“A1.2, step 2: one alphabet in an unknown order on both sides of the table (Quagmire III), a repeating key of length 1-26, with up to three of the 24 known letters carved wrong, decided exactly, against scrambled copies, with planted slipped messages”
A1.2 (Stage 2, step 2, issue #76), second half: one free alphabet. 0119 decided every key length 1-26 exactly with all 24 known letters required (added form dead everywhere, Beaufort alive only at 23 and 26); 0147 (F1.6) allowed one carved letter wrong. Here 2 and 3 of the 24 equations may be dropped as slips: 0119's exact decider, unchanged, on every drop set of 1-3 known places (2,324 per length and form, supersets of a fitting set skipped as not minimal), K4 against 20 scrambled copies, with four planted slipped messages in the same run.
K1the earlier messages The three solved parts of the sculpture. and K2 were made on a table built from one scrambled alphabet. Experiment 119 asked whether any scrambled alphabet at all, with any repeating key wordthe secret word The word or string a cipher needs besides the method., could turn the carving into 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., and found almost none. But if Sanborn made two or three slipsa slip A mistake made while enciphering by hand: a wrong row or column of the table at one letter, so the carved letter is not the one the method gives there. at those places, that test would have thrown the right answer away. Here we let two or three of the 24 disagree, every possible choice of which ones, and ask the same question again, on the carving and on scrambled copiesa shuffled control The same experiment run on the carved letters shuffled into a random order. of it, with planted messages to show the search can find a slipped one.
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
Experiment 119 found that almost no scrambled alphabet fits the 24 known letters with a repeating key. A hand slip at two or three of those places would have hidden the right one.
the check
Every choice of up to 3 slipped places (2,325 per cell), every key length 1-26, both table forms, decided exactly with 119's search, unchanged; K4 against 20 scrambled copies, four planted slipped messages.
the bound
No alphabet fits K4 at lengths 1-7 or 9 even with 3 slips; at 8 and from 10 up K4 fits as the copies do (K4 ahead of every copy in 3 of 52 cells, copies 0-4). Plants at length 7 are found; at 12 they fit without their slips, as their copies do: the reach ends near length 9.
2
Experiment lineage
What this experiment was built on, from the lineage page: 8 experiments before it, back to the root of its line. Its number is in bold.
keptreverted, in its family's colour×error or timeoutreverted under today's rulesbuilt on a parent in the same familybuilt on a parent in another family
3
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 under Result and bounds below.
4
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Choose a key length, say 7, and write the 24 known places with their slot (place mod 7).
For each place write: position of the carved letter in the alphabet = position of the known letter + key value of the slot.
Cross out two of the 24 places, as if Sanborn had slipped there.
Try to place the letters in an alphabet and choose key values so every remaining line holds, backing up when two letters need the same position.
If it works, the length fits with those two slips; if every branch backs up, it does not.
Repeat for every pair and triple of crossed-out places, every length, both table forms, and on scrambled copies.
the worksheet: the first steps by hand, on squared paper
5
Worked example, generated
The same steps run by the code on the real letters.
key length 7, form V: A(carved) = A(known) + k[slot]
slips at none: dead
slips at [66]: dead
slips at [66, 73]: dead
(66 R and 73 C both sit under carved P in slot 2: no alphabet fits without a slip at one of them, and at length 7 none fits even with one.)
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
A1 (knowledge/12-assumptions.md); C27(c) (Sanborn's own tableau 'slightly different from the one on Kryptos', NYT 2010, 02-sanborn-said.md) and C27(a) (the PALIMPCEST worksheet: a slip in his hand); Sanborn 2005, 'Saying I was sloppy is a good way of saying it'; Scheidt 2015, an encoding error by Jim would not have been known to him (03-scheidt-said.md). Model as 0119: one unknown bijection A from letters to Z26 on plain and cipher sides; key k of length L; form V: A(c_i) = A(p_i) + k[i mod L]; form B: A(c_i) = k[i mod L] - A(p_i). A slip at known place q removes q's equation (the carved letter is not what the key gives). For every set D of 1-3 known places, decide exactly (0119's decide(): mod 13 and mod 2, enumeration or pruned search, witness re-checked) whether an alphabet and key satisfy the other 24 - |D| equations. Count the minimal fitting sets per (L, form) by size; a cell is closed at k when K4 has no fitting set of size k or less. Where K4 fits, the reading is not determined (the alphabet is free), so the judge cannot read it; such a cell can be decided only by F1.5's decider (20-30 min per cell) and is left blocked by cost unless K4's count stands outside the copies'. Cost of naming the slips: 4.7 k + log2 C(24, k) bits (17.5 for k = 2, 25.1 for k = 3).
7
Prediction
Written before the run; the result decides it.
Short key lengths (about 1-6) stay dead on K4 even with 3 slips; from there up K4 fits at some lengths with 2 or 3 slips, at the same rate as the copies, cell by cell inside the copies' range (K4 above every copy in no more cells than a copy is above the rest). Planted messages at lengths 7 and 12 are found: the planted slips fit, and the plant's count stands above all of its own copies at k = 2; at k = 3 the copies may fit as well (the reach). Zero candidates: a free alphabet leaves the reading undetermined.
8
Method
For each text, key length L 1-26 and form: the 24 equations (slot, known letter, carved letter).
Decide the system with no drop, then every drop set of 1, 2, 3 known places in order of size, skipping supersets of a set already fitting.
Count the minimal fitting sets by size; record undecided ones.
K4 and 20 copies on every cell; four planted messages and 10 copies of each on their own cell.
Compare K4's count with the copies' per cell and per k.
9
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
K4 above every copy in 3 cells ['B22', 'V25', 'V26']; copies above the rest 0-4; plants found [True, True, False, False]
What was searched, as the harness recorded it.
family
one unknown alphabet on both sides (Quagmire III), periodic key 1-26, with 0-3 known letters slipped (A1.2)
key lengths
1, 26
forms
V
A(c) = A(p) + k
B
A(c) = k - A(p)
drop sets
every set of 0-3 of the 24 known places (2,325 per cell), supersets of a fitting set skipped; 0119's decide() unchanged