Experiment 127

“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.

revertedstatus
experiment 126parents
9 mintime, all phases
0candidates
0/24known letters
unclocked
1

The experiment method

What the method does with the 24 known letters.

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.

OBKRUOXOGHULBSOLIFBBWEASTNORTHEASTOTWTQSJQSSEKZZWATJKLUDIAWINFBBERLINCLOCKWGDKZXTJCDIGKUHUAUEKCARIN: THE COPPER, ALL 97 LETTERS, THE 24 KNOWN PLACES IN ACCENTskip rule, all 3,729 distinct groupings of the known places, on K4 and 30 fresh scrambled coppers: how many fit?settings that fit; the 0126 excess (461 vs 305-449 on 10 coppers) was the small control settwo free alphabets01,300the copper: 1,135one free alphabet, Beaufort form0698the copper: 373one free alphabet, added form01,068the copper: 12030 fresh scrambled coppersthe copperK4 ranks 6 of 31 with two free alphabets; on the standard or KRYPTOS tables no setting fits K4OUT: NO READING; THE SKIP RULE'S EXCESS DOES NOT REPLICATEEASTNORTHEASTBERLINCLOCK
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.

  1. Take a key word of 7 letters and the rule 'skip one key letter every 5 letters'.
  2. 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.
  3. Group the known places by that number.
  4. Ask whether two unknown alphabets and the word can make every group agree.
  5. 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
SKIP RULE AGAINST 30 FRESH SCRAMBLED COPPERS: SETTINGS THAT FIT, BY TESTtestK4coppersK4's rank of 31tables00-12free V12052-106826free B373162-69815two free1135681-13006two free with 28,66 together1430-1231Only the row that counts settings with 28 and 66 together puts K4 first: that is the R-under-P coincidence byconstruction, not a new signal.
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

  1. Build the 3,729 distinct partitions of the skip schedule.
  2. For K4 and 30 shuffles, test each with the 8 tables, one free alphabet (V, B) and two free alphabets.
  3. Count consistent partitions per test; K4's rank.
  4. 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
partitions
3,729
partitions with 28 66 together
635
coverage
k4
3,729
coppers min
3,729
undecided
k4
0
coppers total
70
summary
tables
k4
0
coppers min
0
coppers max
1
k4 rank
2
coppers
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
free V
k4
120
coppers min
52
coppers max
1,068
k4 rank
26
coppers
52, 53, 65, 87, 112, 138, 170, 191, 204, 211, 215, 229, 235, 260, 279, 307, 613, 625, 664, 736, 776, 832, 861, 872, 881, 918, 949, 960, 1,004, 1,068
free B
k4
373
coppers min
162
coppers max
698
k4 rank
15
coppers
162, 241, 247, 248, 271, 287, 289, 297, 300, 304, 318, 320, 338, 357, 368, 373, 390, 396, 420, 432, 500, 525, 525, 549, 559, 581, 602, 625, 631, 698
two free
k4
1,135
coppers min
681
coppers max
1,300
k4 rank
6
coppers
681, 722, 742, 754, 791, 806, 810, 838, 863, 866, 895, 900, 925, 990, 1,003, 1,005, 1,025, 1,028, 1,030, 1,031, 1,033, 1,049, 1,076, 1,114, 1,124, 1,196, 1,208, 1,227, 1,252, 1,300
two free with 28,66 together
k4
143
coppers min
0
coppers max
123
k4 rank
1
coppers
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 111, 123
controls
30 fresh shuffled coppers (seed 127), identical tests

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 2026-10-18 12:26 UTC; updated after every experiment.
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