An efficient and effective hop-based approach for influence maximization in social networks

Abstract

Influence maximization in social networks is a classic and extensively studied problem that targets at selecting a set of initial seed nodes to spread the influence as widely as possible. However, it remains an open challenge to design fast and accurate algorithms to find solutions in large-scale social networks. Prior Monte Carlo simulation-based methods are slow and not scalable, while other heuristic algorithms do not have any theoretical guarantee and they have been shown to produce poor solutions for quite some cases. In this paper, we propose hop-based algorithms that can be easily applied to billion-scale networks under the commonly used Independent Cascade and Linear Threshold influence diffusion models. Moreover, we provide provable data-dependent approximation guarantees for our proposed hop-based approaches. Experimental evaluations with real social network datasets demonstrate the efficiency and effectiveness of our algorithms.

Appendix

Proof of Theorem 1

To consider the outgoing edges from u one at a time, we first disable all the edges from u to its neighbors except for one edge $\langle u,w_1\rangle$. Then, for each neighbor v of $w_1$, all of v’s inverse neighbors other than $w_1$ have their one-hop activation probabilities unchanged by adding $\langle u,w_1\rangle$. Let $\pi _2^{{S}\cup \{u\}}(v|w_1)$ denote the new two-hop activation probability of v. Then, we have

$$\begin{aligned} \frac{1-\pi _2^{{S}\cup \{u\}}(v|w_1)}{1-\pi _2^{{S}}(v)}=\rho (S,u,v,w_1), \end{aligned}$$

(16)

where $\rho (S,u,v,w)=\frac{1-p_{w,v}\cdot \pi _1^{{S}\cup \{u\}}(w)}{1-p_{w,v}\cdot \pi _1^{{S}}(w)}$. Next, we enable the second edge $\langle u,w_2\rangle$. Let $\pi _2^{{S}\cup \{u\}}(v|w_1,w_2)$ denote the new two-hop activation probability of v. Following similar arguments, for each neighbor v of $w_2$, we have

$$\begin{aligned} \frac{1-\pi _2^{{S}\cup \{u\}}(v|w_1,w_2)}{1-\pi _2^{{S}\cup \{u\}}(v|w_1)} =\rho (S,u,v,w_2). \end{aligned}$$

(17)

We continue to enable the outgoing edges of u sequentially. In general, when an edge $\langle u,w_i\rangle$ is enabled after edges $\langle u,w_1\rangle, \langle u,w_2\rangle, \ldots , \langle u,w_{i-1}\rangle$, for each neighbor v of $w_i$, we have

$$\begin{aligned} \frac{1-\pi _2^{{S}\cup \{u\}}(v|w_1,\dots ,w_i)}{1-\pi _2^{{S}\cup \{u\}}(v|w_1,\dots ,w_{i-1})}=\rho (S,u,v,w_i). \end{aligned}$$

(18)

Therefore, we can initialize $\pi _2^{{S}\cup \{u\}}(v)$ with $\pi _2^{{S}}(v)$ and iteratively update $\pi _2^{{S}\cup \{u\}}(v)$ with

$$\begin{aligned} 1-\left(1-\pi _2^{{S}\cup \{u\}}(v)\right)\cdot \rho (S,u,v,w), \end{aligned}$$

(19)

for all the nodes $w\in {N}_u\setminus {S}$ and $v\in {N}_w\setminus {S}$. Moreover, for the direct neighbors of u, their two-hop activation probabilities also need to be adjusted because u’s one-hop activation probability has changed from $\pi _1^{{S}}(u)$ to 1. For each neighbor v of u, the adjustment can be made in a similar way by updating $\pi _2^{{S}\cup \{u\}}(v)$ with

$$\begin{aligned} 1-\left(1-\pi _2^{{S}\cup \{u\}}(v)\right)\cdot \rho (S,u,v,u). \end{aligned}$$

(20)

Then, the final two-hop activation probability $\pi _2^{{S}\cup \{u\}}(v)$ by the iterative updates (19) and (20) is

$$\begin{aligned} \pi _2^{{S}\cup \{u\}}(v) = 1-\left(1-\pi _2^{{S}}(v)\right)\cdot \prod _{w\in ({M}_{u,v}\cup \{u\})}\rho (S,u,v,w). \end{aligned}$$

