Accurate relational reasoning in edge-labeled graphs by multi-labeled random walk with restart

Abstract

Given an edge-labeled graph and two nodes, how can we accurately infer the relation between the nodes? Reasoning how the nodes are related is a fundamental task in analyzing network data, and various relevance measures have been suggested to effectively identify relevance between nodes in graphs. Although many random walk based models have been extensively utilized to reveal relevance between nodes, they cannot distinguish how those nodes are related in terms of edge labels since the traditional surfer does not consider edge labels for estimating relevance scores. In this paper, we propose MuRWR (Multi-Labeled Random Walk with Restart), a novel random walk based model that accurately identifies how nodes are related with, considering multiple edge labels. We introduce a labeled random surfer whose label indicates the relation between starting and visiting nodes, and change the surfer’s label during random walks for multi-hop relational reasoning. We also learn appropriate rules on changing the surfer’s label from the edge-labeled graph to accurately infer relations. We develop an iterative algorithm for computing MuRWR, and prove the convergence guarantee of the algorithm. Through extensive experiments, we show that our model MuRWR provides the best inference performance.

Appendices

Appendix A: Lemma for convexity and solution of label transition probabilities

Lemma 2

The estimator in (2) maximizes the weighted log-likelihood function $WL(\mathbf {S}_{k};\mathcal {D}_{k})$ in (1).

Proof

Our goal is to find S_k that maximizes $WL(\mathbf {S}_{k};\mathcal {D}_{k})$, which is equivalent to minimizing $-WL(\mathbf {S}_{k};\mathcal {D}_{k})$. The probability P(x_kh; S_k) is written as follows:

$$ P(\mathbf{x}_{kh}; \mathbf{S}_{k}) = \mathbf{S}_{k\mathbf{x}_{khs}\mathbf{x}_{kht}}=\prod\limits_{i=1}^{K}\prod\limits_{j=1}^{K}(\mathbf{S}_{kij})^{\mathbf{1}(\mathbf{x}_{khs}=i, \mathbf{x}_{kht}=j)} $$

Then $-WL(\mathbf {S}_{k};\mathcal {D}_{k})$ is represented as follows:

$$ \begin{array}{@{}rcl@{}} -WL(\mathbf{S}_{k};\mathcal{D}_{k}) &=& -\sum\limits_{h=1}^{n_{k}} w_{\mathbf{x}_{kht}}\log{P(\mathbf{x}_{kh};\mathbf{S}_{k})}\\ &=& -\sum\limits_{h=1}^{n_{k}}\sum\limits_{i=1}^{K}\sum\limits_{j=1}^{K}w_{j}\mathbf{1}(\mathbf{x}_{khs} = i, \mathbf{x}_{kht} = j)\log{\mathbf{S}_{kij}}\\ &=& -\sum\limits_{i=1}^{K}\sum\limits_{j=1}^{K}w_{j}\left( \sum\limits_{h=1}^{n_{k}}\mathbf{1}(\mathbf{x}_{khs}=i, \mathbf{x}_{kht}=j)\right)\log\mathbf{S}_{kij}\\ &=& -\sum\limits_{i=1}^{K}\sum\limits_{j=1}^{K}w_{j}N_{kij}\log\mathbf{S}_{kij} \end{array} $$

where $N_{kij}={\sum }_{h=1}^{n_{k}}\mathbf {1}(\mathbf {x}_{khs} = i, \mathbf {x}_{kht} = j)$ is the count of the label transition observations. Then the minimization problem is represented as follows:

$$ \begin{array}{@{}rcl@{}} \underset{\mathbf{S}_{kij}}{\text{minimize}} -WL(\mathbf{S}_{k};\mathcal{D}_{k}) &=& -\sum\limits_{i=1}^{K}\sum\limits_{j=1}^{K}w_{j}N_{kij}\log\mathbf{S}_{kij}\\ \text{subject to} \mathbf{S}_{kij} &\geq& 0 \text{ for } 1\leq i,j \leq K, \\ \sum\limits_{j=1}^{K}\mathbf{S}_{kij}&=&1 \text{ for } 1\leq i \leq K. \end{array} $$

(7)

Note that the above problem is convex (see Lemma 3); thus, the optimization problem is solved by the KKT theorem [5], and the solution of the problem is represented as (2) (details in Lemma 4. □

Lemma 3

The optimization problem in (7) is convex.

