Optimal Parallel Quantum Query Algorithms

Abstract

We study the complexity of quantum query algorithms that make p queries in parallel in each timestep. This model is in part motivated by the fact that decoherence times of qubits are typically small, so it makes sense to parallelize quantum algorithms as much as possible. We show tight bounds for a number of problems, specifically $\Theta ((n/p)^{2/3})$ p-parallel queries for element distinctness and $\Theta ((n/p)^{k/(k+1)})$ for $k$-sum. Our upper bounds are obtained by parallelized quantum walk algorithms, and our lower bounds are based on a relatively small modification of the adversary lower bound method, combined with recent results of Belovs et al. on learning graphs. We also prove some general bounds, in particular that quantum and classical p-parallel query complexity are polynomially related for all total functions f when p is small compared to f’s block sensitivity.

Appendices

Appendix 1: Proof of Fact 1

We use the $\gamma _2$-norm for matrices, which is defined as

$$\begin{aligned} \gamma _2(A)=\min _{X,Y:A=XY} r(X)c(Y), \end{aligned}$$

where r(X) denotes the maximum squared length among the rows of X, and c(Y) denotes the maximum squared length among the columns of Y. Note that the identity and the all-1 matrix both have $\gamma _2$-norm equal to 1 (the latter by taking X and Y to be the all-1 column and row, respectively). Also, $\gamma _2(A\otimes B)=\gamma _2(A)\gamma _2(B)$. Since $\Delta _J$ can be written as $\mathcal{J}-\mathcal{I}_J\otimes \mathcal{J}_{[n]\setminus J}$, where $\mathcal{J}$ is the $[q]^n\times [q]^n$ all-1 matrix, $\mathcal{J}_{[n]\setminus J}$ is the $[q]^{[n]\setminus J}\times [q]^{[n]\setminus J}$ all-1 matrix, and $\mathcal{I}_J$ is the $[q]^J\times [q]^J$ identity matrix, applying the triangle inequality gives $\gamma _2(\Delta _J)\le 2$. The $\gamma _2$-norm satisfies $||{A\circ B}||\le ||{A}||\gamma _2(B)$ by [29, Lemma A.1]. Observe that $\Gamma \circ \Delta _J= (\Gamma \circ \Delta _K)\circ \Delta _J$. Hence we have

$$\begin{aligned} ||{\Gamma \circ \Delta _J}||&=||{(\Gamma \circ \Delta _K)\circ \Delta _J}|| \le ||{\Gamma \circ \Delta _K}||\gamma _2(\Delta _J)\le 2||{\Gamma \circ \Delta _K}||. \end{aligned}$$

Appendix 2: Proof of Lemma 4

The proof is a straightforward adaptation of the proof of [15, Theorem 9], but we repeat it here for completeness. Let $\{w_{S,J}:(S,J)\in \mathcal {E}_p\}$ and $\{\theta _{S,J}(M):(S,J)\in \mathcal {E}_p, M\in \mathcal {C}\}$ be an optimal solution to the primal formulation of $\mathrm{LGC}^{p\parallel }(\mathcal {C})$.

We will use this solution to construct a feasible solution to the dual expression of our p-parallel adversary of Eq. (2), which is the following:

(11)

The dimension k of the vectors $|u_{x,J}\rangle$ can be anything, and is implicitly minimized over.

For each $x\in f^{-1}(1)$, let $M_x\in \mathcal {C}$ be such that for every $y\in [q]^n$, $x_{M_x}=y_{M_x}$ implies $f(y)=1$. For every $x\in \mathcal{D}$ and $J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right)$, define the following state in span$\{|S\rangle |\alpha \rangle :S\subseteq [n],\alpha \in [q]^S\}$:

$$\begin{aligned} |u_{x,J}\rangle :=\left\{ \begin{array}{ll} \sum \nolimits _{S\subseteq [n]\setminus J} \sqrt{w_{S,J}} |S,x_S\rangle &{} \text{ if } f(x)=0\\ \sum \nolimits _{S\subseteq [n]\setminus J} \frac{\theta _{S,J}(M_x)}{\sqrt{w_{S,J}}} |S,x_S\rangle &{} \text{ if } f(x)=1\\ \end{array}\right. \end{aligned}$$

