Experiment 128

“The hand schedules 0126 never reached: stutter, back and forth, triangle and square steps, restart at each carved line, and the tableau rows as keys, tested exactly against 30 coppers”

F1.4, completion. 0126 ran out of its time budget inside the skip schedule, so schedules C (stutter), D (back and forth), E (triangle, square) and F (restart per carved line) were never tested (their zeros in 0126 mean untested), and the tableau rows as fixed keys (A) were compared with only 3 coppers. Here all 1,325 distinct partitions of C-F on K4 and 30 shuffles (seed 128), the four exact tests of 0126/0127, coverage recorded; and A against all 30.

revertedstatus
experiment 127parents
3 mintime, all phases
0candidates
0/24known letters
unclocked
1

The experiment method

What the method does with the 24 known letters.

The rest of the list of hand rules for using a key wordthe secret word The word or string a cipher needs besides the method. (use each letter twice or more, go back and forth through the word, jump by growing steps, start again at each carved line), and the rows of the sculpture's big table as keys, each checked exactly against 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. and against thirty scrambled copies.

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 ACCENTthe rest of the list, every grouping tested exactly on K4 and 30 scrambled coppers: settings that fit with two free alphabetshollow dots: the lowest and highest of the 30 coppersC stutter (207)040the copper: 0E triangle (350)061the copper: 42E square (229)046the copper: 22F line restart (30)018the copper: 17D back and forth (509)0213the copper: 130lowest and highest of 30 coppersthe copperon the standard or KRYPTOS tables no setting of any rule fits K4; the tableau rows as keys reproduce 6 of 24 known letters, coppers 5-8OUT: NO READING; EVERY HAND SCHEDULE ON THE LIST IS DEAD ON THE SCULPTURE'S TABLESEASTNORTHEASTBERLINCLOCK
the idea

0126 never reached these rules: use each key letter r times, go back and forth through the word, step by triangle or square numbers, restart at each carved line; and the tableau rows as keys.

the check

Every distinct grouping of the known places each rule makes, tested exactly on the sculpture's tables and on alphabets in any order, against 30 scrambled copies.

the bound

Dead on the standard and KRYPTOS tables for every rule. On free alphabets K4 fits about as often as the coppers in every cell (one cell lower). The tableau rows as keys match no better than chance.

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 6-letter key word and the rule 'use each key letter twice': key letter numbers 0 0 1 1 2 2 3 3 4 4 5 5, then again.
  2. Group the known places by key letter number.
  3. On the sculpture's table, every group must share one key value (carved place minus known place in KRYPTOS order).
  4. With unknown tables, ask whether any alphabet or pair of alphabets makes every group agree.
  5. Repeat for every rule in the list and on thirty scrambled copies of K4.
the worksheet: the first steps by hand, on squared paper
THE KEY LETTER NUMBER EACH RULE GIVES TO PLACES 1-12 (KEY WORD OF 6 LETTERS, PHASE 0)ruleplaces 1 to 12C use each letter twice0 0 1 1 2 2 3 3 4 4 5 5D forward then back0 1 2 3 4 5 4 3 2 1 0 1E triangle numbers0 1 3 0 4 3 3 4 0 3 1 0E square numbers0 1 4 3 4 1 0 1 4 3 4 1F restart each carved line3 4 5 0 0 1 2 3 4 5 0 1Each rule is tested the same way: known places with the same number must share a key value. K4 starts atcolumn 28 of its carved line, so the line rule restarts after place 4.
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.

C20(c)-(e), C5, C3(a). Models as written in 0126: C f = floor((i + phi)/r), r 2-6; D keyword forward then backward, ends repeated or not, n 2-26, every phase; E f = T(i + phi) or (i + phi)^2 mod n, phi 0-96; F f = column on the carved line minus 1 (K4 fills columns 28-31 of line 11, then lines 12-14), from the line start or from K4's first letter, n 2-31. Key value at place i = W[f(i) mod n], W unknown. Exact tests per distinct partition of the 24 known places: 8 standard/KRYPTOS tables; one free alphabet V and B; two free alphabets (CRT decider, node limit 100,000, undecided counted). A: 26 carved tableau rows with and without the index letter as repeating keys from every start, 16 table settings, known letters reproduced.