Proof

The objective function is convex since the negative log functions $-\log {\mathbf {S}_{kij}}$ are convex, and the sum of non-negatively weighted convex functions is convex (i.e., w_jN_kij ≥ 0) [5]. Let C be a set of S_k satisfying the constraints, i.e., C = {S_k|S_k1 = 1,S_kij ≥ 0, for 1 ≤ i,j ≤ K}. For $\mathbf {S}_{k_{1}},\mathbf {S}_{k_{2}}\!\in \!\mathbf {C}$ and 𝜃₁ + 𝜃₂ = 1 such that 𝜃₁,𝜃₂ ≥ 0, let $\mathbf {S}_{k_{3}}=\theta _{1}\mathbf {S}_{k_{1}} + \theta _{2}\mathbf {S}_{k_{2}}$. Then $\mathbf {S}_{k_{3}}\mathbf {1} = (\theta _{1}\mathbf {S}_{k_{1}} + \theta _{2}\mathbf {S}_{k_{2}})\mathbf {1} = \mathbf {1}$ indicating $\mathbf {S}_{k_{3}}\!\in \!\mathbf {C}$. Thus C is convex by the definition of convex set [5]. □

Lemma 4

The solution of the optimization problem in (7) is represented as (2).

Proof

The lagrangian ${\mathscr{L}}(\cdot )$ of the objective function in (7) is represented as follows:

$$ \begin{array}{@{}rcl@{}} \mathcal{L}(\mathbf{S}_{k}, \mathbf{\lambda}, \mathbf{\nu}) &=& -\sum\limits_{i=1}^{K}\sum\limits_{j=1}^{K}w_{j}N_{kij}\log\mathbf{S}_{kij} + \sum\limits_{i=1}^{K}\sum\limits_{j=1}^{K}-\lambda_{ij}\mathbf{S}_{kij} \\ &+& \sum\limits_{i=1}^{K}\nu_{i}\sum\limits_{j=1}^{K}\left( \mathbf{S}_{kij}-1\right) \end{array} $$

where λ and ν are inequality and equality lagrange multipliers, respectively. Let $\hat {\mathbf {S}}_{kij}$ be the solution that minimizes (7). λ^∗ and ν^∗ denote the optional points for λ and ν, respectively. The stationarity condition $\nabla _{\mathbf {S}_{k}}{\mathscr{L}}(\hat {\mathbf {S}}_{k},\lambda ^{*},\nu ^{*})=0$ implies the following equation:

$$ \frac{\partial \mathcal{L}(\hat{\mathbf{S}}_{k}, \mathbf{\lambda^{*}}, \mathbf{\nu^{*}})}{\partial \mathbf{S}_{kij}} = - \frac{w_{j}N_{kij}}{\hat{\mathbf{S}}_{kij}} - \lambda^{*}_{ij} + \nu^{*}_{i} = 0 \Leftrightarrow \hat{\mathbf{S}}_{kij} = \frac{w_{j}N_{kij}}{\nu^{*}_{i}-\lambda^{*}_{ij}} $$

