Numerical Methods Combining Symmetry and Sparsity for the Calculation of Homogeneous Polynomials Defined by Tensors

Abstract

The homogeneous polynomial defined by a tensor, $\mathcal{A}{\mathbf{x}}^{m-1}$ for $\mathbf{x}\in {\mathbb{R}}^n$, has been used in many recent problems in the context of tensor analysis and optimization, including the tensor eigenvalue problem, tensor equation, tensor complementary problem, tensor eigenvalue complementary problem, tensor variational inequality problem, and least element problem of polynomial inequalities defined by a tensor, among others. However, conventional computation methods use the definition directly and neglect the structural characteristics of homogeneous polynomials involving tensors, leading to a high computational burden (especially when considering iterative algorithms or large-scale problems). This motivates the need for efficient methods to reduce the complexity of relevant algorithms. First, considering the symmetry of each monomial in the canonical basis of homogeneous polynomials, we propose a calculation method using the merge tensor of the involved tensor to replace the original tensor, thus reducing the computational cost. Second, we propose a calculation method that combines sparsity to further reduce the computational cost. Finally, a simplified algorithm that avoids duplicate calculations is proposed. Extensive numerical experiments demonstrate the effectiveness of the proposed methods, which can be embedded into algorithms for use by the tensor optimization community, improving computational efficiency in magnetic resonance imaging, n-person non-cooperative games, the calculation of molecular orbitals, and so on.

1. Introduction

For any positive integer n, we use $\langle n\rangle$ to denote the set $\{1,2,\dots ,n\}$. Throughout this paper, we assume that l, m, and n are positive integers with $m,n\ge 2$, unless otherwise specified. We denote the set of all n-dimensional real vectors by ${\mathbb{R}}^n$, the set of all n-dimensional non-negative vectors by ${\mathbb{R}}_{+}^n$, and the set of all $m\times n$-dimensional real matrices by ${\mathbb{R}}^{m\times n}$. We use the bold lowercase letter $\mathbf{x}:=\left(x_i\right)\in {\mathbb{R}}^n$ to denote an n-dimensional vector with components $x_i\in \mathbb{R}$, where $i\in \langle n\rangle$, and the uppercase letter $A:=\left(a_{ij}\right)\in {\mathbb{R}}^{m\times n}$ to denote an $m\times n$-dimensional matrix of entries $a_{ij}\in \mathbb{R}$, where $i\in \langle m\rangle$ and $j\in \langle n\rangle$.

A real tensor $\mathcal{A}$ of order m and dimension $n_1\times \cdots \times n_m$ means

$$\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\mathrm{where}{a}_{i_1{i}_2\cdots i_m}\in \mathbb{R}\mathrm{for}\mathrm{any}{i}_j\in \langle n_j\rangle \mathrm{with}j\in \langle m\rangle .$$

We use ${\mathbb{R}}^{l\times [m-1,n]}$ to denote the set of m-order $l\times n_2\times \cdots \times n_m$-dimensional real tensors with $n_2=\cdots =n_m=n$. In particular, if $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ with $l=n$, then $\mathcal{A}$ is called an m-th order n-dimensional real tensor, and the set of all m-th order n-dimensional real tensors is denoted by ${\mathbb{R}}^{[m,n]}$. A tensor is said to be non-negative if all its entries are non-negative. For any $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$, $a_{i_1{i}_2\cdots i_m}$ is said to be a diagonal entry when $i_1=i_2=\cdots =i_m$ and is an off-diagonal entry otherwise. If $\mathcal{A}\in {\mathbb{R}}^{[m,n]}$, then it is said to be an identity tensor if all its diagonal entries are one and all its off-diagonal entries are zero, and we denote it by $\mathcal{E}$.

Tensors and homogeneous polynomials are two closely related concepts in mathematics, particularly in algebraic geometry, differential geometry, and physics. A tensor can define a homogeneous polynomial; conversely, a homogeneous polynomial can be defined by a tensor but not uniquely (i.e., it can be defined by different tensors). If the tensor is required to be symmetric, then a homogeneous polynomial is defined by a unique symmetric tensor. Given $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$, for any $\mathbf{x}\in {\mathbb{R}}^n$, the vector of $(m-1)$-degree homogeneous polynomials defined by tensor $\mathcal{A}$, denoted by $\mathcal{A}{\mathbf{x}}^{m-1}\in {\mathbb{R}}^l$, can be defined with its i-th component being given by

$$\begin{array}{c}\hfill {\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i:=\sum _{i_2,\dots ,i_m\in \langle n\rangle }{a}_{i{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m},\forall i\in \langle l\rangle ;\end{array}$$

(1)

when $l=n$, the m-degree homogeneous polynomial defined by the tensor $\mathcal{A}$, denoted by $\mathcal{A}{\mathbf{x}}^m\in \mathbb{R}$, can be defined as

$$\begin{array}{c}\hfill \mathcal{A}{\mathbf{x}}^m:=\sum _{i_1,\dots ,i_m\in \langle n\rangle }{a}_{i_1\cdots i_m}{x}_{i_1}\cdots x_{i_m}.\end{array}$$

Over the past decade, tensors and their related problems have been studied extensively, many of which involve the calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$. Some of these problems are given as follows.

(a)

Tensor eigenvalue problem [1,2,3,4,5]. Given $\mathcal{A}\in {\mathbb{R}}^{[m,n]}$, an H-eigenvalue of $\mathcal{A}$ is a real number $\lambda$, for which

$$\mathcal{A}{\mathbf{x}}^{m-1}=\lambda {\mathbf{x}}^{[m-1]}$$

admits a non-zero solution $\mathbf{x}\in {\mathbb{R}}^n$, where ${\mathbf{x}}^{[m-1]}:={(x_1^{m-1},\dots ,x_n^{m-1})}^{\top }$. Meanwhile, a Z-eigenvalue of $\mathcal{A}$ is a real number $\lambda$, for which

$$\mathcal{A}{\mathbf{x}}^{m-1}=\lambda \mathbf{x}$$

admits a non-zero solution $\mathbf{x}\in {\mathbb{R}}^n$.

(b)

Tensor equation [6,7,8,9,10,11,12,13,14]. Given $\mathcal{A}\in {\mathbb{R}}^{[m,n]}$ and $\mathbf{b}\in {\mathbb{R}}^n$, this problem involves finding a vector $\mathbf{x}\in {\mathbb{R}}^n$ such that

$$\mathcal{A}{\mathbf{x}}^{m-1}=\mathbf{b}.$$

(c)

Tensor complementarity problem [15,16,17,18,19,20,21]. Given $\mathcal{A}\in {\mathbb{R}}^{[m,n]}$ and $\mathbf{q}\in {\mathbb{R}}^n$, the tensor complementarity problem involves finding a vector $\mathbf{x}\in {\mathbb{R}}^n$ such that

$$\mathbf{0}\le \mathbf{x}\perp (\mathcal{A}{\mathbf{x}}^{m-1}+\mathbf{q})\ge \mathbf{0}.$$

See also the survey papers [22,23,24].

(d)

Tensor eigenvalue complementarity problem [25,26,27,28,29,30]. Given $\mathcal{A}\in {\mathbb{R}}^{[m,n]}$, the tensor eigenvalue complementarity problem involves finding a real number $\lambda$ and a vector $\mathbf{x}\in {\mathbb{R}}^n$ such that

$$\mathbf{0}\le \mathbf{x}\perp (\lambda \mathcal{E}{\mathbf{x}}^{m-1}-\mathcal{A}{\mathbf{x}}^{m-1})\ge \mathbf{0},$$

where $\mathcal{E}$ is an identity tensor.

(e)

Tensor variational inequality problem [31]. Given $\mathcal{A}\in {\mathbb{R}}^{[m,n]}$, $\mathbf{q}\in {\mathbb{R}}^n$, and a non-empty set $\Omega \subseteq {\mathbb{R}}^n$, the tensor variational inequality problem involves finding a vector $\mathbf{x}\in {\mathbb{R}}^n$ such that

$$\langle \mathbf{y}-\mathbf{x},\mathcal{A}{\mathbf{x}}^{m-1}+\mathbf{q}\rangle \ge \mathbf{0}\mathrm{for}\mathrm{all}\mathbf{y}\in \Omega .$$

(f)

Least element problem for the set defined by polynomial inequalities [32]. Let $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ be a Z-tensor (i.e., all its off-diagonal entries are non-positive) and $\mathbf{q}\in {\mathbb{R}}^n$. Then, there exists an element $\mathbf{u}\in \mathcal{S}:=\{\mathbf{x}\in {\mathbb{R}}^n:\mathbf{x}\ge \mathbf{0},\mathcal{A}{\mathbf{x}}^{m-1}\ge \mathbf{q}\}$ such that $\mathbf{u}\le \mathbf{x}$ for all $\mathbf{x}\in \mathcal{S}$. This problem is related to the design of algorithms to find the least element $\mathbf{u}$.

For any $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$, $\mathcal{A}$ has $l{n}^{m-1}$ entries and, hence, as n or m becomes large, the computational complexity of $\mathcal{A}{\mathbf{x}}^{m-1}$ with $\mathbf{x}\in {\mathbb{R}}^n$ grows rapidly; therefore, large-scale problems invoke huge computational challenges. Moreover, in the iterative algorithms used to solve the above problems (a)–(f), it is necessary to repeatedly calculate the values of $\mathcal{A}{\mathbf{x}}^{m-1}$ at different points $\mathbf{x}\in {\mathbb{R}}^n$. Therefore, determining how to calculate $\mathcal{A}{\mathbf{x}}^{m-1}$ effectively is an important problem in the context of tensor analysis and optimization.

The purpose of this study is to determine how to effectively calculate the vector of $(m-1)$-degree homogeneous polynomials $\mathcal{A}{\mathbf{x}}^{m-1}$ for $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ and $\mathbf{x}\in {\mathbb{R}}^n$. We aim to reduce the associated computational cost by dissecting the symmetry of each monomial in the canonical basis of homogeneous polynomials and leveraging the sparsity of the involved tensors. Our main contributions are as follows:

• Firstly, taking into account the symmetry of the monomials composing homogeneous polynomials, we replace the original tensors with their merge tensors, significantly reducing computational costs.
• Secondly, considering sparsity, we design algorithms for the calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$ by finding non-zero elements and corresponding positions. When searching for non-zero elements and their corresponding positions in the merge tensor, we also utilize symmetry.
• Thirdly, a simplified algorithm is proposed to avoid duplicate calculations, which further reduces the computational cost.

The rest of this paper is organized as follows. In Section 2, we introduce some symbols, concepts, and results that will be used in the subsequent analyses. In Section 3, we determine the computational complexity of calculating $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ without considering sparsity. In Section 4, we design algorithms for calculating $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ when taking sparsity into consideration and present a simplified algorithm for the calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$. Preliminary numerical results are reported in Section 5, and the concluding remarks are given in Section 6.

2. Preliminaries

2.1. Notations

Let the index set $T\subseteq \langle n\rangle$ be non-empty. We use $\left|T\right|$ to denote the cardinality of the set T. For any $\mathbf{x}\in {\mathbb{R}}^n$, ${\mathbf{x}}_T\in {\mathbb{R}}^{\left|T\right|}$ denotes a sub-vector of $\mathbf{x}$ containing components corresponding to the indices in T. For any $A\in {\mathbb{R}}^{m\times n}$, $A_{T•}=\left(a_{ij}\right)$ denotes a sub-matrix of A composed of its rows corresponding to the indices in T; that is, the entries $a_{ij}$ of $A_{T•}$ satisfy $i\in T$ and $j\in \langle n\rangle$.

