“Every move of the type-1 library, then a table built on two alphabets in unknown orders (Quagmire IV), key lengths 1-26, decided exactly”
F6.2 (two free alphabets after or before each move), first exact test, order A (move, then the table); order B follows as 0132. 0097 tested two unknown alphabets only by a relaxation at key lengths 3, 7 and 21; 0129 did one alphabet exactly. Here 0123's exact two-alphabet decider runs on every one of the 27,852 moves at key lengths 1-26. Controls: 10 scrambled coppers on a fixed sample of 700 moves, K4 counted on the same sample.
The same question as the last experiment, with one more unknown: 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. may use one scrambled alphabet to look up the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. letter and a second, different scrambled alphabet to write the carved letter. For every one of about 28,000 ways of moving the letters first and every key length up to 26, arithmetic on 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. decides whether any such pair of alphabets and any key could work, and the same is done 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..
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
A table with two scrambled alphabets, one to look up the message letter and one to write the carved letter, after every move of the type-1 library: the two-alphabet form of 0129.
the check
The same exact algebra as 0123, run at every key length 1-26. At long keys almost every setting fits and each fit takes a long search, so the run spent its budget there.
the bound
None: 10,625 moves done on K4 only, no scrambled copies finished. On those, fits from key length 3. Rerun as 0132 and 0133.
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.
Pick a move, say a clockwise quarter turn of a grid 7 wide, and write where each known letter lands in the carved text.
Pick a key length, say 9: each known letter's slot is its landing place mod 9.
Write one line per known letter: place of the carved letter in the cipher alphabet = place of the known letter in the plain alphabet + key value of the slot.
If one slot has the same known letter under two carved letters, or two known letters under one carved letter, stop: dead.
Otherwise solve the lines mod 13 and mod 2; if two letters of one alphabet are always forced onto one place, dead.
Else look for two alphabets and a key that satisfy every line; if found, check it: the setting fits.
Repeat for every move and key length 1-26, and on scrambled copies.
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.
move: turn cw 7 (write the message in rows of 7, read the columns bottom to top); then the table, key length 9
place known lands at carved slot (landing place mod 9)
22 E 11 U 1
23 A 25 V 6
24 S 39 Q 2
25 T 53 K 7
26 N 67 V 3
27 O 81 T 8
28 R 94 K 3
29 T 10 H 0
30 H 24 R 5
31 E 38 T 1
32 A 52 J 6
33 S 66 P 2
34 T 80 X 7
64 B 5 U 4
65 E 19 B 0
66 R 33 S 5
67 L 47 Z 1
68 I 61 N 6
69 N 75 W 2
70 C 88 U 6
71 L 4 R 3
72 O 18 F 8
73 C 32 K 4
74 K 46 K 0
a slot sends one known letter to two carved letters (or two to one): dead
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
C1(a) (Sanborn, CNN 2005, knowledge/sources/cnn-2005-sanborn-interview-transcript.txt: 'I used a matrix code... and then went back to this table to do the ultimate'), C27(c) (a table 'slightly different from the one on Kryptos'), C2(a). Model: message p moved by a permutation pi of the type-1 library (as 0097 and 0129: K3's rule at all 9,312 settings, quarter turns, two cw turns, plain and keyed columnar, forward and inverse; 27,852 maps), then Quagmire IV: C(c_j) = P(m_j) + k[j mod L], P and C independent unknown bijections, slot = carved place j (order A). The Beaufort form C(c) = k - P(p) is the same family with P reflected, and a key alphabet drops out, so one form covers all. 24 equations over Z26, exact decision as 0123 (mod 13 and mod 2, each alphabet one-to-one; enumerate at rank <= 5, else depth-first search, 200,000 nodes, 'undecided' reported); every fit re-checked. A per-slot one-to-one filter runs first; it is implied by the model. Zekios (2026, research note combined_methods_and_sevens.md) did any-alphabet keys only to period 12 and only for double rotations.
6
Prediction
Written before the run; the result decides it.
With two free alphabets more settings fit than with one (0129), from key length 6 or 7 up, at the coppers' rate: K4 is above all 10 coppers in no more cells (key lengths 1-26) than the most any single copper manages against the rest. Key lengths 1-4 have no fit. If K4 beats that null count the queue stops for a replication. Zero candidates; fitting settings are blocked (only an English climb could judge them, powerless per 0120).
7
Method
Build the 27,852 moves of the type-1 library.
For each move and key length 1-26, write the 24 equations (slot = landing place mod L, known letter, carved letter there).
Drop settings that fail the per-slot test; decide the rest exactly (dead, fit with a re-checked witness, or undecided).
The same on 10 scrambled coppers over a fixed random sample of 700 moves.
Per key length: K4's fits on the sample against the coppers'; count lengths where K4 is above every copper and compare with the same count for each copper against the rest.
Validate on planted Quagmire IV texts (random move, length, alphabets, key must decide 'fit').
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
K4 on 10,625 of 27,852 moves; smallest key length with a fit {'AIV': 3}; fits at L<=8: 1827. Sample of 0 moves: K4 above every copper in 0 of 26 cells, coppers against the rest 0-0; cells where only K4 fits: []. Stop queue: False.
What was searched, as the harness recorded it.
family
two unknown alphabets (Quagmire IV, Beaufort included by reflection), periodic key 1-26, after a type-1 move (order A)
moves
27,852
k4 moves done
10,625
cells
order A (move, then table) x key lengths 1-26
exhaustive
exact decision per setting (0123's two-alphabet decider; undecided counted where the node limit was hit)