“Replication: the skip schedule on two unknown alphabets, where K4 had more consistent settings than ten coppers (0126), against 30 fresh coppers with every setting covered”
F1.4, replication (order-of-work rule 5). 0126 ran out of time: the skip schedule (B) was tested on part of its 3,729 distinct partitions and schedules C-F not at all (their zeros mean untested). In what was done, B with two free alphabets gave K4 461 consistent partitions against 305-449 on ten coppers. Here all 3,729 B partitions are tested on K4 and 30 fresh shuffles (seed 127), all four tests, with the search's node limit lowered to 100,000 so no case can stall, undecided cases counted, coverage recorded, texts run in parallel.
In the last run, one hand rule (skip a key letter every so often) fitted K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. with two unknown alphabets in more ways than it fitted ten scrambled copies. That run was cut short, so here the rule is tested completely, against thirty new scrambled copies, to see whether K4 really stands out.
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
In 0126 the skip rule with two unknown alphabets fitted K4 in more ways than ten scrambled copies. The rule says: replicate before anything else.
the check
All 3,729 groupings of the known places, the four exact tests, K4 and 30 new scrambled copies, nothing cut short.
the bound
K4 1,135 against 681-1,300: rank 6 of 31. The settings that put places 28 and 66 (both R under P) together are the only ones where K4 leads, as 0124 explained.
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 a key word of 7 letters and the rule 'skip one key letter every 5 letters'.
Number the key letter each place uses: 0 1 2 3 4, skip 5, 6 0 1 2 3, skip, and so on, modulo 7.
Group the known places by that number.
Ask whether two unknown alphabets and the word can make every group agree.
Do it for every skip interval, phase and word length, on K4 and on thirty scrambled copies, and compare the counts.
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.
skip schedule: key word of 7 letters, one key letter skipped every 5 letters; KRYPTOS on both sides, Vigenere
place known carved key letter no. key value
22 E F 4 1
23 A L 5 10
24 S R 6 21
25 T V 0 18
26 N Q 2 1
27 O Q 3 15
28 R P 4 2
29 T R 5 23
30 H N 6 5
31 E G 1 2
32 A K 2 19
33 S S 3 0
34 T S 4 2
64 B N 5 11
65 E Y 6 17
66 R P 1 2
67 L V 2 5
68 I T 3 15
69 N T 4 11
70 C M 5 9
71 L Z 0 8
72 O F 1 7
73 C P 2 20
74 K K 3 0
key letter numbers with clashing values: [0, 1, 2, 3, 4, 5, 6]: dead
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Follow-up of 0126 (C20, C5). Same model and exact tests: key value at place i = W[f(i) mod n], f(i) = i + floor((i + phi)/m), m 2-30, phi 0..m-1, n 2-26; distinct partitions of the 24 known places; tests: 8 standard/KRYPTOS tables, one free alphabet V and B, two free alphabets (CRT decider of 0119/0123). The partitions are heavily nested (different m, phi, n giving related slot patterns), so counts are strongly correlated and a single coincidence in the known letters can lift many at once, as R under P at 28 and 66 did in 0124 (that pair shares a slot whenever f(65) - f(27) is a multiple of n). Report K4's rank among 30 fresh coppers and, after the run, whether K4's excess partitions are those that put 28 and 66 in one slot.
6
Prediction
Written before the run; the result decides it.
K4's count for B with two free alphabets falls inside the range of 30 fresh coppers (rank no better than 3rd), so the 0126 excess was the small control set; the standard/KRYPTOS tables give 0 consistent partitions on K4. If K4 tops all 30, stop and report.
7
Method
Build the 3,729 distinct partitions of the skip schedule.
For K4 and 30 shuffles, test each with the 8 tables, one free alphabet (V, B) and two free alphabets.
Count consistent partitions per test; K4's rank.
Among K4's two-alphabet partitions, count those with 28 and 66 in one slot, and the same on the coppers.
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
B skip, two free alphabets: K4 1135, coppers 681-1300, K4 rank 6 of 31. Tables: K4 0. Coverage 3729/3729; undecided K4 0.
What was searched, as the harness recorded it.
family
skip schedule (one keyword letter skipped every m letters), unknown keyword, 4 exact tests; replication of 0126