(21)

Hence, the theorem is proven. $\square$

Proof of Theorem 2

Consider a single seed $\{u\}$. Let ${A}_u\subseteq {N}_u$ denote a subset of a node u’s neighbors. Let $p({A}_u)$ denote the probability that all the nodes in ${A}_u$ are activated directly by u under the IC and LT models, while all the nodes in ${N}_u\setminus {A}_u$ are not directly activated by u (they may not even be activated eventually). Since each of u’s neighbors is activated by u independently, we have

$$\begin{aligned} p({A}_u)=\left(\prod _{v\in {A}_u} p_{u,v}\right)\cdot \left(\prod _{v\in {N}_u\setminus {A}_u}(1-p_{u,v})\right). \end{aligned}$$

(22)

Furthermore, with h hops of propagation, for each node $w\in {V}\setminus \{u\}$, w can only be activated by a propagation path starting from a node $v\in {A}_u$ whose path length is no longer than $h-1$ hops. In other words, the probability for w to be activated by ${A}_u$ is $\pi _{h-1}^{{A}_u}(w)$. Considering all the possible node sets ${A}_u$ activated directly by u, we have

$$\begin{aligned}&\sigma _h(\{u\})\nonumber \\&\quad = 1+\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot \sum _{w\in {V}\setminus \{u\}}\pi _{h-1}^{{A}_u}(w)\right)\nonumber \\&\quad \le 1+\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot \sum _{w\in {V}}\pi _{h-1}^{{A}_u}(w)\right)\nonumber \\&\quad = 1+\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot \sigma _{h-1}({A}_u)\right)\nonumber \\&\quad \le 1+\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot \sum _{v\in {A}_u}\sigma _{h-1}(\{v\})\right)\nonumber \\&\quad = 1+\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot \sum _{v\in {N}_u}\left(\sigma _{h-1}(\{v\})\cdot p(v\in {A}_u)\right)\right)\nonumber \\&\quad = 1+\sum _{{A}_u\subseteq {N}_u}\left(\sum _{v\in {N}_u}\left(p({A}_u)\cdot \sigma _{h-1}(\{v\})\cdot p(v\in {A}_u)\right)\right)\nonumber \\&\quad = 1+\sum _{v\in {N}_u}\left(\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot \sigma _{h-1}(\{v\})\cdot p(v\in {A}_u)\right)\right)\nonumber \\&\quad = 1+\sum _{v\in {N}_u}\left(\sigma _{h-1}(\{v\})\cdot \sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot p(v\in {A}_u)\right)\right). \end{aligned}$$

(23)

The second “$\le$” is due to the submodularity of $\sigma _{h}(\cdot )$ (see Theorem 3) such that $\sigma _{h-1}({A}_u)\le \sum _{v\in {A}_u}\sigma _{h-1}(\{v\})$. In the third “=”, $p(v\in {A}_u)$ is such a binary value that $p(v\in {A}_u)=1$ if and only if $v\in {A}_u$. Meanwhile, we have

$$\begin{aligned}&\sum _{{A}_u\subseteq {N}_u}\left(p({A}_u)\cdot p(v\in {A}_u)\right)\nonumber \\&\quad =\sum _{{A}_u\subseteq {N}_u\setminus \{v\}}\left(p({A}_u)\cdot p(v\in {A}_u)\right)\nonumber \\&\qquad +\sum _{{A}_u\subseteq {N}_u\setminus \{v\}}\left(p({A}_u\cup \{v\})\cdot p(v\in {A}_u\cup \{v\})\right)\nonumber \\&\quad =\sum _{{A}_u\subseteq {N}_u\setminus \{v\}}p({A}_u\cup \{v\}). \end{aligned}$$

(24)

