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Approximate Identification of the Optimal Epidemic Source in Complex Networks

Part of the Springer Proceedings in Complexity book series (SPCOM)

Abstract

We consider the problem of identifying the source of a network epidemic from a complete snapshot of the infected nodes. We take a fully statistical approach and derive novel recursions to compute the Bayes optimal solution, under a heterogeneous susceptible-infected (SI) epidemic model. Our analysis is time and rate independent, and holds for general network topologies. We then provide two highly scalable algorithms for solving these recursions, a mean-field approximation and a greedy approach, and evaluate their performance on real and synthetic networks. Previous work on the problem has mostly focused on tree-like network topologies. Real networks are far from tree-like and an emphasis will be given to networks with high transitivity, such as social networks and those with communities. We show that on such networks, our approaches significantly outperform popular geometric and spectral centrality measures, most of which perform no better than random guessing.

Keywords

  • Network epidemics
  • Source recovery
  • Approximate inference
  • Dynamic programming

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Acknowledgement

We would like to thank Mason A. Porter for providing the Facebook-100 dataset.

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Correspondence to S. Jalil Kazemitabar .

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Appendices

Appendix 1: Multi-Source Extension

The inference problem discussed in Sect. 2.2 immediately extends to the multi-source situations. Consider the case were more than one independent source, denoted by I , initiate the infection dynamics. Due to the Markovian nature of the dynamics, the infection path that leads to some set I does not influence the value of ρ IO. Hence, Proposition 1 also describes the likelihood of the transition from the source set I to a snapshot O.

If we know that there are s original sources, e.g. |I | = s, with a uniform prior on the patient zeros, the Bayesian solution would be characterized by the optimization

$$\displaystyle \begin{aligned} I^*_{\text{MAP}} = \operatorname*{\mathrm{argmax}}_{I\subset O,\, |I|=s}\, \rho_{I \to O} \end{aligned} $$
(8)

To compute this MAP estimate, we can still use the DP solution in Proposition 1, but we do not need to compute ρ IO for |I| < s. Thus, the multi-source problem is in a sense “easier”, especially when s ≈|O|, since one can terminate the recursion earlier (i.e., the case s = 1 is the hardest).

Appendix 2: Proofs

1.1 6 Proof of Proposition 1

Let us first recall a known fact about the exponential distribution:

Lemma 1

Let \(T_i \sim \operatorname {\mathrm {Exp}}(\beta _i)\) be a collection of independent exponential variables. Then,

$$\displaystyle \begin{aligned} \mathbb{P}\Big(T_i < \min_{j \neq i} T_j\Big) = \frac{\beta_i}{ \sum_{j} \beta_j}. \end{aligned} $$

For a proof of Lemma 1, see [45]. The forward programming (2) is an application of the law of total probability in the following sense: The event that nodes in O ∖ I are infected before any other node in I c splits into sub-events that each node in O ∖ I is infected before those in O c and we have

$$\displaystyle \begin{aligned} \rho_{I \to O} = \sum_{j \in O \setminus I} \rho_{I \to I \cup j} \cdot \rho_{I \cup j \to O} \end{aligned} $$

where we have also used the Markov property of SI dynamics to split the probabilities on the RHS into the products. The ratio in (2) corresponds to the transition probability from I to I ∪ j, that is ρ IIj. Indeed, given that I is infected, we run exponential clocks \(T_j \sim \operatorname {\mathrm {Exp}}( \beta \operatorname {\mathrm {vol}}(I,j))\) and the first to expire determines the next infected node. By Fact 1, this happens for any node j ∈ I c with probability \(\propto _j \beta \operatorname {\mathrm {vol}}(I,j)\). Thus,

$$\displaystyle \begin{aligned} \rho_{I \to I \cup j} = \frac{\beta \operatorname{\mathrm{vol}}(I,j) }{\sum_{j^{\prime}}\beta \operatorname{\mathrm{vol}}(I,j^{\prime})} = \frac{\operatorname{\mathrm{vol}}(I, j)}{\operatorname{\mathrm{vol}}(I, I^c)}. \end{aligned} $$

This proves the forward programming. The backward programming, on the other hand, connects ρ IO to ρ IOj and is proved similarly. Basically, the event of visiting O can be divided into sub-events based on the last node in O that is infected.

