“F6.22: 0114's AMSCO excess at period 23 (K4 21 live settings against 1-15 on 10 shuffles): the same exact count against 200 fresh shuffles, and K4's live settings' forced keys checked against every 23-letter window of the named texts”
F6.22 (Alexis, 2026-10-28, #100: a result that beats its copies gets a row to chase it). 0114 counted AMSCO routes keyed by every dictionary word (3-15 columns, either first cell), then a periodic tableau key on the 8 table classes: at period 23 with the move first, K4 had 21 consistent settings where its 10 shuffles had 1-15, noted and not chased. Here the same exact count at periods 22-24 for K4 and 200 fresh shuffles (vectorised over texts, 0114's routes unchanged), and every setting K4 fits at 23 has its forced key (the shift at each key slot the known letters reach) compared with every 23-letter window of 0210's named texts, both directions, numbered on the standard and KRYPTOS alphabets: a period-23 key taken from a text Sanborn had is the only reading these settings could give.
An old run found that K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page. fits one kind of two-step cipher (a grida grid Writing the message into rows of some width and reading it out in another order, column by column for instance, which shuffles the letters without changing them. shuffle called AMSCOAMSCO A grid shuffle from the American Cryptogram Association's list: the message is written across a grid in cells holding one and two letters alternately, and the columns are read out in the order a keyword gives., then the cipher table with a 23-letter key) in 21 ways where scrambled copiesa shuffled control The same experiment run on the carved letters shuffled into a random order. fit in at most 15. That count is more about the letters than about a message, so we check it against 200 copies, and for each of K4's 21 fits we ask the question that matters: is the 23-letter key it needs a passage from a text Sanborn had to hand?
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
0114 saw K4 agree with AMSCO then a 23-letter key in 21 settings where 10 shuffles reached 15. A result that beats its copies gets a row.
the check
The same exact count on 200 fresh shuffles at key lengths 22-24; every K4 setting at 23 has its forced key compared with every 23-letter passage of K1-K3, the sculpture's texts, Carter and Sanborn's letters.
the bound
At 23 K4's count is in the top 2.5% (p about 0.025, three key lengths looked at); no forced key is a named passage, though a planted one is found. Chased in F6.22b: are the forced keys English at all?
2
Experiment lineage
What this experiment was built on, from the lineage page: 7 experiments before it, back to the root of its line. Its number is in bold.
keptreverted, in its family's colour×error or timeoutreverted under today's rulesbuilt on a parent in the same familybuilt on a parent in another family
3
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 under Result and bounds below.
4
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Write the message into a grid under a keyword, alternating cells of one and two letters (AMSCO), and read the columns in the keyword's alphabetical order.
Encipher the result on the cipher table with a key of 23 letters, repeating.
Each known letter then fixes the key letter at its place mod 23; two known letters at the same key place must agree.
Count the grid keywords and tables where they all agree, for K4 and for scrambled copies.
For K4's agreeing settings, write out the key letters the known letters force and look for them as a passage of 23 letters in K1-K3, Carter's book or Sanborn's letters.
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.
F6.22: AMSCO keyed by a word, then the table with a 23-letter key.
AMSCO: write the message in rows of cells holding 1 and 2 letters alternately, read the columns in keyword order
the table: each moved letter shifted by the key letter at its place mod 23
the known letters force the key letter at their places mod 23; two known letters in one key place must agree
count the agreeing settings for K4 and 200 shuffles; for K4's, look for the forced key as a 23-letter passage.
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
F6 (C1(a): a matrix code, then the table). AMSCO (cells of one and two letters alternately) keyed by a word, then a periodic key of period 23 on the sculpture's table classes (standard or KRYPTOS plain and cipher alphabets, Beaufort or not). A count of consistent settings is a property of a text (0254, 0255: each text has its own level) and a true setting adds only one, so the count cannot single out a message; what can is the key: a period-23 key forced at the known slots that matches a window of K1-K3, the sculpture's texts, Carter or Sanborn's letters. Plant: a Carter message through a random AMSCO route and a 23-letter key from K3's plaintext; its setting must be live and its key window found.
7
Prediction
Written before the run; the result decides it.
K4's count at 23 inside the 200 shuffles' range (rank under the top 5%); no forced key matches a named-text window; the plant's setting live and its key window found. Zero candidates.
8
Method
0114's settings unchanged: every dictionary word's keyword and full-word column order (3-15 columns), first cell 1 or 2, the route and its inverse, deduplicated.
For K4, 200 shuffles (seed 2620) and one plant: per route and table class, the shifts at the 24 known places (move first: the key slot is the carved place mod p); live at p when every pair of known places in one slot agrees; counted at p = 22, 23, 24.
K4's live settings at 23: the forced shift per slot, matched against every 23-letter window of 0210's named texts (both directions), key letters numbered on the standard or KRYPTOS alphabet.
K4's count ranked among the shuffles'; matches listed.
9
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
F6.22: K4 live at 23 in 21 settings; 200 shuffles 1-31 (median 8), 4 at or above K4; forced keys matching a named-text window: 0; plant live True, key found True
What was searched, as the harness recorded it.
item
F6.22 (0114's AMSCO excess at period 23)
routes
891,360
classes
8
periods
22, 23, 24
order
A (move, then the table)
counts
22
k4
2
copies min median max
0, 1, 16
copies at or above
81
23
k4
21
copies min median max
1, 8, 31
copies at or above
4
24
k4
4,184
copies min median max
3,962, 4,197.5, 4,537
copies at or above
111
k4 live at 23
21
k4 key matches
none
n k4 key matches
0
named texts
38
plant
route
MARTINETISM, 11, 2, inverse
class
kry/std/beau
key
YBREACHINTHEUPPERLEFTHA
numbering
std
live
yes
key window found
yes
copies
200 fresh shuffles (seed 2620)
exhaustive
yes
compute seconds
64
modal cost estimate
891,360 routes x 8 classes, vectorised over 202 texts: about 1-2 core-hours (laptop)
How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.
Rendered from the record on 2026-11-05 17:14 UTC; updated after every experiment.