We now verify that $\{|u_{x,J}\rangle \}_{x,J}$ is a feasible solution to the dual formulation of $\mathrm{ADV}^{p\parallel }(f)$:

$$\begin{aligned} \sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) :x_J\ne y_J}\langle {u_{x,J}}|{u_{y,J}}\rangle= & {} \sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) :x_J\ne y_J}\sum _{S\subseteq [n]\setminus J:x_S=y_S}\frac{\theta _{S,J}(M_x)}{\sqrt{w_{S,J}}}\sqrt{w_{S,J}}\end{aligned}$$

(12)

$$\begin{aligned}= & {} \sum _{S\subseteq [n]:x_S=y_S}\sum _{J\in \left( {\begin{array}{c}[n]\setminus S\\ \le p\end{array}}\right) : x_J\ne y_J}\theta _{S,J}(M_x). \end{aligned}$$

(13)

To see that this expression is equal to 1, we need only notice that Eq. (13) is the sum of the flow on all edges across the cut induced by the set $\{S\subseteq [n]:x_S=y_S\}$, and the total flow across a cut is always 1, since $\theta (M_x)$ is a unit flow. Thus the constraint from (11) is satisfied and $\{|u_{x,J}\rangle \}_{x,J}$ is a feasible solution.

We can now bound $\mathrm{ADV}^{p\parallel }(f)$ by the objective value of the feasible solution $\{|u_{x,J}\rangle \}_{x,J}$. First note that for any $x\in f^{-1}(1)$, by constraint (4), we have:

$$\begin{aligned} \sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) }||{|u_{x,J}\rangle }||^2 =\sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) }\sum _{S\subseteq [n]\setminus J}\frac{\theta _{S,J}(M_x)^2}{w_{S,J}}\le 1. \end{aligned}$$

Second, for any $x\in f^{-1}(0)$ we have

$$\begin{aligned} \sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) }||{|u_{x,J}\rangle }||^2 =\sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) }\sum _{S\subseteq [n]\setminus J}w_{S,J}. \end{aligned}$$

We can therefore bound the objective value as:

$$\begin{aligned} \mathrm{ADV}^{p\parallel }(f)\le & {} \sqrt{\max _{x\in [q]^n}\sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) }||{|u_{x,J}\rangle }||^2} \le \sqrt{\max \left\{ 1,\sum _{J\in \left( {\begin{array}{c}[n]\\ \le p\end{array}}\right) }\sum _{S\subseteq [n]\setminus J}w_{S,J}\right\} }\\\le & {} \sqrt{\sum _{e\in \mathcal{E}_p}w_e} = \mathrm{LGC}^{p\parallel }(\mathcal {C}), \end{aligned}$$

where $\sum _e w_e\ge 1$ follows from the Cauchy-Schwarz inequality and the fact that $\sum _e\frac{\theta _e(M_x)^2}{w_e}\le 1$, as follows:

$$\begin{aligned} 1=\left( \sum _e \theta _e(M_x)\right) ^2=\left( \sum _e \frac{\theta _e(M_x)}{\sqrt{w_e}}\sqrt{w_e}\right) ^2\le \sum _e \frac{\theta _e(M_x)^2}{w_e}\sum _e w_e\le \sum _e w_e. \end{aligned}$$

Appendix 3: Proof of Theorem 5

For the upper bound, we immediately obtain $Q^{p\parallel }(f)=O(\mathrm{LGC}^{p\parallel }(\mathcal {C}))$ by Theorem 2 and Lemma 4.

For the lower bound, our proof will be similar to that of [16, Theorem 5], and we will omit parts of the proof that are identical to theirs. Just as in [16, Theorem 5], our proof will start with an optimal feasible solution $\{\alpha _S(M)\}_{M\in \mathcal{C}, S\subseteq [n]}$ to the dual (8). Therefore $\mathrm{LGC}^{p\parallel }(\mathcal {C})=\sqrt{\sum _{M\in \mathcal{C}} \alpha _{\emptyset }(M)^2}$ and moreover $\sum _{M\in \mathcal{C}} (\alpha _S(M)-\alpha _{S\cup J}(M))^2\le 1$. Then we will construct an adversary matrix $\Gamma$ for f such that $||{\Gamma }||\ge \sqrt{\frac{1}{2}\sum _{M\in \mathcal{C}} \alpha _{\emptyset }(M)^2}$ (as proven in [16]) and for every $J\subseteq [n]$,

$$\begin{aligned} ||{\Gamma \circ \Delta _J}||\le 2\max _{S\subseteq [n]} \sqrt{\sum _{M\in \mathcal{C}} (\alpha _S(M)-\alpha _{S\cup J}(M))^2} \end{aligned}$$

(proven below). Then Theorem 2 will imply $Q^{p\parallel }(f)=\Omega (\sqrt{\sum _{M\in \mathcal{C}} \alpha _{\emptyset }(M)^2})=\Omega (\mathrm{LGC}^{p\parallel }(\mathcal {C}))$.

