“The two key lengths a free alphabet survives (Beaufort form, 23 and 26): climb the alphabet toward English with the 24 known letters fixed, against scrambled coppers at equal effort”
F1.1, second step (English climb). 0119 left one unknown alphabet on both sides alive only in the Beaufort form at key lengths 23 (7 checks) and 26 (1 check). Consistency cannot decide those, so here the alphabet and the free key values are climbed by simulated annealing toward English on the 73 unknown letters, with the 24 known letters held by a penalty that must reach zero. The same climb, same effort, runs on 12 scrambled coppers per length that also fit at that length (drawn by 0119's exact test), and on 6 planted English texts per length to measure power.
The last experiment showed that with one scrambled alphabet on both sides of 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., only two key lengths, 23 and 26, can produce 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. at all, and only with the table used the BeaufortVigenere The three rules of arithmetic a table can use: add the key letter to the message letter, subtract the message letter from the key letter, or subtract the key letter from the message letter. way (key minus letter). For those two lengths we now let a computer shuffle the alphabet and the keythe secret word The word or string a cipher needs besides the method. bit by bit, keeping the known letters right, and push the other 73 letters toward English. A climb like this always makes some English-looking fragments, so the same climb is run on scrambled copies of the carved textthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page.; K4 only counts if it does better than all of them. To know whether the climb could find a real answer, we also hide real English under random alphabets and see whether the climb digs it 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
0119 left one scrambled alphabet alive only in the Beaufort form at key lengths 23 and 26. Consistency cannot decide those, so the alphabet is climbed toward English with the known letters held.
the check
The same climb, same effort, on 12 scrambled coppers per length. K4 scores -5.81 at 23 (coppers -5.90 to -5.45) and -5.29 at 26 (-5.33 to -5.01): inside the range both times, low at 23.
the bound
Hidden English beat every copper in 3 of 6 trials at 23 and 1 of 6 at 26, and was never recovered. At 23 the test is weak; at 26, with one check, it cannot decide.
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?WRYZ
NNIPNMUPTOMEORJZEEASTNORTHEASTR
BYEASILETFIPLSNGESMUCHASTDCEBER
LINCLOCKHASANDWLANDWOHLUCEKSENP
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 -5.81; English text scores -3.6
2.21
Key
198 bits to write it down (key words are cheap, arbitrary letters 4.7 bits each); each 60 beyond 60 costs a point
2.31
Loss
of 240, the worst possible
4.52 1.9%
4
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Take key length 23 and write K4 in rows of 23, so each column is one key slot.
Guess an order for the alphabet (26 letters in some sequence) and number its places 0-25.
In a column holding a known letter, the key value is (place of the carved letter + place of the known letter) mod 26; every known letter in that column must give the same value.
In the six columns with no known letter, guess a key value.
Decipher every letter: known-letter place = key value minus carved-letter place, mod 26, read back through the alphabet.
Score the 73 unknown letters by how English their four-letter groups are.
Swap two letters of the alphabet or change one free key value; keep the change if the score improves (sometimes even if not, less often as the search cools); repeat many thousands of times.
Do exactly the same with scrambled copies of K4 and with hidden English, and compare the best scores.