1.2 6 Proof of Proposition 2

We prove the following alternative expressions for \(S = (S_{jj^{\prime }})^{|O| \times |O|}\) and z = (z j)|O|,

$$\displaystyle \begin{aligned} S_{jj^{\prime}} &:= \begin{cases} d^{in}_{O \setminus j^{\prime}}(j) d^{in}_{O \setminus j}(j^{\prime}) + \sum_{i\in O} A_{ij} A_{ij^{\prime}} & j \neq j^{\prime} \\ 2 \big[d^{in}_{O}(j)^2 + \sum_{i\in O} A^2_{ij}\big] & j = j^{\prime} \end{cases} \\ {} z_j &:= \Big[ \operatorname{\mathrm{vol}}(O_{\setminus j}) + 2 \operatorname{\mathrm{vol}}(O{ \setminus\, j}, (O{\setminus \,j})^c)\Big]\, d^{in}_{O}(j) \\ & \qquad + \sum_{i\in O} (d^{out}_{O_{\setminus j}}(i)- d^{in}_{O_{\setminus j}}(i)) A_{ij} + 2 \sum_{i\in O} d^{out}_{(O\setminus \,j)^c}(i)\, A_{ij}. \end{aligned} $$

Here, \(d^{out}_O(i) := \sum _{j \in O} A_{ij}\) is the out-degree of node i in O, \(d^{in}_O(i) := \sum _{j \in O} A_{ji}\) is the in-degree of node i in O, and \( \operatorname {\mathrm {vol}}^{(2)}(i,j) := \sum _{r \in O} A_{ir} A_{rj}\) is the number of paths of length 2 between nodes i and j that pass through O. It is not hard to verify that these expressions are equivalent to the matrix form presented in (2).

Recall that \( \operatorname {\mathrm {vol}}(I, I^c) = \sum _{i,k} A_{ik} 1\{i \in I, k \notin I \}\) and similarity \( \operatorname {\mathrm {vol}}(I, j) = \sum _{r} A_{rj} 1\{r \in I \}\). Here, the indices, i, k and r run over all nodes in the network, i.e. i, k, r ∈ [n]. We have

$$\displaystyle \begin{aligned} (Q^T \boldsymbol r)_j &= \sum_{I\subset O} 1\{j\not\in I\}\, \operatorname{\mathrm{vol}}(I, j)\cdot \operatorname{\mathrm{vol}}(I, I^c) \\ &= \sum_{I \subset O \setminus \{j\}}\, \operatorname{\mathrm{vol}}(I, I^c) \cdot\operatorname{\mathrm{vol}}(I, j) \\ &= \sum_{I \subset O \setminus \{j\}}\, \sum_{i,k,r} A_{ik} A_{rj} \, 1\{i \in I, \,k \notin I, \,r \in I\} \\ &= \sum_{i,k,r} A_{ik} A_{rj} \gamma_{ikr} \end{aligned} $$

where the last equality follows by interchanging the order of summations and defining

$$\displaystyle \begin{aligned} \gamma_{ikr} := \sum_{I \subset O \setminus \{j\}}1\{i \in I, \,k \notin I, \,r \in I\} \end{aligned} $$

If i or r do not belong to O ∖{j}, or k ∈{i, r}, then γ ikr = 0. Thus, it what follows assume that i, r ∈ O j := O ∖{j} and k∉{i, r}. Then,

$$\displaystyle \begin{aligned} \gamma_{ikr} = 0 \begin{cases} 2^{|O|-4} & i \neq r,\; k \in O_{\setminus\, j}\\ 2^{|O|-3} & i=r, \; k \in O_{\setminus\, j}\\ 2^{|O|-3} & i \neq r,\; k \notin O_{\setminus\, j}\\ 2^{|O|-2} & i=r, \; k \notin O_{\setminus\, j}\\ \end{cases} \end{aligned} $$

To see the second equality, note that we are counting subsets of the set O ∖{j} (of cardinality |O|− 1) that contain or exclude certain elements. For example, when k, i, r are pairwise distinct, and k ∈ O ∖{j}, looking at the binary representation of I, we have two ones in the positions i and r and a zero in position k, and the rest of |O|− 1 − 3 positions are free to be zero or one.

