General Rumor Blocking: An Efficient Random Algorithm with Martingale Approach

Abstract

Rumor Blocking, an important optimization problem in social network, has been extensively studied in the literature. Given social network $G=(V,E)$ and rumor seed set A, the goal is asking for k protector seeds that protect the largest expected number of social individuals by truth. However, the source of rumor is always uncertain, rather than being predicted or being known in advance in the real situations, while rumor spreads like wildfire on the Internet.

This paper presents General Rumor Blocking with unpredicted rumor seed set (randomized A) and various personal profits while being protected (weights of nodes in V). We first show that the objective function of this problem is non-decreasing and submodular, and thus a $(1-1/e)$ approximate solution can be returned by greedy approach. We then propose an efficient random algorithm R-GRB which returns a ($1-1/e-\varepsilon$) approximate solution with at least $1-n^{-\ell }$ probability. We show that it runs in $O\left( m(n-r)(k\log (n-r) + \ell \log n)/\varepsilon ^{2}\right)$ expected time, where $m=|E|$, $n=|V|$, $r=|A|$ and k is the number of protector seeds.

This work is supported by NSFC (No. 11271341 and No. 11201439).

Appendix

Proof of Lemma 2. First, we show the monotonicity. For any node set B and any protector seed $u \notin B$, we have

$$\begin{aligned} f_{\mathcal {R}}&(B\cup \{u\}) - f_{\mathcal {R}}(B) = \frac{1}{\rho } \sum _{R_{i}\in \mathcal {R}}w(\overline{A_{i}})\cdot \left( x(B\cup \{u\}, R_{i})-x(B,R_{i})\right) . \end{aligned}$$

Since $x(B, R_{i}) = 1$ leads to $x(B\cup \{u\}) =1$, $f_{\mathcal {R}}(B\cup \{u\}) - f_{\mathcal {R}}(B) \ge 0$. It implies the monotonicity.

Then, we show the submodularity. For any pair of $B_{1}, B_{2}$ with $B_{1} \subseteq B_{2}$ and $u\notin B_{2}$, we have

Observe that if $x(B_{2} \cup \{u\}, R_{i}) - x(B_{2} ,R_{i}) = 1$, $B_{2} \cap R_{i} = \emptyset$ and $u \in R_{i}$. Then $B_{1} \cap R_{i} = \emptyset$ and $u\in R_{i}$, implying that $x(B_{1}\cup \{u\}) -x(B_{1},R_{i}) =1$. Thus,

$$\begin{aligned} \left( f_{\mathcal {R}}(B_{1} \cup \{u\})-f_{\mathcal {R}}(B_{1})\right) - \left( f_{\mathcal {R}}(B_{2} \cup \{u\}) - f_{\mathcal {R}}(B_{2})\right) \ge 0. \end{aligned}$$

The submodularity follows.   $\square$

Proof of Lemma 4.  We first show that the sequence $Z_{1}, \ldots , Z_{\rho }$ is a martingale. Since $Z_{i} = \sum _{j=1}^{i}(w(\overline{A_{j}}) \cdot x(B, R_{j})) - q)$, we have $\mathbb {E}[Z_{i}] = 0$ and $\mathbb {E}[|Z_{i}|] < +\infty$. Based on the process of generating random RP sets, we can observe that the value of $x(B, R_{i})$ is independent of $x(B,R_{1}), \ldots , x(B, R_{i-1})$. Therefore,

$$\begin{aligned} \mathbb {E}[Z_{i} ~| ~Z_{1}, \cdots , Z_{i-1}]&=\mathbb {E}[Z_{i-1} + w(\overline{A_{i}}) \cdot x(B,R_{i}) -q~ |~ Z_{1}, \cdots , Z_{i-1} ]\\&= Z_{i-1} + \mathbb {E}[w(\overline{A_{i}}) \cdot x(B,R_{i})] -q = Z_{i-1}, \end{aligned}$$

implying that $Z_{1}, \cdots , Z_{\rho }$ is a martingale.

