Robust Budget Allocation Via Continuous Submodular Functions

Abstract

The optimal allocation of resources for maximizing influence, spread of information or coverage, has gained attention in the past years, in particular in machine learning and data mining. But in applications, the parameters of the problem are rarely known exactly, and using wrong parameters can lead to undesirable outcomes. We hence revisit a continuous version of the Budget Allocation or Bipartite Influence Maximization problem introduced by Alon et al. (in: WWW’12 - Proceedings of the 21st Annual Conference on World Wide, ACM, New York, 2012) from a robust optimization perspective, where an adversary may choose the least favorable parameters within a confidence set. The resulting problem is a nonconvex–concave saddle point problem (or game). We show that this nonconvex problem can be solved exactly by leveraging connections to continuous submodular functions, and by solving a constrained submodular minimization problem. Although constrained submodular minimization is hard in general, here, we establish conditions under which such a problem can be solved to arbitrary precision $\varepsilon$.

Appendices

Worst-Case Approximation Ratio Versus True Worst-Case

Consider the function $f(x;\theta )$ defined on $\{0,1\} \times \{0,1\}$, with values given by:

$$\begin{aligned} f(x;0) = {\left\{ \begin{array}{ll} 1 &{} x=0 \\ 0.6 &{} x=1, \end{array}\right. } \quad f(x;1) = {\left\{ \begin{array}{ll} 1 &{} x=0 \\ 2 &{} x=1. \end{array}\right. } \end{aligned}$$

(65)

We wish to choose x to maximize $f(x;\theta )$ robustly with respect to adversarial choices of $\theta$. If $\theta$ were fixed, we could directly choose $x_\theta ^*$ to maximize $f(x;\theta )$. In particular, $x^*_0 = 0$ and $x^*_1 = 1$. Of course, we want to deal with worst-case $\theta$. One option is to maximize the worst-case approximation ratio:

$$\begin{aligned} \max _x \min _\theta \frac{f(x;\theta )}{f(x^*_\theta ;\theta )}. \end{aligned}$$

(66)

One can verify that the best x according to this criterion is $x=1$, with worst-case approximation ratio 0.6 and worst-case function value 0.6. In this paper, we optimize the worst-case of the actual function value:

$$\begin{aligned} \max _x \min _\theta f(x;\theta ). \end{aligned}$$

(67)

This criterion will select $x=0$, which has a worse worst-case approximation ratio of 0.5, but actually guarantees a function value of 1, significantly better than the 0.6 achieved by the other formulation of robustness.

DR-Submodularity and $L^\natural$-Convexity

A function is $L^\natural$-convex if it satisfies a discrete version of midpoint convexity, i.e. for all x, y it holds that

$$\begin{aligned} f(x) + f(y) \ge f\left( \left\lceil \frac{x+y}{2}\right\rceil \right) + f\left( \left\lfloor \frac{x+y}{2}\right\rfloor \right) , \end{aligned}$$

(68)

where the floor $\lfloor \cdot \rfloor$ and ceiling $\lceil \cdot \rceil$ functions are interpreted elementwise.

Remark 1

An $L^\natural$-convex function need not be DR-submodular, and vice-versa. Hence algorithms for optimizing one type may not apply for the other.

Proof

Consider $f_1(x_1,x_2) = -x_1^2 - 2x_1 x_2$ and $f_2(x_1,x_2) = x_1^2 + x_2^2$, both defined on $\{0,1,2\} \times \{0,1,2\}$. The function $f_1$ is DR-submodular but violates discrete midpoint convexity for the pair of points (0, 0) and (2, 2), while $f_2$ is $L^\natural$-convex but does not have diminishing returns in either dimension. $\square$

Intuitively-speaking, $L^\natural$-convex functions look like discretizations of convex functions. The continuous objective function ${\mathcal {I}}(x,y)$ we consider need not be convex, hence its discretization need not be $L^\natural$-convex, and we cannot use those tools. However, in some regimes (namely if each $y(s) \in \{0\} \cup [1,\infty )$), it happens that ${\mathcal {I}}(x,y)$ is DR-submodular in x.

Constrained Continuous Submodular Function Minimization

Solving the Optimization Problem

Here, we describe how to solve the convex problem (17) to which we reduced the original constrained submodular minimization problem. Bach [4], at the beginning of Section 5.2, states that this surrogate problem can be optimized via the Frank–Wolfe method and its variants. However, [4] only elaborates on the simpler version of Problem (17) without the extra functions $a_{i x_i}$. Here we detail how Frank–Wolfe algorithms can be used to solve the more general parametric regularized problem. Our aim is to spell out very clearly the applicability of Frank–Wolfe to this problem, for the ease of practitioners.

Bach [4] notes that by duality, Problem (17) is equivalent to:

$$\begin{aligned}&\min _{\rho \in \prod _{i=1}^n {\mathbb {R}}_\downarrow ^{k_i - 1}} h_\downarrow (\rho ) - H(0) + \sum _{i=1}^n \sum _{x_i=1}^{k_i-1} a_{i x_i}[\rho _i(x_i)] \\&\quad = \min _{\rho \in \prod _{i=1}^n {\mathbb {R}}_\downarrow ^{k_i - 1}} \max _{w \in B(H)} \langle \rho , w \rangle + \sum _{i=1}^n \sum _{x_i=1}^{k_i-1} a_{i x_i}[\rho _i(x_i)] \\&\quad = \max _{w \in B(H)} \left\{ \min _{\rho \in \prod _{i=1}^n {\mathbb {R}}_\downarrow ^{k_i - 1}} \langle \rho , w \rangle + \sum _{i=1}^n \sum _{x_i=1}^{k_i-1} a_{i x_i}[\rho _i(x_i)] \right\} \\&\quad := \max _{w \in B(H)} f(w). \end{aligned}$$

Here, the base polytope B(H) happens to be the convex hull of all vectors w which could be output by the greedy algorithm in [4].