In what follows, i and r range over O ∖{j} (otherwise γ ikr = 0). Also, condition k∉{i, r} can be replaced with k ≠ r, since the k ≠ i is implicitly enforced by A ik = 0 if k = i (no self-loops). We have

$$\displaystyle \begin{aligned} (Q^T \boldsymbol r)_j &= \sum_{i,r} \sum_{k \neq r} A_{ik} A_{rj} \big[ 2^{|O|-4}(1+1\{i=r\})1\{k \in O_{\setminus\, j}\} \\ &\quad + 2^{|O|-3}(1+1\{i=r\})1\{k \notin O_{\setminus\, j}\}\big] \\ &= 2^{|O|-4} \sum_{i,r} d^{out}_{O\setminus\{j,r\}}(i) A_{rj} (1+1\{i=r\}) \\ &\quad + 2^{|O|-3}\sum_{i,r} d^{out}_{(O\setminus \,j)^c}(i) A_{rj} (1+1\{i=r\}) \end{aligned} $$

where in the second term, we used the fact that if kO j then we automatically have k ≠ r since r ranges over O j. We have

$$\displaystyle \begin{aligned} \sum_{r} d^{out}_{O\setminus\{j,r\}}(i) A_{rj} &= \sum_{r} (d^{out}_{O_{\setminus j}}(i) - A_{ir}) A_{rj} \\ &= d^{out}_{O_{\setminus j}}(i) d^{in}_{O_{\setminus j}}(j) - \operatorname{\mathrm{vol}}_{O_{\setminus j}}^{(2)}(i,j) \end{aligned} $$

where \( \operatorname {\mathrm {vol}}_{O_{\setminus j}}^{(2)}(i,j) := \sum _{r \in O_{\setminus j}} A_{ir} A_{rj}\) is the number of paths of length two between i and j in O j. Note that \( \operatorname {\mathrm {vol}}_{O_{\setminus j}}^{(2)}(i,j) = \operatorname {\mathrm {vol}}_{O}^{(2)}(i,j)\) and similarly \(d_{O_{\setminus j}}(j) = d_{O}(j)\) since A jj = 0. Thus,

$$\displaystyle \begin{aligned} \sum_{i,r} d^{out}_{O\setminus\{j,r\}}(i)\, A_{rj} \big( 1+1\{i=r\} \big) &= \sum_i \Big[ d^{out}_{O_{\setminus j}}(i) d^{in}_{O}(j) - \operatorname{\mathrm{vol}}_{O}^{(2)}(i,j) + d^{out}_{O_{\setminus j}}(i) A_{ij} \Big] \\ &= \sum_i d^{out}_{O_{\setminus j}}(i) d^{in}_{O}(j) + (d^{out}_{O_{\setminus j}}(i)- d^{in}_{O_{\setminus j}}(i)) A_{ij} \\ &= \operatorname{\mathrm{vol}}(O_{\setminus j}) d^{in}_{O}(j) + \sum_i (d^{out}_{O_{\setminus j}}(i)- d^{in}_{O_{\setminus j}}(i)) A_{ij} \end{aligned} $$

where \( \operatorname {\mathrm {vol}}(O_{\setminus j}) = \operatorname {\mathrm {vol}}(O_{\setminus j},O_{\setminus j})\) and the third equality follows since we have

$$\displaystyle \begin{aligned} \sum_{i\in A} \operatorname{\mathrm{vol}}_{A}^{(2)}(i,j) = \sum_{i \in A} \sum_{r \in A} A_{ir} A_{rj} = \sum_{r \in A} d^{in}_A(r) A_{rj} \end{aligned} $$

which was used with A = O j. Similarly, we have

$$\displaystyle \begin{aligned} \sum_{i,r} d^{out}_{(O\setminus \,j)^c}(i) A_{rj} (1+1\{i=r\}) &= \sum_{i} d^{out}_{(O\setminus \,j)^c}(i) \big (d^{in}_{{O \setminus j}}(j)+A_{ij} \big) \\ &=\operatorname{\mathrm{vol}}(O{ \setminus\, j}, (O{\setminus \,j})^c) \,d^{in}_O(j) \\ &\quad + \sum_{i} d^{out}_{(O\setminus \,j)^c}(i)\, A_{ij} \end{aligned} $$