2.2. Merge Tensor

For any $i_1\in \langle l\rangle$ and $i_2,\dots ,i_m\in \langle n\rangle$, we define

$$\begin{array}{c}\hfill {\delta }_{i_1{i}_2\cdots i_m}=\left\{\begin{array}{c}1\mathrm{if}{i}_1=i_2=\cdots =i_m,\hfill \\ 0\mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

For any given indices $i_2,\dots ,i_m\in \langle n\rangle$, we use ${\mathcal{P}}_{i_2\cdots i_m}$ to denote the set of all permutations of the index arrangement $i_2\cdots i_m$. For any $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$, $R_i\left(\mathcal{A}\right)\in {\mathbb{R}}^{[m-1,n]}$ denotes the i-th row-subtensor of $\mathcal{A}$, with entries ${\left(R_i\left(\mathcal{A}\right)\right)}_{i_2\cdots i_m}=a_{i{i}_2\cdots i_m}$ for any $i\in \langle l\rangle$.

The following concept was introduced in [32].

Definition 1.

For an arbitrary given tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$, we call $\widehat{\mathcal{A}}\in {\mathbb{R}}^{l\times [m-1,n]}$ the merge tensor under the permutation of $\mathcal{A}$ if, for any $i_1\in \langle l\rangle$ and $i_2,\dots ,i_m\in \left[n\right]$,

$$\begin{array}{c}\hfill {\widehat{a}}_{i_1{i}_2\cdots i_m}=\left\{\begin{array}{cc}{a}_{i_1{i}_2\cdots i_m}\hfill & \mathrm{if}{\delta }_{i_1{i}_2\cdots i_m}=1,\hfill \\ \sum _{j_2\cdots j_m\in {\mathcal{P}}_{i_2\cdots i_m}}{a}_{i_1{j}_2\cdots j_m}\hfill & \mathrm{if}{\delta }_{i_1{i}_2\cdots i_m}=0\mathrm{and}{i}_2\le \cdots \le i_m,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

In the following, we use the term merge tensor instead of merge tensor under permutation for the sake of simplicity.

It is easy to see that, through utilizing the symmetry of $x_{i_2{i}_3\cdots i_m}$, the entries of the merge tensor $\widehat{\mathcal{A}}\in {\mathbb{R}}^{l\times [m-1,n]}$ are the sum of entries of the corresponding original tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ under any permutation ${\mathcal{P}}_{i_2{i}_3\cdots i_m}$. From Definition 1, for any given tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$, it is obvious that

$$\begin{array}{c}\hfill \mathcal{A}{\mathbf{x}}^{m-1}=\widehat{\mathcal{A}}{\mathbf{x}}^{m-1},\forall \mathbf{x}\in {\mathbb{R}}^n.\end{array}$$

(2)

We use the following simple example to illustrate this point.

Example 1.

Let $\mathcal{A}=\left(a_{i_1{i}_2{i}_3}\right)\in {\mathbb{R}}^{[3,2]}$, where $a_{111}=5$, $a_{112}=-3$, $a_{121}=1$, $a_{122}=-1$, $a_{211}=-2$, $a_{212}=2$, $a_{221}=-4$, and $a_{222}=4$.

Let $\widehat{\mathcal{A}}=\left({\widehat{a}}_{i_1{i}_2{i}_3}\right)\in {\mathbb{R}}^{[3,2]}$, where ${\widehat{a}}_{111}=5$, ${\widehat{a}}_{112}=-2$, ${\widehat{a}}_{121}=0$, ${\widehat{a}}_{122}=-1$, ${\widehat{a}}_{211}=-2$, ${\widehat{a}}_{212}=-2$, ${\widehat{a}}_{221}=0$, and ${\widehat{a}}_{222}=4$.

It is easy to see that $\mathcal{A}{\mathbf{x}}^2=\widehat{\mathcal{A}}{\mathbf{x}}^2$ for all $\mathbf{x}\in {\mathbb{R}}^2$. Specifically, we have

$$\begin{array}{cc}\mathcal{A}{\mathbf{x}}^2& =\left(\begin{array}{c}5x_1^2-3x_1{x}_2+x_2{x}_1-x_2^2\\ -2x_1^2+2x_1{x}_2-4x_2{x}_1+4x_2^2\end{array}\right)\hfill \\ & =\left(\begin{array}{c}5x_1^2-2x_1{x}_2-x_2^2\\ -2x_1^2-2x_1{x}_2+4x_2^2\end{array}\right)=\widehat{\mathcal{A}}{\mathbf{x}}^2.\hfill \end{array}$$

For any given tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ and $\mathbf{x}\in {\mathbb{R}}^n$, it follows from (2) that we may obtain the value of $\mathcal{A}{\mathbf{x}}^{m-1}$ by computing $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ for any $\widehat{\mathcal{A}}\in {\mathbb{R}}^{l\times [m-1,n]}$ satisfying (2). As the merge tensor $\widehat{\mathcal{A}}$ given in Definition 1 is one of sparsest tensors among all the tensors satisfying (2), we compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ instead of $\mathcal{A}{\mathbf{x}}^{m-1}$, where $\widehat{\mathcal{A}}$ is the merge tensor of $\mathcal{A}$ as this greatly reduces the cost of calculation. In addition, we combine the sparsity of $\widehat{\mathcal{A}}$ to further reduce the computational cost. These aspects are further investigated in the following sections.

3. Calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ Without Considering Sparsity

For any given $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$ and $\mathbf{x}\in {\mathbb{R}}^n$, let $\widehat{\mathcal{A}}=\left({\widehat{a}}_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$ be the merge tensor of $\mathcal{A}$. By (2), we have $\mathcal{A}{\mathbf{x}}^{m-1}=\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ for any $\mathbf{x}\in {\mathbb{R}}^n$. In this section, we show the difference between $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ when they are computed directly, according to their definitions.

On one hand, for any $i\in \langle l\rangle$, as there are at most $n^{m-1}$ non-zero entries in sub-row tensor $R_i\left(\mathcal{A}\right)\in {\mathbb{R}}^{[m-1,n]}$, the computational complexity of ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i$ is $O\left(n^{m-1}\right)$ when we compute ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i$ directly using (1).

On the other hand, we define

$$\begin{array}{c}\hfill r:=C_{n+(m-1)-1}^{m-1}\end{array}$$

(3)

and, for any $\mathbf{x}\in {\mathbb{R}}^n$, we define the monomial vector ${\left[\mathbf{x}\right]}^{\langle m-1\rangle }\in {\mathbb{R}}^r$ as

$$\begin{array}{cc}\hfill {\left[\mathbf{x}\right]}^{\langle m-1\rangle }=& (x_1^{m-1},x_1^{m-2}{x}_2,\dots ,x_1^{m-2}{x}_n,x_1^{m-3}{x}_2^2,x_1^{m-3}{x}_2{x}_3,\dots ,x_1^{m-3}{x}_2{x}_n,x_1^{m-3}{x}_3^2,\hfill \\ & \dots ,x_1^{m-3}{x}_n^2,\dots ,x_1{x}_2^{m-2},x_1{x}_2^{m-3}{x}_3,\dots ,x_1{x}_2^{m-3}{x}_n,\dots ,x_1{x}_3^{m-2},\dots ,x_1{x}_n^{m-2},\hfill \\ & \dots ,x_2^{m-1},x_2^{m-2}{x}_3,\dots ,x_2^{m-2}{x}_n,\dots ,x_2^{m-3}{x}_3^2,x_2^{m-3}{x}_3{x}_4,\dots ,x_3^{m-1},\dots ,x_n^{m-1}{)}^{\top }\hfill \end{array}$$

which represents the canonical basis of the vector space of $(m-1)$-degree homogeneous polynomials in $\mathbf{x}$ with real coefficients. For any $i\in \langle l\rangle$, ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ can be written as

$$\begin{array}{c}\hfill {\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i=\langle \widehat{\mathbf{a}},{\left[\mathbf{x}\right]}^{\langle m-1\rangle }\rangle ,\end{array}$$

(4)

where $\widehat{\mathbf{a}}\in {\mathbb{R}}^r$ denotes the coefficient vector corresponding to the basis vector ${\left[\mathbf{x}\right]}^{\langle m-1\rangle }$. For any $i\in \langle l\rangle$, it is worth noting that there are at most $C_{n+(m-1)-1}^{m-1}$ non-zero entries in the sub-row tensor $R_i\left(\widehat{\mathcal{A}}\right)\in {\mathbb{R}}^{[m-1,n]}$, such that $\widehat{\mathbf{a}}$ has at most $C_{n+(m-1)-1}^{m-1}$ non-zero elements. Hence, the computational complexity of ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ is $O\left(C_{n+(m-1)-1}^{m-1}\right)$ when we compute ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ directly using (4).

The difference between ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i$ and ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ when they are computed using the above mentioned methods is intuitively presented in the following Figure 1.

It can be seen that, while the computational cost of ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i$ and ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ increases with n, the gap between the above two gradually widens; in particular, when m is larger, this phenomenon can be clearly observed. From the above analysis, we can see that, for any given tensor $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$, if its merge tensor $\widehat{\mathcal{A}}=\left({\widehat{a}}_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$ is not itself, then calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ to obtain the value of $\mathcal{A}{\mathbf{x}}^{m-1}$ for any $\mathbf{x}\in {\mathbb{R}}^n$ can greatly reduce the computational cost, when compared to calculating $\mathcal{A}{\mathbf{x}}^{m-1}$ directly.

4. Calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ When Considering Sparsity

In many practical calculations, the tensors considered are usually sparse. In this section, let the tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ be given, and $\widehat{\mathcal{A}}$ be the merge tensor of $\mathcal{A}$. For $\mathbf{x}\in {\mathbb{R}}^n$, we propose methods to calculate $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ by combining the sparsity of the tensors $\mathcal{A}$ and $\widehat{\mathcal{A}}$. It is obvious that the calculation of both $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ is trivial when $\mathbf{x}$ is a zero vector or a vector of all ones. Thus, in the following, we assume that $\mathbf{x}$ is neither a zero vector nor a vector of all ones.

4.1. Calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$

For any $i\in \langle l\rangle$, the i-th component of $\mathcal{A}{\mathbf{x}}^{m-1}$ is equivalent to

$$\begin{array}{c}\hfill {\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i=\left\{\begin{array}{cc}\sum _{(i_2,\dots ,i_m)\in {\Delta }_i}{a}_{i{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m}\hfill & \mathrm{if}{\Delta }_i\ne \varnothing ,\hfill \\ 0\hfill & \mathrm{if}{\Delta }_i=\varnothing ,\hfill \end{array}\right.\end{array}$$

(5)

where ${\Delta }_i:=\{(i_2,\dots ,i_m):a_{i{i}_2\cdots i_m}\ne 0,\forall i_j\in \langle n\rangle ,\forall j\in \langle m\rangle \}$. We divide this section into two parts.

Part 1. Identification of non-zero elements and their positions in tensor $\mathcal{A}$.