By the complementary slackness $\lambda ^{*}_{ij}\hat {\mathbf {S}}_{kij} = 0$, primal feasibility $\hat {\mathbf {S}}_{kij} \geq 0$, and dual feasibility $\lambda ^{*}_{ij} \geq 0$,

$$ \begin{array}{@{}rcl@{}} \bullet \text{ } \hat{\mathbf{S}}_{kij} > 0 \Rightarrow \lambda^{*}_{ij} &=& 0 \Leftrightarrow \hat{\mathbf{S}}_{kij} = \frac{w_{j}N_{kij}}{\nu^{*}_{i}} > 0 \Leftrightarrow w_{j}N_{kij} \neq 0\\ \bullet \text{ } \lambda^{*}_{ij} > 0 \Rightarrow \hat{\mathbf{S}}_{kij} &=& 0 \Leftrightarrow \hat{\mathbf{S}}_{kij} = \frac{w_{j}N_{kij}}{\nu^{*}_{i}\! -\! \lambda^{*}_{ij}} = 0 \Leftrightarrow w_{j}N_{kij} = 0 \end{array} $$

For the case that $\hat {\mathbf {S}}_{kij} > 0$, $\nu ^{*}_{i}$ is obtained from the equality constraint ${{\sum }_{z=1}^{K}\hat {\mathbf {S}}_{kiz}=1}$ as follows:

$$ \begin{array}{@{}rcl@{}} \sum\limits_{z=1}^{K}\hat{\mathbf{S}}_{kiz} &=& \!\!\!\!\!\underset{ \{z | \hat{\mathbf{S}}_{kiz} > 0 \}}{\sum}\!\!\!\!\! \hat{\mathbf{S}}_{kiz} = \!\!\!\!\!\underset{ \{z | \hat{\mathbf{S}}_{kiz} > 0 \}}{\sum}\!\!\!\!\! \frac{w_{z}N_{kiz}}{\nu^{*}_{i}} = 1 \Leftrightarrow\\ \nu^{*}_{i} &=& \!\!\!\!\!\!\! \underset{ \{z | \hat{\mathbf{S}}_{kiz} > 0 \}}{\sum} \!\!\!\!\! w_{z}N_{kiz} = \!\!\!\!\!\!\!\! \underset{ \{z | \hat{\mathbf{S}}_{kiz} > 0 \}}{\sum}\!\!\!\!\! w_{z}N_{kiz} + \!\!\!\!\!\!\!\!\underset{ \{z | \hat{\mathbf{S}}_{kiz} = 0 \}}{\sum}\!\!\!\!\! w_{z}N_{kiz} \! = \!\! \sum\limits_{z=1}^{K}w_{z}N_{kiz} \end{array} $$

Hence, $\hat {\mathbf {S}}_{kij} = w_{j}N_{kij}/\nu ^{*}_{i} = w_{j}N_{kij}/({\sum }_{z=1}^{K}w_{z}N_{kiz})$. □

Appendix B: Lemma for recursive equation of MuRWR score matrix

Lemma 5

Equation (3) is represented as (4).

Proof

In (3), let l_p denote l_vu. For edge $v \rightarrow u$,

$$ l_{i}) = \sum\limits_{j=1}^{K}\mathbf{R}_{vj}P(l_{j} \xrightarrow{l_{p}} l_{i}) = {\sum}_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{pji} $$

where S_pji is the label transition probability $P(l_{j} \! \xrightarrow {l_{p}} \! l_{i})$. By Definition 6, $\tilde {\mathbf {A}}_{pvu} = |\overrightarrow {\mathbf {N}}_{v}|^{-1} = {{\tilde {\mathbf {A}}}^{\top }}_{puv}$ for all p when $\tilde {\mathbf {A}}_{puv}$ is non-zero. Hence,

$$ \underset{v \in \overleftarrow{\mathbf{N}}_{u}}{\sum} \left( \frac{1}{\lvert \overrightarrow{\mathbf{N}}_{v} \rvert} \sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{pji}\right) = \underset{v \in \overleftarrow{\mathbf{N}}_{u}}{\sum} {\tilde{\mathbf{A}}^{\top}}_{puv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{pji} $$

(8)

Let ${\overleftarrow {\mathbf {N}}}_{u}^{(i)}$ be the set of in-neighbors of node u such that node $v \in \overleftarrow {\mathbf {N}}_{u}^{(i)}$ is connected to node u with edge label l_i. Then $\overleftarrow {\mathbf {N}}b_{u}$ is represented as $\overleftarrow {\mathbf {N}}_{uh} = \overleftarrow {\mathbf {N}}_{u}^{(1)} \cup {\cdots } \cup \overleftarrow {\mathbf {N}}_{u}^{(K)}$.