First we use a variation of the adversary bound from [17] that allows the duplication of row and column indices. Concretely, rows and columns of $\Gamma$ are now indexed by (x, a) and (y, b), respectively, where $x\in f^{-1}(1)$, $y\in f^{-1}(0)$, and a and b belong to some finite sets. Then, with slight abuse of notation, $\Delta _J$ is now defined such that $\Delta _J[(x,a),(y,b)]=1$ if $x_J\ne y_J$, and $\Delta _J[(x,a),(y,b)]=0$ otherwise. Specifically, in our case rows of $\Gamma$ will be indexed by (x, M) for some $x\in f^{-1}(1)$ and $M\in \mathcal C$, and columns will simply be indexed by $y\in f^{-1}(0)$.

Second, $\Gamma$ will be the submatrix of a larger matrix $\widetilde{\Gamma }$ (defined below), whose rows are indexed by the elements of $[q]^n\times \mathcal {C}$ and whose columns are indexed by $[q]^n$. Then $\Delta _J$ is naturally extended to all $x,y\in [q]^n$ and $M\in \mathcal {C}$ by $\tilde{\Delta }_J[(x,M),y]=1$ if $x_J\ne y_J$, and $\tilde{\Delta }_J[(x,M),y]=0$ otherwise. Since $\Gamma \circ \Delta _J$ is a submatrix of $\widetilde{\Gamma }\circ \tilde{\Delta }_J$, we will have $||{\Gamma \circ \Delta _J}||\le ||{\widetilde{\Gamma }\circ \tilde{\Delta }_J}||$. Hence it suffices to upper bound the latter norm.

We now define $\widetilde{\Gamma }$. Consider the Hilbert space $\mathbb {C}^q$. Let $E_0$ denote the orthogonal projector onto the vector $\frac{1}{\sqrt{q}}(1,1,\ldots ,1)$, and $E_1=\mathrm {I}-E_0$ its orthogonal complement. For every $S\subseteq [n]$, let $E_S=\otimes _{j\in [n]}E_{s_j}$ (acting on $\mathbb {C}^{q^n}$), where $s_j=1$ if $j\in S$, and $s_j=0$ otherwise. Note that $E_SE_{S'}=E_S$ if $S=S'$, and $E_SE_{S'}=0$ otherwise. Define $\widetilde{\Gamma }$ as

$$\begin{aligned} \widetilde{\Gamma }=\left[ \begin{array}{c} \vdots \\ G_M \\ \vdots \end{array}\right] _{M\in \mathcal {C}},\quad \text {with } G_M=\sum _{S\subseteq [n]}\alpha _S(M) E_S, \end{aligned}$$

where the $\alpha _S(M)$ come from a feasible solution to the dual (8). We then define $\Gamma$ as the submatrix of $\widetilde{\Gamma }$ obtained by keeping only those columns indexed by y such that $f(y)=0$; and only those rows indexed by (x, M) such that M is a 1-certificate for x (i.e., for all $z\in [q]^n$, $z_M=x_M$ implies $f(z)=1$).

Fact 14

$\displaystyle ||{\Gamma }||\ge \sqrt{\frac{1}{2}\sum _{M\in \mathcal{C}} \alpha _{\emptyset }(M)^2}.$

Proof

Belovs and Rosmanis [16, Lemma 17] prove this result for any matrix $\Gamma$ constructed as above assuming that $q\ge 2|\mathcal{C}|$, and for each $M\in \mathcal{C}$, (1) $\alpha _S(M)=0$ whenever $M\subseteq S$, and (2) M is equipped with an orthogonal array $T_M$ of length |M|. Those two assumptions are satisfied in our case too. $\square$

