Quantum context-aware recommendation systems based on tensor singular value decomposition

Abstract

In this paper, we propose a quantum algorithm for recommendation systems which incorporates the contextual information of users to the personalized recommendation. The preference information of users is encoded in a third-order tensor of dimension N which can be approximated by the truncated tensor singular value decomposition (t-svd) of the subsample tensor. Unlike the classical algorithm that reconstructs the approximated preference tensor using truncated t-svd, our quantum algorithm obtains the recommended product under certain context by measuring the output quantum state corresponding to an approximation of a user’s dynamic preferences. The algorithm achieves the time complexity $\mathcal {O}(\sqrt{k}N\mathrm{polylog}(N))$, compared to the classical counterpart with complexity $\mathcal {O}(kN^3)$, where k is the truncated tubal rank.

Appendices

A The proof of Theorem 4

Before proving Theorem 4, we would like to sketch the proof first. According to Assumption 1, each frontal slice $T^{(k)}$ is stored in the binary tree structure. Hence, based on the proof of Lemma 3 in [18], the states $| \mathcal {T}(i,:,k) \rangle$, corresponding to the i-th row of $T^{(k)}$, can be prepared efficiently by operators $P_k$. Based on these operators, two new isometries $\hat{P}_m$ and $\hat{Q}_m$ are constructed in order to perform QSVE on $\hat{T}^{(m)}$. The detail of the proof is given below.

Proof

Since every $T^{(k)}$, $k=0,\ldots ,N-1$, is stored in the binary tree structure, the quantum computer can perform the following mapping in $\mathcal {O}(\mathrm{polylog}(N))$ time, as shown in Theorem 5.1 in [18]:

$$\begin{aligned}&U_{P_k}: | i \rangle | 0 \rangle \nonumber \\&\quad \rightarrow | i \rangle | \mathcal {T}(i,:,k) \rangle =\frac{1}{||\mathcal {T}(i,:,k)||_2}\sum _{j=0}^{N-1}\mathcal {T}_{ijk}| i \rangle | j \rangle , \end{aligned}$$

(29)

where $\mathcal {T}(i,:,k)$ is the i-th row of $T^{(k)}$.

Define the degenerate operator $P_k \in \mathbb {R}^{N^2 \times N}$ related to $U_{P_k}$ as

$$\begin{aligned} P_k: | i \rangle \rightarrow | i \rangle | \mathcal {T}(i,:,k) \rangle . \end{aligned}$$

(30)

That is,

$$\begin{aligned} P_k=\sum _{i=0}^{N-1}| i \rangle | \mathcal {T}(i,:,k) \rangle |\langle i |. \end{aligned}$$

(31)

Based on the efficiently implemented operator $U_{P_k}$, we define another operator

$$\begin{aligned} U_{\hat{P}_m}\triangleq \frac{1}{\sqrt{N}}\sum _{k=0}^{N-1}\omega ^{km}U_{P_k.} \end{aligned}$$

(32)

It can be easily seen that the operator $U_{\hat{P}_m}$ achieves the state preparation of the rows of the matrix $\hat{T}^{(m)}$, i.e., $U_{\hat{P}_m}=\frac{1}{\sqrt{N}}\sum _{k=0}^{N-1}\sum _{i=0}^{N-1}\omega ^{km}| i \rangle | \mathcal {T}(i,:,k) \rangle |\langle i ||\langle 0 |=\sum _{i}| i \rangle | \hat{\mathcal {T}}(i,:,m) \rangle |\langle i ||\langle 0 |$, where $| \hat{\mathcal {T}}(i,:,m) \rangle$ is the state of the i-th row of $\hat{T}^{(m)}$. Similarly, the isometry corresponding to $U_{\hat{P}_m}$ is $\hat{P}_m=\sum _{i}| i \rangle | \hat{\mathcal {T}}(i,:,m) \rangle |\langle i |$. It can be easily shown that $\hat{P}_m$ is an isometry since $\hat{P}_m^{\dagger }\hat{P}_m=I_{N}$.

Since $U_{P_k}$ can be implemented in time $\mathcal {O}(\mathrm{polylog}(N))$, $U_{\hat{P}_m}$ can be implemented in time $\mathcal {O}(N\mathrm{polylog}N)$ using the linear combination of unitaries (LCU) technique [6, 23, 24, 43, 44].