If there is no l_i-labeled edge to node u from any in-neighbor node, Thus (8) is represented as follows:

$$ \begin{array}{@{}rcl@{}} &&\underset{v \in \overleftarrow{\mathbf{N}}_{u}}{\sum}{\tilde{\mathbf{A}}^{\top}}_{puv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{pji} \\ &=& \!\!\!\!\underset{v \in \overleftarrow{\mathbf{N}}_{u}^{(1)}}{\sum}{\tilde{\mathbf{A}}^{\top}}_{1uv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{1ji}+ {\cdots} + \!\!\!\underset{v \in \overleftarrow{\mathbf{N}}_{u}^{(K)}}{\sum}{\tilde{\mathbf{A}}^{\top}}_{Kuv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{Kji} \\ &=& \sum\limits_{k=1}^{K}\left( \underset{v \in \overleftarrow{\mathbf{N}}_{u}^{(k)}}{\sum}{\tilde{\mathbf{A}}^{\top}}_{kuv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{kji}\right) \end{array} $$

(9)

Let (⋅)_ij be (i,j)-th entry of a matrix. Then, ${\sum }_{v \in \overleftarrow {\mathbf {N}}_{u}^{(k)}}{\tilde {\mathbf {A}}^{\top }}_{kuv}{\sum }_{j=1}^{K}\mathbf {R}_{vj}\mathbf {S}_{kji}$ in the above equation is written as:

$$ \underset{v \in \overleftarrow{\mathbf{N}}_{u}^{(k)}}{ \sum}{\tilde{\mathbf{A}}^{\top}}_{kuv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{kji} = \underset{v \in \overleftarrow{\mathbf{N}}_{u}^{(k)}}{\sum}{\tilde{\mathbf{A}}^{\top}}_{kuv}(\mathbf{R}\mathbf{S}_{k})_{vi} = ({\tilde{\mathbf{A}}^{\top}}_{k}\mathbf{R}\mathbf{S}_{k})_{ui} $$

Then (9) is represented as follows:

$$ \sum\limits_{k=1}^{K}\left( \underset{v \in \overleftarrow{\mathbf{N}}_{u}^{(k)}}{\sum}{\tilde{\mathbf{A}}^{\top}}_{kuv}\sum\limits_{j=1}^{K}\mathbf{R}_{vj}\mathbf{S}_{kji}\right) = \sum\limits_{k=1}^{K}({\tilde{\mathbf{A}}^{\top}}_{k}\mathbf{R}\mathbf{S}_{k})_{ui} = \left( \sum\limits_{k=1}^{K}{\tilde{\mathbf{A}}^{\top}}_{k}\mathbf{R}\mathbf{S}_{k}\right)_{\!\!ui} $$

Thus R_ui in (3) is written as follows:

$$ \mathbf{R}_{ui} = (1-c) \left( \sum\limits_{k=1}^{K}{\tilde{\mathbf{A}}^{\top}}_{k}\mathbf{R}\mathbf{S}_{k}\right)_{\!\!ui} \!\!\! + c\mathbf{1}(u=s, l_{i}=l_{d}) $$

For 1 ≤ u ≤ n and 1 ≤ i ≤ K where n is the number of nodes, the above equation is represented as (4). □

Appendix C: Lemma for spectral radius in convergence theorem

Lemma 6

Suppose $\tilde {\mathbf {B}}^{\top } = {\sum }_{k=1}^{K}\mathbf {S}^{\top }_{k} \otimes {\tilde {\mathbf {A}}^{\top }}_{k}$ where $\tilde {\mathbf {A}}_{k}$ is k-th labeled semi-row-normalized matrix, and S_k is k-th label transition probability matrix. Then, $\| \tilde {\mathbf {B}}^{\top } \|_{1} \leq 1$, and the spectral radius of $\tilde {\mathbf {B}}^{\top }$ is bounded as follows: $\rho (\tilde {\mathbf {B}}^{\top }) \leq 1$.