Suppose that $\mathcal{A}$ has p non-zero entries with $p\le l{n}^{m-1}$. We denote the vector of all non-zero entries of $\mathcal{A}$ and the corresponding index matrix by

$$\begin{array}{c}\hfill \mathbf{v}:=\left(\begin{array}{c}{v}_1\\ ⋮\\ v_p\end{array}\right)\in {\mathbb{R}}^p\mathrm{and}S:=\left(\begin{array}{ccc}{\left(i_1\right)}_1& \dots & {\left(i_m\right)}_1\\ ⋮& & ⋮\\ {\left(i_1\right)}_p& \dots & {\left(i_m\right)}_p\end{array}\right)\triangleq \left(\begin{array}{c}{\mathbf{s}}_1\\ ⋮\\ {\mathbf{s}}_p\end{array}\right)\in {\mathbb{R}}^{p\times m},\end{array}$$

(6)

where ${\mathbf{s}}_j\in {\mathbb{R}}^{1\times m}$ is the index corresponding to the non-zero entry $v_j$ of $\mathcal{A}$ (i.e., $v_j=a_{{\mathbf{s}}_j}$ for any $j\in \langle p\rangle$).

The above discussion can be summarized according to the following procedure:

Procedure 1

(Calculation of vector $\mathbf{v}$ and matrix S).

(S0) 

Input: tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$.

(S1) 

Compute vector $\mathbf{v}$ and matrix S by (6) (for any given tensor $\mathcal{A}$, we can easily obtain the vector $\mathbf{v}$ of all its nonzero entries and the corresponding index matrix S by using the Matlab R2023b’s command ‘find’).

Part 2. Calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$ based on vector $\mathbf{v}$ and matrix S.

For any fixed $i\in \langle l\rangle$, we select the rows of the index matrix S satisfying the first component being equal to i and denote the set of numbers corresponding to these rows as ${\mathbf{I}}_i$; that is,

$$\begin{array}{c}\hfill {\mathbf{I}}_i:=\{j\in \langle p\rangle :{\left({\mathbf{s}}_j\right)}_1=i\}.\end{array}$$

(7)

If ${\mathbf{I}}_i=\varnothing$ for some $i\in \langle l\rangle$, then ${\Delta }_i=\varnothing$ obviously; hence, we have ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i=0$. Next, we consider the case of ${\mathbf{I}}_i\ne \varnothing$ for $i\in \langle l\rangle$. In this case, for ${\mathbf{s}}_j\in {\mathbb{R}}^{1\times m}$ with $j\in {\mathbf{I}}_i$ and $i\in \langle l\rangle$, we use ${\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}\in {\mathbb{R}}^{1\times (m-1)}$ to denote a sub-vector of ${\mathbf{s}}_j$ by discarding its first component.

Then, we have

$$\begin{array}{c}\hfill {\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i=\sum _{j\in {\mathbf{I}}_i}{v}_j{\mathbf{x}}_{{\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}}:=\sum _{j\in {\mathbf{I}}_i}{v}_j{x}_{{\left({\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}\right)}_1}\cdots x_{{\left({\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}\right)}_{m-1}},\forall i\in \langle l\rangle .\end{array}$$

Now, for any $i\in \langle l\rangle$, (5) turns to

$$\begin{array}{c}\hfill {\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i=\left\{\begin{array}{cc}\sum _{j\in {\mathbf{I}}_i}{v}_j{\mathbf{x}}_{{\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}}\hfill & \mathrm{if}{\mathbf{I}}_i\ne \varnothing ,\hfill \\ 0\hfill & \mathrm{if}{\mathbf{I}}_i=\varnothing .\hfill \end{array}\right.\end{array}$$

(8)

We can obtain the following algorithm:

Algorithm 1. Calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$ based on vector $\mathbf{v}$ and matrix S

(S0) Input: vector $\mathbf{v}$ and matrix S for tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$. (S1) For all $i\in \langle l\rangle$, compute ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i$ by (8).

4.2. Calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$

In this section, based on the vector $\mathbf{v}$ and matrix S defined according to (6) in Section 4.1, we find the vector of all non-zero entries of $\widehat{\mathcal{A}}$ and the corresponding index matrix and denote them by $\widehat{\mathbf{v}}$ and $\widehat{S}$, respectively. Then, we provide a method to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$. We divide this section into two parts.

Part 1. Identification of non-zero elements and their positions in the tensor $\widehat{\mathcal{A}}$.

Suppose that $\widehat{\mathcal{A}}$ has q non-zero entries with $q\le lr$, where r is defined by (3). Then, $\widehat{\mathbf{v}}\in {\mathbb{R}}^q$ and $\widehat{S}\in {\mathbb{R}}^{q\times m}$. To obtain $\widehat{\mathbf{v}}$ and $\widehat{S}$, we execute a simple operation on vector $\mathbf{v}\in {\mathbb{R}}^p$ and matrix $S\in {\mathbb{R}}^{p\times m}$, as defined by (6).

For any given $i\in \langle l\rangle$, ${\mathbf{I}}_i=\{j\in \langle p\rangle :{\left({\mathbf{s}}_j\right)}_1=i\}$ is defined by (7); that is, the set of numbers of rows in matrix S satisfies that the first element is to i.

First, for any given $i\in \langle l\rangle$, for simplicity, we use

$$\begin{array}{c}\hfill S^i:=S_{{\mathbf{I}}_i•}\in {\mathbb{R}}^{|{\mathbf{I}}_i|\times m}\mathrm{and}{\mathbf{v}}^i:={\mathbf{v}}_{{\mathbf{I}}_i}\in {\mathbb{R}}^{|{\mathbf{I}}_i|}\end{array}$$

(9)

to denote the sub-matrix composed of rows of matrix S and sub-vector composed of elements of the vector $\mathbf{v}$ corresponding to the set ${\mathbf{I}}_i$ (to reduce storage, we delete all the rows corresponding to the index set ${\mathbf{I}}_i$ from matrix S and vector $\mathbf{v}$ after obtaining the sub-matrix $S^i$ and sub-vector ${\mathbf{v}}^i$). Here, we have ${\sum }_{i=1}^l\left|{\mathbf{I}}_i\right|=p$.

Second, for each row of the matrix $S^i$ with $i\in \langle l\rangle$, we rearrange the last $m-1$ elements in increasing order. We denote the matrix obtained through this partial rearrangement as ${\ddot{S}}^i$.

Third, for any $i\in \langle l\rangle$, we assume that there are $r_i$ different rows in ${\ddot{S}}^i\in {\mathbb{R}}^{|{\mathbf{I}}_i|\times m}$ and denote the matrix composed of these different rows as ${\overline{S}}^i\in {\mathbb{R}}^{r_i\times m}$. Moreover, we denote ${\overline{\mathbf{v}}}^i\in {\mathbb{R}}^{r_i}$, with its t-th component being defined by

$$\begin{array}{c}\hfill {\left({\overline{\mathbf{v}}}^i\right)}_t:=\sum _{j\in {\mathbf{K}}_t^i}{v}_j,\forall t\in \langle r_i\rangle ,\end{array}$$

where ${\mathbf{K}}_t^i:=\{j\in \langle |{\mathbf{I}}_i|\rangle :{\ddot{\mathbf{s}}}_j={\overline{\mathbf{s}}}_t\}$ with ${\ddot{\mathbf{s}}}_j$ being the j-th row of matrix ${\ddot{S}}^i$ and ${\overline{\mathbf{s}}}_t$ being the t-th row of matrix ${\overline{S}}^i$ (note that we can add these together utilizing symmetry).

Fourth, for any $i\in \langle l\rangle$, we use ${\widehat{\mathbf{v}}}^i$ to denote the sub-vector of ${\overline{\mathbf{v}}}^i$ obtained by deleting all zero elements of ${\overline{\mathbf{v}}}^i$ and ${\widehat{S}}^i$ to denote the sub-matrix of ${\overline{S}}^i$ obtained by deleting all rows of ${\overline{S}}^i$ corresponding to the zero elements of ${\overline{\mathbf{v}}}^i$.

In summary, we can obtain the matrix $\widehat{S}$ and vector $\widehat{\mathbf{v}}$ as

$$\begin{array}{c}\hfill \widehat{S}:=\left(\begin{array}{c}{\widehat{S}}^1\\ ⋮\\ {\widehat{S}}^l\end{array}\right)\in {\mathbb{R}}^{q\times m}\mathrm{and}\widehat{\mathbf{v}}:=\left(\begin{array}{c}{\widehat{\mathbf{v}}}^1\\ ⋮\\ {\widehat{\mathbf{v}}}^l\end{array}\right)\in {\mathbb{R}}^q,\end{array}$$

(10)

as summarized in Procedure 2:

Procedure 2

(Compute vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$ for merge tensor $\widehat{\mathcal{A}}$).

(S0) 

Input: vector $\mathbf{v}$ and matrix S for tensor $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$.

(S1) 

Compute vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$ for merge tensor $\widehat{\mathcal{A}}$ by (10).

Part 2. Calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ based on vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$.

Based on the above discussions, we provide the following algorithm to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ for any given tensor $\mathcal{A}$ and vector $\mathbf{x}$.    

Algorithm 2. Calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ based on vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$

(S0) Input: vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$ for tensor $\widehat{\mathcal{A}}\in {\mathbb{R}}^{l\times [m-1,n]}$. (S1) For all $i\in \langle l\rangle$, compute ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ by Algorithm 1 with $\mathcal{A}$ being replaced by $\widehat{\mathcal{A}}$ and $\mathbf{v}$ and S being replaced by $\widehat{\mathbf{v}}$ and $\widehat{S}$, respectively.

4.3. Simplification of Calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$

To further reduce computational costs, we propose a simplified algorithm for calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ in this section. From (8), for any $i\in \langle l\rangle$ and ${\mathbf{I}}_i\ne \varnothing$, instead of calculating ${\sum }_{j\in {\mathbf{I}}_i}{v}_j{\mathbf{x}}_{{\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}}$ directly, we can also calculate $z_j={\mathbf{x}}_{{\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}}$ for all $j\in {\mathbf{I}}_i$ first and then calculate ${\sum }_{j\in {\mathbf{I}}_i}{v}_j{z}_j$. These two calculation methods are equivalent, but the difference lies in the order of multiplication.

Now, we further consider the latter case. For any $i,k\in \langle l\rangle$ with ${\mathbf{I}}_i,{\mathbf{I}}_k\ne \varnothing$ and $i<k$, $S^i$ and $S^k$ are defined by (9). Let $S_{•\langle m\rangle \setminus \left\{1\right\}}^i$ (or $S_{•\langle m\rangle \setminus \left\{1\right\}}^k$) be a sub-matrix of $S^i$ (or $S^k$) by getting rid of its first column; thus, for $j\in {\mathbf{I}}_i\bigcup {\mathbf{I}}_k$, ${\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}$ defined in Section 4.1 is a row of matrices $S_{•\langle m\rangle \setminus \left\{1\right\}}^i$ or $S_{•\langle m\rangle \setminus \left\{1\right\}}^k$.

Generally, the matrices $S_{•\langle m\rangle \setminus \left\{1\right\}}^i$ and $S_{•\langle m\rangle \setminus \left\{1\right\}}^k$ are considered as sets of row vectors, and their intersection may not be empty, which means that duplicate rows may appear. Let ${\mathbf{z}}^i\in {\mathbb{R}}^{|{\mathbf{I}}_i|}$ (or ${\mathbf{z}}^k\in {\mathbb{R}}^{|{\mathbf{I}}_k|}$) be a vector whose j-th element is defined as $z_j^i={\mathbf{x}}_{{\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}}(j\in {\mathbf{I}}_i)$ (or $z_j^k={\mathbf{x}}_{{\left({\mathbf{s}}_j\right)}_{\langle m\rangle \setminus \left\{1\right\}}}(j\in {\mathbf{I}}_k)$), and ${\mathbf{v}}^i$ (or ${\mathbf{v}}^k$) is defined by (9). For any $i,k\in \langle l\rangle$ with ${\mathbf{I}}_i,{\mathbf{I}}_k\ne \varnothing$ and $i<k$, we first calculate ${\mathbf{z}}^i$ and obtain ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i={\left({\mathbf{v}}^i\right)}^{\top }{\mathbf{z}}^i$. Then, we calculate ${\mathbf{z}}^k$ and obtain ${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_k={\left({\mathbf{v}}^k\right)}^{\top }{\mathbf{z}}^k$. However, when we calculate ${\mathbf{z}}^k$, some of its elements may have already been calculated in ${\mathbf{z}}^i$. Therefore, we consider utilizing symmetry, such that each monomial in the canonical basis is computed at most once.

