Efficiently spotting the starting points of an epidemic in a large graph

Abstract

Given a snapshot of a large graph, in which an infection has been spreading for some time, can we identify those nodes from which the infection started to spread? In other words, can we reliably tell who the culprits are? In this paper, we answer this question affirmatively and give an efficient method called NetSleuth for the well-known susceptible-infected virus propagation model. Essentially, we are after that set of seed nodes that best explain the given snapshot. We propose to employ the minimum description length principle to identify the best set of seed nodes and virus propagation ripple, as the one by which we can most succinctly describe the infected graph. We give an highly efficient algorithm to identify likely sets of seed nodes given a snapshot. Then, given these seed nodes, we show we can optimize the virus propagation ripple in a principled way by maximizing likelihood. With all three combined, NetSleuth can automatically identify the correct number of seed nodes, as well as which nodes are the culprits. Experimentation on our method shows high accuracy in the detection of seed nodes, in addition to the correct automatic identification of their number. Moreover, NetSleuth scales linearly in the number of nodes of the graph.

Appendix 1

1.1 Proofs

We formally prove various Lemmas used in the paper in this section.

Proof of Lemma 1

The graph $G_I$ is connected (as we assume the set of infected nodes are connected, otherwise we are just dealing with separate problems, one for each connected component). Now, the Laplacian matrix $L(G)$ has all entries except the diagonal elements as nonpositive and all the diagonal elements as positive. In addition, $L(G)$ is a symmetric matrix (as $A(G)$ is symmetric). The matrix $L_A$ is a principal submatrix (i.e., it has been formed by removing matching rows and columns) of size $N_I \times N_I$ of $L(G)$. As a result, $L_A$ is symmetric.

Consider matrix $M = (I - \frac{L_A}{\sigma })$. Clearly, it is symmetric (due to the above). Consider some diagonal element $M_{ii}$ (for some index i):

$$\begin{aligned} M_{ii}&= 1 - \frac{d_i}{1+\sigma } \nonumber \\&= 1 - \frac{d_i}{1+d_{\mathrm{max}}} \end{aligned}$$

(14)

$$\begin{aligned}&> 0 \end{aligned}$$

(15)

Any off-diagonal element $M_{ij}$ is

$$\begin{aligned} M_{ij} = {\left\{ \begin{array}{ll} 0, &{} \mathrm{if }\,\{i, j\} \notin E_I\\ \frac{1}{1+d_{\mathrm{max}}}, &{} \mathrm{if }\,\{i, j\} \in E_I \end{array}\right. } \end{aligned}$$

Hence, first, $M$ is a nonnegative matrix. Further, the structure of the matrix $M = (I - \frac{L_A}{\sigma })$ represents the adjacency matrix of a weighted connected graph $G_M$ (with self-loops). This is because its off-diagonal elements are nonzero only when the corresponding edge is present in $G_I$. Hence, as $G_I$ is connected, so is $G_M$. Now, as $G_M$ is connected, we have that $M$ is irreducible.

Finally, applying the well-known Perron–Frobenius theorem [23] on the nonnegative irreducible matrix $M = (I - \frac{L_A}{\sigma })$, we get that the first (largest) eigenvalue $\lambda _1$ and the corresponding eigenvector $\vec {u}_1$ are all positive and real. $\square$

Proof of Lemma 2

First, note that both matrices $M= (I - \frac{L_A}{\sigma })$ and $L_A$ are symmetric (see proof of Lemma 1 above). Hence, it follows that all their eigenvalues are real [34].

Now, consider the Laplacian matrix $L(G)$ of graph $G$. It is well known that its smallest eigenvalue is 0 [6]. Hence, all its eigenvalues are nonnegative. Let its eigenvalues be

$$\begin{aligned} 0 \le \mu _2 \le \mu _3 \le \cdots \le \mu _N \end{aligned}$$

Consider any cofactor matrix $C_{LG}$ of $L(G)$. Recall that a cofactor matrix is a principal submatrix resulting after the removal of one matching row and column. Clearly, $C_{LG}$ is also symmetric and has all real eigenvalues. Let them be

$$\begin{aligned} \nu _1 \le \nu _2 \le \nu _3 \ldots \le \nu _{N-1} \end{aligned}$$

We can now apply the Cauchy eigenvalue interlacing theorem [34] to $L(G)$ and $C_{LG}$. Applying it, we get that

$$\begin{aligned} 0 \le \nu _1 \le \mu _2 \le \nu _2 \cdots \le \nu _{N-1} \le \mu _N \end{aligned}$$

Hence, all the eigenvalues of $C_{LG}$ are nonnegative.