The LCU technique was first proposed by Long in their work [23] in a more general form, and Shao summarized this result in [44]. The problem of LCU can be formulated as follows: Given $\alpha _i \in \mathbb {C}$ and unitary operators $U_i$, $i = 0, 1, \ldots , N-1$, implement linear operator $L=\sum _{j=0}^{N-1}\alpha _jU_j$. The algorithm stated in [6] implements L in time $\mathcal {O}((T_\mathrm{in} +\mathrm{log}N)N\max _j |\alpha _j|/||L| \psi \rangle ||)$, where $| \psi \rangle$ is any given initial state and $T_\mathrm{in}$ is the time to implement $U_0, U_1, \ldots , U_{N-1}$. In our case, $T_\mathrm{in}=N\mathrm{polylog}N$, $\alpha _j=\omega ^{km}/\sqrt{N}$ and the input state is chosen as $| \psi \rangle =\sum _{i=0}^{N}| i \rangle | 0 \rangle$. Thus, $||L| \psi \rangle ||=||\frac{1}{\sqrt{N}}\sum _{k=0}^{N-1}\omega ^{km}U_{P_k}| \psi \rangle ||=||\frac{1}{\sqrt{N}}\sum _{k=0}^{N-1}\sum _{i=0}^{N-1}\omega ^{km}| i \rangle | \mathcal {T}(i,:,k) \rangle ||=\sqrt{N}$. Therefore, the complexity to implement $U_{\hat{P}_m}$ is $\mathcal {O}(N\mathrm{polylog}N)$.

Next, we define the mapping

$$\begin{aligned}&U_{\hat{Q}_m}: | 0 \rangle | j \rangle \rightarrow | \varvec{s}_{\hat{T}^{(m)}} \rangle | j \rangle \nonumber \\&\quad =\frac{1}{||\hat{T}^{(m)}||_F}\sum _{i}||\hat{\mathcal {T}}(i,:,m)||_2| i \rangle | j \rangle , \end{aligned}$$

(33)

where $\varvec{s}_{\hat{T}^{(m)}}$ is a vector whose i-th entry is $\frac{||\hat{\mathcal {T}}(i,:,m)||}{||\hat{T}^{(m)}||_F}$. Similar with $U_{\hat{P}_m}$, the corresponding isometry is defined as $\hat{Q}_m=\sum _{j}| \varvec{s}_{\hat{T}^{(m)}} \rangle | j \rangle |\langle j |$ satisfying $\hat{Q}_m^{\dagger }\hat{Q}_m=I_{N}$, which can be verified easily.

Now we perform QSVE on the matrix $\hat{T}^{(m)}$. First, the factorization $\frac{\hat{T}^{(m)}}{||\hat{T}^{(m)}||_F}=\hat{P}_m^{\dagger }\hat{Q}_m$ can be easily verified.

Second, we can prove that $2\hat{P}_m\hat{P }_m^{\dagger }-I_{N^2}$ is a reflection and it be implemented through $U_{\hat{P}_m}$. Actually,

$$\begin{aligned}&2\hat{P}_m\hat{P }_m^{\dagger }-I_{N^2} \nonumber \\&\quad =2\sum _{i}| i \rangle | \hat{\mathcal {T}}(i,:,m) \rangle |\langle i ||\langle \hat{\mathcal {T}}(i,:,m) |-I_{N^2} \nonumber \\&\quad =U_{\hat{P}_m}\left[ 2\sum _{i}| i \rangle | 0 \rangle |\langle i ||\langle 0 |-I_{N^2} \right] U_{\hat{P}_m}^{\dagger }, \end{aligned}$$

(34)

where $2\sum _{i}| i \rangle | 0 \rangle |\langle i ||\langle 0 |-I_{N^2}$ is a reflection. The similar result holds for $2\hat{Q}_m\hat{Q}_m^{\dagger }-I_{N^2}$.

