A Graph-Based Recommendation Algorithm on Quaternion Algebra

Abstract

This study presents a novel Quaternion-based link prediction method to be used in different recommendation systems. The method performs Quaternion algebra-based computations while making use of expressive and wide-ranged learning properties of the Hamilton products. The proposed key capabilities rely on link prediction to boost performance in top-N recommendation tasks. According to the achieved experimental results, the proposed method allows for highly improved performance according to three quality measurements: (i) hits rate, (ii) coverage, and (iii) novelty; when applied to two datasets, namely the Movielens and Hetrec datasets. To assess the flexibility level of the proposed algorithm in terms of incorporating alternative sources of information, further wide-scale tests are carried out on three subsets of the Amazon dataset. Hence, the effectiveness of Quaternion algebra in graph-based recommendation algorithms is verified. The algorithms suggested here are further enhanced using similarity and dissimilarity factors between users and items, as well as ‘like’ and ‘dislike’ relationships between users and items. It is observed that this approach is adaptable by incorporating different information sources and can successfully overcome the drawbacks of conventional graph-based recommender systems. It is argued that the proposed novel idea of Quaternion-based link prediction method stands as a superior alternative to existing methods.

Appendix

Theorem 1.

Let ${\varvec{A}}$ be a square matrix with size $n$ by $n$, and let $\lambda = {(}\lambda_{1} ,\lambda_{2} ,...,\lambda_{n} {)}$ be all the eigenvalues of ${\varvec{A}}$. For each positive integer $k$, the $k^{th}$ powers of these values, which are $\lambda^{k} = {(}\lambda_{1}^{k} ,\lambda_{2}^{k} ,...,\lambda_{n}^{k} {)}$, correspond to the eigenvalues of ${\varvec{A}}^{k}$, [35].

Proof 1.

By the triangularization (or Jordan canonical form), there exists a nonsingular matrix ${\varvec{S}}$ such that;

$${\varvec{S}}^{{{\varvec{ - 1}}}} {\varvec{AS}} = \left[ {\begin{array}{*{20}l} {\lambda_{1} } \hfill & * \hfill & * \hfill & * \hfill & * \hfill \\ 0 \hfill & {\lambda_{2} } \hfill & * \hfill & * \hfill & * \hfill \\ \vdots \hfill & \cdots \hfill & \ddots \hfill & \ldots \hfill & \vdots \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & {\lambda_{n - 1} } \hfill & * \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill & {\lambda_{n} } \hfill \\ \end{array} } \right].$$

The result is an upper triangular matrix whose diagonal entries are eigenvalues of ${\varvec{A}}$. After the $k^{th}$ power of the Jordan canonical form are taken, then we have

$${\varvec{S}}^{{{\varvec{ - 1}}}} {\varvec{A}}^{k} {\varvec{S}} = ({\varvec{S}}^{{{\varvec{ - 1}}}} {\varvec{AS}})^{k} = \left[ {\begin{array}{*{20}l} {\lambda_{1}^{k} } \hfill & * \hfill & * \hfill & * \hfill & * \hfill \\ 0 \hfill & {\lambda_{2}^{k} } \hfill & * \hfill & * \hfill & * \hfill \\ \vdots \hfill & \cdots \hfill & \ddots \hfill & \ldots \hfill & \vdots \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & {\lambda_{n - 1}^{k} } \hfill & * \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill & {\lambda_{n}^{k} } \hfill \\ \end{array} } \right].$$

The characteristic polynomial of the matrix ${\varvec{A}}^{k}$ is given by;

$$\begin{gathered} f{(}\lambda ) = \det {(}{\varvec{A}}^{k} - \lambda {\varvec{I}}{)} \hfill \\ = \det {(}{\varvec{S}}^{ - 1} )\det {(}{\varvec{A}}^{k} - \lambda {\varvec{I}}{)}\det {(}{\varvec{S}}{)} \hfill \\ = \det {(}{\varvec{S}}^{ - 1} \cdot {(}{\varvec{A}}^{k} - \lambda {\varvec{I}}{)} \cdot {\varvec{S}}{)} \hfill \\ = \det {(}{\varvec{S}}^{ - 1} \cdot {\varvec{A}}^{k} \cdot {\varvec{S}} - \lambda {\varvec{I}}{)} \hfill \\ \end{gathered}$$