Then we find the value of a and b in the conditions of Martingale’s Property respect with $Z_{1},\cdots , Z_{\rho }$. Recall that $w(\overline{A_{i}}) = \sum _{u\in V\setminus A_{i}} w_{u}$ and $w_{u} \in [\frac{1}{n-r}, 1]$ for any u. Since

$$|Z_{1}| = |w(\overline{A_{1}}) \cdot x(B, R_{1}) - q| \le n-r ~~~and ~~~|Z_{j} - Z_{j-1}| = |w(\overline{A_{j}}) \cdot x(B, R_{j}) - q| \le n-r$$

for each $j\in \{2,\ldots ,i\}$, we can set $a=n-r$. Based on the properties of variance and $Z_{\rho } = \sum _{j=1}^{\rho } (w(\overline{A_{j}}) \cdot x(B, R_{j}) - q)$, we can set $b = (n-r) \cdot \rho q$. It is because

where the second inequality from the end holds from the facts that $w(\overline{A_{i}}) \le n-r$ and $x^{2}(B,R_{j}) = x(B,R_{j})$. Applying Martingale’s Property, one can see that inequalities (1) and (2) hold.   $\square$

Proof of Lemma 5.  Let $\tilde{B}$ be the solution of Algorithm 2 (Node Selection). Let $B^{*}$ be the optimal solution. Based on the greedy approach in Algorithm 2, we have

$$f_{\mathcal {R}}(\tilde{B}) \ge (1-1/e) \cdot f_{\mathcal {R}}(B^{*}).$$

In the sequel we show that if $\rho \ge \rho _{1}$, $\Pr [f_{\mathcal {R}}(B^{*}) \le (1-\varepsilon _{1}) \cdot OPT ] \le \delta _{1}$.

The result means that $\Pr [f_{\mathcal {R}} (B^{*})\ge (1-\varepsilon _{1})\cdot OPT ] \ge 1- \delta _{1}$, and thus $f_{\mathcal {R}}(\tilde{B}) \ge (1-1/e)(1-\varepsilon _{1})\cdot OPT$ holds with at least $1-\delta _{1}$ probability when $\rho \ge \rho _{1}$, implying the lemma.

By Lemmas 3 and 4, we can verify that if $\rho > \rho _{1}$, the following inequality holds.

$$\begin{aligned} \Pr [f_{\mathcal {R}}(B^{*}) \le (1-\varepsilon _{1}) \cdot OPT ] \le \delta _{1}. \end{aligned}$$

(6)

Consider the sampling result $\mathcal {R}$ of R-NRB algorithm. First, for any $R_{i} \in \mathcal {R}$, we denote by $x(B^{*}, R_{i})$ a random variable such that $x(B^{*}, R_{i}) =1$ if $B^{*} \cap R_{i} \ne \emptyset$ and $x^(B^{*}, R_{i}) = 0$ otherwise. Based on Lemma 4, the sequence of $Z_{i} = \sum _{j=1}^{i}\left( w(\overline{A_{i}}) \cdot x(B^{*}, R_{i})-q^{*}\right)$, $i \in \{1, \ldots , \rho \}$ is a martingale. Let $q^{*}= \mathbb {E}[f_{\mathcal {R}}(B^{*})]$. By Lemma 3, $q^{*}= \sigma (B^{*}) = OPT$. We have

By the inequality (2) of Lemma 4 and $\rho \ge \rho _{1}$,

Thus the inequality 6 holds. This completes the proof of the lemma.   $\square$

Proof of Lemma 6

Proof

 Let $\mathcal {R} = \{R_{1}, \ldots , R_{\rho }\}$ be an input of Algorithm 2 and let |R| be the number of nodes in a random RP set R. Recall that Algorithm 2 is a greedy process which returns a protector seed set $\tilde{B}$ by maximizing the marginal utility of $f_{\mathcal {R}}(\cdot )$. Due to

$$f_{\mathcal {R}}(B) = \frac{1}{\rho } \sum _{R_{i}\in \mathcal {R}} (w(\overline{A_{i}}) \cdot x(B,R_{i})),$$

it is clear that Algorithm 2 is equivalent to the greedy approach for a maximum weighted coverage problem. We have known that the time complexity of greedy approach for the maximum weighted coverage is $O(\sum _{R\in \mathcal {R}} |R|)$ in [17]. Thus, Algorithm 2 runs in $O(\sum _{R\in \mathcal {R}} |R|)$ time.   $\square$

Proof of Claim 1. We prove Claim 1 by showing that if OPT$<\xi _{i}$,

for any size-k set B.