6

Prediction

Written before the run; the result decides it.

No partition of C-F is consistent on the standard or KRYPTOS tables with K4; on free alphabets K4's counts lie inside the 30 coppers' range in all 16 cells, give or take one; the tableau rows reproduce at most 7 of 24, as on the coppers. A cell with K4 above every copper is replicated first.

7

Method

  1. Distinct partitions of schedules C-F.
  2. For K4 and 30 shuffles, the four exact tests on each.
  3. Counts per schedule and test, K4's rank.
  4. Tableau rows as keys, best known letters reproduced, K4 and every copper.
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

Outside the copper range: ['D back and forth | free V']. Tables consistent on K4: {'C stutter | tables': 0, 'E triangle | tables': 0, 'E square | tables': 0, 'F line restart | tables': 0, 'D back and forth | tables': 0}. Tableau rows: K4 6/24, coppers 5-8.

What was searched, as the harness recorded it.

family
hand key schedules C-F (stutter, back and forth, triangle, square, restart per carved line), unknown keyword, 4 exact tests; tableau rows as fixed keys
coverage
C stutter
partitions
207
k4 done
207
coppers min done
207
E triangle
partitions
350
k4 done
350
coppers min done
350
E square
partitions
229
k4 done
229
coppers min done
229
F line restart
partitions
30
k4 done
30
coppers min done
30
D back and forth
partitions
509
k4 done
509
coppers min done
509
undecided
C stutter
k4
0
coppers total
0
E triangle
k4
0
coppers total
8
E square
k4
1
coppers total
2
F line restart
k4
0
coppers total
0
D back and forth
k4
1
coppers total
16
summary
C stutter | tables
k4
0
coppers min
0
coppers max
0
k4 rank
1
C stutter | free V
k4
0
coppers min
0
coppers max
12
k4 rank
14
C stutter | free B
k4
0
coppers min
0
coppers max
6
k4 rank
12
C stutter | two free
k4
0
coppers min
0
coppers max
40
k4 rank
28
E triangle | tables
k4
0
coppers min
0
coppers max
0
k4 rank
1
E triangle | free V
k4
0
coppers min
0
coppers max
48
k4 rank
27
E triangle | free B
k4
7
coppers min
1
coppers max
18
k4 rank
10
E triangle | two free
k4
42
coppers min
15
coppers max
61
k4 rank
13
E square | tables
k4
0
coppers min
0
coppers max
0
k4 rank
1
E square | free V
k4
0
coppers min
0
coppers max
25
k4 rank
23
E square | free B
k4
6
coppers min
0
coppers max
15
k4 rank
6
E square | two free
k4
22
coppers min
9
coppers max
46
k4 rank
18
F line restart | tables
k4
0
coppers min
0
coppers max
1
k4 rank
2
F line restart | free V
k4
0
coppers min
0
coppers max
17
k4 rank
23
F line restart | free B
k4
7
coppers min
0
coppers max
11
k4 rank
10
F line restart | two free
k4
17
coppers min
0
coppers max
18
k4 rank
4
D back and forth | tables
k4
0
coppers min
0
coppers max
1
k4 rank
11
D back and forth | free V
k4
1
coppers min
2
coppers max
161
k4 rank
31
D back and forth | free B
k4
41
coppers min
13
coppers max
94
k4 rank
24
D back and forth | two free
k4
130
coppers min
91
coppers max
213
k4 rank
23
outside copper range
D back and forth | free V
tableau rows as keys
keys
52
k4 best hits
6
coppers best hits
5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8
controls
30 shuffled coppers (seed 128), 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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