SNOD: a fast sampling method of exploring node orbit degrees for large graphs

Abstract

Exploring small connected and induced subgraph patterns (CIS patterns, or graphlets) has recently attracted considerable attention. Despite recent efforts on computing how frequent a graphlet appears in a large graph (i.e., the total number of CISes isomorphic to the graphlet), little effort has been made to characterize a node’s graphlet orbit degree, i.e., the number of CISes isomorphic to the graphlet that touch the node at a particular orbit, which is an important fine-grained metric for analyzing complex networks such as learning functions/roles of nodes in social and biological networks. Like global graphlet counting, it is computationally intensive to compute node orbit degrees for a large graph. Furthermore, previous methods of computing global graphlet counts are not suited to solve this problem. In this paper, we propose a novel sampling method SNOD to efficiently estimate node orbit degrees for large-scale graphs and quantify the error of our estimates. To the best of our knowledge, we are the first to study this problem and give a fast scalable solution. We conduct experiments on a variety of real-world datasets and demonstrate that our method SNOD is several orders of magnitude faster than state-of-the-art enumeration methods for accurately estimating node orbit degrees for graphs with millions of edges.

Appendix

1.1 Implementation details

We discuss our methods for implementing the functions in Algorithms 1,  2, and 3. We also analyze their computational complexities.

Initialization of$\phi _v$, $\varphi _v$, $\check{\varphi }_v$, $\tilde{\varphi }_v$, $\psi _v$, and$\gamma _v$: For each node v, we store its degree $d_v$ and its neighbors’ degrees in a list. Therefore, O(1) and $O(d_v)$ operations are required to compute $\phi _v$ and $\varphi _v$, respectively. Similarly, one can easily find that $O(N_v)$, $O(N_v)$, $O(\sum _{u\in N_v} d_u)$, and O(1) operations are required to compute $\check{\varphi }_v$, $\tilde{\varphi }_v$, $\psi _v$, and $\gamma _v$, respectively.

$\mathbf RandomVertex (N_v- \{u\})$: We use an array $N_v[1, \ldots , d_v]$ to store the neighbors of v. Let $i_{v,u}$ denote the index of u in the list $N_v[1, \ldots , d_v]$, i.e., $N_v[i_{v,u}]=u$. Then, function $\text {RandomVertex}(N_v- \{u\})$ includes the following steps:

• Step 1. Select a number z from $\{1, \ldots , d_v\}-\{i_{v,u}\}$ at random;

• Step 2. Return $N_v[z]$.

Therefore, the computational complexity of $\text {RandomVertex}(N_v- \{u\})$ is O(1).

$\mathbf RandomVertex (N_v- \{u, w\})$: Similarly, it includes the following steps:

• Step 1. Select a number z from $\{1, \ldots , d_v\}- \{i_{v,u}, i_{v,w}\}$ at random;

• Step 2. Return $N_v[z]$.

Therefore, the computational complexity of $\text {RandomVertex}(N_v- \{u, w\})$ is O(1).

$\mathbf WeightRandomVertex (N_v, \varvec{\alpha }^{(v)})$: We store an array $A_\alpha ^{(v)}$ in memory, where $A_\alpha ^{(v)}[i]$ is defined as $A_\alpha ^{(v)}[i] = \sum _{j=1}^i (d_{N_v[j]} - 1)$, $1\le i\le d_v$. Let $A_\alpha ^{(v)}[0]=0$. Then, $\text {WeightRandomVertex}(N_v, \varvec{\alpha }^{(v)})$ includes the following steps:

• Step 1. Select a number z from $\{1, \ldots , A_\alpha ^{(v)}[d_v]\}$ at random;

• Step 2. Find i such that

  $$\begin{aligned} A_\alpha ^{(v)}[i-1] < z \le A_\alpha ^{(v)}[i], \end{aligned}$$

  which is solved by binary search;

• Step 3. Return $N_v[i]$.

Therefore, the computational complexity of function $\text {WeightRandomVertex}(N_v, \varvec{\alpha }^{(v)})$ is $O(\log d_v)$.

$\mathbf WeightRandomVertex (N_v, \varvec{\beta }^{(v)})$: We store an array $A_\beta ^{(v)}$ in memory, where $A_\beta ^{(v)}[i]$ is defined as

$$\begin{aligned} A_\beta ^{(v)}[i] = \sum _{j=1}^i (\phi _{N_v[j]} - d_{N_v[j]} + 1), \quad 1\le i\le d_v. \end{aligned}$$

Let $A_\beta ^{(v)}[0]=0$. Then, $\text {WeightRandomVertex}(N_v, \varvec{\beta }^{(v)})$ includes the following steps:

• Step 1. Select a number z from $\{1, \ldots , A_\beta ^{(v)}[d_v]\}$ at random;

• Step 2. Find i such that

  $$\begin{aligned} A_\beta ^{(v)}[i-1] < z \le A_\beta ^{(v)}[i], \end{aligned}$$

  which again is solved by binary search;

• Step 3. Return $N_v[i]$.

Therefore, the computational complexity of function $\text {WeightRandomVertex}(N_v, \varvec{\beta }^{(v)})$ is $O(\log d_v)$.

