Exact distributed quantum algorithm for generalized Simon’s problem

Abstract

Simon’s problem is one of the most important problems demonstrating the power of quantum algorithms, as it greatly inspired the proposal of Shor’s algorithm. The generalized Simon’s problem is a natural extension of Simon’s problem and also a special hidden subgroup problem: Given a function $f:\{0,1\}^n \rightarrow \{0,1\}^m$, it is promised that there exists a hidden subgroup $S\le \mathbb {Z}_2^n$ of rank k such that for any $x, y\in {\{0, 1\}}^n$, $f(x) = f(y)$ iff $x \oplus y \in S$. The goal of generalized Simon’s problem is to find the hidden subgroup S. In this paper, we present two key contributions. Firstly, we characterize the structure of the generalized Simon’s problem in distributed scenario and introduce a corresponding distributed quantum algorithm. Secondly, we refine the algorithm to ensure exactness due to the application of quantum amplitude amplification technique. Our algorithm offers exponential speedup compared to the distributed classical algorithm. When contrasted with the quantum algorithm for the generalized Simon’s problem, our algorithm’s oracle requires fewer qubits, thus making it easier to be physically implemented. Particularly, the exact distributed quantum algorithm we develop for the generalized Simon’s problem outperforms the best previously proposed distributed quantum algorithm for Simon’s problem in terms of generalizability and exactness.

Proof of Lemma 3

Lemma 3

Let $\phi =2\arctan {\left( \sqrt{\frac{2^{n-t-r-k_l}}{3\cdot 2^{n-t-r-k_l}-4}}\right) }$, $\varphi = \arccos {\left( \frac{2^{n-t-r-k_l-1}-1}{2^{n-t-r-k_l}-1}\right) }$, where $r=|Y|-1$. Then

$$\begin{aligned} \mathcal {Q}\Big \vert S_l^{\perp },0^{2^tm},S(T)\Big \rangle =\Big \vert \Psi _{X}\Big \rangle . \end{aligned}$$

In line 6 of Algorithm 3, the measurement result z is obtained and satisfies $z\in S_l^{\perp }{\setminus } \langle Y \rangle$ conditions with certainty.

Proof

From Eq. (39), we can write $\mathcal {R}_{0}(\phi )$ as follows.

$$\begin{aligned} \mathcal {R}_{0}(\phi ) =&I^{\otimes n-t + 2^{t+1}m}- \left( 1 - e^{i\phi }\right) \Big \vert 0^{n-t}, 0^{2^{t+1}m}\Big \rangle \Big \langle 0^{n-t}, 0^{2^{t+1}m}\Big \vert . \end{aligned}$$

(A1)

From the definitions of $\mathcal {R}_{\mathcal {A}}(\varphi , Y)$, $\Big \vert \Psi _{X}\Big \rangle$ and $\Big \vert \Psi _{Y}\Big \rangle$, we have

$$\begin{aligned}&\left( \mathcal {R}_{\mathcal {A}}(\varphi , Y)\otimes I^{\otimes {2^{t+1}}m}\right) \Big \vert \Psi _{X}\Big \rangle&= e^{i\varphi }\Big \vert \Psi _{X}\Big \rangle . \end{aligned}$$

(A2)

$$\begin{aligned}&\left( \mathcal {R}_{\mathcal {A}}(\varphi , Y)\otimes I^{\otimes {2^{t+1}}m}\right) \Big \vert \Psi _{Y}\Big \rangle&= \Big \vert \Psi _{Y}\Big \rangle . \end{aligned}$$

(A3)

Let $\mathcal {U}(\mathcal {A}, \phi ) = -\mathcal {A}\mathcal {R}_{0}(\phi )\mathcal {A}^{\dagger }$, then based on Eq. (41), $\mathcal {Q}$ can be written as

$$\begin{aligned} \mathcal {Q} = \mathcal {U}(\mathcal {A}, \phi )\left( \mathcal {R}_{\mathcal {A}}(\varphi , Y)\otimes I^{\otimes {2^{t+1}}m}\right) . \end{aligned}$$

(A4)