We use the following simple example to illustrate the above description.

Example 2.

Let $\mathcal{A}=\left(a_{i_1{i}_2{i}_3}\right)\in {\mathbb{R}}^{[3,3]}$, where $a_{111}=5$, $a_{112}=-3$, $a_{121}=1$, $a_{122}=1$, $a_{113}=-4$, $a_{131}=2$, $a_{123}=2$, $a_{132}=1$, $a_{133}=2$, $a_{211}=2$, $a_{212}=4$, $a_{221}=-3$, $a_{233}=5$, $a_{322}=2$, $a_{313}=4$, $a_{331}=-2$, and $a_{333}=-1$.

It is easy to calculate that

$$\begin{array}{cc}\mathcal{A}{\mathbf{x}}^2& =\left(\begin{array}{c}5x_1^2-3x_1{x}_2+x_2{x}_1+x_2^2-4x_1{x}_3+2x_3{x}_1+2x_2{x}_3+x_3{x}_2+2x_3^2\\ 2x_1^2+4x_1{x}_2-3x_2{x}_1+5x_3^2\\ 2x_2^2+4x_1{x}_3-2x_3{x}_1-x_3^2\end{array}\right),\hfill \end{array}$$

for a given $\mathbf{x}\in {\mathbb{R}}^3$. When we calculate the first component of $\mathcal{A}{\mathbf{x}}^2$, we need to calculate the vector $z^1={(x_1^2,x_1{x}_2,x_2{x}_1,x_2^2,x_1{x}_3,x_3{x}_1,x_2{x}_3,x_3{x}_2,x_3^2)}^{\top }$ first and then take the inner product of the vectors $v^1={(5,-3,1,1,-4,2,2,1,2)}^{\top }$ and $z^1$. However, when we calculate the second component of $\mathcal{A}{\mathbf{x}}^2$, the elements in vector $z^2={(x_1^2,x_1{x}_2,x_2{x}_1,x_3^2)}^{\top }$ have already been calculated in $z^1$, and the same applies to the third component of $\mathcal{A}{\mathbf{x}}^2$.

According to the above analysis, there is a similar discussion regarding the calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$. In the following, we introduce the details of a process for calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ in order to avoid duplicate calculations. Based on the matrix $\widehat{S}$ defined by (10) in Section 4.2, we can obtain ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ by getting rid of its first column and identify the different rows of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ denoted as ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ (for any given matrix A, we can easily obtain the matrix C containing all different rows using the Matlab R2023b’s command ‘unique’). Then, for any $i\in \langle l\rangle$, we can find ${\widehat{\mathbf{v}}}^i$ and the corresponding index matrix ${\widehat{S}}^i$. Based on ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ and ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$, we can obtain the intersection of these two sets and the index of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ in ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ with respect to the rows. Finally, we calculate the vector $\mathbf{z}$ whose t-th element is defined as $z_t={\mathbf{x}}_{{\left({\tilde{\mathbf{s}}}_t\right)}_{\langle m\rangle \setminus \left\{1\right\}}}$, where ${\left({\tilde{\mathbf{s}}}_t\right)}_{\langle m\rangle \setminus \left\{1\right\}}$ is the t-th row of ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$; thus, for any $i\in \langle l\rangle$, ${\mathbf{z}}^i$ can be obtained from $\mathbf{z}$ and the index of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ in ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ directly. Then, ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i={\left({\widehat{\mathbf{v}}}^i\right)}^{\top }{\mathbf{z}}^i$ can be calculated. We divide this section into three parts.

Part 1. Identification of different rows based on $\widehat{S}$ by getting rid of its first column.

Let ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ be the sub-matrix of $\widehat{S}$ by getting rid of its first column. Then, we can find all the different rows of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$, and we denote the matrix composed of these different rows by ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$. Suppose that there are s different rows of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$. Then, ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}\in {\mathbb{R}}^{s\times (m-1)}$.

Part 2. Identification of the intersection of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ and ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ with respect to rows and the corresponding row index vector of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ in ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ for any $i\in \langle l\rangle$.

For any given $i\in \langle l\rangle$, ${\widehat{S}}^i$ and ${\widehat{\mathbf{v}}}^i$ are defined by (10), which represent the sub-matrix composed of rows of matrix $\widehat{S}$ and the sub-vector composed of elements of vector $\widehat{\mathbf{v}}$, respectively. Let ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ be the sub-matrix of ${\widehat{S}}^i$ obtained by discarding its first column. Then, we can obtain the intersection of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ and ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ to obtain the index set of ${\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i$ in ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ with respect to rows. We denote the index set as ${\mathbf{J}}_i$, which is characterized by the following relationship:

$$\begin{array}{c}\hfill {\widehat{S}}_{•\langle m\rangle \setminus \left\{1\right\}}^i={\left({\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}\right)}_{{\mathbf{J}}_i•}.\end{array}$$

Then, we can obtain the index set J as follows

$$\begin{array}{c}\hfill \mathbf{J}:=\left(\begin{array}{c}{\mathbf{J}}_1\\ {\mathbf{J}}_2\\ ⋮\\ {\mathbf{J}}_l\end{array}\right)\in {\mathbb{R}}^q.\end{array}$$

(11)

The above two parts can be summarized through the following procedure:

Procedure 3

(Compute index set $\mathbf{J}$ for tensor $\widehat{\mathcal{A}}$).

(S0) 

Input: matrix $\widehat{S}$ for tensor $\widehat{\mathcal{A}}\in {\mathbb{R}}^{l\times [m-1,n]}$.

(S1) 

Compute $\mathbf{J}$ by (11).

Part 3. Calculation of ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ based on matrix ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$, vector ${\widehat{\mathbf{v}}}^i$, and index set ${\mathbf{J}}_i$ for any $i\in \langle l\rangle$.

Let ${\left({\tilde{\mathbf{s}}}_t\right)}_{\langle m\rangle \setminus \left\{1\right\}}$ be the t-th ($t\in \langle s\rangle$) row of ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$. We calculate the vector $\mathbf{z}$ based on $\mathbf{x}$ and ${\tilde{S}}_{•\langle m\rangle \setminus \left\{1\right\}}$ first, as follows:

$$\begin{array}{c}\hfill z_t={\mathbf{x}}_{{\left({\tilde{\mathbf{s}}}_t\right)}_{\langle m\rangle \setminus \left\{1\right\}}}=x_{{\left({\left({\tilde{\mathbf{s}}}_t\right)}_{\langle m\rangle \setminus \left\{1\right\}}\right)}_1}\cdots x_{{\left({\left({\tilde{\mathbf{s}}}_t\right)}_{\langle m\rangle \setminus \left\{1\right\}}\right)}_{m-1}},\forall t\in \langle s\rangle .\end{array}$$

For any $i\in \langle l\rangle$, we can obtain the corresponding ${\widehat{\mathbf{v}}}^i$ and the index set ${\mathbf{J}}_i$. Thus, ${\mathbf{z}}^i$ can be obtain by $\mathbf{z}$ and ${\mathbf{J}}_i$ directly; that is,

$$\begin{array}{c}\hfill {\mathbf{z}}^i={\mathbf{z}}_{{\mathbf{J}}_i},\forall i\in \langle l\rangle ,\end{array}$$

then ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i={\left({\widehat{\mathbf{v}}}^i\right)}^{\top }{\mathbf{z}}^i$ can be calculated.

Now, the i-th component of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ is equivalent to

$$\begin{array}{c}\hfill {\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i=\left\{\begin{array}{cc}{\left({\widehat{\mathbf{v}}}^i\right)}^{\top }{\mathbf{z}}_{{\mathbf{J}}_i}\hfill & \mathrm{if}{\mathbf{I}}_i\ne \varnothing ,\hfill \\ 0\hfill & \mathrm{if}{\mathbf{I}}_i=\varnothing .\hfill \end{array}\right.\end{array}$$

(12)

Based on the above discussions, we give the following simplified algorithm to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ for any given tensor $\mathcal{A}$ and vector $\mathbf{x}$.

Algorithm 3. Simplified calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ based on vector $\widehat{\mathbf{v}}$ and index set $\mathbf{J}$

(S0) Input: vector $\widehat{\mathbf{v}}$ and index set $\mathbf{J}$ for tensor $\widehat{\mathcal{A}}\in {\mathbb{R}}^{l\times [m-1,n]}$. (S1) For all $i\in \langle l\rangle$, compute ${\left(\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}\right)}_i$ by (12).

5. Numerical Experiments

In this section, we implement the methods proposed in Section 4 to calculate $\mathcal{A}{\mathbf{x}}^{m-1}$ and $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ and conduct a series of experiments to verify the effectiveness of the proposed methods.

All the experiments were performed using MATLAB R2023b on a laptop computer with an Intel Core i7-10710U at 1.1 GHz and 16 GB of RAM.

5.1. Performance of Algorithm 1 (or Algorithm 2) for Computing $\mathcal{A}{\mathbf{x}}^{m-1}$ (or $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$) in the Situation That $\mathbf{v}$ and S (or $\widehat{\mathbf{v}}$ and $\widehat{S}$) Are Known

According to the description in Section 4, we can easily find that the difference between Algorithms 1 and 2 relates to their inputs. In this section, we verify the efficiencies of Algorithms 1 and 2 through a comparison using ‘ttv’ in the tensor toolbox [33]. In addition, $\widehat{\mathbf{v}}$ and $\widehat{S}$ are sometimes easily obtained in practical applications, and the tensor may even be given by $\widehat{\mathbf{v}}$ and $\widehat{S}$; in this case, we can use Algorithm 2 to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ directly without conducting Procedures 1 and 2. In the following example, to demonstrate the performance of Algorithm 1 (or Algorithm 2), we construct a tensor for which the related merge tensor is itself, such that Algorithms 1 and 2 are the same.

Example 3.

