If every repeated ciphertext neighbor corresponds exactly to one fixed plaintext rune, that rune must occupy five different short-word patterns across 20 words. No repeat event lands on a one-rune word. These are useful conditional constraints, but they do not identify a letter, word or key.
A tempting result illustrates the risk: O is the only rune satisfying all five patterns within our small solved-page vocabulary. Ordinary English counterexamples immediately allow other letters, and O's reference frequency is much higher than the observed repeat frequency. We do not treat O as a decipherment.
Hypothesis and scope
The hypothesis is:
C[i] = C[i-1] if and only if P[i] = R
Here R is a placeholder for one fixed plaintext rune, not the rune labelled R. The plaintext position is the right-hand member of the repeated ciphertext pair. The left-hand member need not have the same plaintext value.
This is a shared property of the fixed repeat-channel models in Sessions 46 and 47. It does not describe every feedback cipher. We assume preserved rune-word lengths and boundaries, one common R across all eight substantive sections, and continuous state across words, punctuation, number blocks and pages. The first rune of each section is unknown because its predecessor is not supplied. It is not silently asserted to be different from R.
The inventory uses the corrected lexical parser from Session 42, joining intra-rune apostrophes. It contains 86 repeat events among 12,939 eligible positions. This pass uses the retained transcription; it does not newly authenticate or visually audit all 86 source-image locations. Session 27 includes earlier selected source checks.
What the short words require
An underscore means a plaintext rune other than R under the stated if-and-only-if hypothesis. Lengths count runes, not Latin characters: a rune such as TH still occupies one position.
| Rune length | Required pattern | Observed words |
|---|---|---|
| 2 | R_ |
3 |
| 2 | _R |
1 |
| 3 | R__ |
6 |
| 3 | _R_ |
3 |
| 3 | __R |
7 |
| Total | Five patterns | 20 |
Each listed word contains one repeat event. The complete locations and ciphertext labels are retained in the results file. For example, the previously spot-checked TH–TH ciphertext word on LP2 page 35, starting at rune offset 195, requires _R, not RR.
The 83 audited singleton words contain zero repeat events at their positions. This does not force R to be A or I. Instead, under the model, no singleton plaintext word is R. If R were A, every ordinary A/I singleton would have to be I; if R were I, every such singleton would have to be A. Without knowing the message, neither possibility is a logical contradiction.
The pooled five-pattern requirement uses a common R across sections. Allowing a different repeat rune in each section weakens the constraint; per-section patterns are included in the evidence.
Why the small vocabulary appears to select O
Our five decoded LP1 groups supply only 18 distinct two-rune words and 32 distinct three-rune words. Within that reference vocabulary, O alone supports all five required patterns:
| Pattern | Examples from the solved reference, with R = O |
|---|---|
R_ |
OF, OR, ON |
_R |
GO, SO, TO, DO |
R__ |
OFF, OUR, OWN |
_R_ |
FOR, NOW, NOT, YOU |
__R |
WHO, TWO |
Leaving any one solved group out still leaves O as the only compatible reference-vocabulary rune. That stability does not solve the coverage problem: all five samples are small and related, and they omit many ordinary words.
For a direct counterexample, each following row satisfies all five patterns with familiar words. These are illustrative alternatives, not proposed plaintext assignments:
| Possible R | R_ |
_R |
R__ |
_R_ |
__R |
|---|---|---|---|---|---|
| T | TO | IT | TOP | ATE | NOT |
| S | SO | IS | SAD | ASK | HIS |
| N | NO | ON | NOW | AND | MAN |
These examples use only single-letter rune labels, so their segmentation is unambiguous. The code verifies their masks. They prove that O's apparent uniqueness is limited to our retained vocabulary, not a consequence of ordinary English or the cipher evidence.
Frequency tension
If the hypothesis is true with a shared R, its plaintext frequency at scored positions is exactly the repeat rate:
86 / 12,939 = 0.6647%.
In the known LP1 reference, excluding the first rune of each group to match the adjacency scope, O occurs 190 / 1,958 = 9.7038%. Thus the reference-vocabulary candidate O would need to be far less common in the unknown message than in that sample.
That is a mismatch with the reference, not proof that an O-poor English text is impossible. Nor do the T, S and N examples establish that those letters have plausible overall frequencies. The examples address only the logical short-word patterns. No language probabilities or statistical significance are claimed in this pass.
The repeated six-rune phrase also contains no repeated neighboring ciphertext rune, including its transition from the preceding rune, at either occurrence. Under this same fixed-R hypothesis, all six plaintext positions must therefore be different from R. This is another conditional crib filter; it gives no phrase translation.
Behavioral controls
We encrypted each decoded LP1 group using additive ciphertext feedback with R set in turn to O, T and X. The test checks the actual repeat mask against the known plaintext positions, excluding each group's first position.
All positions agree exactly in all three cases: 190 repeats for O, 128 for T and 2 for X. This validates which side of the pair supplies the plaintext constraint and the handling of boundaries. It is not unknown-key recovery or an independent language test.
Decision
The repeat clue yields a reusable list of forced plaintext positions, but no identified letter. We retain the five short-word patterns and the repeated phrase's exclusion of R as conditional checks for future candidates. We do not promote O, or any other letter, into a key guess on this evidence.
The next substantive model would need to explain the positional constraints and the low overall frequency together, then produce unchanged coherent continuation. A vocabulary match alone is insufficient.
Evidence and reproduction
- All event words, the 20 short-word locations, pattern counts and vocabulary comparisons
- Executable audit and controls
- Earlier repeat-position inventory and image spot checks
- Corrected singleton image audit
- Fixed repeat-channel model
Run python repeat_letter_constraints.py with NumPy and the companion notebook files. The output records script and transcription SHA-256 hashes. Page numbers and rune offsets are zero-based; location fields in this report's inventory identify word starts. No transcription changes were made, and no new plaintext was recovered.