The last “=” follows the fact that $p(v\in {A}_u)=0$ since $v\not \in {A}_u\subseteq {N}_u\setminus \{v\}$ and $p(v\in {A}_u\cup \{v\})=1$ since $v\in {A}_u\cup \{v\}$. Therefore, from (23) and (24), we have

$$\begin{aligned} \sigma _h(\{u\})\le 1+\sum _{v\in {N}_u}\left(\sigma _{h-1}(\{v\})\cdot \sum _{{A}_u\subseteq {N}_u\setminus \{v\}}p({A}_u\cup \{v\})\right). \end{aligned}$$

(25)

Furthermore, by definition,

$$\begin{aligned}&\sum _{{A}_u\subseteq {N}_u\setminus \{v\}}p({A}_u\cup \{v\})\nonumber \\&\quad = \sum _{{A}_u\subseteq {N}_u\setminus \{v\}}\left(\left(\prod _{w\in {A}_u\cup \{v\}} p_{u,w}\right)\cdot \left(\prod _{w\in {N}_u\setminus ({A}_u\cup \{v\})}(1-p_{u,w})\right)\right)\nonumber \\&\quad = \sum _{{A}_u\subseteq {N}_u\setminus \{v\}}\left(p_{u,v}\cdot \left(\prod _{w\in {A}_u} p_{u,w}\right)\cdot \left(\prod _{w\in {N}_u\setminus ({A}_u\cup \{v\})}(1-p_{u,w})\right)\right)\nonumber \\&\quad = p_{u,v}\cdot \sum _{{A}_u\subseteq {N}_u\setminus \{v\}}\left(\left(\prod _{w\in {A}_u} p_{u,w}\right)\cdot \left(\prod _{w\in {N}_u\setminus ({A}_u\cup \{v\})}(1-p_{u,w})\right)\right)\nonumber \\&\quad = p_{u,v}\cdot 1\nonumber \\&\quad = p_{u,v}. \end{aligned}$$

(26)

Thus, by (25) and (26), it holds that

$$\begin{aligned} \sigma _h(\{u\})\le 1+\sum _{v\in {N}_u}\left(\sigma _{h-1}(\{v\})\cdot p_{u,v}\right). \end{aligned}$$

(27)

Inequality (11) can be proved by induction. When $h=1$, the inequality follows directly from Inequality (10). Suppose that it holds for $h-1$ hops of propagation, i.e., $\sigma _{h-1}(\{u\}) \le \hat{\sigma }_{h-1}(\{u\})$. Then, for h hops of propagation, we have

$$\begin{aligned} \sigma _h(\{u\})&\le 1+\sum _{v\in {N}_u}\left(p_{u,v}\cdot \sigma _{h-1}(\{v\})\right)\nonumber \\&\le 1+\sum _{v\in {N}_u}\left(p_{u,v}\cdot \hat{\sigma }_{h-1}(\{v\})\right)\nonumber \\&= \hat{\sigma }_{h}(\{u\}). \end{aligned}$$

(28)

Therefore, for any $h\ge 0$, we have $\sigma _{h}(\{u\}) \le \hat{\sigma }_{h}(\{u\})$. $\square$

Proof of Theorem 3

This can be proved using the live edge approach (Kempe et al. 2003).

• Under the IC model, for each edge $\langle u,v\rangle\in {E}$, we independently flip a coin of bias $p_{u,v}$ to decide whether the edge ⟨u, v⟩ is live or blocked to generate a sample influence propagation outcome X.

• Under the LT model, for each node $v\in V$, it picks at most one of its incoming edge at random—selecting the edge from an inverse neighbor u with probability $p_{u,v}$ and not selecting any incoming edge with probability $1-\sum _{u\in {I}_v}p_{u,v}$.

We use p(X) to denote the probability of a specific outcome X in the sample space. Let ${V}_h^X(v)$ denote the node set that can be reached from a node v within h hops in the sample outcome X. Then, the number of nodes that can be reached from a seed set S within h hops in the outcome X is given by $\sigma _h^X({S})=\Big |\bigcup _{v\in {S}}{V}_h^X(v)\Big |$. Thus,

$$\begin{aligned} \sigma _h({S})=\sum _{X}\left(p(X)\cdot \sigma _h^X({S})\right), \end{aligned}$$

(29)

where the monotonicity of $\sigma _h({S})$ holds since $\sigma _h^X({S})$ increases as S expands.

