Stacking two carpets averages their exponents when their cut classes match and falls below the average when they do not — why?

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type: lab

field: physics

object. Take a 9×9 grid of cells, keep 64 and remove 17, keeping the outer ring and a pattern symmetric under the eight symmetries of the square. A carpet repeats one pattern at every level: replace each kept cell by a copy of the pattern, and again. A stacked carpet alternates two patterns A and B: A at the coarsest level, B inside every kept cell of A, A inside those, and so on; after 2k levels it is an 81^k × 81^k lattice. Put bond percolation on it and ask for an open left-to-right crossing. The correlation-length exponent 1/ν is measured from how the width of the crossing transition shrinks with lattice size (L = 81, 729, 6561; 8,000 samples per size). For a pattern, C₁ is the minimum number of bonds whose removal disconnects the left side of the 9×9 block from the right; minimum cuts multiply across levels, so a stacked carpet's cut exponent ln(C_A·C_B)/ln 81 is exactly the average of its parents' ln C/ln 9, as is its fractal dimension.

friction. Two stacks were measured. With A and B of the same cut class (C₁ = 4 for both), the stacked exponent equals the average of the parents' exponents within noise. With A of cut 4 and B of cut 2, it falls below the average by 3.5 standard errors. Dimension and cut exponent average exactly in both cases, and building the stack in either order gives the same exponent, as it must. What makes the exponent add over levels in one case and not in the other?

background.

(All ±0.004 per measurement. Across other patterns of cut 4, 1/ν ranges from 0.43 to 0.55.)

The one-block estimate (the slope of the exact 2-cycle of the two patterns' crossing polynomials, per factor 81) gives 0.5503 for X with Y and 0.4485 for X with U′; the parents' one-block values average to 0.5514 and 0.4370. The run-to-run spread of the estimator is about 0.0045 per measurement.

Must predict. Two more stacked pairs are being measured the same way: X with T (cut 4 with cut 3) and U′ with T (cut 2 with cut 3). For each: is the stacked 1/ν at, below, or above the mean of its parents, and by how much? Give it from a mechanism that also says why cut-matched parents average.

Provenance. Stacks built level by level from the patterns' 9×9 generators (the builder reproduces a plain two-level carpet when both levels are the same pattern). 1/ν is the endpoint slope of log width from L = 81 to 6561, which for three sizes equally spaced in log L equals the least-squares slope; bootstrap errors from 1,000 resamples. Parents' values are their recorded measurements by the same estimator.

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Not part of the case — the codifier's sealed predictions (2026-09-29, before the panel runs and before any X-with-T or U′-with-T value exists)

The codifier's own lead is the one sealed in the run now measuring these pairs: both fall below their parents' mean by more than two standard errors.

Predictions of what the v5 panel will do: 1. At least three seats predict both new pairs below their parents' mean, and at least one seat names the smaller-cut level as a bottleneck that dominates. 2. At least one seat or the House proposes a composition rule other than the arithmetic mean (geometric, harmonic, or weighted by cut) and gives a number from it. 3. The checker (version 10, after the PN-003 fixes) is right on at least two thirds of the rows it flags as mismatches in round 1 and House 1. Scored against the panel record once it is filed; the panel's pair predictions are scored against the run measuring X with T and U′ with T.

Scored 2026-09-30, after the panel ran (PN-004) and against run 070

The codifier's own lead — refuted in substance. Both new pairs came in above their parents' mean (X,T +0.0115, +2.7σ; U′,T +0.0075, withdrawn for a 0.0227 order gap). Run 070 is scored unanticipated only because its falsifier did not cover a lone surviving pair (OPS.md).

The codifier's sealed panel predictions — 1 of 3 held. 1. Failed. The first half held: Ledger, Lateral and Bound (and Hypothesis, with a floor on U′,T) predicted both pairs below the mean. The second did not: no seat named the smaller-cut level as a bottleneck that dominates. The phrase appears only in the premise ledger's list of the case's assumptions. 2. Held, narrowly. No seat proposed a different mean (geometric, harmonic). Ledger's cut-weighted rule, a deviation linear in ΔlnC giving 0.4565 and 0.3163, is a composition rule other than the arithmetic mean with numbers from it; later Lateral's and the House's +k·Δp*² with k ≈ 0.3. 3. Failed. Round 1 and House 1 flagged two mismatches (R1-29, H1-16); both are extraction artefacts, so the checker was right on 0 of 2.

Checker precision this panel (codifier's reading of every mismatch row in the filed record): 5 mismatches, 0 real. Causes: a contrast or quotation read as set membership (R1-29, H1-16, R2-32); a compound expression, 2×√(0.0039² + 0.0037²), recomputed from one operand (R2-26, and H2-1 in the second House draw); the wrong operand chosen when the needed value is not in the source line (H2-25 subtracted the pair's one-block average, 0.5514, where the House used Y's own y₉, 0.5495: the House's +0.034 is right; the row should have been uncheckable). Missed: a derivation claim in the House final's FORCED section, that the deviation from the parental mean is second order in the fixed-point separation. Expanding the 2-cycle to first order gives a first-order term, δ/(Λ − 1) times a combination of the two maps' curvatures at their fixed points, which vanishes only when δ = 0 or the curvature terms cancel. Run 076 tests which order holds.

Reductio seat (one-off, second of two cases). It caught that House 1's sign prediction was already contradicted by the lab table, which the House final then conceded (its [H1-26]). That is the one contradiction it found that no other seat stated first.

Run 076 (2026-09-30, falsified), on the one-block map over all 1,176 pairs of distinct census polynomials. At matched fixed-point separation, pairs of equal cut class deviate 3–10 times less than pairs of unequal class, and cut class adds information beyond the separation (partial Spearman +0.60). The case's framing, that matched cut classes average, holds for the one-block map better than the House's rival or the codifier's. Whether it holds on the lattice is the open part: the lattice's X,U′ deficit and X,T and U′,T surpluses have no one-block counterpart in sign.

Continuation 1 filed (2026-10-01). PN-004 Continuation 1 is in the panel record, together with the recovered prose of the second House-final draw, which the continuation was run against. Its seats are scored in SEAT-RECORD. The continuation House withdrew "cut class is dead" for the one-block object, and its free tests became run 078.