$$\left[ {\begin{array}{*{20}c} {\lambda_{1}^{k} - \lambda } & * & * & * & * \\ 0 & {\lambda_{2}^{k} - \lambda } & * & * & * \\ \vdots & \ldots & \ddots & \ldots & \vdots \\ 0 & 0 & 0 & {\lambda_{n - 1}^{k} - \lambda } & * \\ 0 & 0 & 0 & 0 & {\lambda_{n}^{k} - \lambda } \\ \end{array} } \right]$$

$$= \prod\limits_{i = 1}^{n} {(} \lambda_{i}^{k} - \lambda {)}$$

(A1)

Since the roots of the characteristic polynomial are all the eigenvalues as mentioned in Proposition 1, it can be seen from Eq. (A.1) that $\lambda^{k} = {(}\lambda_{1}^{k} ,\lambda_{2}^{k} ,...,\lambda_{n}^{k} {)}$ are all the eigenvalues of ${\varvec{A}}^{k}$.

Hence, when the eigenvalue matrix of ${\varvec{A}}$ is denoted as ${{\varvec{\Lambda}}}$, and the eigenvector matrix of ${\varvec{A}}$ is denoted as ${\varvec{U}}$, then the $k^{th}$ power of ${\varvec{A}}$, i.e. ${\varvec{A}}^{k}$ can be written as;

$${\varvec{A}}^{k} = {\varvec{U\Lambda }}^{k} {\varvec{U}}^{T}$$

Theorem 2.

If ${\varvec{A}}$ is a square matrix with eigenvalues.

$\lambda_{i} ,i = 1,2,...,n$,

(a)$\det {(}{\varvec{A}}{) = }\prod\limits_{i = 1}^{n} {\lambda_{i} }$

(b)${\text{trace}} {(}{\varvec{A}}{)} = \sum\limits_{i = 1}^{n} {\lambda_{i} }$

Proof 2. (Part a)

Recall that eigenvalues are roots of the characteristic polynomial.

It follows that this polynomial corresponds to $f_{{\varvec{A}}} {(}\lambda {)} = \det {(}{\varvec{A}} - \lambda {\varvec{I}}_{n} {)}$; and then

$$\begin{gathered} \det {(}{\varvec{A}} - \lambda {\varvec{I}}_{n} {)} = \left| {\begin{array}{*{20}c} {a_{11} - \lambda } & {a_{12} } & \cdots & {a_{1,n} } \\ {a_{21} } & {a_{22} - \lambda } & \cdots & {a_{2,n} } \\ \vdots & \vdots & \ddots & \vdots \\ {a_{n1} } & {a_{m2} } & \cdots & {a_{nn} - \lambda } \\ \end{array} } \right| \hfill \\ = \prod\limits_{i = 1}^{n} {{(}\lambda_{i} - \lambda {)}{\text{.}}} \hfill \\ \end{gathered}$$

(A2)

If $\lambda$ is assigned a value as $\lambda = 0$, it can be seen that

$$\det {(}{\varvec{A}}{) = }\prod\limits_{i = 1}^{n} {\lambda_{i} }$$

and this completes the proof of part (a).

(Part b) Let’s now compare the coefficients of $\lambda^{n - 1}$ of both sides of Eq. (A.2). The coefficient of $\lambda^{n - 1}$ of the determinant on the left side of Eq. (A.2) is, $( - 1)^{n - 1} (a_{11} + a_{22} + \ldots + a_{nn} ) = ( - 1)^{n - 1} trace({\varvec{A}})$.

The coefficient of $\lambda^{n - 1}$ of the determinant on the right side of Eq. (A.2) is

Thus, as the ${( - 1)}^{n - 1} \sum\nolimits_{i = 1}^{n} {\lambda_{i} }$ proof of part (b), we have: ${\text{trace}} {(}{\varvec{A}}{)} = \sum\nolimits_{i = 1}^{n} {\lambda_{i} }$.