For $\mathcal {U}(\mathcal {A}, \phi )$, we have

$$\begin{aligned} \mathcal {U}(\mathcal {A}, \phi )= & {} -\mathcal {A}\mathcal {R}_{0}(\phi )\mathcal {A}^{\dagger }\nonumber \\= & {} -\mathcal {A}\left( I^{\otimes n-t + 2^{t+1}m}\right. - \left. \left( 1 - e^{i\phi }\right) \Big \vert 0^{n-t}, 0^{2^{t+1}m}\Big \rangle \Big \langle 0^{n-t}, 0^{2^{t+1}m}\Big \vert \right) \mathcal {A}^{\dagger }\nonumber \\= & {} \left( 1 - e^{i\phi }\right) \left( \mathcal {A}\Big \vert 0^{n-t}, 0^{2^{t+1}m}\Big \rangle \Big \langle 0^{n-t}, 0^{2^{t+1}m}\Big \vert \mathcal {A}^{\dagger }\right) - I^{\otimes n-t + 2^{t+1}m}\nonumber \\= & {} \left( 1 - e^{i\phi }\right) \Big \vert K^{\perp }, 0^{2^tm}, S(T)\Big \rangle \Big \langle K^{\perp }, 0^{2^tm}, S(T)\Big \vert - I^{\otimes n-t + 2^{t+1}m}\nonumber \\= & {} \left( 1 - e^{i\phi }\right) \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big )\big (\Big \langle \Psi _{X}\Big \vert +\Big \langle \Psi _{Y}\Big \vert \big ) - I^{\otimes n-t + 2^{t+1}m}. \end{aligned}$$

(A5)

Since $| \langle Y \rangle | = 2 ^ {r}$and $| S_l ^ {\perp } | = 2 ^ {n-t - k_l}$, according to the definitions of $\Big \vert \Psi _ {X}\Big \rangle$ and $\Big \vert \Psi _ {Y}\Big \rangle$, we have

$$\begin{aligned} \langle \Psi _{X}|\Psi _{X}\rangle= & {} 1-2^{r+k_l+t-n}.\nonumber \\ \langle \Psi _{Y}|\Psi _{Y}\rangle= & {} 2^{r+k_l+t-n}.\nonumber \\ \langle \Psi _{X}|\Psi _{Y}\rangle= & {} 0. \end{aligned}$$

(A6)

Thus, we can obtain the following equations.

$$\begin{aligned} \mathcal {Q}\Big \vert \Psi _{X}\Big \rangle= & {} \mathcal {U}(\mathcal {A}, \phi )\left( \mathcal {R}_{\mathcal {A}}(\varphi , Y)\otimes I^{\otimes 2^{t+1}m}\right) \Big \vert \Psi _{X}\Big \rangle \nonumber \\= & {} e^{i\varphi }\mathcal {U}(\mathcal {A}, \phi )\Big \vert \Psi _{X}\Big \rangle \nonumber \\= & {} e^{i\varphi }\Big (\left( 1 - e^{i\phi }\right) \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big )\big (\Big \langle \Psi _{X}\Big \vert +\Big \langle \Psi _{Y}\Big \vert \big ) - I^{\otimes n-t + 2^{t+1}m}\Big )\Big \vert \Psi _{X}\Big \rangle \nonumber \\= & {} e^{i\varphi }\left( 1 - e^{i\phi }\right) \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big )\big (\Big \langle \Psi _{X}\Big \vert +\Big \langle \Psi _{Y}\Big \vert \big ) \Big \vert \Psi _{X}\Big \rangle - e^{i\varphi }\Big \vert \Psi _{X}\Big \rangle \nonumber \\= & {} e^{i\varphi }\left( 1 - e^{i\phi }\right) \langle \Psi _{X}|\Psi _{X}\rangle \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big ) - e^{i\varphi }\Big \vert \Psi _{X}\Big \rangle \nonumber \\= & {} e^{i\varphi }\left( 1 - e^{i\phi }\right) \left( 1-2^{r+k_l+t-n}\right) \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big ) - e^{i\varphi }\Big \vert \Psi _{X}\Big \rangle \nonumber \\= & {} e^{i\varphi }\left( (1-e^{i\phi })(1-2^{r+k_l+t-n})-1\right) \Big \vert \Psi _{X}\Big \rangle + e^{i\varphi }(1-e^{i\phi })(1-2^{r+k_l+t-n})\Big \vert \Psi _{Y}\Big \rangle .\nonumber \\ \end{aligned}$$

(A7)

