“F1.4: every hand key schedule K4 fits on two free alphabets, each decided by D1b's four-run decider against kept-place copies one at a time, with 20 shuffled stand-ins through the same procedure”
F1.4, the two-alphabet residue, which waited for D1b. 0253-0255 decided the one-alphabet residue (K4 inside 40 stand-ins, p 0.51). D1b (0257) qualified four runs of 0144's two-alphabet tempering (66 of 96 plants against ten runs' 71). Here 0144's climb is generalised to any slot map (the change 0252 made to 0140's), every 0128 schedule that fits K4 on two free alphabets (4,706 slot maps; the 1,346 in the record were distinct partitions of the 24 known places) is decided with it against kept-place copies climbed one at a time (stop at the first at or above K4, at most 12), and 20 stand-ins (K4 with its 73 unknown places shuffled, so the same settings fit) go through the same procedure on 60 settings each: the verdict is K4's rate against theirs (0254's lesson).
Sanborn may have stepped through his key wordthe secret word The word or string a cipher needs besides the method. with a twist (skipping a letter now and then, going back and forth) and used two scrambled alphabets of his own, one for the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. and one for the cipher. About five thousand such twists agree with 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., so those letters cannot tell them apart. For each one we search for the most English-looking reading of K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. and do the same on scrambled copiesa shuffled control The same experiment run on the carved letters shuffled into a random order.; to know what 'K4 wins often' means, twenty scrambled stand-insa stand-in A scrambled copy of the carved letters put through a whole experiment exactly as K4 is, with scrambled copies of its own. for K4 go through the very same test.
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
Sanborn's hand key schedules (skip, stutter, back and forth...) on two scrambled alphabets of his own: 4,706 schedules fit the 24 known letters.
the check
Every schedule climbed four times on K4 and on copies that keep the 24 known letters; 20 shuffles of K4 through the same test.
the bound
K4 beat all 12 copies in 363 schedules (7.7%, chance 7.7%); 8 of 20 stand-ins at or above. Plants found 8/12 clean, 4/12 Sanborn-style: closed at that power.
2
Experiment lineage
What this experiment was built on, from the lineage page: 19 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.
Pick a schedule, say: a key of 11 letters, skip one key letter after every 7 carved letters.
Write the key slot under each of the 97 carved letters.
With two unknown alphabets, two known letters in one slot must satisfy C(c1) - P(p1) = C(c2) - P(p2); check that some pair of alphabets satisfies all of them.
Search alphabet pairs and keys for the most English-looking reading of K4, four times.
Shuffle the 73 letters at unknown places (the known places keep their carved letters, so step 3 still holds) and search the copy the same way; stop when a copy reads as well as K4, at most 12 copies.
Do steps 4-5 for every schedule, and for 20 shuffled stand-ins of K4; compare how often K4 wins all 12 with how often the stand-ins do.
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.
F1.4: two free alphabets under a hand key schedule, decided setting by setting.
setting skip one key letter every 7 letters, 11 key letters
slot map place 1 2 3 4 5 6 7 8 9 ...
key letter 1 2 3 4 5 6 8 9 10 ... (letter 7 skipped)
cipher C(carved) = P(plain) + key[slot], P and C two unknown alphabets
K4 gets four tempering searches under that slot map; then copies that keep the carved letters at the 24 known
places (so they fit the same checks) get the same, one at a time, until a copy reads as well as K4 (at most 12).
20 shuffled stand-ins for K4 go through the same test: K4's rate of wins must stand out from theirs.
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
F1.4: a keyed table with a hand key schedule (C20(d): tableau rows as keys; 0126-0128's schedules: a key letter skipped every m letters, stutter, back and forth, triangle, square, a restart per carved line), on two free alphabets (Quagmire IV with unknown orders: Sanborn's 'slightly different' tableau on both sides, C27(c)). The added and Beaufort forms are one on two free alphabets (reverse the cipher alphabet). 0128 found the fitting settings at the copies' rate, so the 24 known letters cannot decide them and English must. Decider: D1b's four runs of 0144's tempering (16 x 625 x 1000) from the exact witness (0128's two-alphabet decider), patched only to read the key slot of each carved letter from the schedule; self-checked against 0144's library on periodic slot maps. Copies keep the carved letters at the 24 known places (same equations, same freedom; 0143, 0257). Clue entries: C20(d), C27(c), C5(a).
7
Prediction
Written before the run; the result decides it.
K4's rate of settings above all 12 copies inside the 20 stand-ins' spread (p above 0.05); plants (24, on settings K4 fits) found above all 12 own copies about 14 of 24; no reading English. Zero candidates.
8
Method
Settings: 0252's f14_settings() (every 0128 schedule at every slot count 2-31, deduplicated); each decided on K4 by 0128's exact two-alphabet decider; the fitting ones are the cell (undecided ones counted).
Per setting (one job, own seed): K4 climbed with four runs of the generalised tempering from the exact witness; then kept-place copies (the 73 unknown places shuffled) climbed the same way one at a time, stopping at the first at or above K4, at most 12.
Stand-ins: 20 kept-place shuffles of K4 (seed 2630), each decided the same way on 60 settings drawn from the cell.
Plants: 12 clean and 12 Sanborn-style messages on settings drawn from the cell, two random alphabets and key; found if above all 12 own copies.
Verdict: K4's share of settings above all 12 copies, ranked among the stand-ins' shares.
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
F1.4 two-alphabet residue: 4706 settings; K4 above all 12 kept-place copies in 363 (7.7%); 20 stand-ins 0.0-15.0%, 8 at or above K4 (p 0.43); plants clean [8, 12], Sanborn-style [4, 12].
What was searched, as the harness recorded it.
item
F1.4 (two free alphabets residue, after D1b)
decider
D1b's four-run decider: 4 runs of 0144's tempering (16 x 625 x 1000) generalised to any slot map, from 0128's exact two-alphabet witness; copies keep the carved letters at the 24 known places and shuffle the other 73, climbed one at a time, stop at the first at or above, at most 12
self check against 0144
13, -5.262, -5.262
16, -4.892, -4.892
settings listed
12,640
decider status
dead
7,933
fit
4,706
undecided
1
k4 fitting settings
4,706
k4 above all 12
363
k4 rate
0.077
chance if independent
0.077
stand ins
n
20
settings each
60
rates
s0
0.067
s1
0.017
s2
0.133
s3
0.033
s4
0.05
s5
0.05
s6
0.05
s7
0.033
s8
0.067
s9
0
s10
0.117
s11
0.05
s12
0.017
s13
0.083
s14
0.1
s15
0.1
s16
0.15
s17
0.1
s18
0.083
s19
0.033
at or above k4
8
p
0.429
how
K4 with its 73 unknown places shuffled (seed 2631): the same 24 checks, so every setting fits; each decided as K4 on 60 settings drawn from the cell
every setting of 0128's schedules (slot counts 2-31) K4 fits on two free alphabets; each decided at D1b's four-run power
resumed units
821
cost estimate
4,706 settings (desk count 2026-10-30; the 1,346 in the record counted distinct partitions of the 24 known places, 0127/0128, not slot maps over all 97 places, which is what a climb reads) x about 16.4 runs (4 for K4 plus about 3.1 copies of 4, H_12) plus 20 x 60 stand-in settings and 24 plants: about 98,000 tempering runs at 12-15 core-s (desk timing on the loaded laptop, 12.2 s): about 330-410 core-hours; about $5.6 on Modal, above the credit left, so on the laptop: several days
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 2026-11-05 17:14 UTC; updated after every experiment.