“Two alphabets in unknown orders (Quagmire IV) after or before every move of the type-1 library: exact at key lengths 1-10 on K4 and ten coppers, the per-slot test at 11-26”
F6.2, second attempt. 0131 ran the exact two-alphabet decider at every key length and ran out of its budget after 10,625 of 27,852 moves with no copper comparison done (incomplete; item left open). Timing since: the decider costs about 4 ms per move for lengths 1-10 in both orders but 0.2 s at 11-12 and more above, where 97-99% of settings that pass the per-slot test fit anyway (0131: 7,072 of 7,152 at 26). So here: exact decision at lengths 1-10, both orders, every move, on K4 and on ten scrambled coppers; at 11-26 the per-slot test (exact as a refutation: a setting failing it is dead) on every move and text, and the full decider on K4 for 300 sampled moves to state the fit rate.
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 one to write the carved letter. For every one of about 28,000 ways of moving the letters, before or after the table, and every key length up to 10, 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 could work. For longer keys almost everything that passes a quick check fits, so the quick check is used there. 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. go through the same test.
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
The second attempt at F6.2: decide two unknown alphabets exactly only where it is cheap (key lengths 1-10) and use the per-slot test above, on K4 and ten scrambled coppers.
the check
A move counts only when K4 and every copper are done for it; none was, because a heavy tail of slow searches and a long-key sample filled the budget.
the bound
None. What it measured: at long keys 97-98% of settings passing the per-slot test fit, so the per-slot test is nearly the whole story there.
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, and an order: each known letter's slot is its landing place mod 9 (move first) or its message place mod 9 (table first).
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, both orders and key lengths 1-10, and on scrambled copies; for longer keys stop at step 4.
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: the type-1 library (27,852 maps, as 0097 and 0129) and Quagmire IV: C(c) = P(m) + k[slot], P and C independent unknown bijections, slot = carved place (order A: move, then table) or message place (order B: table, then move). The Beaufort form is the same family with P reflected; a key alphabet drops out. 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. The per-slot test (within a slot, known to carved must be one-to-one) is implied by the model. 0100 and 0102 found Quagmire IV at period 3 after a move marginal (12 distinct check sets against 0-11, p about 0.24 after selection) under 0097's relaxation; this run decides period 3 exactly on all moves and ten coppers.
6
Prediction
Written before the run; the result decides it.
K4's fits per (order, length) at 1-10, and its per-slot survivors at 11-26, lie at the coppers' rate: K4 is above all ten coppers in no more of the 52 cells than the most any single copper manages against the rest. Period 3 after a move (0131 found 6 fits on part of the library) is inside the coppers' range. On the K4 sample at 11-26, at least 90% of per-slot survivors fit. A K4 count beyond the null stops the queue. Zero candidates.
7
Method
Build the 27,852 moves.
For each move, order and key length, write the 24 equations; apply the per-slot test.
Lengths 1-10: decide the survivors exactly, on K4 and ten scrambled coppers, every move.
Lengths 11-26: count per-slot survivors on every text; on K4 for 300 sampled moves, decide them exactly.
Per cell (order, length), K4 against the coppers; cells where K4 is above every copper, against each copper against the rest.
Validate on planted Quagmire IV texts.
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
0 of 27,852 moves on K4 and 10 coppers. Smallest key length with a K4 fit: {'A': None, 'B': None}; fits at L<=5: 0. K4 above every copper in 0 of 52 cells; each copper against the rest 0-0. Long lengths on the K4 sample: {'A': {'per_slot_survivors': 440, 'fit': 430, 'undecided': 0, 'fit_share': 0.9773}, 'B': {'per_slot_survivors': 614, 'fit': 598, 'undecided': 1, 'fit_share': 0.9739}}. 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) or before (B) a type-1 move
moves
27,852
moves done all texts
0
exhaustive
exact decision at key lengths 1-10 (0123's decider); per-slot refutation at 11-26; exact at 11-26 on 64 sampled moves for K4