Experiment 169
F12.3 (C28). Stein found a peak in the average index of coincidence at m = 13 alphabets in K1 (his Part I, key length 10) and in K2's Part II (the 125 letters between the panel's first two question marks, key length 8) and asked: 'Could this be a sign of another code overlaid on top of the others... Or just a coincidence?' Here the question is answered against the right null: shuffles that keep each part's key-period structure (letters permuted only within their own key column), 10,000 per text; also plain shuffles. Statistics: the I.C. at 13 and the local peak (I.C. at 13 minus the larger of 12 and 14). Texts: K1, K2 Part II (Stein's), K2 Part III, all of K2, and the K1 and K2 plaintexts. K4's own value at 13 is in 0168 (p 0.21).
What the method does with the 24 known letters.
When Stein cut the first two messages into 13 interleaved strands, the letters matched a little more often than expected, and he wondered whether a second, 13-letter code lay underneath. We check whether that bump is bigger than what the known keys alone would give, by shuffling the letters only among positions that share a key letter, 10,000 times.
The carved letters go in at the top; the reading comes out at the bottom.
Stein saw the letters match more often when K1 and part of K2 were cut into 13 strands, and wondered whether a second, 13-letter code lay underneath.
Shuffle each text only among the places that share a key letter, which keeps everything the known keys explain, 10,000 times, and see how often the bump at 13 is reached.
K1's bump is unusual (p 0.0075, about 0.02 allowing for the strand counts scanned); Part II's is not (0.24), nor K4's (0.21). A coincidence, closed.
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.
Redo the method on a few letters with pencil and paper.
The same steps run by the code on the real letters.
K1's 63 letters in 13 strands (letter i goes to strand i mod 13):
strand 1: EUGBY
strand 2: MSNQQ
strand 3: UDFVT
strand 4: FJIYM
strand 5: PKVUK
strand 6: HZJVY
strand 7: ZLYLR
strand 8: LDQLD
strand 9: RKTTM
strand 10: FRQRF
strand 11: ANUED
strand 12: XSXV
strand 13: YHQJ
average index of coincidence over the 13 strands: 0.0744; over 10 strands (the key length): 0.0914
Family, key space, the constraint under test, and the clues it reads.
C28 (knowledge/11-clue-register.md); Stein 1999 (knowledge/sources/stein-1999-cracking-the-courtyard-crypto.txt), Figures 7, 17, 26: 'I knew that the peak at m = 10 was caused by the 10-letter keyword, and the peaks at 5 and 15 are expected artifacts... But what was this peak at m = 13? ... it also showed up in the I.C. plot for Part II.' Model of the null: a Quagmire III ciphertext of period L has column multisets fixed by the plaintext and key; permuting letters within each column (place mod L) keeps every period-L property and destroys any other structure. If K1's and Part II's I.C. at 13 sit inside that null, the peak is a coincidence of the plaintext's letters; if outside, something keyed at 13 exists and K4 would be tested for it.
Written before the run; the result decides it.
The m = 13 values sit inside the key-period null for K1 and Part II (p above 0.05): a coincidence of short texts. Zero candidates.
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
K1 ciphertext (Stein's Part I, key 10): I.C.13 0.07436, p within-columns 0.0075, plain 0.0359; K2 Part II (Stein's 125 letters, key 8): I.C.13 0.04231, p within-columns 0.2435, plain 0.331; K2 Part III (key 8): I.C.13 0.04103, p within-columns 0.4051, plain 0.4586; K2 ciphertext without '?' (key 8): I.C.13 0.04837, p within-columns 0.0894, plain 0.1486; K1 plaintext: I.C.13 0.04359, p within-columns -, plain 0.8613; K2 plaintext: I.C.13 0.06619, p within-columns -, plain 0.3845
What was searched, as the harness recorded it.
How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.