A fractional memory-efficient approach for online continuous-time influence maximization

Abstract

Influence maximization (IM) under a continuous-time diffusion model requires finding a set of initial adopters which when activated lead to the maximum expected number of users becoming activated within a given amount of time. State-of-the-art approximation algorithms applicable to solving this intractable problem use reverse reachability influence samples to approximate the diffusion process. Unfortunately, these algorithms require storing large collections of such samples which can become prohibitive depending on the desired solution quality, properties of the diffusion process and seed set size. To remedy this, we design an algorithm that allows the influence samples to be processed in a streaming manner, avoiding the need to store them. We approach IM using two fractional objectives: a fractional relaxation and a multi-linear extension of the original objective function. We derive a progressively improved upper bound to the optimal solution, which we empirically find to be tighter than the best existing upper bound. This enables instance-dependent solution quality guarantees that are observed to be vastly superior to the theoretical worst case. Leveraging these, we develop an algorithm that delivers solutions with a superior empirical solution quality guarantee at comparable running time with greatly reduced memory usage compared to the state-of-the-art. We demonstrate the superiority of our approach via extensive experiments on five real datasets of varying sizes of up to 41M nodes and 1.5B edges.

Appendix A : Omitted proofs

Proof of Lemma 4

Starting from $\Pr [X \ge (1 + \epsilon ) \chi ] < \exp \left( - \frac{\epsilon ^2 \chi }{2 (1 + \epsilon /3)} \right)$, where $\chi {:}{=}\mu t$, a bound violation probability of at most $\delta$ is needed (i.e., $\delta = \exp ( - \frac{\epsilon ^2 \chi }{2 (1 + \epsilon /3)} )$). Rearranging into quadratic form in $\epsilon$,

$$\begin{aligned} 2 (1 + \epsilon /3) \log (\delta )&= - \epsilon ^2 \chi \\ \chi \epsilon ^2 + 2/3 \log (\delta ) \epsilon + 2 \log (\delta )&= 0 \end{aligned}$$

By solving the quadratic it can be determined that this is ensured if,

$$\begin{aligned} \epsilon&= \left( -2/3\log (\delta ) + \sqrt{4/ 9 \log (\delta )^2 - 8 \chi \log (\delta )}\right) /(2 \chi ) \end{aligned}$$

(30)

$$\begin{aligned}&= \left( -\log (\delta ) + \sqrt{-\log (\delta )(18 \chi - \log (\delta ))}\right) /(3 \chi ) \end{aligned}$$

(31)

Hence,

$$\begin{aligned} \Pr [X \ge \chi + \left( -\log (\delta ) + \sqrt{-\log (\delta )(18 \chi - \log (\delta ))}\right) /3]&\le \delta \end{aligned}$$

(32)

From the inner constraint we have,

$$\begin{aligned}&X \ge \chi + \left( -\log (\delta ) + \sqrt{-\log (\delta )(18 \chi - \log (\delta ))}\right) /3 \end{aligned}$$

(33)

$$\begin{aligned}&\quad 3 X + \log (\delta ) - 3 \chi \ge \sqrt{-\log (\delta )(18 \chi - \log (\delta ))} \end{aligned}$$

(34)

Squaring both sides and rearranging.

$$\begin{aligned} 0&\le \log (\delta )(18 \chi - \log (\delta )) + (3 X + \log (\delta ) - 3 \chi )^2 \end{aligned}$$

(35)

$$\begin{aligned} 0&\le 3 \chi ^2 + (4 \log (\delta ) - 6 X) \chi + 2 X \log (\delta ) + 3 X^2 \end{aligned}$$

(36)

Solving for $\chi$ via the quadratic formula and simplifying.

$$\begin{aligned} \chi&\le X - 2/3 \log (\delta ) - \sqrt{-2/9 \log (\delta )(9 X - 2 \log (\delta ))} \end{aligned}$$

(37)

As such we have,

$$\begin{aligned}&\textstyle \Pr [\chi \le X - \frac{2}{3} \log (\delta ) - \sqrt{-\frac{2}{9} \log (\delta )(9 X - 2 \log (\delta ))}] \le \delta \end{aligned}$$

