Decoder floors: the smallest two-sided case (m = 10, one outside direction per side) costs ≥ 64 on P112 — conjecture K3 of #2832 (no κ claim)

[DreamingOfClouds]

Follow-up to #2832. It proves that conjecture K3 holds. This is a research note: no κ claim, no supplier word, no change to any existing file. Only research/two-sided-ten-term-floor/ is added.

Result

Theorem K3. Every normalised two-sided decoder of N₀ = G/2 on P112 (dense h = 9 table) with m = 10 terms and (s, t) = (1, 1) costs at least 64. Normalised means no two proportional p's or q's (Lemma D of #2338). The decoder may have J_P ∩ J_Q disjoint or not.

• Sharp. Two exact decoders in the class cost exactly 64 (P Qᵀ = N₀ on all 112² entries, no proportional terms): one with J_P ∩ J_Q = ∅ (|J_P| = |J_Q| = 3) and one with |J_P ∩ J_Q| = 2.
• Proof. The bound is cost = 90 − Σ s_k with the exact identity cost = 72 − D̂_P − D̂_Q + ζ_P + ζ_Q + Λ, where D̂ summarises the low-weight outside rows and columns (one low of deficiency D ≥ 3, one or two weight-7 lows, or none). Every (D̂_P, D̂_Q) cell is closed:
  • sum ≤ 8: the identity itself.
  • a side with D̂ ≤ 3: the other side needs D̂ ≥ 6, so at least five lossless frame-row dual terms with outside columns in non-frame classes; the pair systems have no 5-clique. This is exact on all 93 sets I^c with |I^c| ≤ 3; the independent verify_cl.py gives largest cliques 2, 2, 2, 4.
  • both sides ≥ 4: Σ s ≤ 26 when the two lows sit in different terms (swap slots are Ω-orthonormal, and each extra saving costs a slot). If one term is low on both sides, Σ s ≤ 36 − cost_b − (D_P + D_Q); the equality case (cost_b = 0, D_P + D_Q = 9) is impossible. One written proof covers it, with idempotents A + B = I₈ for repeated indices and the Steiner system S(3,4,8) for distinct ones.
• Dependencies. Conditional on the open PR Decoder floors, case (b) for every number of terms: one-sided P112 decoders cost ≥ 64 without corrections (unique optimum) and with ≤ 2 non-full corrections (no κ claim) #2816 only (Research: rank-8 complex decoders are gauge-rigid; E-graph eight-term floor (no new kappa) #1212, Decoder floors, case (b): nine terms outside col M (P112 ≥ 64, #968 P104 ≥ 56; no κ claim) #2724 and Decoder floors, case (b): the six remaining d = 8 multisets of #2816 are dead (≥ 64 with ≤ 3 non-full corrections; E8 lifts ≥ 67; no κ claim) #2826 are not used) (Lemma 3.1, cover facts C1/C3, g(5) = 12, completeness of the direction list), through Decoder floors: universality, projection transfer and batching reductions; C(P112) ≥ 64 reduced to named conjectures, C(P120) ≥ 72 restated on N₀ (no κ claim) #2832 Lemmas 2.3 and 3.7–3.9, Theorem 3.5, Corollary 3.6. Decoder floors, case (b): nine terms outside col M (P112 ≥ 64, #968 P104 ≥ 56; no κ claim) #2724 is not used, and neither is any disjointness hypothesis (the disjoint sub-case is covered again; Decoder floors: universality, projection transfer and batching reductions; C(P112) ≥ 64 reduced to named conjectures, C(P120) ≥ 72 restated on N₀ (no κ claim) #2832 Theorem 3.13 is an independent proof of it).

Consequence for #2832

K3 is now a theorem, conditional as stated. With #2832, the P112 master C(P112) ≥ 64 reduces to K1 (every two-sided decoder of N₀(P112), general m, s, t) and K2 (excess pays for corrections, one-sided) only. Nothing is claimed about K1, K2, P120 or any κ.

Verification

bash verify.sh (Python ≥ 3.10 with numpy, sympy, networkx; about 12 minutes wall unloaded (696 s measured; about 22 minutes under heavy load); at most 4 processes, or 2 with JOBS=2, each under timeout 1200; no SAT solver). It checks the manifest, re-runs the finite checks and asserts every number quoted in the README as PASS/FAIL:

• clique2.py 5 on all 93 sets (no 5-clique passes the joint test), verify_cl.py, sym.py, stab.py.
• diag9.py (280 pairs, no candidates) and caseb4.py (22,020,096 solutions, none valid).
• The second proof chain clique2.py 4 → d4A.py → d4A3.py (16128 → 128 → 0).
• witness16.py (the two cost-64 witnesses).

Caveats: the dependency on #2816 above; the finite parts (clique tests, S(3,4,8) enumeration) are exact computations, not proofs by hand; nothing beyond m = 10, (s, t) = (1, 1) is claimed.

Prepared by DreamingOfClouds with Claude Opus 5.5 (Anthropic), via Claude Code; Apache-2.0.

🤖 Generated with Claude Code