Now denote

$$\begin{aligned} W_m=\left( 2\hat{P}_m\hat{P}_m^{\dagger }-I_{N^2}\right) \left( 2\hat{Q}_m\hat{Q}_m^{\dagger }-I_{N^2}\right) . \end{aligned}$$

(35)

Let $\hat{T}^{(m)} = \sum _{i=0}^{r-1}\hat{\sigma }_i^{(m)}\hat{u}_i^{(m)}\hat{v}_i^{(m)\dagger }$ be the singular value decomposition of $\hat{T}^{(m)}$. We can prove that the subspace spanned by $\{\hat{Q}_m| \hat{v}_i^{(m)} \rangle , \hat{P}_m| \hat{u}_i^{(m)} \rangle \}$ is invariant under the unitary transformation $W_m$:

$$\begin{aligned}&W_m\hat{Q}_m| \hat{v}_i^{(m)} \rangle =\frac{2\hat{\sigma }_i^{(m)}}{||\hat{T}^{(m)}||_F}\hat{P}_m| \hat{u}_i^{(m)} \rangle -Q| \hat{v}_i^{(m)} \rangle , \\&W_m\hat{P}_m| \hat{u}_i^{(m)} \rangle = \\&\quad \left( \frac{4\hat{\sigma }_{i}^2}{||\hat{T}^{(m)}||_F}-1 \right) \hat{P}_m| \hat{u}_i^{(m)} \rangle -\frac{2\hat{\sigma }_{i}}{||\hat{T}^{(m)}||_F}\hat{Q}_m| \hat{v}_i^{(m)} \rangle . \end{aligned}$$

The matrix $W_m$ can be calculated under an orthonormal basis using the Schmidt orthogonalization. It is a rotation in the subspace spanned by its eigenvectors $| \omega ^{(m)}_{i \pm } \rangle$ with correspondent eigenvalues $\mathrm{e}^{\pm i\theta _i^{(m)}}$, where $\theta _i^{(m)}$ is the rotation angle satisfying

$$\begin{aligned} \cos (\theta _i^{(m)}/2)=\frac{\hat{\sigma }_i^{(m)}}{||\hat{T}^{(m)}||_F}, \end{aligned}$$

(36)

that is,

$$\begin{aligned}&\hat{Q}_m| \hat{v}_i^{(m)} \rangle =\sqrt{2}\left( | \omega ^{(m)}_{i +} \rangle +| \omega ^{(m)}_{i -} \rangle \right) , \\&\hat{P}_m| \hat{u}_i^{(m)} \rangle =\sqrt{2}\left( \mathrm{e}^{ i\theta _i/2}| \omega ^{(m)}_{i +} \rangle +\mathrm{e}^{ -i\theta _i/2}| \omega ^{(m)}_{i -} \rangle \right) . \end{aligned}$$

Here, we choose the input state $| \hat{\mathcal {T}}(i,:,m) \rangle$ represented in (12), then

$$\begin{aligned}&\hat{Q}_m| \hat{\mathcal {T}}(i,:,m) \rangle =\sum _{j=0}^{N-1}\sqrt{2}\beta _{j}^{(im)}\left( | \omega ^{(m)}_{i +} \rangle +| \omega ^{(m)}_{i -} \rangle \right) . \end{aligned}$$

(37)

We perform the phase estimation on $W_m$ with running time $\mathcal {O}\left( N\mathrm{polylog}N/{\epsilon _\mathrm{SVE}^{(m)}}\right)$ and compute the estimated singular value of $\hat{T}^{(m)}$ through oracle with a computable function $f(x)=||\hat{T}^{(m)}||_F\cos (x/2)$. According to the relations between $\theta _i^{(m)}$ and $\hat{\sigma }_i^{(m)}$ in (36), we obtain

$$\begin{aligned} \sum _{j=0}^{N-1}\sqrt{2}\beta _{j}^{(im)}\left( | \omega ^{(m)}_{i +} \rangle | \overline{\theta }_i^{(m)} \rangle + | \omega ^{(m)}_{i -} \rangle | -\overline{\theta }_i^{(m)} \rangle \right) | \overline{\sigma }_{i}^{(m)} \rangle . \end{aligned}$$

(38)

We next uncompute the phase estimation procedure and then apply the inverse of $U_{\hat{Q}_m}$ to obtain the desired state (13) in Theorem 4. $\square$