(38)

$$\begin{aligned}&\textstyle \Pr [\chi > X + \frac{2}{3} \log (\frac{1}{\delta }) - \sqrt{\frac{2}{9} \log (\frac{1}{\delta })(9 X + 2 \log (\frac{1}{\delta }))}] \ge 1 - \delta \end{aligned}$$

(39)

Let $\text{ LB }(X,\delta ) {:}{=}X + \frac{2}{3} \log (\frac{1}{\delta }) - \sqrt{\frac{2}{9} \log (\frac{1}{\delta }) (9 X + 2 \log (\frac{1}{\delta })}$ then, $\Pr [\mu > \text{ LB }(X,\delta )/t] \ge 1 - \delta$ $\square$

Proof of Lemma 5

Starting from $\Pr [X \le (1 - \epsilon ) \chi ] \le \exp \left( - \frac{1}{2} \epsilon ^2 \chi \right)$, where $\chi {:}{=}\mu t$, a concentration bound violation probability of at most $\delta$ is needed (i.e., $\delta = \exp ( - \frac{1}{2} \epsilon ^2 \chi )$). This is ensured if $\epsilon = \sqrt{-2 \log (\delta ) / \chi }$. Hence,

$$\begin{aligned} \Pr [X \le (1 - \sqrt{-2 \log (\delta ) / \chi }) \chi ]&\le \delta \end{aligned}$$

(40)

From the inner constraint we have, $0 \le \chi - \sqrt{-2 \log (\delta ) \chi } - X$ Solving for $\sqrt{\chi }$ via the quadratic formula gives, $\sqrt{\chi } \ge \left( \sqrt{-2 \log (\delta )} + \sqrt{4X - 2 \log (\delta )}\right) /2$. As such we have,

$$\begin{aligned} \Pr [\chi \ge \left( \sqrt{-2 \log (\delta )} + \sqrt{4X - 2 \log (\delta )}\right) ^2/4]&\le \delta \end{aligned}$$

(41)

$$\begin{aligned} \Pr [\chi < \left( \sqrt{-2 \log (\delta )} + \sqrt{4X - 2 \log (\delta )}\right) ^2/4]&\ge 1 - \delta \end{aligned}$$

(42)

$$\begin{aligned} \Pr [\chi < X - \log (\delta ) + \sqrt{-\log (\delta ) (2 X - \log (\delta ))}]&\ge 1 - \delta \end{aligned}$$

(43)

Let $\text{ UB }(X,\delta ) {:}{=}X + \log (\frac{1}{\delta }) + \sqrt{\log (\frac{1}{\delta })(2 X + \log (\frac{1}{\delta }))}$ then, $\Pr [\mu < \text{ UB }(X,\delta )/t] \ge 1 - \delta$ $\square$

Proof of Lemma 8

Without lost of generality we will consider the left child of a group. From the applicability condition succeeding we have, $\textstyle T \cdot [\sum _{i \in L \cup R} (1/\lambda _i)^\ell ]^{1/\ell } < 1$. Using this and the fact that the scale parameters are non-negative we have that, $\textstyle T \cdot [\sum _{i \in L} (1/\lambda _i)^\ell ]^{1/\ell } < 1$. What remains to be shown is that increasing $\ell$ can only decrease the left-hand side (the minimum shape of a child group may only be equal or large than the minimum shape of the parent group).

Consider $\ell '$ such that $\ell \le \ell '$ and let $\sum _{i \in L} (1/\lambda _i)^\ell ]^{1/\ell } < c$ for $c > 0$ then, $\sum _{i \in L} (1/\lambda _i)^\ell < c^\ell$. For all i, $(1/\lambda _i)^\ell < c ^ \ell$ must hold, since all terms are positive, and hence $(1/\lambda _i) < c$ must also hold. Now multiplying both sides by $c^{\ell ' - \ell }$ gives, $\sum _{i \in L} (1/\lambda _i)^\ell c^{\ell ' - \ell } < c^{\ell '}$. Since $(1/\lambda _i) < c$ for all i it follows that $(1/\lambda _i) ^{\ell ' - \ell } < c ^ {\ell ' - \ell }$. From this it we have, $\sum _{i \in L} (1/\lambda _i)^{\ell '} < c^{\ell '}$. Which gives, $[\sum _{i \in L} (1/\lambda _i)^{\ell '}]^{1/\ell '} < c$. $\square$