“F1.11: a split key phrase: its first part mixes the plain alphabet, its second the cipher alphabet, the whole phrase is the repeating key”
F1.11 (added 2026-10-16 from Elementary Military Cryptography, 1935/1943, par. 59d: the key phrase CENTRAL AMERICA split into CENTRAL for one alphabet and AMERICA for the other, the whole phrase the key). 0123 showed two unknown alphabets admit K4 only at key lengths 8, 13, 16, 19, 20, 23, 24 and 26; here the alphabets and key are fixed by a phrase, so each phrase is one exact decryption. Phrases: every ordered pair of puzzle words, and every run of 2-4 consecutive words of the K1-K3 plaintexts split at each inner word boundary, of total length 3-30 (all lengths tried, not only 0123's eight, as a check).
A wartime US Army manual describes a key phrase cut in two: the first half reorders the alphabet for the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered., the second half reorders the alphabet for the carved lettersthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page., and the whole phrase is the repeating secret word. We try every pair of the puzzle's wordsthe 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. and every short run of words from the first three messagesthe earlier messages The three solved parts of the sculpture. as such a phrase, and count how many of 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. each one gives.
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
A US Army manual (1935, 1943) cuts a key phrase in two: the first part reorders the message alphabet, the second the carved alphabet, and the whole phrase is the repeating key.
the check
Every pair of the puzzle's words and every short run of words from K1-K3, split at each word boundary: 29,880 phrases, three ways of numbering the key, three table rules, on K4 and ten scrambled copies.
the bound
The best phrase gives 7 of the 24 known letters, as on the copies (7-8). Closed.
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 the phrase LAYER TWO: LAYER mixes the plain alphabet (LAYERBCDFG...), TWO the cipher alphabet (TWOABCDEFG...).
The key is LAYERTWO, eight letters, repeated under the message.
For a message letter, find its place in the plain alphabet, add the key letter's place in the cipher alphabet, and read the cipher alphabet there.
Run it backwards at the 24 known places: does each carved letter give the known letter?
Repeat for every phrase, the three numberings, the three rules, and 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.
phrase LAYER TWO: plain alphabet LAYERBCDFGHIJKMNOPQSTUVWXZ, cipher alphabet TWOABCDEFGHIJKLMNPQRSUVXYZ, key LAYERTWO
place carved key cipher place key value plain place letter known
22 F T 8 0 8 F E
23 L W 14 1 13 K A
24 R O 19 2 17 P S
64 N O 16 2 14 M B
65 Y L 24 14 10 H E
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
C5(b) ('key word/s'), C6 (keys concealed at the sculpture, K1-K3 vocabulary), C7(a); source knowledge/sources (Elementary Military Cryptography par. 59d, per research_notes/Scheidt patents and field ciphers/addendum_downloaded_manuals.md). Model: plain alphabet P = keyword_mixed(part 1), cipher alphabet C = keyword_mixed(part 2), key = part 1 + part 2 repeating; key letters numbered by their place in C (the manual's rule: the key letter in the cipher component is set against the plain component's first letter), in P, or in the standard alphabet; rules Vigenere (C[P(p) + k]), Beaufort (C[k - P(p)]), variant (C[P(p) - k]). Exact: count the known letters each phrase reproduces. Ten scrambled coppers; statistic fixed before the run: phrases reproducing all 24 (a hit), and the best count on K4 against the coppers' best.
6
Prediction
Written before the run; the result decides it.
No phrase reproduces all 24 known letters; K4's best count is inside the coppers' range (about 6-9 of 24). Zero candidates.
7
Method
Build the phrases.
For each phrase, numbering and rule: decrypt the 24 known places and count matches.
Same on 10 coppers; compare best counts; decrypt any full hit.
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
29,880 phrases x 3 numberings x 3 rules: K4 best 7/24 (coppers 7-8); full hits K4 0, coppers [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
What was searched, as the harness recorded it.
family
split key phrase (EMC par. 59d): part 1 mixes the plain alphabet, part 2 the cipher alphabet, the whole phrase the key
phrases
29,880
numberings
cipher alphabet, plain alphabet, standard
rules
vig, beau, var
k4 best hits
7
controls best hits
7, 8, 8, 7, 7, 7, 7, 7, 7, 7
k4 full hits
0
controls full hits
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
k4 top
none
controls
10 shuffled coppers (seed 1411)
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.