Let $\mathbf{x}=2\mathbf{e}\in {\mathbb{R}}^n$ with $\mathbf{e}={(1,\cdots ,1)}^{\top }$ and $\mathcal{A}=\left(a_{i_1{i}_2{i}_3{i}_4}\right)\in {\mathbb{R}}^{[4,n]}$ be a tensor generated with entries such that, for any $i\in [n-10]$,

$$\left\{\begin{array}{cccccc}{a}_{ii(i+1)(i+1)}\hfill & =a_{i(i+1)(i+3)(i+4)}\hfill & =a_{i(i+2)(i+5)(i+5)}\hfill & =a_{i(i+2)(i+5)(i+8)}\hfill & =a_{i(i+3)(i+4)(i+5)}\hfill & ={\displaystyle \frac{1}{4}},\hfill \\ a_{i(i+4)(i+4)(i+6)}\hfill & =a_{i(i+5)(i+8)(i+9)}\hfill & =a_{i(i+6)(i+8)(i+9)}\hfill & =a_{i(i+7)(i+9)(i+10)}\hfill & =a_{i(i+8)(i+9)(i+9)}\hfill & ={\displaystyle \frac{1}{4}};\hfill \\ a_{(i+1)i(i+1)(i+6)}\hfill & =a_{(i+1)(i+1)(i+1)(i+6)}\hfill & =a_{(i+1)(i+2)(i+3)(i+6)}\hfill & =a_{(i+1)(i+3)(i+6)(i+6)}\hfill & =a_{(i+1)(i+5)(i+5)(i+6)}\hfill & ={\displaystyle \frac{1}{10}},\hfill \\ a_{(i+1)(i+6)(i+7)(i+9)}\hfill & =a_{(i+1)(i+7)(i+7)(i+8)}\hfill & =a_{(i+1)(i+8)(i+9)(i+9)}\hfill & =a_{(i+1)(i+8)(i+9)(i+10)}\hfill & ={\displaystyle \frac{1}{10}};\hfill \\ a_{(i+2)i(i+1)(i+6)}\hfill & =a_{(i+2)(i+1)(i+1)(i+6)}\hfill & =a_{(i+2)(i+2)(i+3)(i+6)}\hfill & =a_{(i+2)(i+3)(i+6)(i+6)}\hfill & =a_{(i+2)(i+4)(i+5)(i+6)}\hfill & ={\displaystyle \frac{1}{10}},\hfill \\ a_{(i+2)(i+5)(i+5)(i+6)}\hfill & =a_{(i+2)(i+6)(i+7)(i+9)}\hfill & =a_{(i+2)(i+7)(i+7)(i+10)}\hfill & =a_{(i+2)(i+8)(i+9)(i+9)}\hfill & =a_{(i+2)(i+8)(i+9)(i+10)}\hfill & ={\displaystyle \frac{1}{10}};\hfill \\ a_{(i+3)i(i+2)(i+4)}\hfill & =a_{(i+3)(i+1)(i+1)(i+3)}\hfill & =a_{(i+3)(i+2)(i+2)(i+5)}\hfill & =a_{(i+3)(i+3)(i+4)(i+5)}\hfill & =a_{(i+3)(i+4)(i+5)(i+7)}\hfill & ={\displaystyle \frac{1}{5}},\hfill \\ a_{(i+3)(i+5)(i+5)(i+8)}\hfill & =a_{(i+3)(i+6)(i+7)(i+7)}\hfill & =a_{(i+3)(i+7)(i+7)(i+9)}\hfill & =a_{(i+3)(i+8)(i+9)(i+9)}\hfill & =a_{(i+3)(i+8)(i+9)(i+10)}\hfill & ={\displaystyle \frac{1}{5}};\hfill \\ a_{(i+4)i(i+1)(i+5)}\hfill & =a_{(i+4)i(i+2)(i+5)}\hfill & =a_{(i+4)(i+1)(i+3)(i+5)}\hfill & =a_{(i+4)(i+2)(i+3)(i+5)}\hfill & ={\displaystyle \frac{1}{4}},\hfill & \\ a_{(i+4)(i+3)(i+5)(i+7)}\hfill & =a_{(i+4)(i+5)(i+7)(i+8)}\hfill & =a_{(i+4)(i+6)(i+6)(i+8)}\hfill & =a_{(i+4)(i+8)(i+8)(i+9)}\hfill & ={\displaystyle \frac{1}{4}};\hfill & \\ a_{(i+5)(i+1)(i+2)(i+5)}\hfill & =a_{(i+5)(i+1)(i+3)(i+5)}\hfill & =a_{(i+5)(i+2)(i+3)(i+5)}\hfill & =a_{(i+5)(i+3)(i+4)(i+5)}\hfill & =a_{(i+5)(i+4)(i+4)(i+5)}\hfill & ={\displaystyle \frac{1}{5}},\hfill \\ a_{(i+5)(i+5)(i+5)(i+8)}\hfill & =a_{(i+5)(i+6)(i+8)(i+9)}\hfill & =a_{(i+5)(i+7)(i+8)(i+9)}\hfill & =a_{(i+5)(i+8)(i+9)(i+9)}\hfill & ={\displaystyle \frac{1}{5}};\hfill \\ a_{(i+6)i(i+1)(i+6)}\hfill & =a_{(i+6)(i+1)(i+2)(i+9)}\hfill & =a_{(i+6)(i+2)(i+3)(i+8)}\hfill & =a_{(i+6)(i+2)(i+4)(i+9)}\hfill & =a_{(i+6)(i+3)(i+5)(i+10)}\hfill & ={\displaystyle \frac{1}{5}},\hfill \\ a_{(i+6)(i+4)(i+5)(i+7)}\hfill & =a_{(i+6)(i+5)(i+6)(i+9)}\hfill & =a_{(i+6)(i+6)(i+6)(i+9)}\hfill & =a_{(i+6)(i+6)(i+8)(i+9)}\hfill & =a_{(i+6)(i+7)(i+9)(i+9)}\hfill & ={\displaystyle \frac{1}{5}};\hfill \\ a_{(i+7)i(i+1)(i+4)}\hfill & =a_{(i+7)(i+1)(i+4)(i+5)}\hfill & =a_{(i+7)(i+2)(i+3)(i+6)}\hfill & =a_{(i+7)(i+3)(i+5)(i+6)}\hfill & =a_{(i+7)(i+4)(i+6)(i+6)}\hfill & ={\displaystyle \frac{1}{4}},\hfill \\ a_{(i+7)(i+5)(i+7)(i+7)}\hfill & =a_{(i+7)(i+6)(i+7)(i+7)}\hfill & =a_{(i+7)(i+6)(i+8)(i+9)}\hfill & =a_{(i+7)(i+8)(i+8)(i+9)}\hfill & ={\displaystyle \frac{1}{4}};\hfill & \\ a_{(i+8)ii(i+1)}\hfill & =a_{(i+8)(i+1)(i+2)(i+2)}\hfill & =a_{(i+8)(i+2)(i+4)(i+5)}\hfill & =a_{(i+8)(i+3)(i+4)(i+5)}\hfill & =a_{(i+8)(i+4)(i+6)(i+8)}\hfill & ={\displaystyle \frac{1}{5}},\hfill \\ a_{(i+8)(i+4)(i+7)(i+8)}\hfill & =a_{(i+8)(i+5)(i+6)(i+7)}\hfill & =a_{(i+8)(i+6)(i+7)(i+9)}\hfill & =a_{(i+8)(i+7)(i+7)(i+8)}\hfill & ={\displaystyle \frac{1}{5}};\hfill & \\ a_{(i+9)i(i+2)(i+3)}\hfill & =a_{(i+9)(i+1)(i+1)(i+2)}\hfill & =a_{(i+9)(i+2)(i+4)(i+8)}\hfill & =a_{(i+9)(i+3)(i+3)(i+5)}\hfill & =a_{(i+9)(i+4)(i+5)(i+6)}\hfill & ={\displaystyle \frac{1}{10}},\hfill \\ a_{(i+9)(i+4)(i+7)(i+8)}\hfill & =a_{(i+9)(i+5)(i+6)(i+7)}\hfill & =a_{(i+9)(i+6)(i+7)(i+8)}\hfill & =a_{(i+9)(i+7)(i+8)(i+9)}\hfill & =a_{(i+9)(i+8)(i+9)(i+9)}\hfill & ={\displaystyle \frac{1}{10}};\hfill \end{array}\right.$$

with all other entries being 0.

Obviously, $\mathcal{A}$ given in Example 3 is equal to $\widehat{\mathcal{A}}$ and, so, $\widehat{\mathbf{v}},\widehat{S}$ can be obtained easily via Procedure 1. Next, we apply both Algorithm 2 and ‘ttv’ to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$. Note that, before executing Algorithm 2, we need to use Procedure 1 to obtain $\widehat{\mathbf{v}}$ and $\widehat{S}$, while ‘ttv’ can be directly applied to the tensor $\widehat{\mathcal{A}}$. Moreover, we use ${\mathrm{CPU}}_{\widehat{\mathbf{v}},\widehat{S}}$, ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}^{\widehat{\mathbf{v}},\widehat{S}}$, and ${\mathrm{CPU}}_{\mathtt{ttv}}$ to denote the CPU time cost of Procedure 1, Algorithm 2, and ‘ttv’ in the tensor toolbox, respectively. The numerical experimental results for different values of n are presented in

Table 1. Numerical results for computing $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ in Example 3.

n	p	$\mathit{sr}$	${\mathrm{CPU}}_{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}^{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\mathtt{ttv}}$
50	3078	4.92 × ${10}^{-4}$	0.0156	0.0000	0.0000
100	6878	6.88 × ${10}^{-5}$	1.3750	0.0000	0.3906
120	8398	4.05 × ${10}^{-5}$	1.7344	0.0000	0.7031
150	10,678	2.11 × ${10}^{-5}$	2.1563	0.0000	1.7031
200	14,478	9.05 × ${10}^{-6}$	7.4688	0.0000	5.7969
220	15,998	6.83 × ${10}^{-6}$	22.5781	0.0000	85.1250
250	18,278	4.68 × ${10}^{-6}$	44.3125	0.0313	171.7656
300	22,078	2.73 × ${10}^{-6}$	-	-	-

, where p is the number of non-zero elements and $sr$ represents the sparsity ratio.

From Table 1, we can see that our Algorithm 2 significantly outperforms ‘ttv’ in the tensor toolbox for instances with low sparsity. However, our algorithm requires additional time to find $\widehat{\mathbf{v}}$ and $\widehat{S}$ via Procedure 1, which can be implemented as an off-line process. Moreover, as n increases, Procedure 1 and Algorithm 2 present certain advantages. In the case $n=300$, the MATLAB function ‘tensor’ ran out of memory.

Furthermore, the tensors involved in Table 1 are stored in a general way. When the tensors in Example 3 are stored by $\widehat{\mathbf{v}}$ and $\widehat{S}$, Procedure 1 does not need to be executed. Therefore, we only need to directly compare Algorithm 2 and ‘ttv’. It is worth noting that ‘ttv’ in the tensor toolbox is an overloaded function, where the function ‘ttv’ invoked with general tensor arguments differs from that with sparse tensor arguments. When the tensor is stored by $\widehat{\mathbf{v}}$ and $\widehat{S}$—that is, the tensor is stored in the sparse structure—it is more fair to use ‘ttv’ with sparse tensors as parameters for comparison. The numerical experimental results for different n are shown in

Table 2. Numerical results for computing $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ in Example 3 with sparse structure.

n	p	$\mathit{sr}$	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}^{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\mathbf{s}\text{-}\mathbf{ttv}}$
50	3078	4.92 × ${10}^{-4}$	0.0000	0.0000
100	6878	6.88 × ${10}^{-5}$	0.0000	0.0000
120	8398	4.05 × ${10}^{-5}$	0.0000	0.0000
150	10,678	2.11 × ${10}^{-5}$	0.0000	0.0313
200	14,478	9.05 × ${10}^{-6}$	0.0000	0.0469
220	15,998	6.83 × ${10}^{-6}$	0.0000	0.0000
250	18,278	4.68 × ${10}^{-6}$	0.0000	0.0313
300	22,078	2.73 × ${10}^{-6}$	0.0313	0.0000
500	37,278	5.96 × ${10}^{-7}$	0.0000	0.0156
1000	75,728	7.53 × ${10}^{-8}$	0.0000	0.0156
2000	151,278	9.45 × ${10}^{-9}$	0.0313	0.1094

, where we use ${\mathrm{CPU}}_{\mathrm{s}\text{-}\mathrm{ttv}}$ to denote the CPU time cost of ‘ttv’ with sparse tensors as parameters and the other notation are as same as that in Table 1. Without a loss of generality, in the following description, if the tensors involved are the general structure, we execute the function ‘ttv’ in the tensor toolbox with general tensors as parameters (denoted by ‘ttv’), while, if the tensors involved have sparse structure, we execute the function ‘ttv’ in the tensor toolbox with sparse tensors as parameters (denoted by ‘s-ttv’).