Upper bounding $||{\widetilde{\Gamma }\circ \tilde{\Delta }_J}||$ requires some additional steps compared to [16]. We first review the approach of [16], which is for the special case $J=\{j\}$. Define a linear map $\varphi _j$ on matrix $\widetilde{\Gamma }$ by its action on blocks $E_S$, for every $S\subseteq [n]$. First, let $\varphi$ be such that $\varphi (E_0)=E_0$ and $\varphi (E_1)=-E_0$. Then $\varphi _j(E_S)= E_{s_1}\otimes \ldots \otimes E_{s_{j-1}} \otimes \varphi (E_{s_j}) \otimes E_{s_{j+1}}\otimes \ldots \otimes E_{s_n}$. An alternative but equivalent definition is

$$\begin{aligned} \varphi _j(E_S)={\left\{ \begin{array}{ll} E_S,&{}\text { if }j\not \in S;\\ \displaystyle -E_{S\setminus \{j\}}&{}\text { otherwise.} \end{array}\right. } \end{aligned}$$

The map $\varphi _j$ was introduced because it satisfies $E_S\circ \Delta _{j} =\varphi _j(E_S)\circ \Delta _j$. This comes from the observation that $\varphi (E_1)\circ \Delta _1=E_1\circ \Delta _1$, since $E_1=\mathrm {I}-E_0$ and $\mathrm {I}\circ \Delta _1=0$. The approach of [16] then consists of applying $\varphi _j$ to $\widetilde{\Gamma }$ before computing the norm of $\widetilde{\Gamma }\circ \tilde{\Delta }_j$.

We now generalize $\varphi _j$ to subsets $J\subseteq [n]$ as

$$\begin{aligned} \varphi _J(E_S)={\left\{ \begin{array}{ll} E_S,&{}\text { if }J\not \subseteq S;\\ \displaystyle -\sum _{S':S\setminus J \subseteq S'\subsetneq S} E_{S'},&{}\text { otherwise.} \end{array}\right. } \end{aligned}$$

Then $\varphi _j$ satisfies the following fact, which is an extension of the case $J=\{j\}$ (proved in [16]).

Fact 15

Let $J\subseteq [n]$ be any subset. Then $\widetilde{\Gamma }\circ \tilde{\Delta }_J = \varphi _J(\widetilde{\Gamma })\circ \tilde{\Delta }_J$.

Proof

Since

$$\begin{aligned} \widetilde{\Gamma }\circ \tilde{\Delta }_J=\left[ \begin{array}{c} \vdots \\ G_M\circ \Delta _J \\ \vdots \end{array}\right] _{M\in \mathcal {C}} =\quad \quad \left[ \begin{array}{c} \vdots \\ \sum _{S\subseteq [n]}\alpha _S(M) E_S\circ \Delta _J \\ \vdots \end{array}\right] _{M\in \mathcal {C}}, \end{aligned}$$

by linearity it suffices to prove the fact for $E_S$, i.e., that $E_S\circ {\Delta }_J = \varphi _J (E_S) \circ {\Delta }_J$, where $S,J\subseteq [n]$. This equality is immediate when $J\not \subseteq S$, since then $\varphi _J (E_S)=E_S$.

Assume from now on that $J\subseteq S$. For notational simplicity, assume further and without loss of generality that $J=\{1,2,\ldots ,j\}$, and set $F=E_{s_{j+1}}\otimes \ldots \otimes E_{s_n}$, hence $E_S= E_1^{\otimes j} \otimes F$. Using $\mathrm {I}=E_0+E_1$ we have

$$\begin{aligned} \mathrm {I}^{\otimes j} =(E_0+E_1)^{\otimes j} = \sum _{S':\emptyset \subseteq S'\subseteq J} E_{S'}, \end{aligned}$$

where the notations $E_J$ and $E_{S'}$ stand for the first j bits only. This implies

$$\begin{aligned} E_J= \mathrm {I}^{\otimes j}- \sum _{S':\emptyset \subseteq S'\subsetneq J} E_{S'}. \end{aligned}$$

Since

$$\begin{aligned} (\mathrm {I}^{\otimes j} \otimes F) \circ {\Delta }_J=0, \quad \text {and}\quad \varphi _J(E_S) = - \sum _{S':\emptyset \subseteq S'\subsetneq J} E_{S'}\otimes F, \end{aligned}$$

this concludes the proof. $\square$

Now by [29, Lemma A.1], $||{A\circ B}||\le ||{A}||\gamma _2(B)$, so in particular, $||{\phi _J(\widetilde{\Gamma })\circ \tilde{\Delta }_J}||\le ||{\phi _J(\widetilde{\Gamma })}||\gamma _2(\tilde{\Delta }_J)$. Combining this with Fact 15 and the fact that $\gamma _2(\tilde{\Delta }_J)\le 2$ (see Appendix 1), we can upper bound $||{\widetilde{\Gamma }\circ \tilde{\Delta }_J}||$ by $2||{\varphi _J(\widetilde{\Gamma })}||$. It remains to upper bound the latter norm.