The marginal influence gain

$$\begin{aligned} \sigma _h^X({S}\cup \{u\})-\sigma _h^X({S})=\Big |{V}_h^X(u)\setminus \bigcup _{v\in {S}}{V}_h^X(v)\Big | \end{aligned}$$

(30)

is the number of nodes that are reachable from a node u within h hops but are not reachable from any node in a seed set S within h hops in a sample outcome X. For any two node sets S and T where ${S}\subseteq {T}$, we have $\bigcup _{v\in {S}}{V}_h^X(v)\subseteq \bigcup _{v\in {T}}{V}_h^X(v)$. Thus, ${V}_h^X(u)\setminus \bigcup _{v\in {S}}{V}_h^X(v)\supseteq {V}_h^X(u)\setminus \bigcup _{v\in {T}}{V}_h^X(v)$, which implies that

$$\begin{aligned} \sigma _h^X({S}\cup \{u\})-\sigma _h^X({S})\ge \sigma _h^X({T}\cup \{u\})-\sigma _h^X({T}). \end{aligned}$$

(31)

Since $p(X)\ge 0$ for any X, taking the linear combination, we have

$$\begin{aligned} \sigma _h({S}\cup \{u\})-\sigma _h({S})\ge \sigma _h({T}\cup \{u\})-\sigma _h({T}). \end{aligned}$$

(32)

Thus, $\sigma _h(\cdot )$ is submodular. $\square$

Proof of Theorem 4

Let ${S}_h^*$ denote the optimal seed set for maximizing the influence spread within h hops of propagation, i.e., $\sigma _h({S}_h^*)=\max _{|{S}|=k}\sigma _h({S})$. We have

$$\begin{aligned} \sigma ({S}_h)&\ge \sigma _h({S}_h)\nonumber \\&\ge \left(\frac{1}{\kappa _{\sigma _h}}(1-e^{-\kappa _{\sigma _h}})\right)\sigma _h({S}_h^*)\nonumber \\&\ge \left(\frac{1}{\kappa _{\sigma _h}}(1-e^{-\kappa _{\sigma _h}})\right)\sigma _h({S}^*)\nonumber \\&=\left(\frac{1}{\kappa _{\sigma _h}}(1-e^{-\kappa _{\sigma _h}})\alpha \right)\sigma ({S}^*) \end{aligned}$$

(33)

The first inequality follows from the fact that the exact influence spread is equal to the influence spread without any hop limitation of propagation. The second inequality is because that the greedy algorithm can achieve $\left(\frac{1}{\kappa _f}(1-e^{-\kappa _f})\right)$-approximation for maximizing a monotone submodular function f with a cardinality constraint (Conforti and Cornuéjols 1984), where the submodularity and monotonicity of $\sigma _h(\cdot )$ is given by Theorem 3. The third inequality is because ${S}_h^*$ is the optimal solution for maximizing $\sigma _h(\cdot )$. $\square$

We first introduce some lemmas used to prove Theorem 5.

Lemma 1

For scale-free random graphs with propagation probability $p_{u,v}=p$ for every edge $\langle u,v\rangle\in {E}$, the expected influence spread produced within one hop of propagation from a random seed set S satisfies

$$\begin{aligned} \mathbb {E}[\sigma _1({S})]\ge (p+1)k-pk^2/|{V}|. \end{aligned}$$

(34)

Proof of Lemma 1

With one hop of propagation, for a randomly selected node v, it is not activated if and only if v is not a seed and v is not activated by any of its inverse neighbors. The probability for v to be a non-seed node is $1-\frac{k}{|{V}|}$. The probability for an inverse neighbor of v to be a seed is $\frac{k}{|V|}$, and thus, the probability for it to activate v is $p\cdot \frac{k}{|{V}|}$. Therefore, the probability for all of v’s inverse neighbors to fail to activate v is

$$\begin{aligned} \prod _{u\in {I}_v}\left(1-p\cdot \frac{k}{|{V}|}\right)=\left(1-\frac{pk}{|{V}|}\right)^{|{I}_v|}. \end{aligned}$$

(35)