From Table 2, we can see that when the tensors are stored using $\widehat{\mathbf{v}}$ and $\widehat{S}$ or in the sparse structure, both Algorithm 2 and ‘s-ttv’ can solve problems of larger scale, and Algorithm 2 and ‘s-ttv’ were found to perform equally well. When $n=2000$, Algorithm 2 performed better.

5.2. Verification of the Advantages of Calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ After Merging

In order to verify the advantages of merging, we determined the numerical performances of Algorithm 2 and ‘s-ttv’ based on $\mathbf{v},S$.

Example 4.

Let $\mathbf{x}=2\mathbf{e}\in {\mathbb{R}}^n$ with $\mathbf{e}={(1,\cdots ,1)}^{\top }$ and $\mathcal{A}=\left(a_{i_1{i}_2{i}_3{i}_4{i}_5}\right)\in {\mathbb{R}}^{[5,n]}$ be a tensor generated with entries satisfying

$$\left\{\begin{array}{ccccc}{a}_{iiiii}=1,\hfill & a_{iiii(i+2)}=2,\hfill & a_{iii(i+2)i}=-3,\hfill & a_{ii(i+2)ii}=-1,\hfill & \\ a_{i(i+2)iii}=-1,\hfill & a_{i(i+1)(i+1)(i+2)(i+2)}=1,\hfill & a_{i(i+2)(i+2)(i+1)(i+1)}=-2,\hfill & \forall i\in \{1,4,7,\dots \};\hfill & \\ a_{iiiii}=1,\hfill & a_{i(i-1)(i-1)ii}=2,\hfill & a_{i(i-1)i(i-1)i}=-2,\hfill & a_{ii(i-1)(i-1)i}=-2,\hfill & \\ a_{iii(i-1)(i-1)}=-2,\hfill & a_{iii(i+1)(i+1)}=2,\hfill & a_{ii(i+1)(i+1)i}=-2,\hfill & a_{i(i+1)(i+1)ii}=-2,\forall i\in \{2,5,8,\dots \};\hfill \\ a_{iiiii}=-1,\hfill & a_{i(i-2)(i-2)ii}=3,\hfill & a_{iii(i-2)(i-2)}=-2,\hfill & a_{ii(i-2)(i-2)i}=-3,\hfill \\ a_{i(i-2)(i-2)(i-1)(i-1)}=-{\displaystyle \frac{1}{2}},\hfill & a_{i(i-1)(i-1)(i-2)(i-2)}={\displaystyle \frac{1}{4}},\hfill & \forall i\in \{3,6,9,\dots \};\hfill & \end{array}\right.$$

with all other entries being 0.

Obviously, $\mathcal{A}$ given in Example 4 is very sparse and its merge tensor $\widehat{\mathcal{A}}$ can reduce the number of non-zero elements by more than half. Suppose that the tensors in Example 4 are stored in the sparse structure; then, we can directly apply Procedure 2 and Algorithm 2 to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$, while ‘s-ttv’ is used to compute $\mathcal{A}{\mathbf{x}}^{m-1}$. Moreover, we use ${\mathrm{CPU}}_{\widehat{\mathbf{v}},\widehat{S}}$, ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}^{\widehat{\mathbf{v}},\widehat{S}}$, and ${\mathrm{CPU}}_{\mathrm{s}\text{-}\mathrm{ttv}}$ to denote the CPU time cost of Procedure 2, Algorithm 2, and ‘s-ttv’ in the tensor toolbox, respectively. The numerical experimental results for different values of n are shown in

Table 3. Numerical results for computing $\mathcal{A}{\mathbf{x}}^{m-1}$ in Example 4.

n	p	q	${\mathrm{CPU}}_{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}^{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\mathbf{s}\text{-}\mathbf{ttv}}$
100	694	298	0.0000	0.0000	0.0000
1000	6994	2998	0.0000	0.0000	0.0000
2000	13,992	5997	0.0000	0.0000	0.0000
10,000	69,994	29,998	0.0000	0.0000	0.0313
50,000	349,992	149,997	0.0625	0.0156	0.0938
100,000	699,994	299,998	0.2656	0.0313	0.1094
200,000	1,399,992	599,997	0.5156	0.0313	0.2188

, where p is the number of non-zero elements of $\mathcal{A}$, and q is the number of non-zero elements of $\widehat{\mathcal{A}}$.

From Table 3, it can be seen that both Algorithm 2 and ‘s-ttv’ can solve problems of different scales and have remarkable computational efficiency. For small-scale problems, Algorithm 2 and ‘ttv’ take almost no CPU time, while, for large-scale problems, Algorithm 2 takes less CPU time than ‘s-ttv’. In addition, although Procedure 2 may take additional CPU time for large-scale problems, it can be performed in an offline manner in practical applications.

5.3. Comparison of Algorithms 1 and 2 with s-ttv

For any given $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ and $\mathbf{x}\in {\mathbb{R}}^n$ with different $(l,m,n)$, we used ‘s-ttv’ and Algorithm 1 to compute tensor–vector products $\mathcal{A}{\mathbf{x}}^{m-1}$ and used Algorithm 2 to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$, where ‘s-ttv’ was implemented using the tensor toolbox [33], consistent with the description in the Section 5.1.

Example 5.

Let $\mathbf{x}=2\mathbf{e}\in {\mathbb{R}}^n$ with $\mathbf{e}={(1,\cdots ,1)}^{\top }$ and $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$ be a sparse tensor generated by sptenrand with sparsity ratio $sr$; that is, $\mathcal{A}$ has $sr\times l{n}^{m-1}$ non-zero entries uniformly distributed in $(0,1)$.

To demonstrate the performance of the different algorithms, we calculated the average CPU time cost from 10 random experiments. In particular, we use ${\mathrm{CPU}}_{\mathrm{Axm}1}^{\mathbf{v},S}$, ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}^{\widehat{\mathbf{v}},\widehat{S}}$ and ${\mathrm{CPU}}_{\mathrm{s}\text{-}\mathrm{ttv}}$ to denote the average CPU time cost of Algorithms 1 and 2 and ‘s-ttv’, respectively. Moreover, to present the computation cost of the pre-processing procedures clearly, we use ${\mathrm{CPU}}_{\mathbf{v},S}$ to denote the average CPU time cost for $\mathbf{v}$ and S in Procedure 1, and we use ${\mathrm{CPU}}_{\widehat{\mathbf{v}},\widehat{S}}$ to denote the average CPU time cost for computation of the vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$ for the merge tensor $\widehat{\mathcal{A}}$ in Procedure 2. Here, p is the average number of non-zero elements of original tensors, and q is the average number of non-zero elements of corresponding merge tensors. The numerical results are shown in

Table 4. Numerical results for computing $\mathcal{A}{\mathbf{x}}^{m-1}$ in Example 5 with different values of $sr$.

$(\mathit{l},\mathit{m},\mathit{n}),\mathbf{sr}$	$\mathit{p}/\mathit{q}$	${\mathrm{CPU}}_{\mathbf{s}\text{-}\mathbf{ttv}}$	${\mathrm{CPU}}_{\mathit{v},\mathbf{S}}$	${\mathrm{CPU}}_{\mathbf{Axm}1}^{\mathit{v},\mathit{S}}$	${\mathrm{CPU}}_{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}^{\widehat{\mathit{v}},\widehat{\mathit{S}}}$
(50, 4, 50), 50%	2,459,310/1,027,410.6	0.1750	0.0047	0.1844	1.5047	0.0906
(80, 4, 100)	31,477,918.2/12,909,783.7	3.4578	0.0016	4.5922	35.1703	1.0750
(100, 4, 100)	39,347,945/16,138,614	3.1031	0.0609	2.9375	36.5516	1.2438
(30, 5, 30)	9,561,777.4/1,221,955.8	1.9438	0.0156	0.8719	9.9344	0.1328
(10, 6, 10)	393,401/19,875.4	0.0563	0.0000	0.0484	0.7719	0.0094
(10, 6, 20)	12,590,775.4/424,585	2.4984	0.0000	1.4578	18.3719	0.0391
(50, 4, 50), 20%	1,133,057/740,778.5	0.0688	0.0031	0.0844	0.6859	0.0422
(80, 4, 100)	14,501,746/9,398,356.5	2.2109	0.0000	1.2063	16.5984	0.8078
(100, 4, 100)	18,127,358/11,748,351	1.4609	0.0344	1.4609	16.4421	0.9344
(30, 5, 30)	4,404,639.6/1,171,318.7	1.5125	0.0109	0.4531	4.9031	0.1313
(10, 6, 10)	181,208.3/19,414.9	0.0438	0.0000	0.0844	0.3469	0.0000
(10, 6, 20)	5,800,478.7/422,239.6	1.6516	0.0000	0.6703	7.5078	0.0375
(50, 4, 50), 10%	594,732.1/474,245.9	0.0484	0.0000	0.0688	0.3500	0.0297
(80, 4, 100)	7,613,230.3/6,042,749.7	1.6203	0.0000	0.5828	7.7609	0.4828
(100, 4, 100)	9,516,266/7,553,460.2	0.7781	0.0234	0.7828	8.1047	0.6250
(30, 5, 30)	2,312,471.9/1,017,402	0.9859	0.0031	0.2453	3.2547	0.1125
(10, 6, 10)	95,190.2/18,359.3	0.0094	0.0000	0.0297	0.1750	0.0094
(10, 6, 20)	3,045,135.4/414,376.1	1.3063	0.0000	0.4000	4.0703	0.0578
(50, 4, 50), 5%	304,809.5/271,144.4	0.0156	0.0015	0.0188	0.2203	0.0125
(80, 4, 100)	3,901,636.1/3,463,587.5	1.3219	0.0141	0.3281	4.3328	0.2563
(100, 4, 100)	4,876,798/4,329,426.8	0.4578	0.0078	0.4469	3.8844	0.3609
(30, 5, 30)	1,185,079.8/747,613.9	0.5484	0.0016	0.1344	2.1266	0.0781
(10, 6, 10)	48,769.4/16,144.3	0.0125	0.0000	0.0000	0.0953	0.0094
(10, 6, 20)	1,560,658.4/389,369.2	0.8641	0.0000	0.5484	7.5078	0.0437
(50, 4, 50), 1%	62,188.1/60,713.8	0.0219	0.0000	0.0188	0.1328	0.0094
(80, 4, 100)	796,011.4/776,760	0.2109	0.0000	0.2281	1.5453	0.0859
(100, 4, 100)	995,026.8/971,036.7	0.0688	0.0016	0.0797	0.5922	0.0875
(30, 5, 30)	241,783.1/218,543.7	0.0391	0.0000	0.0266	0.1484	0.0141
(10, 6, 10)	9951.4/7275.2	0.0016	0.0000	0.0000	0.0078	0.0000
(10, 6, 20)	318,414.5/211,318.5	0.1688	0.0000	0.1563	0.7797	0.1047

.