Proof

According to spectral radius theorem [39], $\rho (\tilde {\mathbf {B}}^{\top }) \leq \| \tilde {\mathbf {B}}^{\top } \|_{1}$

where $\| \tilde {\mathbf {B}}^{\top } \|_{1}$ is the maximum absolute column sum of $\tilde {\mathbf {B}}^{\top }$. Since each entry of $\tilde {\mathbf {B}}^{\top }$ is non-negative, $\| \tilde {\mathbf {B}}^{\top } \|_{1}$ is equal to the maximum value of the column sums of the matrix. The column sums are represented as follows:

$$ \begin{array}{@{}rcl@{}} (\mathbf{1}^{\top} \otimes \mathbf{1}^{\top})\tilde{\mathbf{B}}^{\top} &=& (\mathbf{1}^{\top} \otimes \mathbf{1}^{\top})\left( \sum\limits_{k=1}^{K}\mathbf{S}^{\top}_{k} \otimes {\tilde{\mathbf{A}}^{\top}}_{k}\right) \\ & =&\sum\limits_{k=1}^{K}(\mathbf{1}^{\top} \otimes \mathbf{1}^{\top})(\mathbf{S}^{\top}_{k} \otimes {\tilde{\mathbf{A}}^{\top}}_{k}) = \sum\limits_{k=1}^{K}\mathbf{1}^{\top}\mathbf{S}^{\top}_{k} \otimes \mathbf{1}^{\top}{\tilde{\mathbf{A}}^{\top}}_{k} \end{array} $$

According to Definition 3, the sum of each row of S_k is 1, i.e., $\mathbf {S}_{k}\mathbf {1} = \mathbf {1} \Leftrightarrow \mathbf {1}^{\top }\mathbf {S}^{\top }_{k} = \mathbf {1}^{\top }$. Hence,

$$ \begin{array}{@{}rcl@{}} \sum\limits_{k=1}^{K}\mathbf{1}^{\top}\mathbf{S}^{\top}_{k} \otimes \mathbf{1}^{\top}{\tilde{\mathbf{A}}^{\top}}_{k} &=& \sum\limits_{k=1}^{K}\mathbf{1}^{\top} \otimes \mathbf{1}^{\top}{\tilde{\mathbf{A}}^{\top}}_{k} = \mathbf{1}^{\top} \otimes \sum\limits_{k=1}^{K}\mathbf{1}^{\top}{\tilde{\mathbf{A}}^{\top}}_{k} \\ &=& \mathbf{1}^{\top} \otimes \mathbf{1}^{\top}\sum\limits_{k=1}^{K}{\tilde{\mathbf{A}}^{\top}}_{k} \end{array} $$