Fact 16

$\displaystyle ||{\varphi _J(\widetilde{\Gamma })}|| = \max _{S\subseteq [n]} \sqrt{\sum _{M\in \mathcal{C}} (\alpha _S(M)-\alpha _{S\cup J}(M))^2}.$

Proof

We first compute $\varphi _J(G_M)$:

$$\begin{aligned} \varphi _J(G_M)=\sum _{S\subseteq [n]}\beta _S(M) E_S, \quad \text {where }\beta _S(M)=\alpha _S(M)-\alpha _{S\cup J}(M). \end{aligned}$$

Observe that $\beta _S(M)=0$ if $J\subseteq S$. Now rewrite $(\varphi _J(\widetilde{\Gamma }))^*\varphi _J(\widetilde{\Gamma })$ as

$$\begin{aligned} (\varphi _J(\widetilde{\Gamma }))^*\varphi _J(\widetilde{\Gamma }) =\sum _{M\in \mathcal {C}} (\varphi _J(G_M))^*\varphi _J(G_M) = \sum _{S\subseteq [n]} \left( \sum _{M\in \mathcal {C}}\beta _S(M)^2 \right) E_S. \end{aligned}$$

Since the different $E_S$ project onto orthogonal subspaces, we can conclude

$$\begin{aligned} ||{\varphi _J(\widetilde{\Gamma })}|| = \sqrt{||{(\varphi _J(\widetilde{\Gamma }))^*\varphi _J(\widetilde{\Gamma })}||}=\max _{S\subseteq [n]} \sqrt{\sum _{M\in \mathcal {C}}\beta _S(M)^2}. \end{aligned}$$

$\square$

We therefore have

$$\begin{aligned} ||{\Gamma \circ \Delta _J}||\le ||{\widetilde{\Gamma }\circ \tilde{\Delta }_J}||\le 2||{\varphi _J(\widetilde{\Gamma })}||= 2\max _{S\subseteq [n]} \sqrt{\sum _{M\in \mathcal{C}} (\alpha _S(M)-\alpha _{S\cup J}(M))^2}. \end{aligned}$$

When J has size at most p, the right-hand side is at most 2 because of the constraint parallel-(9), applied to edge $(S,J')\in \mathcal {E}_p$ with $J'=J\setminus S$. Therefore

$$\begin{aligned} \mathrm{ADV}^{p\parallel }(f)\ge \frac{||{\Gamma }||}{\max _{J\in {[n]\atopwithdelims ()p}}||{\Gamma \circ \Delta _J}||} \ge \frac{\sqrt{\frac{1}{2}\sum _{M\in \mathcal{C}} \alpha _{\emptyset }(M)^2}}{2}=\frac{1}{2\sqrt{2}}\mathrm{LGC}^{p\parallel }(\mathcal{C}). \end{aligned}$$

Appendix 4: Proof of Theorem 9

Here we prove our parallel lower bound for the ${k\mathrm{-sum}}$ problem. The proof strategy is the same as in Theorem 8. We now use certificate structure $\mathcal{C}={[n]\atopwithdelims ()k}$ with the orthogonal array

$$\begin{aligned} T=\{(v_1,\ldots ,v_k) : \sum _{i=1}^k v_i=0 \mod q\}. \end{aligned}$$

This induces the ${k\mathrm{-sum}}$ problem in the way mentioned in Theorem 5. We define the following solution to the dual for $\mathrm{LGC}^{p\parallel }(\mathcal{C})$:

$\displaystyle \alpha _j=\frac{1}{2n^{k/2}}\max ((n/p)^{k/(k+1)} - j/p, 0)$

$\displaystyle \alpha _S(M)=0$ if $M\subseteq S$

$\displaystyle \alpha _S(M)=\alpha _{|S|}$ otherwise

Fix some $e=(S,J)$ with $S\subseteq [n]$ of size s, and disjoint $J\subseteq [n]$ of size at most p. Let L denote the left-hand side of constraint parallel-(9). In order to establish that the above solution is feasible, we want to show $L\le 1$. With respect to e, we can distinguish different kinds of $M=\{i_1,\ldots ,i_k\}$, depending on $i:=|M\cap S|$ and $j:=|M\cap J|$:

• 1. $i+j<k$. There are ${s\atopwithdelims ()i}{|J|\atopwithdelims ()j}{n-s-|J|\atopwithdelims ()k-i-j}$ such M, and each contributes $\le |\alpha _s-\alpha _{s+|J|}|^2\le 1/4n^k$ to L.

