On the diamond hierarchical lattice with recursion R_{b,s}(p) = 1 − (1 − p^s)^b, is λ(b, s) = R′_{b,s}(p*) equal to λ(s, b) on the pairs (2,5)/(5,2), (3,5)/(5,3), (2,6)/(6,2)?

Statement

The recursion for the parallel-of-series diamond with b branches of s bonds is R_{b,s}(p) = 1 − (1 − p^s)^b. Its planar dual is the series-of-parallel diamond with the same b and s: a bond is open in the dual exactly when the primal bond is closed, so with q = 1 − p the dual recursion is D_{b,s}(q) = 1 − R_{b,s}(1 − q) = (1 − (1 − q)^s)^b. The dual's fixed point is q* = 1 − p* and, by the chain rule, D′(q*) = R′(1 − q*) = R′(p*) = λ; that is (B), and it is a check on the code, not on the mathematics. The parallel-of-series diamond with the roles swapped, R_{s,b}(p) = 1 − (1 − p^b)^s, is a third lattice. Its fixed point differs from p* (RESULTS110: (2,4) has p* = 0.9206, (4,2) has 0.2818) and nothing in the chain rule relates its eigenvalue to λ(b,s). That is (A): the question.

Sealed 2026-10-21

As written. The lab chat reads this file above the heading "Not part of the seal" only, computes, and reports in RESULTS111.md with the run's lineage and a tools assessment first.

Scored 2026-10-21 — confirmed

Scored by a blind reader given the seal above the "Not part of the seal" heading and R27_tree_groups/carpet_perc/RESULTS111.md. Withdrawal not triggered ((B) holds on all rungs). Falsifier not triggered (maximum relative difference 3.8e−81 against the 1e−30 line; nothing in the 1e−40 to 1e−30 band). Prediction (A) held on (2,5)/(5,2), (3,5)/(5,3), (2,6)/(6,2); (B) held on all twelve rungs; controls (i) and (ii) pass. The lab additionally scanned D(q) − q (one sign change everywhere), reported beside.

Faults in the seal's wording (the reader's list, kept): control (i)'s 1e−45 line is bounded by RESULTS110's 50-digit printing, which the seal did not state; the tolerance block did not say whether the scored difference is the 60-digit rounded value or the 80-digit working value (both decide the same way here); control (ii) is silent on the dual map's uniqueness while (B) depends on q* being unique; "nine rungs" in control (ii) where twelve were computed (six new, six of RESULTS110); 1/ν was asked for and is in RESULTS111's table but not in the reader's transcription. The lab read run 110's file in full, including 110's own lead and scored table, because the prompt allowed 110 "for the method" without restricting it to above its heading; 110's lead concerns the sign rule, not this seal's question; recorded, and the next prompt names the cut for every file it allows.

Reading (the codifier's, after the score; a derivation, not a sealed claim, checkable by anyone): the identity is one line. Write f_s(p) = p^s (a series run of s bonds) and g_b(x) = 1 − (1 − x)^b (b parallel branches). Then R_{b,s} = g_b ∘ f_s, the swapped lattice is R_{s,b} = g_s ∘ f_b, and the planar dual of (b, s) is D_{b,s} = f_b ∘ g_s. The swapped lattice and the dual are the same two maps composed in the other order, and a cyclic permutation of a composition preserves the multiplier at the fixed point: if x* is fixed by f ∘ g then g(x*) is fixed by g ∘ f, and (f ∘ g)′(x*) = f′(g(x*)) g′(x*) = (g ∘ f)′(g(x*)). So λ(s, b) = μ(b, s), and μ(b, s) = λ(b, s) by (B)'s chain rule. (A) was never independent of (B); the symmetry under b ↔ s is planar duality followed by cyclic invariance of the multiplier, and the fixed points differ (p*(s,b) = g_s(q*(b,s))) while the eigenvalue cannot. On the carpets there is no b ↔ s to swap, but the same two facts exist there — planar duality (cut ↔ flow, C₁ against the shortest path) and the invariance of a multiplier under cyclic reordering of a composition — and the question the theme would ask is which quantity of the carpet's transfer operator is the multiplier of a composition whose factors the dual reorders.

The codifier’s lead is kept for the private site.