B The proof of Lemma 5

Proof

Let $\sigma _i$ denote the singular value of A and l be the largest integer for which $\sigma _{l} \ge \frac{\epsilon ||A||_F}{\sqrt{k}}.$ By the triangle inequality, $||A-A_{\ge \sigma }||_F \le ||A-A_k||_F +||A_k-A_{\ge \sigma }||_F.$ If $k \le l$, it’s easy to conclude that $||A_k-A_{\ge \sigma }||_F \le ||A-A_k||_F \le \epsilon ||A||_F$. If $k > l$, $||A_k-A_{\ge \sigma }||_F^2=\sum _{i=l+1}^{k}\sigma _i^2 \le k\sigma _{l+1}^2 \le k\left( \frac{\epsilon ||A||_F}{\sqrt{k}}\right) ^2 \le \left( \epsilon ||A||_F\right) ^2$. In either case, we have $||A-A_{\ge \sigma }||_F \le 2\epsilon ||A||_F$. $\square$

The proof of Theorem 5

Proof

Based on Lemma 5 in the main text, if the best rank-k approximation satisfies $||\hat{T}^{(m)}-\hat{T}^{(m)}_{k}||_F \le \epsilon ^{(m)}||\hat{T}^{(m)}||_F,$ then

$$\begin{aligned} ||\hat{T}^{(m)}-\hat{T}^{(m)}_{\ge \tau _m}||_F \le 2 \epsilon ^{(m)}||\hat{T}^{(m)}||_F \le \epsilon _0||\hat{T}^{(m)}||_F \end{aligned}$$

(39)

for $m=0, \ldots , N-1.$ By summing both sides of (39) over m, we get

$$\begin{aligned} ||\hat{\mathcal {T}}-\hat{\mathcal {T}}_{\ge \varvec{\tau }}||_F^2 = \sum _{m=0}^{N-1}||\hat{T}^{(m)}-\hat{T}^{(m)}_{\ge \tau _m}||_F^2 \le \epsilon _0^2||\hat{\mathcal {T}}||_F^2. \end{aligned}$$

(40)

Since the inverse QFT along the third mode of the tensor $\mathcal {T}$ cannot change the Frobenius norm of its horizontal slice, (40) can be rewritten as

$$\begin{aligned} ||\mathcal {T}-\mathcal {T}_{\ge \varvec{\tau }}||_F^2 \le \epsilon _0^2||\mathcal {T}||_F^2. \end{aligned}$$

(41)

Moreover, noticing that $||\mathcal {T}-\mathcal {T}_{\ge \varvec{\tau }}||_F^2=\sum _{i=0}^{N-1}||\mathcal {T}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F^2$, we have ${E}\left( ||\mathcal {T}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F^2\right) \le \frac{\epsilon _0^2||\mathcal {T}||_F^2}{N}.$ Due to Markov’s Inequality [41, Proposition 2.6], for $\delta \in (0,1)$,

$$\begin{aligned}&\Pr \left( ||\mathcal {T}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F^2 > \frac{\epsilon _0^2||\mathcal {T}||_F^2}{\delta N}\right) \nonumber \\&\quad \le \frac{{E}\left( ||\mathcal {T}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F^2\right) \delta N}{\epsilon _0^2||\mathcal {T}||_F^2} \le \delta \end{aligned}$$

(42)

holds. This means at least $(1-\delta )N$ users i satisfy

$$\begin{aligned} ||\mathcal {T}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F^2 \le \frac{\epsilon _0^2||\mathcal {T}||_F^2}{\delta N}. \end{aligned}$$

(43)

Notice ${E}\left( ||\mathcal {T}||_F^2\right) =||\mathcal {A}||_F^2/p$. Using the Chernoff bound, we have $\Pr \left( ||\mathcal {T}||_F^2 >(1+\theta )||\mathcal {A}||_F^2/p\right) \le \mathrm{e}^{-\theta ^2||\mathcal {A}||_F^2/3p}$ for $\theta \in [0,1]$, which is exponentially small. Here, we choose $\theta =1/10$, then $||\mathcal {T}||_F^2 \le 11||\mathcal {A}||_F^2/10p$.

