Maximizing misinformation restriction within time and budget constraints

Abstract

Online social networks have become popular media worldwide. However, they also allow rapid dissemination of misinformation causing negative impacts to users. With a source of misinformation, the longer the misinformation spreads, the greater the number of affected users will be. Therefore, it is necessary to prevent the spread of misinformation in a specific time period. In this paper, we propose maximizing misinformation restriction ($\mathsf {MMR}$) problem with the purpose of finding a set of nodes whose removal from a social network maximizes the influence reduction from the source of misinformation within time and budget constraints. We demonstrate that the $\mathsf {MMR}$ problem is NP-hard even in the case where the network is a rooted tree at a single misinformation node and show that the calculating objective function is #P-hard. We also prove that objective function is monotone and submodular. Based on that, we propose an $1{-}1/\sqrt{e}$-approximation algorithm. We further design efficient heuristic algorithms, named $\mathsf {PR}$-$\mathsf {DAG}$ to show $\mathsf {MMR}$ in very large-scale networks.

Appendix

We let $e=(u, v), f=(u', v')$, $\mathcal {G}^e(G\setminus X)$ is the set of sample graph where incoming edge $e=(u, v)$ is selected for node v, $\mathcal {G}^{\overline{e}}(G\setminus X)$ is the set of sample graph where a different incoming edge $\overline{e}=(y, v)$ is selected for node v and $\mathcal {G}^{\emptyset }(G\setminus X)$ is the set of sample graph where no incoming edge $\overline{e}=(y, v)$ is selected for node v. According to Khalil et al. (2014), we have the following results:

Proposition 2

(Khalil et al. (2014), proposition 1) For every live-edge $g \in \mathcal {G}^{\emptyset }(G \setminus X)$, there exits a corresponding live-edge graph $\widetilde{g} \in \mathcal {G}^e(G \setminus X)$ and vice versa. If $g=(V, E_g)$ then $\widetilde{g}=(V, E_g \cup \{e\} )$.

Proposition 3

(Khalil et al. (2014), proposition 2) $\mathcal {G}(G \setminus (X\cup \{e\})) \subseteq \mathcal {G}(G \setminus X)$ and furthermore $\mathcal {G}(G \setminus (X\cup \{e\}))=\mathcal {G}^{\overline{e}}(G \setminus X) \cup \mathcal {G}^{\emptyset }(G \setminus X)$.

Proposition 4

(Khalil et al. (2014), proposition 3) Given $f=(u', v') \in E \setminus X, v' \ne v$, let $t=|\mathcal {G}^{\emptyset }(G \setminus (X \cup \{f\}))|$ then $\mathcal {G}^{\emptyset }(G \setminus X)$ can be partitioned into t sets $\{ \varPhi _i \}_{i=1}^t$ such that, for every $\varPhi _i$ there exits a corresponding $g_i \in \mathcal {G}^{\emptyset }( G \setminus (X \cup \{f\}))$ and vice versa.

Proposition 5

(Khalil et al. (2014), proposition 4) For every $\varPhi _i \subseteq \mathcal {G}^{\emptyset }(G \setminus X)$ and its associated $g_i \in \mathcal {G}^{\emptyset }(G \setminus (X \cup \{e\}))$, $\Pr [g_i|G \setminus (X \cup \{f \})]= \sum _{H \in \varPhi _i} \Pr [H|G \setminus X]$.

Proof of Lemma 3

We need to show that $\sigma _{d,E}(S, X) \ge \sigma _{d,E}(S, X \cup \{e\})$. The idea of the proof is similar to the theorem 5 in Khalil et al. (2014). Using Proposition 2 and  3, we have:

$$\begin{aligned} \sigma _{d,E}(S, X) - \sigma _{d,E}(S,X \cup \{e\})= \sum _{g \in \mathbb {G}(G\setminus X)}\Pr [g|G \setminus X]f_d(g,S) \\ - \sum _{g \in \mathcal {G}(G \setminus (X \cup \{e\}))}\Pr [g|G \setminus (X \cup \{e\})]f_d(g,S) = \sum _{g \in \mathcal {G}^e(G \setminus X)}{\Pr [g|G \setminus X]\cdot f_d(g,S)} \\ + \sum _{g \in \mathcal {G}^{\emptyset }(G \setminus X)}\Big ( \Pr [g|G \setminus X)]- \Pr [g|G \setminus (X \cup \{e\})]\Big )\cdot f_d(g,S) \\ + \sum _{g \in \mathcal {G}^{\overline{e}}(G \setminus X)}\Big ( \Pr [g|G \setminus X)]- \Pr [g|G \setminus (X\cup \{e\})]\Big )\cdot f_d(g,S) \end{aligned}$$

Recall that $e=(u, v)$, for $g \in \mathcal {G}^{\emptyset }(G \setminus X)$, we have:

$$\begin{aligned} \Pr [g|G \setminus X]-\Pr [g|G \setminus (X \cup \{e\})]= -w(u, v)\prod _{v' \ne v} P(v', g, G \setminus X) \end{aligned}$$

(35)

