An Exact Algorithm for Robust Influence Maximization

Abstract

We propose a Branch-and-Cut algorithm for the robust influence maximization problem. The influence maximization problem aims to identify, in a social network, a set of given cardinality comprising actors that are able to influence the maximum number of other actors. We assume that the social network is given in the form of a graph with node thresholds to indicate the resistance of an actor to influence, and arc weights to represent the strength of the influence between two actors. In the robust version of the problem that we study, the node thresholds are affected by uncertainty and we optimize over a worst-case scenario within a given robustness budget. Numerical experiments show that we are able to solve to optimality instances of size comparable to other exact approaches in the literature for the non-robust problem, but in addition to this we can also tackle the robust version with similar performance.

Proofs

Proof

Proposition 1. It suffices to show that for a given choice of the adversarial threshold modification $\theta _j$, problem $\text {RI}_{x,\theta }(\bar{y})$ computes the total influence spread as if we had applied InfluenceSpread$(\bar{y}, t + \theta )$. Notice that $\text {RI}_{x,\theta }(\bar{y})$ is a minimization problem and each $x_j$ is lower bounded by two quantities only: $\bar{y}_j$, and $\frac{\sum _{i \in \delta ^-(j)} w_{ij} x_{i} - t_j - \theta _j + \epsilon }{\sum _{i \in \delta ^-(j)} w_{ij}}$. The latter quantity is $> 0$ if and only if $\sum _{i \in \delta ^-(j)} w_{ij} x_{i} > t_j + \theta _j$, implying that $x_j = 1$ if and only if its activation rule is triggered by its neighbors. If we apply InfluenceSpread$(\bar{y}, t + \theta )$, it is easy to see by induction over the main loop that for each $x_j \in A_k$ there is an implied lower bound $x_j \ge 1$, and for all nodes $\not \in A_n$ there is no such implied lower bound. It follows that in the optimal solution $x_j = 1$ if and only if $j \in A_n$.

Proof

Proposition 2. Every $x_j$ is lower bounded by $\bar{y}_j$ and by $\sum _{i \in S} x_i - |S| + 1$ for some subset of nodes S adjacent to node j.

We first show by induction for $k=1,\dots ,n$ that for every node $j \in A_k$ in InfluenceSpread$(\bar{y}, t)$, we have an implied lower bound $x_j \ge 1$ in $\text {AS}_x(\bar{y})$.

For $k = 1$ the claim is obvious because of the constraints $x_j \ge \bar{y}_j$. To go from $k-1$ to k, notice that if node j is added to $A_k$ at step k of InfluenceSpread$(\bar{y}, t)$, it must be that $\sum _{i \in \delta ^-(j) : i \in A_{k-1}} w_{ij} \ge t_j$. By the induction hypothesis for all $i \in A_{k-1}$ we have $x_i \ge 1$, hence $\sum _{i \in \delta ^-(j)} w_{ij} x_i \ge t_j$. By definition of minimal activation set, there must exist some $S \in \mathcal {C}_j$, say $\bar{S}$, such that $\bar{S} \subseteq A_{k-1}$. Then the corresponding constraint $\sum _{i \in \bar{S}} x_i - x_j \le |\bar{S}| - 1$ in the formulation $\text {AS}_x(\bar{y})$ reads $|\bar{S}| - x_j \le |\bar{S}| - 1$, implying $x_j \ge 1$.

Finally, for every $j \not \in A_n$, all the constraints $\sum _{i \in S} x_i - x_j \le |S| - 1$ are slack because there does not exist $S \in \mathcal {C}_j, S \subseteq A_k$ for some k. Hence, the implied lower bound for $x_j$ is 0. Since we are minimizing $\sum _{j \in V} x_j$, at the optimum $x_j = 1$ if and only if $j \in A_n$.

Proof

Proposition 3. We show that for a given 0–1 vector $\bar{y}$, the remaining system in (DUAL-$\theta 0$) is total dual integral. This implies that it defines an integral polyhedron [6].

The discussion in this section shows that the dual of (DUAL-$\theta 0$) for fixed $y = \bar{y}$ is the problem $\text {AS}_x(\bar{y})$ defined in (1). To show total dual integrality of the desired system, we need to show that for any integer value of the r.h.s. of the first set of constraints in $\text {AS}_x(\bar{y})$, either $\text {AS}_x(\bar{y})$ is infeasible, or it has an optimal solution that is integer.

Let b be a given vector of integer r.h.s. values for the first set of constraints, which are indexed by $j \in V, S \in \mathcal {C}_j$. First, notice that if $b_{j,S} < 0$ for any j, S, the problem is infeasible; hence, we only need to consider the case $b \ge 0$. We show how to construct an integer optimal solution.