It follows that

$$\displaystyle \begin{aligned} (Q^T \boldsymbol r)_j =2^{|O|-4} &\Big[ \operatorname{\mathrm{vol}}(O_{\setminus j}) d^{in}_{O}(j) + \sum_i (d^{out}_{O_{\setminus j}}(i)- d^{in}_{O_{\setminus j}}(i)) A_{ij} \\ &+ 2 \operatorname{\mathrm{vol}}(O{ \setminus\, j}, (O{\setminus \,j})^c) \,d^{in}_O(j) + 2 \sum_{i} d^{out}_{(O\setminus \,j)^c}(i)\, A_{ij}\Big]. \end{aligned} $$

Calculating Q TQ

Let us first take j ≠ j . Then, similar to the previous argument,

$$\displaystyle \begin{aligned} (Q^T Q)_{jj^{\prime}} &= \sum_{I\subset O\setminus \{j,j^{\prime}\}} \operatorname{\mathrm{vol}}(I, j)\, \operatorname{\mathrm{vol}}(I, j^{\prime}) \\ &=\sum_{I\subset O\setminus \{j,j^{\prime}\}} \sum_{i,r} A_{ij}\, A_{rj^{\prime}} 1\{i \in I,\;r \in I\} \\ &= \sum_{i,r} A_{ij}\, A_{rj^{\prime}} \beta_{ir} \end{aligned} $$

where we have defined

$$\displaystyle \begin{aligned} \beta_{ir} &:= \sum_{I\subset O\setminus \{j,j^{\prime}\}} 1\{i \in I,\;r \in I\} \\ &= 2^{|O|-4} 1\{i\neq r\} + 2^{|O|-3} 1\{i=r\}\\ &= 2^{|O|-4} \big(1+ 1\{i=r\}\big) \end{aligned} $$

assuming i, r ∈ O ∖{j, j }, otherwise β ir = 0. Thus, restricting summations over indices i, r ∈ O ∖{j, j }

$$\displaystyle \begin{aligned} (Q^T Q)_{jj^{\prime}} &= 2^{|O|-4}\Big[ \sum_{i,r} A_{ij}\, A_{rj^{\prime}} + \sum_{i} A_{ij} A_{ij^{\prime}}\Big] \\ &= 2^{|O|-4}\Big[ d^{in}_{O \setminus j^{\prime}}(j) d^{in}_{O \setminus j}(j^{\prime}) + \sum_{i} A_{ij} A_{ij^{\prime}}\Big]. \\ \end{aligned} $$

Now consider the case j = j . Then,

$$\displaystyle \begin{aligned} (Q^T Q)_{jj} &= \sum_{I\subset O\setminus \{j\}} \operatorname{\mathrm{vol}}(I, j)^2 \\ &=\sum_{I\subset O\setminus \{j\}} \sum_{i,r} A_{ij}\, A_{rj} 1\{i \in I,\;r \in I\} \\ &= \sum_{i,r} A_{ij}\, A_{rj} \,2^{|O|-3} \big( 1 + 1\{i=r\} \big), \end{aligned} $$

assuming i, r ∈ O ∖ j. It follows that

$$\displaystyle \begin{aligned} (Q^T Q)_{jj} &= 2^{|O|-3} \Big[ \sum_{i,r} A_{ij}\, A_{rj} + \sum_{i} A^2_{ij} \Big] \\ &= 2^{|O|-3} \big[ d^{in}_{O}(j)^2 + \sum_{i} A^2_{ij} \big]. \end{aligned} $$

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Kazemitabar, S.J., Amini, A.A. (2020). Approximate Identification of the Optimal Epidemic Source in Complex Networks. In: Masuda, N., Goh, KI., Jia, T., Yamanoi, J., Sayama, H. (eds) Proceedings of NetSci-X 2020: Sixth International Winter School and Conference on Network Science. NetSci-X 2020. Springer Proceedings in Complexity. Springer, Cham. https://doi.org/10.1007/978-3-030-38965-9_8

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