For $g \in \mathcal {G}^{\overline{e}}(G \setminus X)$ we have $P(v, g, G \setminus X)=P(v, g, G \setminus (X\cup \{e\}))=w(\overline{e})$ which leads to $\Pr [g|G \setminus X]=\Pr [g|G \setminus (X\cup \{e\})]$, it infers:

$$\begin{aligned} \sigma _{d,E}(S, X) - \sigma _{d,E}(S, X \cup \{e\})= \sum _{g \in \mathcal {G}^e(G \setminus X)}{\Pr [g|G \setminus X]\cdot f_d(g,S)} \\ + \sum _{g \in \mathcal {G}^{\emptyset }(G \setminus X)}{-w(u, v)\prod _{v' \ne v}p(v', g, G \setminus X)\cdot f_d(g,S)} \end{aligned}$$

Using prop. 2, for $g\in \mathcal {G}^{\emptyset }(G \setminus X)$ there exits a corresponding $\widetilde{g} \in \mathcal {G}^e(G \setminus X)$ and $\Pr [\widetilde{g}|G \setminus X]=w(u, v)\prod _{v' \ne v}p(v', \widetilde{g}, G \setminus X)$. Therefore,

$$\begin{aligned} \sigma _{d,E}(S, X) - \sigma _{d,E}(S, X \cup \{e\})= \sum _{g \in \mathcal {G}^{\emptyset }(G \setminus X)} \Pr [\widetilde{g}|G \setminus X] \Big ( f_d(\widetilde{g}, S) - f_d(g, S) \Big ) \end{aligned}$$

(36)

We can see that g is a subgraph of $\widetilde{g}$, the set of vertices which can reach from S in g is subset of the set of vertices which can reach from S in $\widetilde{g}$. Hence, $f_d(\widetilde{g}, S) - f_d(g, S) \ge 0$, which completes the proof. $\square$

Proof of Lemma 4

the idea of the proof is similar to that of theorem 6 in Khalil et al. (2014). For edge $f \in E$, let $t=|\mathcal {G}(G \setminus (X\{f\}))|$. From Proposition 4, we can partition $\mathcal {G}^{\emptyset }(G \setminus X)$ into t sets $\{\varPhi \}_{i=1}^t$, rewrite (36) as:

$$\begin{aligned} \sigma _{d,E}(S, X) - \sigma _{d,E}(S, X \cup \{e\})= \sum _{g \in \mathcal {G}^{\emptyset }(G \setminus X)} \Pr [\widetilde{g}|G \setminus X] \Big ( f_d(\widetilde{g}, S) - f_d(g, S) \Big ) \nonumber \\ =\sum _{i=1}^t\sum _{g \in \varPhi _i} \Pr [\widetilde{g}|G \setminus X] \Big ( f_d(\widetilde{g}, S) - f_d(g, S) \Big ) \end{aligned}$$

(37)

Using similar reasoning to that in Eq. (36) in the proof of lemma 1 for $G \setminus (X \cup \{f \})$, we have:

$$\begin{aligned}&\sigma _{d,E}(S, X \cup \{f\}) - \sigma _{d,E}(S, X \cup \{f, e\}) \nonumber \\&=\sum _{g \in \mathcal {G}^{\emptyset }(G \setminus (X \cup \{f \}))} \Pr [\widetilde{g}|G \setminus (X \cup \{f\})] \Big ( f_d(\widetilde{g}, S) - f_d(g, S) \Big ) \end{aligned}$$

(38)

We will compare two Eqs. (37) and (38) term by term for each $g_i \in \mathcal {G}^{\emptyset }(G \setminus X), i=1,\ldots ,t$. It can be divided into two cases: (1) $\varPhi _i =\{X_i\}$ in case in $g_i$ has another incoming edge to $v'$ not f, now the terms in two equations are equal; (2) in the case in $g_i$ has only incoming edge to $v'$ is f, then $\varPhi _i=\{g_i, g_i'\}$, we need to prove:

$$\begin{aligned} \Pr [\widetilde{g_i}|G \setminus X] \Big ( f_d(\widetilde{g_i}, S) - f_d(g_i, S) \Big )+\Pr [\widetilde{g'_i}|G \setminus X] \Big ( f_d(\widetilde{g'_i}, S) - f_d(g'_i, S) \Big ) \nonumber \\ \ge \Pr [\widetilde{g_i}|G \setminus (X \cup \{f\})] \Big ( f_d(\widetilde{g_i}, S) - f_d(g_i, S) \Big ) \end{aligned}$$

(39)

Using prop. 5 in Khalil et al. (2014), we have:

$$\begin{aligned} \Pr [\widetilde{g_i}|G \setminus (X \cup \{f\})] = \Pr [\widetilde{g_i}|G \setminus X] + \Pr [\widetilde{g'_i}|G \setminus X] \end{aligned}$$

(40)

Hence, inequality (39) is true if:

$$\begin{aligned} f_d(\widetilde{g'_i}, S) - f_d(g'_i, S) \ge f_d(\widetilde{g_i}, S) - f_d(g_i, S) \end{aligned}$$

(41)

Note that $\widetilde{g'_i}=(V, E_{g_i} \cup \{f\})$ and live-edge graphs are constructed in a way that each node has at most one incoming edge. We can see that: a reachability path in $\widetilde{g_i}$ is clearly presented in $\widetilde{g'_i}$, hence if removing edge e from $\widetilde{g_i}$ results in unreachability of some nodes in $g_i$. Similarly, some nodes become unreachable when removing e from $\widetilde{g'_i}$. Removing edge e from $\widetilde{g'_i}$ may disconnect some additional nodes whose paths derived from the source including edge f. Therefore, the reduction in reachable nodes when removing edge e from $\widetilde{g'_i}$ is the same or larger than the reduction when removing e from $\widetilde{g_i}$, it implies (41) is true. $\square$