Based on the second assumption in Assumption 1, we sum both sides of (6) for m and i, respectively, obtaining

$$\begin{aligned} \frac{1}{1+\gamma }\frac{||\mathcal {A}||_F^2}{N} \le ||\mathcal {A}(i,:,:)||_F^2 \le (1+\gamma )\frac{||\mathcal {A}||_F^2}{N}, \end{aligned}$$

(44)

and

$$\begin{aligned} \frac{1}{1+\gamma }\frac{||\mathcal {A}||_F^2}{N} \le ||A^{(m)}||_F^2 \le (1+\gamma )\frac{||\mathcal {A}||_F^2}{N}. \end{aligned}$$

(45)

Then, (43) becomes

$$\begin{aligned} ||\mathcal {T}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F^2 \le \frac{11\epsilon _0^2(1+\gamma )}{10\delta p}||\mathcal {A}(i,:,:)||_F^2. \end{aligned}$$

(46)

Meanwhile, since

$$\begin{aligned} {E}\left( ||\mathcal {A}(i,:,:)-\mathcal {T}(i,:,:)||_F^2\right) =\left( \frac{1}{p}-1\right) ||\mathcal {A}(i,:,:)||_F^2, \end{aligned}$$

then

$$\begin{aligned}&\Pr \left( ||\mathcal {A}(i,:,:)-\mathcal {T}(i,:,:)||_F^2 > \nu ||\mathcal {A}(i,:,:)||_F^2\right) \nonumber \\&\quad \le \mathrm{e}^{-\zeta ^2\left( \frac{1}{p}-1\right) \frac{||\mathcal {A}||_F^2}{3N(1+\gamma )}}, \end{aligned}$$

(47)

where $\nu =(1+\zeta )\left( \frac{1}{p}-1\right)$ and $\zeta \in (0,1)$. This means with probability at least $p_1$,

$$\begin{aligned} ||\mathcal {A}(i,:,:)-\mathcal {T}(i,:,:)||_F^2 \le \nu ||\mathcal {A}(i,:,:)||_F^2. \end{aligned}$$

(48)

Combining (46) and (48) together and by triangle inequality, we obtain

$$\begin{aligned} ||\mathcal {A}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F \le \epsilon ||\mathcal {A}(i,:,:)||_F. \end{aligned}$$

(49)

According to Lemma 4, the probability that sampling according to $\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)$ provides a bad recommendation is

$$\begin{aligned} \Pr \limits _{t\sim \mathcal {U}_N, j\sim \mathcal {T}_{\ge \varvec{\tau }}(i,:,t)} [(i,j,t) \mathrm{bad}] \le \left( \frac{\epsilon }{1-\epsilon }\right) ^2. \end{aligned}$$

(50)

$\square$

D. The proof of Theorem 6

Proof

According to Theorem 4, the modified QSVE algorithm performed on the frontal slice $\hat{T}^{(m)}$ takes time $\mathcal {O}\left( N\mathrm{polylog}(N)/{\epsilon _\mathrm{SVE}^{(m)}}\right)$.

In Step 5 of Algorithm 2, we need to repeat the measurement $\mathcal {O}\left( \frac{||\mathcal {T}(i,:,:)||_F}{||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F}\right)$ times in order to ensure the probability of getting the outcome $| 0 \rangle$ in this step is close to 1. For most users, we can prove that $\frac{||\mathcal {T}(i,:,:)||_F}{||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F}$ is bounded and the upper bound is a constant for appropriate parameters. The proof is in the following.

Since ${E}\left( ||\mathcal {T}(i,:,:)||_F^2\right) =\frac{||\mathcal {T}(i,:,:)||_F^2}{p} \le (1+\gamma )\frac{||\mathcal {A}||_F^2}{pN},$ then by Chernoff bound,

$$\begin{aligned} ||\mathcal {T}(i,:,:)||_F^2 \le \frac{2(1+\gamma )||\mathcal {A}||_F^2}{pN} \end{aligned}$$

(51)

holds with probability close to 1. Moreover, by Theorem 5, there are at least $(1-\delta )N$ users satisfying $||\mathcal {A}(i,:,:)-\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F\le \epsilon ||\mathcal {A}(i,:,:)||_F$, then $(1+\epsilon )||\mathcal {A}(i,:,:)||_F \le ||\mathcal {T}_{\ge \varvec{\tau }}(i,:,:)||_F \le (1+\epsilon )||\mathcal {A}(i,:,:)||_F.$ Since the Frobenius norm is unchanged under the Fourier transform, we get

$$\begin{aligned} (1+\epsilon )||\hat{T}_{(i)}||_F \le ||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F \le (1+\epsilon )||\hat{T}_{(i)}||_F. \end{aligned}$$

(52)

Therefore,

$$\begin{aligned} ||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F^2 \ge (1+\epsilon )^2||\hat{T}_{(i)}||_F^2\ge \frac{(1+\epsilon )^2}{1+\gamma }\frac{||\mathcal {A}||_F^2}{N}. \end{aligned}$$

