“F1.6, the revived cell: K1 and K2's own system (one free alphabet, added form) at key length 19 with the carved S at place 33 taken as a slip; design 2's power on planted texts and K4 against its copies”
F1.6. 0147 found that replacing the carved S at 33 (S under S) revives K1 and K2's system with a free alphabet at key length 19, for 25 of the possible replacements, the one cell where K4 stood above every scrambled copy in that model (25 against at most 23; inside the multiple-comparison null overall). The exact test cannot judge it further. Here place 33's carved letter is treated as unknown (its equation dropped, the message letter S kept), and 0139's English decider, rewritten for the added form, judges the cell: power on 5 planted texts against copies of themselves, and K4 against 12 copies of itself, in one run.
K1the earlier messages The three solved parts of the sculpture. and K2 were made with the tablethe table The right-hand half of the sculpture, 26 rows of the alphabet each shifted one place further, starting from the word KRYPTOS. and a scrambled alphabet; the same method on K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. is ruled outdead Said of a period, grid or method when the known letters contradict it: no secret word of that kind can produce the carved letters. only because two carved letters equal the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. letters above them. If the S at place 33 is a slip of the chisel, the method comes back to life at a key length of 19. We let the search that tells hidden Englisha planted message A real English passage, with the 24 known letters written in at their places, enciphered by the method under test with a random key. from scrambled letters judge that one case, after checking on planted messages that it can.
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
0147 found one slip of the chisel, the S at 33, would bring K1 and K2's own method back at key length 19. That case is judged by English, after checking the search on planted messages with the same slip.
the check
Five planted texts with a slip at 33, each against four copies keeping the known places; K4 against twelve. One planted text came back whole (97 of 97); the others got 32-38 letters right.
the bound
Only 3 of 5 planted texts beat all their copies (bar 4): the search is too weak here, so the cell stays blocked whatever K4 does (seventh of 13).
2
Best candidate against the known letters
Laid out as the copper carries K4: OBKR ends the line that finishes K3, then three lines of 31. Left, as carved, the 24 known places in accent. Right, the candidate: a known place it gets right is an accent cell, one it misses is outlined. This is the candidate with most known letters right; none passed every gate.
in: the copper, always the same
as carved, 97 letters; the 24 known places in accent
ECDMRIPFEIMEHNLSSTTRTVDOHW?OBKR
UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO
TWTQSJQSSEKZZWATJKLUDIAWINFBNYP
VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
OBKR ends the line that finishes K3, then three lines of 31
out: the candidate
every place read; known places in accent, misses outlined
ECDMRIPFEIMEHNLSSTTRTVDOHW?YADK
ARZAMESURECOINGOREASTNORTHEASTE
TENAITEPETJFBGLIONOUCHERYTHIBER
LINCLOCKEELHURMOFTSSAYMCHTELZEP
checked: the known letters in place, and the reading turns back into the copper
3
How the score is made
The loss of the best candidate, part by part. Zero is solved.
Known letters missed
0 of 24, ten points each
0.00
English
the 73 free letters score -5.43; English text scores -3.6
1.83
Key
186 bits to write it down (key words are cheap, arbitrary letters 4.7 bits each); each 60 beyond 60 costs a point
2.10
Loss
of 240, the worst possible
3.93 1.6%
4
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Write K4 in rows of 19; each column is one key slot.
Leave place 33 out: its carved S is taken as a slip.
For every other known letter: place(carved) = place(known) + key value of its column, in one unknown alphabet.
Search alphabets that keep these equations, sixteen searches side by side, scoring the whole text as English, with S at 33.
Do the same on copies of K4 that keep the known places' carved letters and shuffle the rest, and on planted English.
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.
key length 23, Beaufort form: K4 in rows of 23; each column is one key slot
O B K R U O X O G H U L B S O L I F B B W F L
R V Q Q P R N G K S S O T W T Q S J Q S S E K
Z Z W A T J K L U D I A W I N F B N Y P V T T
M Z F P K W G D K Z X T J C D I G K U H U A U
E K C A R
the seven checks every move keeps (place(carved) + place(known) equal within a column):
column 0: R+S (24) = M+C (70)
column 1: V+T (25) = Z+L (71)
column 2: Q+N (26) = F+O (72)
column 3: Q+O (27) = P+C (73)
column 4: P+R (28) = K+K (74)
column 21: F+E (22) = T+I (68)
column 22: L+A (23) = T+N (69)
best alphabet on K4: OVLIGSMJAZBFDXNCYPQRTWUHEK
key values: [10, 2, 13, 20, 14, 7, 4, 18, 24, 25, 17, 6, 17, 7, 11, 2, 0, 23, 6]
place 22: carved F at 11, key 13: 13 - 11 = 2 -> L
place 64: carved N at 14, key 4: 4 - 14 = 16 -> Y
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
C17(b), family G (a slip of the chisel, as K2's dropped letter and UNDERGRUUND), C27(c), C1(b): K1 and K2 are Quagmire III on the KRYPTOS alphabet (knowledge/01-facts.md); with a free alphabet the added form x(c) = x(p) + k[i mod L] dies at every length 1-26 only because S under S (33) and K under K (74) force key 0 in shared slots (0119). Model here: same, L 19, place 33's carved letter unknown (23 equations; its message letter is S). Search: 0139's compiled parallel tempering, check-keeping moves, added form, 10 runs x 16 replicas x 625 rounds x 1,000 moves per text. Power: 5 planted texts (Carter windows, random alphabet and key, added form, the carved letter at 33 then replaced by a random other letter; seed 148), each against 4 kept-place copies (carved letters kept at the 24 known places, other 73 shuffled; seed 1481); bar fixed before the run: at least 4 of 5 beat all their copies (luck 0.2 each). K4 against 12 kept-place copies (seed 1482); K4 above all 12 stops the queue for a replication; otherwise, if the bar is met, the cell closes sampled at that power, and if not, it stays blocked.
7
Prediction
Written before the run; the result decides it.
The decider has power here (at least 4 of 5 planted texts beat their copies: 23 equations, 4 checks); K4 sits inside its 12 copies, so the cell closes. No reading.
8
Method
Compile the added-form search; check its decryption against Python.
5 planted texts with a slip at 33, 4 kept-place copies each; K4 and 12 kept-place copies.
The search on all 38 texts, place 33 left out of the equations and read as S.
Plants beating all their copies; K4's rank among its copies.
9
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
1candidates returned
0passed every gate
24/24known letters, best
1.6%lowest loss
nonechampion before
All 1 candidates, lowest loss first.
1failed at baseline24/24 known lettersloss 1.6%
ECDMRIPFEIMEHNLSSTTRTVDOHW?YADK
ARZAMESURECOINGOREASTNORTHEASTE
TENAITEPETJFBGLIONOUCHERYTHIBER
LINCLOCKEELHURMOFTSSAYMCHTELZEP
added form, one alphabet OVLIGSMJAZBFDXNCYPQRTWUHEK, key length 19, carved letter at 33 read as a slip
power 3/5 (bar 4, null 1.00); K4 -5.111 rank 7 of 13 (-5.309 to -5.009); stop False
What was searched, as the harness recorded it.
family
one free alphabet, added form (K1/K2's system), key length 19, carved letter at place 33 taken as a slip (equation dropped, S kept)
search
parallel tempering (0139) in the added form, 10 runs x 16 replicas x 625 rounds x 1000 moves per text