$$\begin{aligned} \mathcal {Q}\Big \vert \Psi _{Y}\Big \rangle= & {} \mathcal {U}(\mathcal {A}, \phi )\left( \mathcal {R}_{\mathcal {A}}(\varphi , Y)\otimes I^{\otimes 2^{t+1}m}\right) \Big \vert \Psi _{Y}\Big \rangle \nonumber \\= & {} \mathcal {U}(\mathcal {A}, \phi )\Big \vert \Psi _{Y}\Big \rangle \nonumber \\= & {} \Big (\left( 1 - e^{i\phi }\right) \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big )\big (\Big \langle \Psi _{X}\Big \vert +\Big \langle \Psi _{Y}\Big \vert \big ) - I^{\otimes n-t + 2^{t+1}m}\Big )\Big \vert \Psi _{Y}\Big \rangle \nonumber \\= & {} \left( 1 - e^{i\phi }\right) \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big )\big (\Big \langle \Psi _{X}\Big \vert +\Big \langle \Psi _{Y}\Big \vert \big )\Big \vert \Psi _{Y}\Big \rangle - \Big \vert \Psi _{Y}\Big \rangle \nonumber \\= & {} \left( 1 - e^{i\phi }\right) \langle \Psi _{Y}|\Psi _{Y}\rangle \big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big ) - \Big \vert \Psi _{Y}\Big \rangle \nonumber \\= & {} \left( 1 - e^{i\phi }\right) 2^{r+k_l+t-n}\big (\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle \big ) - \Big \vert \Psi _{Y}\Big \rangle \nonumber \\= & {} (1-e^{i\phi })2^{r+k_l+t-n}\Big \vert \Psi _{X}\Big \rangle -\left( (1-e^{i\phi })(1-2^{r+k_l+t-n})+e^{i\phi } \right) \Big \vert \Psi _{Y}\Big \rangle . \end{aligned}$$

(A8)

By making sure the resulting superposition $\mathcal {Q}(\Big \vert \Psi _{X}\Big \rangle +\Big \vert \Psi _{Y}\Big \rangle )$ has inner product zero with $\Big \vert \Psi _{Y}\Big \rangle$, then based on Eqs. (A7) and (A8), we can obtain the following equation.

$$\begin{aligned} e^{i\varphi }(1-e^{i\phi })(1-2^{r+k_l+t-n}) =(1-e^{i\phi })(1-2^{r+k_l+t-n})+e^{i\phi }. \end{aligned}$$

(A9)

Denote by

$$\begin{aligned} b=1-2^{r+k_l+t-n}. \end{aligned}$$

(A10)

Then according to Eq. (A9), we have

$$\begin{aligned} b= & {} e^{-i\varphi }\left( b + \frac{1}{e^{-i\phi } - 1}\right) \nonumber \\= & {} e^{-i\varphi }\left( b + \frac{1}{\cos \phi - 1 - i\sin \phi }\right) \nonumber \\= & {} e^{-i\varphi }\left( b + \frac{\cos \phi - 1}{\left( \cos \phi - 1\right) ^2 + \sin ^2\phi }\right. \left. + i\frac{\sin \phi }{\left( \cos \phi - 1\right) ^2 + \sin ^2\phi }\right) \nonumber \\= & {} e^{-i\varphi }\left( b - \frac{1}{2} + i\frac{\sin \phi }{2 - 2\cos \phi }\right) . \end{aligned}$$

(A11)

Taking the square of b, we can further have

$$\begin{aligned} b^2 = \left( b - \frac{1}{2}\right) ^2 + \frac{\sin ^2\phi }{4\left( 1 - \cos \phi \right) ^2}. \end{aligned}$$

(A12)

Arrange Eq. (A12) to get

$$\begin{aligned} 4b - 1= & {} \frac{\sin ^2\phi }{\left( 1 - \cos \phi \right) ^2}\nonumber \\= & {} \cot ^2\frac{\phi }{2}. \end{aligned}$$

(A13)

From Eq. (A13), we can obtain

$$\begin{aligned} \phi= & {} 2\arctan {\left( \sqrt{\frac{1}{4b-1}}\right) }\nonumber \\= & {} 2\arctan {\left( \sqrt{\frac{2^{n-t-r-k_l}}{3\cdot 2^{n-t-r-k_l}-4}}\right) }. \end{aligned}$$

(A14)

Since b is a real number, Eq. (A11) also needs to be a real number, and we can further obtain

$$\begin{aligned} \varphi= & {} \arccos {\left( \frac{b-\frac{1}{2}}{b}\right) }\nonumber \\= & {} \arccos {\left( \frac{2^{n-t-r-k_l-1}-1}{2^{n-t-r-k_l}-1}\right) }. \end{aligned}$$

(A15)

Thus, from the above derivation, it follows that if $\phi =2\arctan {\left( \sqrt{\frac{2^{n-t-r-k_l}}{3\cdot 2^{n-t-r-k_l}-4}}\right) }$, $\varphi = \arccos {\left( \frac{2^{n-t-r-k_l-1}-1}{2^{n-t-r-k_l}-1}\right) }$, where $r=|Y|-1$, we have $\mathcal {Q}\Big \vert S_l^{\perp },0^{2^tm},S(T)\Big \rangle =\Big \vert \Psi _{X}\Big \rangle$, and $z\in X={S_l^{\perp }}{\setminus } \langle Y \rangle$ is obtained in line 6 of Algorithm 3, after measuring the first register of the quantum state $Q\Big \vert S_l^{\perp },0^{2^tm}, S(T)\Big \rangle$. $\square$