(53)

Combining (51) and (53) together, we can conclude that for at least $(1-\delta )N$ users, $\frac{||\mathcal {T}(i,:,:)||_F}{||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F}$ is bounded, that is,

$$\begin{aligned} \frac{||\mathcal {T}(i,:,:)||_F}{||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F} \le \left( \frac{(1+\gamma )\frac{2||\mathcal {A}||_F^2}{pN}}{\frac{(1+\epsilon )^2}{1+\gamma }\frac{||\mathcal {A}||_F^2}{N}}\right) ^{1/2} = \frac{\sqrt{2}(1+\gamma )}{(1+\epsilon )\sqrt{p}}. \end{aligned}$$

(54)

The precision for the singular value estimation algorithm on the matrix $||\hat{T}^{(m)}||_F$ can be chosen as $\epsilon _\mathrm{SVE}^{(m)}=\frac{\tau _m}{||\hat{T}^{(m)}||_F}.$ Therefore, the total time complexity of Algorithm 2 is

$$\begin{aligned}&({\mathrm{log}N})^4 \cdot \frac{N\mathrm{polylog}(N)}{\min \limits _{m}{\epsilon _\mathrm{SVE}^{(m)}}} \cdot \frac{||\mathcal {T}(i,:,:)||_F}{||\hat{\mathcal {T}}_{\ge \varvec{\tau }}(i,:,:)||_F} \\&\quad \le ({\mathrm{log}N})^4{N\mathrm polylog}(N) \max \limits _{m}{\frac{||\hat{T}^{(m)}||_F}{\tau _m}}\cdot \frac{\sqrt{2}(1+\gamma )}{(1+\epsilon )\sqrt{p}} \\&\quad \approxeq \frac{\sqrt{k}N\mathrm{polylog}(N) (1+\gamma )}{\min \limits _{m}{\epsilon ^{(m)}(1+\epsilon )\sqrt{p}}}, \end{aligned}$$

where $\epsilon ^{(m)}$ and $\epsilon$ are defined in Theorem 5. $\square$

E The quantum cost

Defining the cost of quantum circuits is not an easy task due to the fact that each quantum computer model may have a different cost for a given quantum gate. Here, the quantum cost of a reversible gate is defined to be the number of $1\times 1$ and $2\times 2$ reversible gates or quantum logic gates required in its design. The quantum costs of all reversible $1\times 1$ and $2\times 2$ gates are taken as unity [4, 42, 49, 50]. The cost of a circuit is calculated by summing up the costs of the gates composing the circuit. First, we analyze the quantum cost of QFT and QSVE circuits and then we analyze the cost of each gate in Fig. 3.

1.1 E.1 The cost of QFT

The QFT under an orthonormal basis $|x\rangle \in \{|0\rangle , \ldots ,|N-1\rangle \}$ is defined as the linear operator with the following action on the basis vectors:

$$\begin{aligned} \mathrm {QFT}:|x\rangle \rightarrow \frac{1}{\sqrt{N}} \sum _{k=0}^{N-1} \omega ^{x \cdot k}|k\rangle , \end{aligned}$$

where $\omega =\mathrm{e}^{\frac{2\pi \mathrm {i}}{N}}$. The inverse QFT is then defined as

$$\begin{aligned} \mathrm {QFT}^{\dagger }:|k\rangle \rightarrow \frac{1}{\sqrt{N}} \sum _{x=0}^{N-1} \omega ^{-k \cdot x}|x\rangle . \end{aligned}$$

The circuit of QFT, shown in Fig. 7, is composed of a total number of $\mathcal {O}\left( \lceil \mathrm{log}N \rceil ^2\right)$ H gates, CNOT gates and controlled phase gate $R_{m}$ (see [27, Section 5.1]), where

$$\begin{aligned} H=\frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 &{} 1 \\ 1 &{} -1 \end{array}\right) , \quad \quad R_{m}=\left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} \mathrm{e}^{\frac{2 \pi \mathrm {i}}{2^{m}}} \end{array}\right) \end{aligned}$$

with $m=2, \ldots , n$, $n=\lceil \mathrm{log}N \rceil$.

Fig. 7

The circuit of QFT

1.2 E.2 The cost of QSVE

The QSVE algorithm, originally proposed in [18], is introduced in the preliminary part of the manuscript. The QSVE algorithm and its circuit are, respectively, given in Algorithm 3 and Fig. 8. In what follows, we focus on the quantum cost of this algorithm.

