“F12.5: K4 as two parts with their own repeating keys, split between the two known runs, on the sculpture's tables: exact, with the checks each part leaves”
F12.5 (assumption A7: one key for the whole text). Stein found K1 and K2 used different keys in different parts of the panel. F12.2 and F12.4 found no break inside K4, so the split is at the known stretches: a first part holding EASTNORTHEAST and a second holding BERLINCLOCK, each with its own repeating key of length 1-26 on one of the 8 standard/KRYPTOS table classes. With separate keys the split point does not matter for consistency (a key's slots inside one run depend only on distances), and each run checks only keys shorter than itself: 13 letters check lengths 1-12, 11 letters check 1-10. The checks each length leaves are stated; ten scrambled coppers.
K1the earlier messages The three solved parts of the sculpture. and K2 used different secret words. Perhaps K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page.'s two halves did too: one word for the part with EAST NORTHEAST, another for the part with BERLIN CLOCK. Then each known word can only test its own half's secret word, and only word lengths shorter than itself. We check every length that can be checked, on the sculpture's tables, and on scrambled copies.
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 idea
K1 and K2 used different secret words. If K4's half with EAST NORTHEAST and its half with BERLIN CLOCK had their own words, each known word checks only its own half's word.
the check
Every word length a known run can check (up to 12 and up to 10), on the eight ways of using the sculpture's tables, on K4 and ten scrambled copies.
the bound
No length fits either half of K4; copies manage 0-1 cells. Longer words leave no checks at all, so this is all the known letters can say.
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.
Take EASTNORTHEAST at places 22-34 and a first secret word of length 5.
Places 22 and 27 share a slot (27 - 22 = 5): on the table, E over F and O over Q must ask for the same key letter.
Check every pair five apart inside the run, and the same for every length up to 12.
Do the same for BERLINCLOCK with the second word, lengths up to 10.
Repeat on scrambled copies.
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.
part 1, sculpture's table (KRYPTOS alphabets), Vigenere: key value forced at each place of EASTNORTHEAST
place 22: E under F -> key 1, slot at length 5: 0
place 23: A under L -> key 10, slot at length 5: 1
place 24: S under R -> key 21, slot at length 5: 2
place 25: T under V -> key 18, slot at length 5: 3
place 26: N under Q -> key 1, slot at length 5: 4
place 27: O under Q -> key 15, slot at length 5: 0
place 28: R under P -> key 2, slot at length 5: 1
place 29: T under R -> key 23, slot at length 5: 2
place 30: H under N -> key 5, slot at length 5: 3
place 31: E under G -> key 2, slot at length 5: 4
place 32: A under K -> key 19, slot at length 5: 0
place 33: S under S -> key 0, slot at length 5: 1
place 34: T under S -> key 2, slot at length 5: 2
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Assumption A7 (knowledge/12-assumptions.md, for information), Stein 1999 (parts with their own keys: ABSCISSA and PALIMPSEST), C5(b) ('key word/s'). Model: c_i = T(p_i, k1[i mod a]) for i in part 1, T(p_i, k2[(i - s) mod b]) in part 2, T one of 8 classes (plain and cipher alphabets standard or KRYPTOS, Beaufort or not). Checks for part 1 at length a: 13 - (distinct slots of places 22-34 mod a); part 2: 11 - (distinct slots of 64-74 mod b). A length is live if no slot holds two different forced key letters. Statistic: live (class, length) cells per part with at least one check, K4 against the coppers' range.
6
Prediction
Written before the run; the result decides it.
No length below 11-13 is live on K4 in either part at more than the coppers' rate (the periodic sweep 0005 already fails at these lengths with one key; separate keys loosen it only by the cross-run checks). Zero candidates.
7
Method
For each text, table class and part: the forced key value at each known place of that part.
For lengths 1-12 (part 1) and 1-10 (part 2): live if each slot holds one value; count checks.
Compare live cells on K4 with ten coppers.
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
part 1 (EASTNORTHEAST, 22-34): K4 live lengths [] (0 cells), coppers 0-1; part 2 (BERLINCLOCK, 64-74): K4 live lengths [] (0 cells), coppers 0-1
What was searched, as the harness recorded it.
family
two parts with separate repeating keys (split between the runs), standard/KRYPTOS tables, lengths up to the run length - 1
summary
part 1 (EASTNORTHEAST, 22-34)
k4 live lengths
none
k4 live cells
0
coppers live cells
0, 1, 0, 0, 0, 1, 0, 0, 0, 0
checks by length
1
12
2
11
3
10
4
9
5
8
6
7
7
6
8
5
9
4
10
3
11
2
12
1
part 2 (BERLINCLOCK, 64-74)
k4 live lengths
none
k4 live cells
0
coppers live cells
0, 0, 0, 1, 1, 0, 1, 0, 0, 0
checks by length
1
10
2
9
3
8
4
7
5
6
6
5
7
4
8
3
9
2
10
1
k4 live
none
controls
10 shuffled coppers (seed 1615)
exhaustive
yes
How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.
Rendered from the record on 2026-10-18 12:26 UTC; updated after every experiment.