Note that ${\tilde {\mathbf {A}}^{\top }}_{k}={\mathbf {A}}_{k}^{\top }\mathbf {D}^{-1}$ according to Definition 5. Suppose $\mathbf {A}^{\prime }={\sum }_{k=1}^{K}\mathbf {A}_{k}$ is the adjacency matrix of the graph G without edge labels. Then, $\mathbf {1}{\sum }_{k=1}^{K}{\tilde {\mathbf {A}}^{\top }}_{k} = \mathbf {1}{\sum }_{k=1}^{K}{\mathbf {A}}_{k}^{\top }\mathbf {D}^{-1} = (\mathbf {A}^{\prime }\mathbf {1})^{\top }\mathbf {D}^{-1}$. The u-th row of $\mathbf {A}^{\prime }\mathbf {1}$ indicates the out-degree of node u, denoted by deg_u. If node u is a deadend node, then deg_u = 0. Otherwise, deg_u > 0. Note that D is the out-degree diagonal matrix of G and ${\mathbf {D}}^{-1}_{uu} = 1/deg_{u}$ if node u is not a deadend. Otherwise, $\mathbf {D}^{-1}_{uu} = 0$. Thus, $(\mathbf {A}^{\prime }\mathbf {1})^{\top }\mathbf {D}^{-1} = \mathbf {b}^{\top }$ where u-th entry of b is 1 if node u is non-deadend, or 0 otherwise. Hence, $(\mathbf {1} \otimes \mathbf {1})\tilde {\mathbf {B}}^{\top } \!= \mathbf {1} \otimes \mathbf {b}^{\top }$ which is the column sum vector of $\tilde {\mathbf {B}}^{\top }$, and the maximum value of the vector is less than or equal to 1. Therefore, $\| \tilde {\mathbf {B}}^{\top } \|_{1} \leq 1$, implying $\rho (\tilde {\mathbf {B}}^{\top }) \leq \| \tilde {\mathbf {B}}^{\top } \|_{1} \leq 1$. □

Appendix D: Lemma for Complexity Analysis

Lemma 7

The time complexity of Algorithms 1 and 2 is O(m^1.5 + K³) where m is the number of edges, and K is the number of edge labels.

Proof

In Algorithm 1, it takes O(m^1.5) time to enumerate all transitive triangles in the given graph G using a triangle enumeration algorithm [17] (line 1 in Algorithm 1). Also, estimating S_k requires O(K³) time (lines 2 $\sim$ 8 in Algorithm 1). Algorithm 2 takes O(m) time for counting out-degrees of nodes (line 2 in Algorithm 2) and computing $\tilde {\mathbf {A}}_{k} = \mathbf {D}^{-1}\mathbf {A}_{k}$ for 1 ≤ k ≤ K (line 3 in Algorithm 2).

□

Lemma 8

The time complexity of Algorithm 3 is O(T(Km + K³n)) where $T=\log _{(1-c)}\frac {\epsilon }{2}$ indicates the number of iterations for convergence, 𝜖 is an error tolerance, m is the number of edges, n is the number of nodes, and K is the number of edge labels.

Proof

Let m_k denote the number of non-zeros in k-th semi-row normalized matrix $\tilde {\mathbf {A}}_{k}$ stored in a sparse matrix format such as compressed column storage (CCS). For each iteration, it takes O(Km_k + K²n) time to compute ${{\tilde {\mathbf {A}}}^{\top }}_{k}\mathbf {R}^{(t-1)}\mathbf {S}_{k}$ since the sparse matrix product ${\tilde {\mathbf {A}}^{\top }}_{k}\mathbf {R}^{(t-1)}$ requires O(Km_k) time, and the dense matrix product $({\tilde {\mathbf {A}}^{\top }}_{k}\mathbf {R}^{(t-1)})\mathbf {S}_{k}$ takes O(K²n) time. Thus, computing ${\sum }_{k=1}^{K}({\tilde {\mathbf {A}}^{\top }}_{k}\mathbf {R}^{(t-1)}\mathbf {S}_{k})$ takes $O({\sum }_{k=1}^{K}(Km_{k} + K^{2}n)) = O(Km + K^{3}n)$ where ${\sum }_{k=1}^{K}m_{k} = m$ (line 6). Note that when 2(1 − c)^t ≤ 𝜖, R^(t) converges since δ^(t) ≤ 2(1 − c)^t by Theorem 1. Hence, for $t \geq \log _{(1-c)}\frac {\epsilon }{2}$, the iteration is necessarily terminated. Thus, the number of iterations for convergence is estimated at $\log _{(1-c)}\frac {\epsilon }{2}$, and the total time complexity is $O((\log _{(1-c)}\frac {\epsilon }{2})(Km + K^{3}n))$. □