Define $x^0 := \bar{y}$. Apply the following algorithm: for $k=1,\dots ,n$, (i) set $x^k_j \leftarrow 0 \, \forall j$; (ii) for $j \in V, S \in \mathcal {C}_j$, set $x^k_j \leftarrow \max \{x^k_j, \sum _{i \in S} x^{k-1}_i - b_{j,S}\}$. It is clear that this defines an integral vector $x^n$. We now show that this solution is optimal. Let $x^*$ be an optimal solution for the problem with r.h.s. b. We first show by induction that $x^k \le x^*$. For $k = 0$ this is obvious as $x \ge \bar{y} = x^0$ is among the constraints. Assume $x^{k-1} \le x^*$ and suppose $x^k_h > x^*_h$ for some h. Since $x^k_h$ is initially 0, it must be that for some S, $x^k_h$ is set to $\sum _{i \in S} x^{k-1}_i - b_{h,S} > x^*_h$ for the first time. But $\sum _{i \in S} x^{k-1}_i - b_{h,S} \le \sum _{i \in S} x^*_i - b_{h,S} \le x^*_h$, because $x^*$ satisfies the constraints; this is a contradiction. It follows that $x^k \le x^*$ for all $k=1,\dots ,n$. It is easy to check that $x^n$ is feasible by construction, and therefore it must be optimal.

Proof

Proposition 4. Let $\pi ^*, y^*, \mu ^*, z^*$ be the optimal solution of (R-IMP-LAZY). Consider the following LP, obtained by fixing $y = y^*$, keeping only one of the constraints involving z for a given value $\bar{y}$ in (R-IMP-LAZY), and eliminating the z variable (which is unnecessary if it appears in one constraint only):

$$\begin{aligned} \left. \begin{array}{rrcl} \max _{\pi , \mu , y} &{} \sum \nolimits _{j \in V} \sum \nolimits _{S \in \mathcal {C}_j^{\bar{y}} } (|S|-1) \pi _S + \\ {} &{}\sum \nolimits _{j \in V} \sum \nolimits _{S \in \mathcal {C}^e_j \setminus \mathcal {C}_j^{\bar{y}} } |S| \pi _S + \sum \nolimits _{j \in V} \mu _j &{} &{} \\[2pt] \forall j \in V &{} \sum \nolimits _{k \in \delta ^+(j)} \sum \nolimits _{S \in \mathcal {C}^e_k: j \in S} \pi _S - \sum \nolimits _{S \in \mathcal {C}^e_j} \pi _S + \mu _j &{}\le &{} 1 \\[2pt] \forall j \in V, y^*_j = 0 &{} \mu _j &{}=&{} 0 \\[2pt] \forall j \in V, \forall S \in \mathcal {C}^e_j &{} \pi _S &{}\le &{} 0 \\[2pt] \forall j \in V &{} \mu _j &{}\ge &{} 0. \end{array} \right\} \end{aligned}$$

(2)

This problem is feasible as the all-zero solution is feasible. Using the reverse of the transformations discussed in Sect. 3, we can show that the dual of the above problem is equivalent to the following LP:

$$\begin{aligned} \left. \begin{array}{rrcl} \min &{} \sum \nolimits _{j \in V} x_j &{} &{} \\ \forall j \in V, \forall S \in \mathcal {C}_j^{\bar{y}} &{} \sum \nolimits _{i \in S} x_i - x_j &{}\le &{} |S| - 1 \ \\[2pt] \forall j \in V, \forall S \in \mathcal {C}^e_j \setminus \mathcal {C}_j^{\bar{y}} &{} \sum \nolimits _{i \in S} x_i - x_j &{}\le &{} |S| \\[2pt] \forall j \in V &{} x_j &{}\ge &{} y^*_j \\[2pt] \forall j \in V &{} x_j &{}\le &{} 1 \\[2pt] \forall j \in V &{} x_j \ge 0. \end{array} \right\} \end{aligned}$$

(3)

The constraints with r.h.s. value |S| are redundant and can be dropped. As a result, with the same argument used for Proposition 2, the optimum value of (3) and therefore (2) is equal to $\text {RI}_{x, \theta =\theta ^{\bar{y}}}(y^*)$, i.e., the influence spread with seeds determined by $y^*$ and node thresholds equal to $t + \theta ^{\bar{y}}$.

Now notice that the objective function of (R-IMP-LAZY) corresponding to $\pi ^*, y^*, \mu ^*, z^*$ is equal to the minimum objective function of all the problems (2), for all $\bar{y} \in \{0,1\}^n$. In other words, (R-IMP-LAZY) yields the following value:

$$\begin{aligned} \min _{\bar{y} \in \{0,1\}^n} \text {RI}_{x, \theta =\theta ^{\bar{y}}}(y^*). \end{aligned}$$

By definition of $\theta ^{\bar{y}}$, this minimum is attained for $\bar{y} = y^*$. Hence, the optimum of (R-IMP-LAZY) has value $\text {RI}_{x, \theta =\theta ^{y^*}}(y^*)$, i.e., the influence spread with seeds $y^*$, when the node thresholds are chosen to be the worst possible within the allowed set of node thresholds. Since we are maximizing, this is equivalent to solving (R-IMP).