Correction: Session 42 joins three apostrophe suffixes and recalculates this test. The shared maximum remains 575/12,947; the independently keyed maximum changes from 779 to 780/12,947 (6.0246%). The historical calculation below uses the original segmentation. Read Session 42.
Result: No new plaintext. Resetting the previous two ciphertext runes at every word, with any fixed reset state, still cannot produce the peak rune frequency of our solved LP benchmark. The exact maximum is 4.4412% with shared coefficients and reset state, or 6.0168% when every section chooses its own coefficients and reset state. The benchmark is 12.6337%.
This closes a specific gap in Session 39, which carried memory across word boundaries. It does not exclude all word-based ciphers.
The hypothesis and scope
We keep the same arithmetic but reset the memory before each word:
p[i] = (A*c[i] + B*c[i-1] + C*c[i-2] + K) mod 29
A and C: 1..28; B and K: 0..28
Before every word: c[-2] = s2, c[-1] = s1
s2 and s1: independently 0..28
The same coefficients and initial pair apply to every word in the shared-rule test. The independently keyed section test permits a different complete set in each section. Natural Gematria index order is used throughout. The output has one rune per input rune, preserving word lengths.
All 12,947 runes in 2,897 words from the established eight substantial unsolved sections are scored. Unlike Session 39, this calculation includes every first and second rune: initialization is optimized explicitly, not discarded. The nine-rune section 3 remains excluded.
The existing transcription parser treats rune separators as word boundaries. Spaces, newlines, slashes and page breaks do not themselves end a word, allowing words to continue across layout breaks. Other non-rune delimiters end a word. This is a rune-only interpretation of the retained transcription, not a new image-level punctuation audit.
The repeated phrase changes its implication
The two copies are D J U · B E I, at LP2 page 27 rune offset 28 and page 55 offset 70. Each comprises two three-rune words, as confirmed in Session 32.
| Memory rule | Consequence under the same key and relevant initial state |
|---|---|
| Session 39: continuous two-rune memory | Last four plaintext runes match; second differs; first may differ |
| Session 40: reset before each word | All six plaintext runes match |
The reset erases the different external lead-ins, B F versus K X. It also resets at the separator inside the phrase. The script verifies this behavior with a constructed key. Its arbitrary output is a fixture, not a proposed translation.
This illustrates why guessing an English phrase before choosing the cipher mechanism can mislead us. Identical ciphertext implies different plaintext constraints under different mechanisms. The shared-rule conclusions do not apply if the two sections use different keys or reset states.
Checking every reset state without a large key search
Normalize A out of the equation:
a = B/A mod 29; b = C/A mod 29
p[i] = A*q[i] + K mod 29
A and K only permute histogram bins. Split each word into three position classes:
| Word position | Normalized output |
|---|---|
| First | c[0] + u |
| Second, where present | c[1] + a*c[0] + v |
| Third onward | c[i] + ac[i-1] + bc[i-2] |
Here u = a*s1 + b*s2 and v = b*s1. Since b is nonzero modulo the prime 29, any desired pair u,v corresponds to exactly one reset pair s2,s1. The first-position and second-position histograms can therefore be shifted independently to align their peaks with the interior histogram's peak.
For each a,b, the exact largest possible count of one output rune is:
max(first-position histogram)
+ max(second-position histogram for a)
+ max(interior histogram for a,b)
We evaluate all 812 normalized a,b pairs. This covers all permitted coefficients and all 841 reset pairs. The calculation includes 2,897 first positions, 2,811 second positions and 7,239 interior positions. No word guesses, fitted language model or selected candidate continuation is involved.
Results
| Scope | Maximum count of one output rune | Maximum frequency |
|---|---|---|
| Shared coefficients and reset pair | 575 / 12,947 | 4.4412% |
| Independently chosen coefficients and reset pair per section | 779 / 12,947 | 6.0168% |
| Solved LP1 benchmark, all positions | 248 / 1,963 | 12.6337% |
| Solved LP2 page 57, all positions | 15 / 95 | 15.7895% |
These maxima are attainable, not just generous upper bounds. For the shared case, A=1, B=4, C=6, K=0, with reset pair s2=19, s1=6, produces 575 occurrences of output bin 22. The script decodes the complete corpus and verifies the count. This maximizing output is not claimed to be readable.
For the independently keyed case, the script constructs each section's maximizing rule and shifts its peak to output bin zero. The resulting combined histogram contains exactly 779 zeros. This verifies that the sum is attainable under the relaxed assumptions.
| Section | Runes | Words | Maximum count | Maximum frequency |
|---|---|---|---|---|
| 0 | 729 | 158 | 59 | 8.0933% |
| 1 | 1,145 | 256 | 77 | 6.7249% |
| 2 | 1,729 | 383 | 99 | 5.7259% |
| 4 | 1,894 | 436 | 108 | 5.7022% |
| 5 | 1,021 | 224 | 70 | 6.8560% |
| 6 | 1,524 | 349 | 97 | 6.3648% |
| 7 | 1,589 | 355 | 96 | 6.0415% |
| 8 | 3,316 | 736 | 173 | 5.2171% |
The benchmark percentage differs slightly from Session 39 because all positions are included here. Session 39 excluded the first two runes of each section or solved group. Neither report establishes a universal required frequency for plaintext; both compare against the stated solved-text sample.
Controls
Two constructed encryptions use the solved LP1 groups and solved LP2 page 57 with their original word boundaries. Different nonzero reset states and coefficient sets are used, including a B=0 case. Both complete plaintexts recover exactly. The calculated frequency limits admit their true plaintext peaks:
| Constructed control | True plaintext peak | Permitted maximum | Full recovery |
|---|---|---|---|
| Solved LP1 groups | 12.6337% | 16.1488% | Exact |
| Solved LP2 page 57 | 15.7895% | 18.9474% | Exact |
An independent sequential decoder exhausts 10,000 coefficient-and-reset settings over a five-symbol alphabet, using words of lengths 1, 2, 3, 4 and 6. Its maximum count is 10, agreeing exactly with the analytical calculation. This checks the normalization and handling of short words, rather than merely reusing the histogram implementation.
Controls verify known-key inversion, bound admissibility and the exact maximum calculation. They do not demonstrate unknown-key language recovery. All real-text analysis remains retrospective on an already studied corpus.
Reproducibility
- Executable audit
- Full results and all 812 pooled histogram calculations
- Word parser and benchmark definitions
- Rune handling
- Retained transcription
Run the audit alongside its inputs with Python 3 and NumPy:
python3 -B word_reset_feedback_audit.py
The script verifies transcription SHA-256:
b0b2e13c3cd35512ee5955a2d9a56c22d7410e283d35dd9c90abbaa7b99279eb
What this changes
Allowing arbitrary fixed word resets does not make this linear feedback family resemble the solved text's uneven rune distribution. We can deprioritize this particular extension without choosing a speculative keyword.
The result is an exact empirical frequency limit, not a p-value or proof that no conceivable message could use the rule. Nonlinear rules, arbitrary input alphabet permutations, changing per-word states, extra key streams, skipped symbols and intermediate encoded material remain outside scope. A fixed substitution applied after the decoder preserves its peak frequency and would not remove this limit.
The next distinct question is whether repeated whole-word frequencies support any deterministic word-by-word cipher with a common reset state. That would test a broader structural property before choosing another arithmetic formula. No new plaintext or previously unknown community discovery is claimed here.