Note that if v is selected as a seed, it must be activated. Hence, the overall activation probability of v is

$$\begin{aligned} \pi _1^{S}(v)=1-\left(1-\frac{k}{|{V}|}\right)\cdot \left(1-\frac{pk}{|{V}|}\right)^{|{I}_v|}. \end{aligned}$$

(36)

As a result, the expectation of the activation probability of a random node v is given by

$$\begin{aligned} \mathbb {E}[\pi _1^{S}(v)]&=\mathbb {E}\left [1-\left(1-\frac{k}{|{V}|}\right )\cdot \left(1-\frac{pk}{|{V}|}\right )^{|{I}_v|}\right ]\nonumber \\&=1-\left (1-\frac{k}{|{V}|}\right )\cdot \sum _{|{I}_v|=1}^{\infty }\left (P_0(|{I}_v|)\cdot \left (1-\frac{pk}{|{V}|}\right )^{|{I}_v|}\right )\nonumber \\&\ge 1-\left (1-\frac{k}{|{V}|}\right )\cdot \left (1-\frac{pk}{|{V}|}\right )\cdot \sum _{|{I}_v|=1}^{\infty }P_0(|{I}_v|)\nonumber \\&=1-\left (1-\frac{k}{|{V}|}\right )\cdot \left (1-\frac{pk}{|{V}|}\right )\nonumber \\&=\frac{(1+p)k}{|{V}|}-\frac{pk^2}{|{V}|^2}. \end{aligned}$$

(37)

Therefore, it holds that $\mathbb {E}[\sigma _1({S})]=|{V}|\cdot \mathbb {E}[\pi _1^{S}(v)]\ge (p+1)k-pk^2/|{V}|$. This completes the proof. $\square$

Lemma 2

(Li et al. 2012) For an infinite random power law graph, the expected fraction of nodes activated $\phi ({S})=\mathbb {E}[\sigma ({S})]/|{V}|$ can be computed by

$$\begin{aligned} {\left\{ \begin{array}{ll} 1-\varphi ({S})=\left(1-\frac{k}{|{V}|}\right)\sum _{d=0}^{\infty }P_1(d+1)\left (1-p\varphi ({S})\right )^d,\\ 1-\phi ({S})=\left (1-\frac{k}{|{V}|}\right )\sum _{d=1}^{\infty }P_0(d)\left (1-p\varphi ({S})\right )^d, \end{array}\right. } \end{aligned}$$

(38)

where $P_1(d)=\frac{d^{1-\gamma }}{\sum _{d=1}^{\infty }d^{1-\gamma }}$ is the probability of a node connecting to a neighbor whose degree is d, and $\varphi ({S})$ is an instrumental variable.

Lemma 3

The expected fraction of nodes activated $\phi ({S})$ is bounded by

$$\begin{aligned} \mathbb {E}[\sigma ({S})]\le |{V}|\cdot \left(1-\left(1-\frac{k}{|{V}|}\right)P_0(1)(1-pA)\right), \end{aligned}$$

(39)

where $A=1-\left (1-\frac{k}{|{V}|}\right )P_1(1)$.

Proof of Lemma 3

From (38) in Lemma 2, we have

$$\begin{aligned} 1-\varphi ({S})\ge \left (1-\frac{k}{|{V}|}\right )P_1(1)\left (1-p\varphi ({S})\right )^0=1-A, \end{aligned}$$

(40)

and

$$\begin{aligned} 1-\phi ({S})\ge \left (1-\frac{k}{|{V}|}\right )P_0(1)\left (1-p\varphi ({S})\right ). \end{aligned}$$

(41)

Hence, by (40) and (41), the lemma follows. $\square$

Proof of Theorem 5

Lemma 1 indicates that

$$\begin{aligned} \mathbb {E}[\sigma _h({S})]\ge \mathbb {E}[\sigma _1({S})]\ge (p+1)k-pk^2/|{V}|. \end{aligned}$$

(42)

Lemma 3 indicates that

$$\begin{aligned} E[\sigma ({S})]\le |{V}|\cdot \left (1-\left (1-\frac{k}{|{V}|}\right )P_0(1)(1-pA)\right ). \end{aligned}$$

(43)

Putting (42) and (43) together, the theorem follows. $\square$