“F6.3b: every mixed triple of matrix readings with a width of 25 or more, enumerated exactly, then or before a repeating key on the table”
F6.3b. 0134 enumerated every triple of the four column readings at widths 2-24 and every pair at 2-96; triples with any width from 25 to 96 rested on 0037's sample of 200,000. Here they are enumerated exactly: 63 reading triples (all-clockwise is 0083) x every width triple in 2-96 with at least one width of 25 or more, forward and inverse, deduplicated within each first stage. The count (about 95 million distinct routes on 11 texts) does not fit Python in 30 minutes, so the inner loop is compiled C, written out and compiled by this file and checked inside the run against a Python reference on random routes.
Sanborn described writing the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. in rows and copying it out down the columns, sometimes two or three times, before using 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.. We already tried every way of doing that three times with rows up to 24 letters wide. Here we try every three-step combination where at least one of the grids is wider, 25 to 96 letters, which gives about 95 million different scrambles. After each one, a repeating secret wordthe secret word The word or string a cipher needs besides the method. of 1 to 26 letters on the sculpture's tables either has to fit 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. or it does not, and we compare how often that happens with 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
Sanborn's double and triple matrixes: write in rows, copy out down the columns, up to three times, then use the table. Every triple with all widths up to 24 was done (0134); here every triple with a grid 25 to 96 wide.
the check
105,548,253 distinct routes, each with a repeating key of 1-26 on the sculpture's tables after or before, against the 24 known letters and ten scrambled copies, in compiled code checked against Python.
the bound
Shortest key that fits K4: 18 after the stages, 24 before; copies 19 and 17. K4 above every copy in 3 of 52 cells, a copy 0-3: 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.
Write the 97 letters in rows of 30: three full rows and a row of 7.
Read the columns left first, top to bottom (Sanborn's matrix).
Write the result in rows of 9 and read the columns right first, bottom up.
Write that in rows of 41 and read the columns left first, bottom up (a quarter turn).
Note where each of the 24 known letters has landed and the carved letter there.
With a repeating key of length p on the table, two known letters whose landing places differ by a multiple of p use the same key letter, so they must force the same one.
Do this for every reading and width of the three grids, the key before or after, 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.
three matrix stages: rows of 30 read down the columns, left first; rows of 9 read up, right first; rows of 41 read up, left first
place letter after 1 after 2 after 3 copper there
22 E 71 13 39 Q
23 A 74 78 88 U
24 S 77 45 11 U
64 B 15 41 97 R
65 E 19 95 37 W
74 K 49 59 50 A
with a key of length p on the table, known letters whose final places differ by a multiple of p share a key letter
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: 'you write a line of plain text out, right, in successive rows. And then basically you write it going from top to bottom... sometimes I used double matrixes and triple matrixes... and then went back to this table'), C2(a), C11. Stage: write the 97 symbols in rows of w (last row short) and read the columns one of four ways: cw (left column first, bottom up: the K3 turn), ccw (right first, top down), col (left first, top down: Sanborn's description), colr (right first, bottom up). Three stages, widths w1, w2, w3 in 2-96, at least one >= 25 (0134 did all <= 24), 63 reading triples; the composed permutation and its inverse. Key: periodic tableau 1-26 after (order A, slot = landing place mod p) or before (order B, slot = message place mod p), 8 classes (plain and cipher alphabets standard or KRYPTOS, Beaufort or not); a period is live if no slot holds two different forced key letters. Exact. Ten scrambled coppers (seed 133, as 0134); statistic fixed before the run: cells (order, period) where K4's live count is above every copper's, against the same count for each copper against the rest.
6
Prediction
Written before the run; the result decides it.
As in 0134 and 0083: no key shorter than 20 fits K4 after any route in either order beyond what the coppers show, and K4 is above every copper in no more cells than the most any copper manages against the rest. A K4 count outside that null stops the queue for a replication. Zero candidates.
7
Method
Build the where-map of one stage for every width 2-96 and reading.
Compile the C counter; check it against a Python reference on 25 random routes (every order, period, class and text).
For each first stage (reading, width), compose every second and third stage with at least one width >= 25, skip all-clockwise, forward and inverse; skip permutations already seen within that first stage.
For each route, class and text: the forced key letter at each known letter's slot; period p live if no slot holds two different letters; count per (order, period, text); list K4's settings live below 20.
Compare K4 with the ten coppers cell by cell.
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
105,548,253 distinct routes (380 of 380 first stages) x 8 classes x 2 orders, periods 1-26. Shortest live key on K4: A 18, B 24; coppers A 19, B 17. K4 settings live below 20: A 5, B 0. K4 above every copper in 3 of 52 cells; null 0-3. Stop queue: False.
What was searched, as the harness recorded it.
family
three matrix stages (rows of w, columns read in one of four ways), at least one width 25-96, then or before a periodic tableau key
readings
cw
left column first, bottom up
ccw
right first, top down
col
left first, top down
colr
right first, bottom up
triples
63 reading triples (all-cw skipped, 0083) x width triples in 2-96 with max >= 25 (all <= 24 was 0134), forward and inverse