• 2. $i+j=k$. There are ${s\atopwithdelims ()i}{|J|\atopwithdelims ()j}$ such M, each contributes $\alpha _s^2$ to L if $i<k$, and 0 if $i=k$.

  Over all such choices of i and j, at most $|J|\left( {\begin{array}{c}s+|J|-1\\ k-1\end{array}}\right)$ of these M have $j\ge 1$ (i.e., $\alpha _S(M)\ne 0$), since this counts the number of ways of choosing one index from J, and $k-1$ more from $J\cup S$.

Note that $\alpha _s=0$ if $s\ge p(n/p)^{k/(k+1)}$, so below we may assume $s+p-1\le 2p(n/p)^{k/(k+1)}$. Also $\alpha _s\le \alpha _0=(n/p)^{k/(k+1)}/2n^{k/2}$. Hence we can bound L as

$$\begin{aligned} L&\le \sum _{i=0}^{k-1} \sum _{j=0}^{k-1-i} {s\atopwithdelims ()i}{|J|\atopwithdelims ()j}{n-s-|J|\atopwithdelims ()k-i-j}|\alpha _s-\alpha _{s+|J|}|^2 + \sum _{i=0}^{k-1} {s\atopwithdelims ()i}{|J|\atopwithdelims ()k-i}\alpha _s^2\\&\le \sum _{i=0}^{k-1}{s\atopwithdelims ()i}{n-s\atopwithdelims ()k-i}|\alpha _s-\alpha _{s+|J|}|^2 + |J|{s+|J| - 1\atopwithdelims ()k-1}\alpha _s^2\\&\le {n\atopwithdelims ()k}\frac{1}{4n^k} + \frac{p(s+p-1)^{k-1}}{(k-1)!}\frac{(n/p)^{2k/(k+1)}}{4n^k}\\&\le \frac{1}{4 k!}+\frac{2^{k-1}}{(k-1)!}p^k(n/p)^{(k-1)k/(k+1)}\frac{(n/p)^{2k/(k+1)}}{4n^k}\\&\le \frac{1}{4k!}+2p^k\frac{(n/p)^k}{4n^k} = \frac{1}{4k!}+\frac{1}{2}\le 1. \end{aligned}$$

Hence our solution is feasible. Its objective value is $\displaystyle \sqrt{{n \atopwithdelims ()k} \alpha _0^2}=\Omega \left( (n/p)^{k/(k+1)}\right)$.

Appendix 5: Parallel Quantum Oracle Interrogation

The following quantum algorithm recovers the complete input $x\in \{0,1\}^n$ with high probability, using roughly n / 2p p-parallel queries:

• 1. With $T=n/2+O(\sqrt{n\log (1/\varepsilon )})$ and $B=\sum _{i=0}^{T}{n\atopwithdelims ()i}$ being the number of $y\in \{0,1\}^n$ with weight $|y|\le T$, set up the n-qubit superposition $\frac{1}{\sqrt{B}}\sum _{y\in \{0,1\}^n:|y|\le T}|y\rangle .$

• 2. Apply the unitary $|y\rangle \mapsto (-1)^{x\cdot y}|y\rangle$. We can implement this using $\lceil {T/p}\rceil$ p-parallel queries for $|y|\le T$: the first batch of p queries would query the first p positions where y has a one and put the answer in the phase; the second batch queries the next p positions, etc.

• 3. Apply a Hadamard transform to all qubits and measure.

To see the correctness of this algorithm, note that the fraction of n-bit strings y that have weight $>T$ is $\ll \varepsilon$. Hence the state obtained in step 2 is very close to the state $\frac{1}{\sqrt{2^n}}\sum _{y\in \{0,1\}^n}(-1)^{x\cdot y}|y\rangle$, whose Hadamard transform is exactly $|x\rangle$.