“Two alphabets in unknown orders, one for the message and one for the carved letters (Quagmire IV), no move: which key lengths 1-26 can any pair fit, and is that unusual against scrambled coppers?”
F1.2. 0097 tested two unknown alphabets only at key lengths 3, 7 and 21, by a relaxation. Here every length 1-26 is decided exactly with 0119's method generalised to two independent alphabets (each one-to-one), on K4 and 100 scrambled coppers. Power is stated before running: at each length, the share of coppers on which the exact test fails is the test's power to reject a random text there; lengths where K4 fits are then compared with that rate.
K1the earlier messages The three solved parts of the sculpture. and K2 used one scrambled alphabet for the whole table. A variant uses two: one scrambled order to look up the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. letter and another to write the carved letter. With both orders unknown there are vastly more tables. For each key length 1 to 26 we ask, by exact arithmetic on 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., whether any pair of orders and any key could produce them, and how often the same is true of scrambled copies of the carved textthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page..
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 variant of K1 and K2's table uses two scrambled alphabets, one to look up the message letter and one to write the carved letter. Both may be any order; the known letters decide exactly which key lengths any pair can fit.
the check
Per length, the 24 equations are solved modulo 13 and 2 and searched for two one-to-one alphabets. The share of scrambled coppers that fail is the test's power at that length.
the bound
K4 fits at 8, 13, 16, 19, 20, 23, 24 and 26 and nowhere else. Every length but 19 is inside the coppers' range; 19 (3 of 100) was replicated in 0124.
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.
Choose a key length, say 8, and group the known places by slot (place mod 8).
For each known place write: place of the carved letter in the cipher alphabet = place of the known letter in the plain alphabet + key value of the slot.
Two known places in one slot give one check with the key removed: C(c1) - P(p1) = C(c2) - P(p2).
If one slot holds the same known letter over two different carved letters, those carved letters would share a place: dead.
Otherwise solve the checks modulo 13 and modulo 2 and search the solutions for one where each alphabet keeps its letters apart.
Repeat for every length 1-26 and for 100 scrambled copies of the carved text.
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.
Family, key space, the constraint under test, and the clues it reads.
C27(c) (Sanborn's 'slightly different' tableau), C1(b). Model: plain alphabet P and cipher alphabet C, unknown bijections letters -> Z26, key k of length L: C(c_i) = P(p_i) + k[i mod L] (Quagmire IV; the Beaufort form C(c) = k - P(p) is the same family with P reflected, so one form covers both; a key alphabet drops out). 24 linear equations over Z26; exact decision by CRT: solve mod 13 and mod 2, each alphabet's letters at most two per residue class mod 13 with each pair differing mod 2; refute at once on forced coincidences; otherwise enumerate the mod-13 letter space up to the two translations and scaling (rank <= 5) or search it depth first with pruning, a node limit reporting 'undecided'. The generalised decider agreed with 0119's on 780 one-alphabet cases and recovered planted Quagmire IV texts at lengths 3-26 in a dry run, which also showed K4 fitting at lengths 8, 13, 16, 19, 20, 23, 24 and 26. Power and copper comparison are what this run adds. Two alphabets have more freedom than one, so where both K4 and most coppers fit, 0120's finding (a climb on one alphabet already has weak power at 23 and none at 26) means an English climb here could not decide; such lengths are recorded blocked, not run weakly.
6
Prediction
Written before the run; the result decides it.
Power: the exact test rejects nearly all coppers (at least 90%) at lengths 1-7 and fewer than half at lengths with 7 or fewer checks (17, 18, 23-26). K4's 8 fitting lengths lie inside the middle 90% of the coppers' counts, and at every length where K4 fits at least 5% of coppers fit too; if K4 fits at a length where fewer than 5% of coppers do (length 8, 16 checks, is the candidate), that is outside the range and stops the queue for a replication. Zero candidates.
7
Method
For each key length 1-26 write the 24 equations.
Decide exactly whether a pair of one-to-one alphabets and a key satisfy them; re-check every witness.
Do the same on 100 scrambled coppers (seed 121).
Report per length the checks, K4's verdict, the share of coppers that fit (the power is one minus that share), and K4's count of fitting lengths against the coppers'.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.