Let $A \in \mathbb {R}^{N \times N}$ be a matrix with singular value decomposition $A=\sum _{i} \sigma _{i} u_{i} v_{i}^\mathrm{T}$ stored in the binary tree data structure introduced in Lemma 2. In Step 2 of Algorithm 3, the unitary $U_Q$ corresponding to Q is defined as

$$\begin{aligned} U_Q: \left| 0 \right\rangle \left| j \right\rangle \rightarrow |\widetilde{A}, j\rangle =\frac{1}{\Vert A\Vert _{F}} \sum _{i \in [N]}\left\| A_{i}\right\| |i, j\rangle \end{aligned}$$

(55)

for $j =0,\ldots ,N-1$, where $\widetilde{A}$ is a vector whose i-th row is $\widetilde{A}_{i}=\left\| A_{i}\right\|$ for $i \in [N].$ Since $\widetilde{A}$ is stored in a classical binary tree with depth $\lceil \mathrm{log}N \rceil$, $U_Q$ can be implemented by performing $\lceil \mathrm{log}N \rceil$ controlled rotations on $| 0 \rangle ^{\otimes \lceil \mathrm{log}N \rceil }$ (see [18, Appendix A]). To implement each controlled rotation $\sum _{\tilde{\theta } \in \{0,1\}^{\otimes \lceil \mathrm{log}N \rceil }}|\widetilde{\theta }\rangle \langle \widetilde{\theta }| \otimes \mathrm{e}^{-\mathrm {i} Y \tilde{\theta }}$ with Pauli Y matrix $Y=\begin{bmatrix} 0 &{} -\mathrm {i} \\ \mathrm {i} &{} 0 \\ \end{bmatrix}$, we use one rotation controlled on each qubit of the first register which can be implemented with cost ${{\mathcal {O}}}(\lceil \mathrm{log}N \rceil )$ (see [52, Lemma 2]). Therefore, the cost of implementing $U_Q$ in Step 2 is $\sum _{\lceil \mathrm{log}N \rceil }{\lceil \mathrm{log}N \rceil }^2$, i.e., $\mathcal {O}({\lceil \mathrm{log}N \rceil }^3)$.

Fig. 8

The circuit of QSVE algorithm. $U_{f}$ is a unitary operator implemented through oracle with a computable function $f(x)=||A||_F\cos (x/2)$. $w=\mathrm{log}N-1$

In Step 3, we can prove that the quantum cost of implementing the unitary W is $\mathcal {O}(N{\lceil \mathrm{log}N \rceil }^3)$. To be specific, notice that $W=U \cdot V,$ where $V=2 Q Q^\mathrm{T}-I_{N^2}$, $U=2 P P^\mathrm{T}-I_{N^2}=U_{P}\left( 2 \sum _{i}|i\rangle |0\rangle \langle i|\langle 0|-I_{N^2}\right) U_{P}^{\dagger }$, and $U_P: \left| i\right\rangle \left| 0\right\rangle \rightarrow \left| i, A_{i}\right\rangle$ for $i \in [N]$.

The unitary operator $2 \sum _{i}|i\rangle |0\rangle \langle i|\langle 0|-I_{N^2}$ can be realized by the circuit shown in Fig. 9. Since each row of the matrix A is stored in the classical binary tree structure, $U_P$ can be implemented with cost $\mathcal {O}(N{\lceil \mathrm{log}N \rceil }^3)$, as compared to the cost $\mathcal {O}({\lceil \mathrm{log}N \rceil }^3)$ of implementing $U_Q$. To summarize, a total cost of $\mathcal {O}(N{\lceil \mathrm{log}N \rceil }^3)$ is required to implement W if we ignore less significant cost. Next, we perform phase estimation on W. When the singular values of W is precise to the $\lceil \mathrm{log}N \rceil$-th bit, the total number of rotation gate invocations is $\mathcal {O}\left( \lceil \mathrm{log}N \rceil ^2\right)$ (See [5, Section 5.2]). The sequence of controlled-$W^{2^j}$, $j=0, \ldots , \lceil \mathrm{log}N \rceil -1$ operations in the phase estimation procedure can be implemented using $\mathcal {O}(\lceil \mathrm{log}N \rceil ^3)$ gates by the modular exponentiation technique (see [27, Box 5.2]). Thus, the cost of Step 3 is $\mathcal {O}(N{\lceil \mathrm{log}N \rceil }^3)$, which dominates the cost of QSVE algorithm.