From Table 4, we can see that when we calculate $\mathcal{A}{\mathbf{x}}^{m-1}$, Algorithm 1 is faster than ‘s-ttv’ in the tensor toolbox for high-order examples while, for low-order examples, Algorithm 1 and ‘s-ttv’ perform equally well. In general, Algorithm 2 is faster than Algorithm 1 and ‘s-ttv’. In our examples, we generated a series of sparse tensors, such that Procedure 1 used to calculate $\mathbf{v},S$ is basically not time-consuming due to the special storage structure of sparse tensors, while Procedure 2 to calculate $\widehat{\mathbf{v}},\widehat{S}$ takes a relatively long time. As discussed in Section 4, we need to implement Procedures 1 and 2 to obtain $\widehat{\mathbf{v}},\widehat{S}$ before executing Algorithm 2.

In many practical applications, we need to compute $\mathcal{A}{\mathbf{x}}^{m-1}$ repeatedly if we use ‘s-ttv’ in the tensor toolbox. However, in our proposed methods, we only use Procedures 1 and 2 once to obtain $\widehat{\mathbf{v}},\widehat{S}$ and then call Algorithm 2 repeatedly, thus greatly shortening the computation time. Therefore, our proposed methods may not have advantages in a single calculation but instead have significant advantages in repeated calculations.

5.4. Comparison of Algorithms 2 and 3

In this section, in order to demonstrate the effectiveness of the simplified strategy presented in Section 4.3, we demonstrate the numerical performance of Algorithms 2 and 3 based on $\widehat{\mathbf{v}},\widehat{S}$.

Example 6.

Let $\mathbf{x}=2\mathbf{e}\in {\mathbb{R}}^n$ with $\mathbf{e}={(1,\cdots ,1)}^{\top }$ and $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{l\times [m-1,n]}$ be a sparse tensor generated by sptenrand with sparsity ratio $sr$. Then, we can use Procedure 2 to obtain the vector of all non-zero entries $\widehat{\mathbf{v}}$ and the corresponding index matrix $\widehat{S}$.

In the following, for any given $\mathcal{A}\in {\mathbb{R}}^{l\times [m-1,n]}$ and $\mathbf{x}\in {\mathbb{R}}^n$ with different $(l,m,n)$, we use Algorithms 2 and 3 to compute tensor–vector products $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ based on $\widehat{\mathbf{v}},\widehat{S}$. In addition, we use ‘s-ttv’ to compute $\mathcal{A}{\mathbf{x}}^{m-1}$ as a baseline for comparison.

To demonstrate the performance of the different algorithms, we show the average CPU time cost for 10 random experiments. We use ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}^{\widehat{\mathbf{v}},\widehat{S}}$, ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}^{\mathbf{J}}$, and ${\mathrm{CPU}}_{\mathtt{s}-\mathtt{ttv}}$ to denote the average CPU time cost for Algorithms 2, 3, and ‘s-ttv’, respectively. As calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ with Algorithm 3 requires an additional procedure, in order to determine the computation cost associated with the pre-processing procedure clearly, we use ${\mathrm{CPU}}_{\mathbf{J}}$ to denote the average CPU time cost for computing the index set $\mathbf{J}$ based on $\widehat{\mathbf{v}}$ and $\widehat{S}$ for the merge tensor $\widehat{\mathcal{A}}$ via Procedure 3. Here, q is the average number of non-zero elements of corresponding merge tensors (i.e., the average number of rows of $\widehat{S}$), and $q^{\prime }$ is the average number of different rows for $\widehat{S}$ obtained by discarding the first column. The numerical results are presented in

Table 5. Numerical results for computing $\mathcal{A}{\mathbf{x}}^{m-1}$ in Example 6 based on $\widehat{\mathbf{v}}$ and $\widehat{S}$.

$(\mathit{l},\mathit{m},\mathit{n}),\mathbf{sr}$	$\mathit{q}/{\mathit{q}}^{\prime }$	${\mathrm{CPU}}_{\mathbf{s}\text{-}\mathbf{ttv}}$	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}^{\widehat{\mathit{v}},\widehat{\mathit{S}}}$	${\mathrm{CPU}}_{\mathbf{J}}$	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}^{\mathbf{J}}$
(50, 4, 50), 50%	1,027,344.4/22,100	0.7016	0.0688	1.2125	0.0313
(80, 4, 100)	12,909,783.7/171,700	3.4922	1.1484	20.8891	0.8984
(100, 4, 100)	16,137,621.3/171,700	4.2016	1.4516	28.4625	1.2314
(30, 5, 30)	1,221,928.7/40,920	2.0297	0.1375	2.0484	0.0594
(10, 6, 10)	19,878.7/2002	0.0547	0.0031	0.0547	0.0000
(10, 6, 20)	424,585/42,503.8	2.3375	0.0531	0.7375	0.0188
(50, 4, 50), 20%	740,674.6/22,100	0.3063	0.0406	1.0297	0.0281
(80, 4, 100)	9,398,356.5/171,700	2.1141	0.6859	12.6781	0.5781
(100, 4, 100)	11,748,329.8/171,700	2.4484	0.8672	16.8688	0.6141
(30, 5, 30)	1,171,271.9/40,920	1.4000	0.1016	1.7031	0.0391
(10, 6, 10)	19,399.1/2000.5	0.0281	0.0031	0.0266	0.0000
(10, 6, 20)	422,239.6/42,501.2	1.7156	0.0422	0.7250	0.0188
(50, 4, 50), 10%	474,356.7/22,099.8	0.1406	0.1172	1.0281	0.0156
(80, 4, 100)	6,042,749.7/171,700	1.7094	0.5109	11.2156	0.4125
(100, 4, 100)	7,553,286.2/171,700	1.7328	0.5656	11.9125	0.3469
(30, 5, 30)	1,017,688.9/40,918.4	0.8391	0.0891	1.5969	0.0375
(10, 6, 10)	18,358.6/1997.7	0.0156	0.0031	0.0625	0.0000
(10, 6, 20)	414,376.1/42,493.6	1.3891	0.0547	0.6438	0.0031
(50, 4, 50), 5%	271,201.6/22,093.4	0.0453	0.0328	0.9266	0.0094
(80, 4, 100)	3,463,587.5/171,698	1.1625	0.2719	7.1453	0.1609
(100, 4, 100)	4,329,140.1/171,699	1.3859	0.3125	8.8484	0.2391
(30, 5, 30)	747,763.8/40,911.3	0.4813	0.0625	1.1188	0.0297
(10, 6, 10)	16,135.6/1988	0.0156	0.0000	0.0406	0.0000
(10, 6, 20)	389,369.2/42,456.2	0.7875	0.0328	0.5656	0.0125
(50, 4, 50), 1%	60,719.4/20,555.9	0.0094	0.0031	0.4250	0.0000
(80, 4, 100)	776,760/169,413.8	0.2031	0.1656	4.7703	0.0438
(100, 4, 100)	970,963.6/170,773.7	0.2500	0.0734	5.7016	0.0344
(30, 5, 30)	218,543.7/40,215.5	0.0859	0.0594	1.0797	0.0063
(10, 6, 10)	7275.2/1837.3	0.0016	0.0000	0.0141	0.0000
(10, 6, 20)	211,318.5/41,432.7	0.1063	0.0859	0.8656	0.0313

.

The total number of multiplications is $q(m-1)$ when we calculate $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ using Algorithm 2, while the total number is $q^{\prime }(m-2)+q$ when we calculate $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ with Algorithm 3. From Table 5, we can see that Algorithm 3 can avoid a significant number of multiplications. Regarding the CPU time cost, Algorithm 3 is faster than Algorithm 2 with different $(l,m,n)$ and sparse ratio $sr$.

Compared to Algorithm 2, Algorithm 3 takes additional time to find the index set $\mathbf{J}$ using Procedure 3 before execution. Similar to the analysis in Section 5.3, when we need to compute $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ iteratively, Procedure 3 only needs to be executed once to obtain $\mathbf{J}$, following which Algorithm 3 is called repeatedly, thus greatly reducing the computational time.

5.5. Finding the Least Element for the Set Defined by Polynomial Inequalities

In this section, we consider the least element problem for the set defined by polynomial inequalities [32] in which the proposed iterative algorithm for finding the least element of the considered set involves the calculation of $\mathcal{A}{\mathbf{x}}^{m-1}$. In order to verify the advantage of our Algorithm 2 over ttv and s-ttv in the tensor toolbox [33], we conducted the numerical experiments using Example 5.3 from [32]:

Example 7

(Example 5.3 in [32]). Consider the set $S:=\{\mathbf{x}\in {\mathbb{R}}^{2n+10}:\mathbf{x}\ge \mathbf{0}\mathrm{and}\mathcal{A}{\mathbf{x}}^3\ge \mathbf{b}\}$, where

$$\begin{array}{c}\hfill \mathbf{b}=\left(\begin{array}{c}{\mathbf{b}}_1\\ {\mathbf{b}}_2\\ {\mathbf{b}}_3\end{array}\right)\in {\mathbb{R}}^{3n+10}\mathrm{with}{b}_i=\left\{\begin{array}{cc}{i}^3& \mathrm{if}i\in \{1,\dots ,n\},\hfill \\ -2i^3& \mathrm{if}i\in \{n+11,\dots ,2n+10\},\hfill \\ -i^3& \mathrm{if}i\in \{2n+11,\dots ,3n+10\},\hfill \end{array}\right.\end{array}$$

with ${\mathbf{b}}_2={(-10,-12,-20,-11.0279,-20.5699,-10,-10,-10,-10,-10)}^{\top }\in {\mathbb{R}}^{10}$, and $\mathcal{A}=\left(a_{i_1{i}_2{i}_3{i}_4}\right)\in {\mathbb{R}}^{(3n+10)\times [3,2n+10]}$ with diagonal entries

$$\begin{array}{c}\hfill a_{iiii}=\left\{\begin{array}{cc}1& \mathrm{if}i\in \{1,\dots ,n+10\}\setminus \{n+2\},\hfill \\ 2& \mathrm{if}i=n+2,\hfill \\ -1& \mathrm{if}i\in \{n+11,\dots ,2n+10\},\hfill \end{array}\right.\end{array}$$

and off-diagonal entries

$$\left\{\begin{array}{ccc}{a}_{1(n+1)(n+1)(n+2)}=4,\hfill & a_{1(n+1)(n+2)(n+1)}=-2,\hfill & a_{1(n+2)(n+1)(n+1)}=-2;\hfill \\ a_{2(n+2)(n+3)(n+3)}=-1;\hfill & & \\ a_{i1(n+11)(n+11)}=a_{i(n+2)(n+4)(n+4)}=2,\hfill & a_{i(n+11)(n+11)1}=a_{i(n+4)(n+2)(n+4)}=-3,\hfill & \mathrm{for}\mathrm{any}i\in \{1,2,5,10,\dots ,\lfloor {\displaystyle \frac{n}{5}}\rfloor \};\hfill \\ a_{(n+1)112}=a_{(n+1)122}=-2,\hfill & a_{(n+1)(n+1)(n+1)(n+3)}=-1,\hfill & a_{(n+1)(n+1)(n+3)(n+3)}=-1;\hfill \\ a_{(n+2)111}=a_{(n+2)223}=-1,\hfill & a_{(n+2)(n+2)(n+2)(n+3)}=-1,\hfill & a_{(n+2)113}=-4;\hfill \\ a_{(n+3)111}=-1,\hfill & a_{(n+3)223}=-1,\hfill & a_{(n+3)112}=-4;\hfill \\ a_{(n+4)112}=a_{(n+4)113}=-2,\hfill & a_{(n+4)(n+4)(n+4)(n+5)}=-1,\hfill & a_{(n+4)(n+4)(n+5)(n+5)}=-1;\hfill \\ a_{(n+5)11n}=-{\displaystyle \frac{4}{n}},\hfill & a_{(n+5)12n}=-1;\hfill & \\ a_{(n+6)111}=-2,\hfill & a_{(n+6)(n+6)(n+6)(n+7)}=-3;\hfill & \\ a_{(n+7)112}=-1,\hfill & a_{(n+7)(n+6)(n+7)(n+7)}=-1;\hfill & \\ a_{(n+8)11n}=-{\displaystyle \frac{1}{n}},\hfill & a_{(n+8)(n+6)(n+8)(n+8)}=-2;\hfill & \\ a_{(n+9)12n}=-{\displaystyle \frac{1}{n}},\hfill & a_{(n+9)(n+8)(n+8)(n+9)}=-1;\hfill & \\ a_{(n+10)112}=-1,\hfill & a_{(n+10)(n+7)(n+7)(n+8)}=-2;\hfill & \\ a_{i11n}=-{\displaystyle \frac{1}{n}},\hfill & a_{i(n+6)(n+6)i}=-2,\hfill & \mathrm{for}\mathrm{any}i\in \{n+11,\dots ,2n+10\};\hfill \\ a_{i11(n+1)}=a_{i(n+1)(n+1)(i-n-5)}=-1,\hfill & a_{i(i-n-5)(i-n-5)(i-n)}=-2,\hfill & \mathrm{for}\mathrm{any}i\in \{2n+11,\dots ,3n+10\};\hfill \end{array}\right.$$

whille all other entries are 0. Here, we use $\lfloor x\rfloor$ to denote the floor operation of $x\in \mathbb{R}$.