Now, recall that according to the famous Kirchoff matrix tree theorem [6], the determinant of any cofactor of the Laplacian matrix $L(G)$ of graph $G$ is equal to the number of spanning trees of $G$. As the number can not be zero, the determinant of $C_{LG}$ is also nonzero, i.e.,

$$\begin{aligned} \mathrm{det}(C_{LG}) > 0 \end{aligned}$$

Further, it is well known that the determinant of any matrix is equal to the product of its eigenvalues [34]. So,

$$\begin{aligned} \mathrm{det}(C_{LG}) = \prod _{i=1}^{N-1} \nu _i > 0 \Rightarrow \, \forall _i~~\nu _i > 0, \end{aligned}$$

(16)

i.e., none of the eigenvalues of $C_{LG}$ are zero.

Recall that $L_A$ is a principal submatrix of $L(G)$—hence, it is a cofactor of some other larger principal submatrix, which is a cofactor of some other still larger principal submatrix and so on till we reach $C_{LG}$. Hence, there is a sequence of submatrices which can lead us from $C_{LG}$ to $L_A$, in which each submatrix is a cofactor of the one before it. Hence, applying Eq. 16 above successively to this sequence, we get that all the eigenvalues of $L_A$ are strictly positive (nonzero).

Finally, note that any eigenvalue of $(I - \frac{L_A}{\sigma }), \mathrm{eig}((I - \frac{L_A}{\sigma })) = 1 - eig(\frac{L_A}{\sigma })$. Hence, as all the eigenvalues of $L_A$ are nonzero, we get

$$\begin{aligned} \lambda _1 \left( 1 - \frac{L_A}{\sigma }\right) = 1 - \frac{\lambda _{N_I}(L_A)}{\sigma }, \end{aligned}$$

where $\lambda _{N_I}(L_A)$ is the smallest eigenvalue of $L_A$. $\square$

Proof of Lemma 3

We keep finding $\mathcal S _k$ for each seed set size until MDL tells us to stop. Hence, the running time is $O(k^* (\mathcal E _{I} + T_{\mathrm{RIPPLE}} + T_{\mathrm{MDL}}))$ if $k^*$ is the optimal seed set size and $T_{\mathrm{MDL}}$ is the running time of computing the MDL score given the seed set size is $k^*$. Here, we used the fact that calculating the eigenvector using the Lanczos method is approximately $O(E)$ (# edges) for sparse graphs.

The worst-case complexity $T_{\mathrm{MDL}}$ of calculating $\mathcal{L }(G_I, \mathcal{S }, R)$ for a given $G_I, \mathcal{S }$, and $R$ is $O(\mathcal E _I + \mathcal E _{F} + \mathcal V _I)$. The $\mathcal{L }(\mathcal{S })$ term is $O(1)$. For the $\mathcal{L }(R\mid \mathcal{S })$ term, we need to iterate over the ripple, which is at most $\mathcal V _{I}$ steps long. We only have to update the frontier set $\mathcal F$ when one or more nodes got infected, for which we then have to update the attack degrees of the nodes connected to the nodes infected at time $t$. Hence, we traverse every edge in $\mathcal E _{I}+\mathcal E _F$ and every node in $\mathcal V _{I}$, which gives it the complexity of $O(\mathcal E _I + \mathcal E _{F} + \mathcal V _{I})$.

Finally, the running time $T_{\mathrm{RIPPLE}}$ of computation of the MLE ripple for a given $\mathcal S _k$ is also $O(\mathcal E _I + \mathcal E _{F} + \mathcal V _{I})$.

So, the overall complexity of NetSleuth is $O(k^* (\mathcal E _I + \mathcal E _F + \mathcal V _I))$. $\square$