1.2 Proof of Theorem 7

According to Theorems 1 and 4, we have

$$\begin{aligned} \text {Var}(\hat{d}^{(1)}_v) = \frac{d^{(1)}_v}{k}\left( \frac{1}{\pi _{1, v}} - d^{(1)}_v\right) . \end{aligned}$$

According to Theorems 1 and 5, we have

$$\begin{aligned} \text {Var}(\hat{d}^{(i)}_v) = \frac{d^{(i)}_v}{\check{k}}\left( \frac{1}{\check{\pi }_{i, v}} - d^{(i)}_v\right) ,\quad i\in \{5, 8, 11\}. \end{aligned}$$

According to Theorems 1 and 6, we have

$$\begin{aligned} \text {Var}(\hat{d}^{(i)}_v) = \frac{d^{(i)}_v}{\tilde{k}}\left( \frac{1}{\tilde{\pi }_{i, v}} - d^{(i)}_v\right) ,\quad i\in \{6, 9\}. \end{aligned}$$

By Theorem 2 and the definition of $\hat{d}^{(3)}_v$, $\hat{d}^{(10)}_v$, $\hat{d}^{(12)}_v$, $\hat{d}^{(13)}_v$, and $\hat{d}^{(14)}_v$ in Eqs. (7) and (10), we have

$$\begin{aligned} \begin{aligned} \text {Var}(\hat{d}^{(i)}_v)&= \text {Var}\left( \frac{\text {Var}(\tilde{d}^{(i)}_v) \check{d}^{(i)}_v + \text {Var}(\check{d}^{(i)}_v) \tilde{d}^{(i)}_v}{\text {Var}(\check{d}^{(i)}_v) + \text {Var}(\tilde{d}^{(i)}_v)}\right) \\&= \frac{\text {Var}(\tilde{d}^{(i)}_v) \text {Var}(\check{d}^{(i)}_v)}{\text {Var}(\check{d}^{(i)}_v) + \text {Var}(\tilde{d}^{(i)}_v)}, \qquad i\in \{3, 10, 12, 13, 14\}. \end{aligned} \end{aligned}$$

In the above derivation, the last equation holds because $\check{d}^{(i)}_v$ and $\tilde{d}^{(i)}_v$ are independent, which can be easily obtained from their definition in Eqs. (5), (6), (8), and (9). For $\text {Var}(\hat{d}^{(2)}_v)$, we have

$$\begin{aligned} \text {Var}(\hat{d}^{(2)}_v) = \text {Var}(\phi _v - \hat{d}^{(3)}_v) = \text {Var}(\hat{d}^{(3)}_v). \end{aligned}$$

By the definition of $\hat{d}^{(4)}_v$ and $\hat{d}^{(7)}_v$, we easily prove that their variances are

$$\begin{aligned} \begin{aligned} \text {Var}(\hat{d}^{(4)}_v) =&\sum _{j\in \{3, 8, 9, 10, 12, 13, 14\}} \chi _j^2 \text {Var}(\hat{d}^{(j)}_v)\\&+\sum _{j, l\in \{3, 8, 9, 10, 12, 13, 14\}\wedge j\ne l} \chi _j \chi _l \text {Cov}(\hat{d}^{(j)}_v, \hat{d}^{(l)}_v),\\ \text {Var}(\hat{d}^{(7)}_v) =&~ \text {Var}(\hat{d}^{(11)}_v) + \text {Var}(\hat{d}^{(13)}_v) + \text {Var}(\hat{d}^{(14)}_v)\\&+ \sum _{j, l\in \{11,13,14\}\wedge j\ne l} \text {Cov}(\hat{d}^{(j)}_v, \hat{d}^{(l)}_v) + \sum _{j, l\in \{11,13,14\}\wedge j\ne l} \text {Cov}(\hat{d}^{(j)}_v, \hat{d}^{(l)}_v). \end{aligned} \end{aligned}$$

The covariances in the above formulas of $\text {Var}(\hat{d}^{(4)}_v)$ and $\text {Var}(\hat{d}^{(7)}_v)$ are computed based on the following properties: (1) The covariance of estimates given by different sampling methods (e.g., Path2, Path3, and Star3) is zero; (2) the covariance of estimates given by the same sampling method is computed based on Theorem 1. For example, when $j, l\in \{10, 12, 13, 14\}$ and $j\ne l$, by the definition of $\hat{d}^{(j)}_v$ in Eq. (10), we have $\text {Cov}(\hat{d}^{(j)}_v, \hat{d}^{(l)}_v)= \text {Cov}(\lambda ^{(j,1)}_v \check{d}^{(j)}_v + \lambda ^{(j,2)}_v \tilde{d}^{(j)}_v, \lambda ^{(l,1)}_v \check{d}^{(l)}_v + \lambda ^{(l,2)}_v \tilde{d}^{(l)}_v)$. By the definition of $\check{d}^{(j)}_v$ and $\tilde{d}^{(j)}_v$ in Eqs. (8) and (9), we easily find that $\check{d}^{(j)}_v$ and $\check{d}^{(l)}_v$ given by sampling method Path3 are independent with $\tilde{d}^{(j)}_v$ and $\tilde{d}^{(l)}_v$ given by sampling method Star3. Moreover, we have $\text {Cov}(\check{d}^{(j)}_v,\check{d}^{(l)}_v) = -\frac{d^{(j)}_v d^{(l)}_v}{\check{k}}$ and $\text {Cov}(\tilde{d}^{(j)}_v,\tilde{d}^{(l)}_v) = -\frac{d^{(j)}_v d^{(l)}_v}{\tilde{k}}$ from Theorem 1. Thus, we have $\text {Cov}(\hat{d}^{(j)}_v, \hat{d}^{(l)}_v) = -\frac{\lambda ^{(j,1)}_v \lambda ^{(l,1)}_v d^{(j)}_v d^{(l)}_v}{\check{k}}-\frac{\lambda ^{(j,2)}_v \lambda ^{(l,2)}_v d^{(j)}_v d^{(l)}_v}{\tilde{k}}$.