From the definition of the tensor $\mathcal{A}$ in Example 7, the merge tensor $\widehat{\mathcal{A}}$ can be easily obtained. We used the algorithm proposed in [32] to find the least element of the set $\mathcal{S}$ involving the merge tensor $\widehat{\mathcal{A}}$, where we replaced ttv used in the fixed point iteration process of the algorithm proposed in [32] with s-ttv and our Algorithm 2. We compared the original algorithm in [32] with those replaced by s-ttv and Algorithm 2. The numerical results are shown in

Table 6. Numerical results of Example 7 for different values of n.

n	It	It(fp)	${\mathrm{CPU}}_{\mathbf{ttv}}$/${\mathrm{CPU}}_{\mathbf{s}\text{-}\mathbf{ttv}}$/${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}$	${\text{CPU(fp)}}_{\mathbf{ttv}}$/${\text{CPU(fp)}}_{\mathbf{s}\text{-}\mathbf{ttv}}$/${\text{CPU(fp)}}_{\widehat{\mathbf{A}}\mathbf{xm}1}$	${\mathit{x}}^{\mathit{k}+1}$	Val
10	3	915	4.5313/2.1719/0.2969	4.3281/1.9688/0.0938	+	−2.07 × ${10}^{-11}$
20	3	914	4.8750/2.3281/0.3750	4.5938/2.0781/0.1875	+	−2.46 × ${10}^{-11}$
50	3	913	34.1094/3.9063/2.1563	32.4219/2.1250/0.3750	+	−2.62 × ${10}^{-10}$
100	3	913	468.9375/39.5156/36.2969	433.4063/0.9375/0.5000	+	−1.75 × ${10}^{-9}$
120	3	913	995.7344/84.4375/79.2813	913.2969/1.4063/0.7969	+	−2.79 × ${10}^{-9}$
150	3	907	2809.6875/258.2656/256.9063	2552.2188/5.3750/4.1719	+	−5.59 × ${10}^{-9}$

, where It represents the number of iterations, and It(fp) represents the number of iterations in the fixed point method. ${\mathrm{CPU}}_{\mathtt{ttv}}$, ${\mathrm{CPU}}_{\mathtt{s}-\mathtt{ttv}}$, and ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}$ represent the CPU times for the respective algorithms. ${\text{CPU(fp)}}_{\mathtt{ttv}}$, ${\text{CPU(fp)}}_{\mathtt{s}-\mathtt{ttv}}$, and ${\text{CPU(fp)}}_{\widehat{\mathrm{A}}\mathrm{xm}1}$ represent the CPU time for the fixed point methods used in the respective algorithms. Here, ${\mathbf{x}}^{k+1}$ is the least element, Val$:=min\{{(\mathcal{A}{\left({\mathbf{x}}^{k+1}\right)}^{m-1}-\mathbf{b})}_i:i\in [3n+10]\}$, and the symbol $‘‘+"$ means ${\mathbf{x}}^{k+1}\ge 0$. It is worth noting that ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}$ and ${\text{CPU(fp)}}_{\widehat{\mathrm{A}}\mathrm{xm}1}$ include the CPU time associated with Procedure 1 for calculating the vector $\widehat{\mathbf{v}}$ and matrix $\widehat{S}$.

From Table 6, it can be seen that replacing ttv with s-ttv and Algorithm 2 greatly reduced the CPU time of the iterative algorithm due to the sparsity of the example. Moreover, Algorithm 2 always performed better than s-ttv for different values of n, especially when n is relatively small.

5.6. Tensor Complementarity Problems with the Implicit Z-Tensors

To further validate the effectiveness of embedding our Algorithm 2 in iterative algorithms, we consider tensor complementarity problems with the implicit Z-tensors [20] in which the proposed fixed point iterative algorithm involves the calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$. We conducted numerical experiments using Example 4 in [20]:

Example 8

(Example 4 in [20]). Consider TCP$(\mathcal{A},\mathbf{q})$, where $\mathcal{A}=\left(a_{i_1{i}_2{i}_3{i}_4{i}_5}\right)\in {\mathbb{R}}^{[5,n]}$ with diagonal entries $a_{iiiii}=1$ for any $i\in \left[n\right]\setminus \{3,6,9,\dots \}$ and $a_{iiiii}=-1$ for any $i\in \{3,6,9,\dots \}$, non-diagonal entries

$$\left\{\begin{array}{cccc}{a}_{iiii(i+2)}=2,\hfill & a_{i(i+2)iii}=-5,\hfill & a_{i(i+1)(i+1)(i+2)(i+2)}=-1,\hfill & \forall i\in \{1,4,7,\dots \},\hfill \\ a_{i(i-1)(i-1)ii}=2,\hfill & a_{iii(i-1)(i-1)}=-6,\hfill & a_{iii(i+1)(i+1)}=-2,\hfill & \forall i\in \{2,5,8,\dots \},\hfill \\ a_{i(i-2)(i-2)ii}=3,\hfill & a_{iii(i-2)(i-2)}=-5,\hfill & a_{i(i-2)(i-2)(i-1)(i-1)}=-{\displaystyle \frac{1}{4}},\hfill & \forall i\in \{3,6,9,\dots \},\hfill \end{array}\right.$$

with all other entries being 0; furthermore, $\mathbf{q}\in {\mathbb{R}}^n$ with $q_i=-1$ for $i\in \{1,4,7,\dots \}$, $q_i=0$ for $i\in \{2,5,8,\dots \}$ and $q_i=16$ for $i\in \{3,6,9,\dots \}$.

The tensor $\mathcal{A}$ in Example 8 is an implicit Z-tensor, and we can easily obtain the merge tensor $\widehat{\mathcal{A}}$. We embedded our Algorithm 2 into the fixed point iterative algorithm in [20] and compared it with the algorithm presented in [20], where the calculation of $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ uses the tensor toolbox [33]. The numerical results are shown in

Table 7. Numerical results for computing $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ in Example 8.

n	${\mathit{x}}^0$	Iteration	${\mathrm{CPU}}_{\widehat{\mathbf{A}}\mathbf{xm}1}$	${\mathrm{CPU}}_{\mathbf{ttv}}$	Res
10	${(5,3,0,5,3,0,\dots )}^{\top }$	54	0.0781	0.1719	6.65 × ${10}^{-13}$
20	${(5,3,0,5,3,0,\dots )}^{\top }$	55	0.1094	1.3281	5.26 × ${10}^{-13}$
50	${(5,3,0,5,3,0,\dots )}^{\top }$	55	0.2656	9.8281	8.20 × ${10}^{-13}$
80	${(5,3,0,5,3,0,\dots )}^{\top }$	56	0.5625	88.3906	5.54 × ${10}^{-13}$
10	${(5,0,0,5,0,0,\dots )}^{\top }$	42	0.0312	0.1250	9.27 × ${10}^{-13}$
20	${(5,0,0,5,0,0,\dots )}^{\top }$	43	0.0625	0.9375	6.13 × ${10}^{-13}$
50	${(5,0,0,5,0,0,\dots )}^{\top }$	43	0.1094	7.7344	9.56 × ${10}^{-13}$
80	${(5,0,0,5,0,0,\dots )}^{\top }$	44	0.3438	81.5156	6.00 × ${10}^{-13}$

, where $x^0$ represents the initial solution, Iteration represents the number of iterations, and ${\mathrm{CPU}}_{\widehat{\mathrm{A}}\mathrm{xm}1}$ and ${\mathrm{CPU}}_{\mathtt{ttv}}$ represent the CPU time for the whole algorithm embedded with our Algorithm 2 and that presented in [20], respectively. Furthermore, Res represents the natural residual.

It can be seen that Algorithm 2 outperformed the algorithm in [20] for different initial solutions and values of n. Algorithm 2 greatly reduced the computational time for the fixed point iteration algorithm, especially when n is larger.

In order to demonstrate the difference in calculating $\widehat{\mathcal{A}}{\mathbf{x}}^{m-1}$ between the fixed point iterative algorithm in [20] in comparison with our Algorithm 2, using the tensor toolbox more intuitively, we took the initial solution $x^0=(5,0,0,5,0,0,\dots )$ as an example and determined the total computational time when using iterative algorithms adopting the above two techniques, as shown in Figure 2. The bar chart shows the computational time with n equal to $10,20,30,40,$ and 50, while the line chart shows the difference in computational time between the algorithms.

From Figure 2, we can see that our Algorithm 2 and tensor toolbox have similar and shorter computational times when n is relatively small. Our algorithm does not significantly increase the computational time with increasing n, while the tensor toolbox algorithm increases rapidly. Therefore, the proposed algorithm has a better application value in large-scale problems.

6. Concluding Remarks

Many tensors and their related problems involve a calculation of the vector of a $(m-1)$-degree homogeneous polynomial defined by a tensor $\mathcal{A}{\mathbf{x}}^{m-1}$. In this study, taking symmetry and sparsity properties into consideration, we proposed efficient algorithms that avoid the large amount of computation inherent to existing algorithms. Specifically, utilizing the symmetry of monomials in the canonical basis of homogeneous polynomials, we proposed a method to calculate $\mathcal{A}{\mathbf{x}}^{m-1}$ using the merge tensor of the involved tensor to replace the original tensor, thus reducing the computational cost. Then, an algorithm was designed to calculate this tensor, combining sparsity to further reduce the computational cost. Moreover, through analysis of the calculation details, a simplified algorithm that avoids duplicate calculations was further proposed. Finally, the results of extensive numerical experiments verified the effectiveness of the proposed methods.

There are still some aspects worthy of further study. First, although Procedure 2 or Procedure 3 only need to be executed once (or offline) for iterative algorithms, they take a relatively long time; as such, further accelerating the speed of these procedures would be desirable. Second, it is clear that more entries can be merged for sparser tensors, thus making our algorithms faster; therefore, can we determine the range of sparsity that is suitable for our algorithms? Finally, in addition to considering computation time, storage space may also be taken into account. Future investigations may help to address these outstanding questions.