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
columns holding two known letters (each gives one check: place(c1)+place(p1) = place(c2)+place(p2)):
column 0: S under R (24), C under M (70)
column 1: T under V (25), L under Z (71)
column 2: N under Q (26), O under F (72)
column 3: O under Q (27), C under P (73)
column 4: R under P (28), K under K (74)
column 21: E under F (22), I under T (68)
column 22: A under L (23), N under T (69)
best climbed alphabet: JCRBOQYZPXVWHSMGAUKNFTIELD
key values: [15, 5, 24, 9, 10, 23, 5, 12, 8, 0, 8, 6, 24, 17, 18, 21, 0, 22, 3, 10, 8, 17, 14]
place 22: carved F at 20, key 17: 17 - 20 = 23 -> E
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
C27(c) (a tableau 'slightly different from the one on Kryptos'), C1(b) (the table alone). Model: one unknown bijection A on both sides, key k of length L in {23, 26}, Beaufort form A(c_i) = k[i mod L] - A(p_i) (decryption p_i = A^-1(k[i mod L] - A(c_i))); the 0119 exact test shows only these two (form, length) pairs admit any alphabet for 1-26. Given A, the key at every slot holding a known letter is forced (majority value; disagreements are violations); the remaining slots (6 at 23, 3 at 26) are free. Search: simulated annealing over A (swap two letters' places) and the free key values, objective = mean log10 quadgram score of the 73 non-crib letters (the harness's english_score of free_text) minus 1.0 per violation, linear cooling from 0.05 to 0.002, 20 restarts of 200,000 moves; only zero-violation states count. These settings were chosen in dry runs on planted texts before the run (penalty 0.4 and hotter schedules did worse); in those dry runs no setting got a planted text back, and the climb could not even return to the true alphabet from two swaps away, because one swap changes letters in every slot and the forced key values with them. Controls: shuffles of K4 that fit exactly at the same length and form (0119's decider), 12 per length, same restarts and iterations. Power: 6 planted texts per length, 97-letter windows of Carter's Tomb of Tut-ankh-Amen vol. 1 (knowledge/sources) with EASTNORTHEAST and BERLINCLOCK written in at 22 and 64, enciphered under a random alphabet and key of that length; a plant counts as found when the climb's best scores above every copper at that length. A key from such a climb costs about 88 bits for the alphabet plus 4.7 per key letter.
7
Prediction
Written before the run; the result decides it.
Power: the climb finds at least 4 of 6 planted texts at length 23 (best score above every copper) and at least 3 of 6 at 26. K4: its best zero-violation score lies inside the range of the 12 coppers at both lengths, so F1.1 closes as sampled at stated power. If plants are found in fewer than half the trials at a length, that length is recorded as not decidable by this climb and the item stays open there. A K4 score above every copper stops the queue for a replication against fresh coppers.
8
Method
Draw 12 shuffles of K4 that fit at length 23 in the Beaufort form and 12 that fit at 26 (exact test from 0119).
Build 6 planted texts per length from Carter's English.
For K4, each copper and each plant, run the same annealing climb (restarts x iterations fixed in the bounds) over the alphabet and the free key values, the known letters held by a penalty.
Keep the best zero-violation English score per text.
Compare K4 with the coppers per length; report the plants found and how many of their letters the climb got right.
9
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
2candidates returned
0passed every gate
24/24known letters, best
1.8%lowest loss
nonechampion before
All 2 candidates, lowest loss first.
1failed at baseline24/24 known lettersloss 1.8%
ECDMRIPFEIMEHNLSSTTRTVDOHW?QBHE
FEULCLATLYEPHUVALEASTNORTHEASTP
OILTYNESTUCOOTEVELHOURKMRTISBER
LINCLOCKTBYINAROMTHEENEIGNORGHF
Beaufort form, one alphabet NDPEAFCIXWRMUHGYZKOBTJVSQL on both sides, key length 26
2failed at baseline24/24 known lettersloss 1.9%
ECDMRIPFEIMEHNLSSTTRTVDOHW?WRYZ
NNIPNMUPTOMEORJZEEASTNORTHEASTR
BYEASILETFIPLSNGESMUCHASTDCEBER
LINCLOCKHASANDWLANDWOHLUCEKSENP
Beaufort form, one alphabet JCRBOQYZPXVWHSMGAUKNFTIELD on both sides, key length 23
L23: K4 -5.811, coppers -5.899 to -5.446, plants found 3/6 (letters right [37, 41, 34, 34, 32, 36]); L26: K4 -5.289, coppers -5.326 to -5.008, plants found 1/6 (letters right [29, 37, 38, 38, 43, 36])
What was searched, as the harness recorded it.
family
one unknown alphabet on both sides, Beaufort form, periodic key 23 or 26, no move (the only pairs 0119 left)
search
simulated annealing, 20 restarts x 200,000 moves per text, penalty 1.0 per known-letter violation, T 0.05->0.002