Based on Lemma 4, the sequence $Z_{d} = \sum _{j=1}^{d}\left( w(\overline{A_{j}}) \cdot x(B, R_{j})\right)$, $d\in \{1,\ldots ,\rho _{i}\}$ is a martingale. Denote $q= \mathbb {E}[f_{\mathcal {R}}(B)]$. It is clear that

$$\begin{aligned} \Pr [&f_{\mathcal {R}}(B) \ge (1+\varepsilon _{0}) \cdot \xi _{i}]\\&= \Pr \left[ \rho _{i} \cdot f_{\mathcal {R}}(B) -\rho _{i} q \ge \left( (1+\varepsilon _{0}) \cdot \xi _{i})/q) -1\right) \cdot \rho _{i} q \right] \\&\le \exp \left( -\frac{\zeta ^{2}}{(2+\frac{2}{3}\zeta )(n-r)} \cdot \rho _{i} q\right) , \end{aligned}$$

where $\zeta =(1+\varepsilon _{0}) \cdot \xi _{i} /q - 1$ and the last inequality holds from the inequality (1) of Lemma 4. By Lemma 3, $q= \sigma (B) \le OPT < \xi _{i}$, we have $\zeta = (1+\varepsilon _{0}) \cdot \xi _{i}/q -1> \varepsilon _{0} \cdot \xi _{i} / q > \varepsilon _{0}$. Then the right side of above inequality

Since $\rho \ge \mu _{0} / \xi _{i}$, the right side of above inequality

Thus the Claim 1 is proved.   $\square$

Proof of Claim 2. We prove Claim 2 by showing that if OPT $\ge \xi _{i}$,

for any size-k set B. Based on Lemma 4, the sequence $Z_{d} = \sum _{j=1}^{d}(w(\overline{A_{j}}) \cdot x(B, R_{j}))$, $d \in \{1, \ldots ,\rho \}$ is a martingale. Recall that $f_{\mathcal {R}}(B) = \frac{1}{\rho }\sum _{R_{i}\in \mathcal {R}} \left( w(\overline{A_{i}}) \cdot x(B, R_{i})\right)$ and denote $q= \mathbb {E}[f_{\mathcal {R}}(B)] < OPT$. It is clear that

By inequality (1) of Lemma 4 and $q < OPT$,

By $\rho \ge \mu _{0} / \xi _{i}$ and $OPT \ge \xi _{i}$, the right side of above inequality

Thus, the Claim 2 holds.   $\square$

Proof of Lemma 7.  Denote by $\hat{v}$ a random node is subjected to some probability distribution $\mathcal {V}$ with $\Pr ^{*}[\hat{v}]$. Let $\Pr ^{*}[\hat{v}]= \frac{d(\hat{v})}{2m}$, where $d(\hat{v})$ is the in-degree of $\hat{v}$ in G and m is the number of edges in G. For any random node $\hat{v}$, denote by $p_{\hat{v}}$ the probability that $\hat{v}$ is covered by a random RP set. Let $\varphi (\hat{v}, R)$ be a function as follows:

$$\varphi (\hat{v}, R)=\left\{ \begin{array}{ll} 1,&{}{\hat{v} \in R};\\ 0,&{}{otherwise}. \end{array}\right. $$

Recall that for any protector seed set B, $x(B, R)=1$ if $B \cap R \ne \emptyset$ and $x(B, R)=0$ otherwise. $\mathcal {P}$ is a collection consisting of all possible RP sets. Then we obtain that

where the last inequality holds from the fact that $\varphi (\hat{v}, R) =1$ if and only if $x(\hat{v}, R) =1$. Recall the Definition 1 (Random RP Set), assume that the sampled rumor seed set is A, then the target v is selected with $\Pr [v] = \frac{w_{v}}{w(\overline{A})}$. Since $w(\overline{A_{i}})=\sum _{u\in V\setminus A_{i}} w_{u} \ge (n-r) \cdot \frac{1}{n-r}=1$, the right side of above inequality

where the first and second equalities can be derived from the proof of Lemma 3. Then the last inequality holds by $\sum _{\hat{v} \in V} d(\hat{v}) = m$ and $\sigma (\{\hat{v}\}) \le$ OPT, Therefore the lemma holds.   $\square$