Quantum counting error and boundary cases

For $m=1$, the $M=2$ procedure is the one-qubit instance of [F22]; for $m\ge 2$ it is [F22] specialized by the amplitude-estimation output map. The $n$ Hadamards turn $\mathrm{\mid }{0}^n⟩$ into the uniform superposition; the definition of ${\mathrm{\Pi }}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}}$ makes its squared projection norm $t\mathrm{/}N$ by [F21] and [F24]. The eigenphase relation [F23] applies to this projector, and the amplitude-estimation instructions of [F22] use the corresponding ${\sin }^2$ output map. Thus the procedure estimates precisely this fraction and multiplying by $N$ gives the count estimate.

For $M\ge 4$, substitute $a=t\mathrm{/}N$ in [F2] and multiply its error bound by $N$. Since $N\sqrt{a(1-a)}=\sqrt{t(N-t)}$ and $N(a)=t$, the result is the displayed inequality with unchanged event probability.

In repeated-unitary mode, each controlled $Q$ call includes one explicitly supplied or implemented controlled membership reflection. In power-query mode, each supplied controlled power carries its declared cost vector, including any membership-reflection implementation it uses; neither mode infers controlled access from a bare phase query. The exact algebraic result has separate postprocessing cost by [F25].

The output remains the real-valued pair $(\tilde{a},\tilde{t})$ unless an integer is explicitly requested. Preparation uses exactly $n$ ordinary Hadamards by [F24]. The access-mode-specific query costs are those declared in [F26], while ordinary gate and bit costs remain separate by [F24]–[F25].

If $t=0$, the phase is the exact grid phase zero, so QPE returns $y=0$ and $\tilde{t}=0$. If $t=N$, the phase is $1\mathrm{/}2$; the even grid size has exact label $M\mathrm{/}2$, which returns $\tilde{t}=N$.