“Words a riddle would use, dragged across K4 at every repeating-key length the known letters allow”
Earlier periodic experiments at the live lengths (27-29 and 53 and up) filled the free key letters by climbing toward English, which cannot beat its own controls. Here nothing is climbed: each word of a fixed riddle lexicon (spelled numbers, units, compass words, the puzzle's own words, Sanborn's themes) is placed at every position, and a placement counts only if the key letters it forces agree with the ones the known letters already fix. Random letter strings of the same lengths are the control.
revertedstatus
a new rootparents
25 stime, all phases
0candidates
0/24known letters
unclocked
1
The experiment method
What the method does with the 24 known letters.
If the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. is a riddle like K2the earlier messages The three solved parts of the sculpture., which spelled out numbers and directions, it probably uses words such as THIRTY, DEGREES, NORTH or SHADOW. For every length of repeating secret word that fits 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., we slide each such word along the message and ask whether it agrees with the keythe secret word The word or string a cipher needs besides the method. letters the known letters already fix. Random strings of letters, slid the same way, show how often agreement happens by luck.
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 drag
Lay a word Sanborn might have used, a number, a unit, a compass point or a theme word, at every place, and see whether a repeating key consistent with the 24 known letters can carry it.
the check
Under a 60-letter key the word's letters that fall on already fixed slots must need the same key letters. DEGREES at place 80 fails at its first fixed slot.
the bound
93 words, 376 key lengths and tables that fit the known letters: no word does better than random strings. The test has little power: a word meets few fixed slots, so a miss proves little.
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 sculpture's alphabet, Vigenere, and a key of length 60; the known letters fix the key letters at slots 21-33 and 3-13.
Place the word DEGREES so that it starts at place 80.
Its letters fall on slots 19-25 (place mod 60); slots 21-25 are already fixed by EASTNORTHEAST's letters at places 22-26.
For each of those, the key letter that turns the word's letter into the carved letter must equal the fixed one.
All agree: a consistent placement with that many checks. One disagreement: rejected.
Repeat for every word, start, period and table rule, then for random strings.
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.
sculpture's alphabet, Vigenere, key length 60: slots fixed by the known letters: [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33]
DEGREES starting at place 80
place letter copper slot needed fixed
80 D X 19 14
81 E T 20 19
82 G J 21 3 1
83 R C 22 8 10
84 E D 23 25 21
85 E I 24 4 18
86 S G 25 7 1
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Type 8 of directions.md, 'a plaintext that is itself a riddle'. K2 spells out its coordinates (THIRTY EIGHT DEGREES FIFTY SEVEN MINUTES...) and ends LAYER TWO (01-facts.md); Sanborn 2025: the plaintext joins the 1986 Egypt trip, the 1989 Wall and 'delivering a message' (site_texts.json); issue #11 asks for credit for spelled numbers and the puzzle's words. Classical probable-word attack: for every single-alphabet periodic key the cribs allow (each of 8 tableau classes, every consistent period 27-96), each lexicon word w at every start x forces key values at slots (x + i) mod p; it is consistent if they agree with the slots fixed by the 24 known letters; placements overlapping the known stretches are skipped. The number of such agreements is its check count. Null: 40 random strings per word length on the same text and keys. A real word that is the plaintext would show up as a placement with many checks, or as an excess of 3+-check placements over the decoys.
6
Prediction
Written before the run; the result decides it.
Lexicon words reach consistent placements with three or more checks at the same rate as random strings of the same length (within the decoys' spread), and no placement has five or more checks. Zero candidates.
7
Method
For each class and period 27-96, test crib consistency and keep the consistent pairs with their fixed key slots.
For each lexicon word and start clear of the known stretches, count agreements with fixed slots; reject on any disagreement.
Do the same for 40 random strings per length.
Compare counts of placements with 2, 3 and 4+ checks between words and decoys, per word length.
List the real-word placements with 3 or more checks.
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
93 words x 376 consistent (class, period) keys, every start. Most checks: real words 2, decoys 4. Words above every decoy of their length at 3+ checks: []; at 4+: [].
What was searched, as the harness recorded it.
family
probable-word drag under every crib-consistent single-alphabet periodic key