“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.
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.
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.
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.
Group the known places by key letter number.
On the sculpture's table, every group must share one key value (carved place minus known place in KRYPTOS order).
With unknown tables, ask whether any alphabet or pair of alphabets makes every group agree.
Repeat for every rule in the list and on thirty scrambled copies of K4.
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.
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
Distinct partitions of schedules C-F.
For K4 and 30 shuffles, the four exact tests on each.
Counts per schedule and test, K4's rank.
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