Fig. 9

Quantum circuit to run $2 \sum _{i}| i \rangle ^{\otimes n}| 0 \rangle |\langle i |^{\otimes n}\langle 0|-I^{\otimes (n+1)}$ for the input state with $n+1$ qubits. The block Z is the Pauli Z gate

To summarize, the quantum cost of QSVE is $\mathcal {O}(N{\lceil \mathrm{log}N \rceil }^3)$.

1.3 E.3 The cost of the circuit in Fig. 3 for Algorithm 2

As discussed in Appendix E.1, the circuit of QFT in Step 1 of Algorithm 2 is composed of $\mathcal {O}(\lceil \mathrm{log}N \rceil ^2)$ H gates, CNOT gates and two-qubit controlled phase gates.

In Step 2, we give a rough estimation of the cost of the controlled-unitary operator U defined in (10). Current standard methods for realizing controlled-unitary gates rely on the decomposition of $U_\mathrm{SVE}^{(m)}$ into a set of $1\times 1$ and $2\times 2$ reversible gates. Specifically, based on the method proposed in [61], the number of additional operations required to add a control to each $U_{\mathrm {SVE}}^{(m)}$ will generally be far less than the cost of the constructed $U_\mathrm{SVE}^{(m)}$, so we next focus on estimating the cost of achieving $U_\mathrm{SVE}^{(m)}, m =0,\ldots ,N-1.$ The most costly step of this process lies in constructing the operator $U_{\hat{P}_m}$ in (32), $m =0,\ldots ,N-1$. According to the analysis in the paragraph above (33), implementing all $U_{P_k}$ in (29) for $k=0, \ldots , N-1$ occupies the major cost of achieving $U_{\hat{P}_m}, m =0,\ldots ,N-1$, which takes a total of $\mathcal {O}(N^2{\lceil \mathrm{log}N \rceil }^3)$ $1\times 1$ and $2\times 2$ reversible gates, since the cost of each $U_{P_k}$ is $\mathcal {O}(N{\lceil \mathrm{log}N \rceil }^3)$ based on the cost of $U_P$ analyzed in Appendix E.2. To summarize, the quantum cost of implementing the operator U is $\mathcal {O}(N^2{\lceil \mathrm{log}N \rceil }^3)$.

In conclusion, the quantum cost of implementing Algorithm 2 is $\mathcal {O}(N^2{\lceil \mathrm{log}N \rceil }^3)$, which is mainly concentrated in achieving the operator U if we ignore the insignificant cost, such as the cost of implementing the operator V in (29). Actually, it is reasonable that the cost of Algorithm 2 is N times more expensive than that of the QSVE algorithm proposed in [18]. The cost of the QSVE algorithm on matrix $A \in \mathbb {C}^{N \times N}$ mainly concentrates on the cost of quantum access to classical data with N rows of A stored in N binary trees. By contrast, our algorithm deals with tensor $\mathcal {T}\in \mathbb {R}^{N \times N \times N}$ whose $N^2$ tubes are assumed to be stored in $N^2$ binary trees, so our algorithm needs N times more gates than the QSVE algorithm to quantum access to the data structure.

Since implementing the physical operations depends on many factors, such as the initial quantum circuit, the quantum computer to be performed on, or the Hamiltonian of the system, the cost presented here is not necessarily the true cost but it provides a reference value. Also, it is not necessarily minimal because it can be decreased by finding more efficient quantum circuits.

The number of $1\times 1$ and $2\times 2$ reversible gates required by our quantum algorithm scales polynomially with the dimension of the preference tensor, which presents a big obstacle to practical realization. In fact, not only our algorithm and the QSVE algorithm, but also some benchmark algorithms, such as the HHL algorithm [13], appear too expensive to be executed efficiently by a quantum computer and are not likely to be feasible in the Noisy Intermediate-Scale Quantum (NISQ) era; see [30, Sections 6.6 and 6.7], [44, Section 1.10]. Finally, we point out most quantum machine learning algorithms focus on the time complexity rather than their quantum cost; see, e.g., [6, 13, 17, 36, 37].