“A1.2, step 3: two alphabets in unknown orders (Quagmire IV), 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), last part: two free alphabets, run because the cost allows (about 1 s per length per text). 0123 decided every key length 1-26 exactly with all 24 known letters required (K4 fits at 8, 13, 16, 19, 20, 23, 24, 26; dead at the rest). Here 1-3 of the 24 equations may be dropped as slips: 0123's exact decider, unchanged, on every drop set (2,325 per length, supersets of a fitting set skipped), K4 against 20 scrambled copies, planted slipped messages in the same run (three pre-registered at lengths 4, 5, 7; three more at 3 and 9 added after a dry run showed copies fit at 4-8, to locate the reach). 0208 found the reach of one free alphabet near length 9; two alphabets have 25 more free places, so the reach is expected lower.
The same question as experiment 208, with one more freedom: the alphabet used to look up the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. letter and the one used to write the carved letter may be two different scrambles. That is so much freedom that scrambled copiesa shuffled control The same experiment run on the carved letters shuffled into a random order. of the carving often fit already; we want to know up to which word length a search allowing 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. can still tell anything apart, and whether the carving fits anywhere the copies do not.
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
With two scrambled alphabets, one to look up the message letter and one to write the carved letter, a hand slip at the known places is easier still to hide; the question is how far a search allowing slips can still tell anything.
the check
Every choice of up to 3 slipped places, every key length 1-26, decided exactly with 123's search, unchanged; K4 against 20 scrambled copies and six planted slipped messages.
the bound
No pair fits K4 at lengths 1-3 or 9 even with 3 slips; elsewhere K4 fits as the copies do (ahead of every copy in 2 of 26 lengths, copies 0-2). Plants at 3 and 9 are found, at 4, 5 and 7 they are not told from their copies: that is the reach.
2
Experiment lineage
What this experiment was built on, from the lineage page: 9 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 4, and write the 24 known places with their slot (place mod 4).
For each place write: position of the carved letter in the cipher alphabet = position of the known letter in the plain alphabet + key value of the slot.
Cross out up to three of the 24 places, as if Sanborn had slipped there.
Try to place the letters in two alphabets and choose key values so every remaining line holds, backing up when two letters of one alphabet need the same position.
If it works, the length fits with those slips; if every branch backs up, it does not.
Repeat for every choice of crossed-out places, every length, 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 4, two free alphabets: C(carved) = P(known) + k[slot]
slips at none: dead
slips at [66]: dead
slips at [66, 73]: dead
slips at [28, 66, 73]: dead
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', 02-sanborn-said.md), C27(a) (the PALIMPCEST slip), Sanborn 2005 'Saying I was sloppy is a good way of saying it', Scheidt 2015 on Jim's encoding errors (03-scheidt-said.md). Model as 0123: two unknown bijections P, C from letters to Z26 (plain and cipher sides), key k of length L: C(c_i) = P(p_i) + k[i mod L] (the Beaufort form is the same family with P negated). A slip at known place q removes its equation. For every set D of 0-3 known places decide exactly (0123's decide(two=True), witness re-checked) whether alphabets and key satisfy the rest. Count minimal fitting sets per length by size, K4 against each copy. Where K4 fits, the reading is undetermined (two free alphabets), so the judge cannot read it; only F1.5's decider could, and two alphabets are above 0136's English bound from length 19. Cost of naming the slips: 17.5 bits for 2, 25.1 for 3.
7
Prediction
Written before the run; the result decides it.
K4 has no fitting pair of alphabets at lengths 1-4 even with 3 slips; from about 5 the copies fit as well, and K4 fits where they do, inside their range (K4 above every copy in no more cells than a copy is above the rest). A plant at length 4 with 2 slips is found; plants at 5 and 7 with 3 slips may not be told from their copies (the reach). Zero candidates.
8
Method
For each text and key length L 1-26: the 24 equations (slot, known letter, carved letter).
Decide the system with no drop, then every drop set of 1, 2, 3 known places, skipping supersets of a set already fitting.
Count the minimal fitting sets by size; record undecided ones.
K4 and 20 copies on every length; three planted messages and 10 copies of each on their own length.
Compare K4's count with the copies' per length 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 2 cells ['V18', 'V22']; copies above the rest 0-2; plants found [False, False, False, True, True, True]
What was searched, as the harness recorded it.
family
two unknown alphabets (Quagmire IV), periodic key 1-26, with 0-3 known letters slipped (A1.2)
key lengths
1, 26
forms
V
C(c) = P(p) + k (Beaufort is the same family with P negated)
drop sets
every set of 0-3 of the 24 known places (2,325 per cell), supersets of a fitting set skipped; 0123's decide(two=True) unchanged