
29 T R 11 0 4
30 H N 1 1 5
64 B N 6 0 7
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Two passes through the same tableau compose to t[i] = x[i mod a] + s y[i mod b] mod 26 (0014). Under keyword-mixed plain and cipher alphabets (C5, C29) the forced totals t change, so the weighted union-find test is rerun over 63 x 63 alphabet pairs, both plaintext signs, both s, and every length pair with a < b, a + b <= 24 and lcm outside the dead periods. With a + b - 1 effective unknowns and 24 equations, each case has 25 - a - b redundant checks at 1/26 each; chance survival per case is below 1/26 and the number of cases is about a million, so a few thousand chance survivors at a + b = 23 and 24 are expected and none below 20. Survivors are enumerated and X, Y matched against the vocabulary and K1-K3 windows under both key alphabets. Bounds: complete for these alphabets and lengths.
6
Prediction
Written before the run; the result decides it.
No length pair with a + b below 20 survives under any alphabet pair. Survivors at 20 to 24 appear at chance rates, and none yields a word from the vocabulary or K1-K3. Zero candidates pass.
7
Method
Build the alphabet pool.
For each plain and cipher alphabet and plaintext sign, compute the forced total at each crib.
For each length pair and combiner sign, add the 24 difference constraints to a weighted union-find; reject on contradiction.
Count survivors by a + b.
Enumerate survivors with at most three components; match X and Y against the vocabulary and K1-K3 windows; decrypt and submit matches.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.