It is the dual problem, where we maximize over w, which is amenable to Frank–Wolfe. For Frank–Wolfe methods, we need two oracles: an oracle which, given w, returns $\nabla f(w)$; and an oracle which, given $\nabla f(w)$, produces a point s which solves the linear optimization problem $\max _{s \in B(H)} \langle s, \nabla f(w) \rangle$.

Per [4], an optimizer of the linear problem can be computed directly from the greedy algorithm. For the gradient oracle, recall that we can find a subgradient of $g(x) = \min _y h(x,y)$ at the point $x_0$ by finding $y(x_0)$ which is optimal for the inner problem, and then computing $\nabla _x h(x,y(x_0))$. Moreover, if such $y(x_0)$ is the unique optimizer, then the resulting vector is indeed the gradient of g(x) at $x_0$. Hence, in our case, it suffices to first find $\rho (w)$ which solves the inner problem, and then $\nabla f(w)$ is simply $\rho (w)$ because the inner function is linear in w. Since each function $a_{i x_i}$ is strictly convex, the minimizer $\rho (w)$ is unique, confirming that we indeed get a gradient of f, and that f is differentiable.

Of course, we still need to compute the minimizer $\rho (w)$. For a given w, we want to solve

$$\begin{aligned} \min _{\rho \in \prod _{i=1}^n {\mathbb {R}}_\downarrow ^{k_i - 1}} \langle \rho , w \rangle + \sum _{i=1}^n \sum _{x_i=1}^{k_i-1} a_{i x_i}[\rho _i(x_i)] \end{aligned}$$

There are no constraints coupling the vectors $\rho _i$, and the objective is similarly separable, so we can independently solve n problems of the form

$$\begin{aligned} \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} \langle \rho , w \rangle + \sum _{j=1}^{k-1} a_{j}(\rho _j). \end{aligned}$$

Recall that each function $a_{i y_i}(t)$ takes the form $\frac{1}{2} t^2 r_{i y_i}$ for some $r_{i y_i} > 0$. Let $D = {{\,\mathrm{{\mathrm {diag}}}\,}}(r)$, the $(k-1)\times (k-1)$ matrix with diagonal entries $r_j$. Our problem can then be written as

$$\begin{aligned} \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} \langle \rho , w \rangle + \frac{1}{2} \sum _{j=1}^{k-1} r_j \rho _j^2&= \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} \langle \rho , w \rangle + \frac{1}{2} \langle D \rho , \; \rho \rangle \\&= \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} \langle D^{1/2}\rho , \; D^{-1/2} w \rangle + \frac{1}{2} \langle D^{1/2} \rho , \; D^{1/2}\rho \rangle . \end{aligned}$$

Completing the square, the above problem is equivalent to

$$\begin{aligned} \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} ||D^{1/2} \rho + D^{-1/2} w ||_2^2&= \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} \sum _{j=1}^{k-1} \Big (r_j^{1/2} \rho _j + r_j^{-1/2} w_j\Big )^2 \\&= \min _{\rho \in {\mathbb {R}}_\downarrow ^{k - 1}} \sum _{j=1}^{k-1} r_j \Big (\rho _j + r_j^{-1} w_j\Big )^2. \end{aligned}$$

This last expression is precisely the problem which is called weighted isotonic regression: we are fitting $\rho$ to ${{\,\mathrm{{\mathrm {diag}}}\,}}(r^{-1}) w$, with weights r, subject to a monotonicity constraint. Weighted isotonic regression is solved efficiently via the Pool Adjacent Violators algorithm of [12].

Runtime

Frank–Wolfe returns an $\varepsilon$-suboptimal solution in $O(\varepsilon ^{-1} D^2 L)$ iterations, where D is the diameter of the feasible region, and L is the Lipschitz constant for the gradient of the objective [42]. Our optimization problem is $\max _{w\in B(H)} f(w)$ as defined in the previous section. Each $w \in B(H)$ has $O(n\delta ^{-1})$ coordinates of the form $H^\delta (x+e_i)-H^\delta (x)$. Since $H^\delta$ is an expected influence in the range [0, T], we can bound the magnitude of each coordinate of w by T and hence $D^2$ by $O(n\delta ^{-1} T^2)$. If $\alpha$ is the minimum derivative of the functions $R_i$, then the smallest coefficient of the functions $a_{ix_i}(t)$ is bounded below by $\alpha \delta$. Hence the objective is the conjugate of an $\alpha \delta$-strongly convex function, and therefore has $\alpha ^{-1}\delta ^{-1}$-Lipschitz gradient. Combining these, we arrive at the $O(\varepsilon ^{-1} n\delta ^{-2} \alpha ^{-1} T^2)$ iteration bound. The most expensive step in each iteration is computing the subgradient, which requires sorting the $O(n\delta ^{-1})$ elements of $\rho$ in time $O(n\delta ^{-1} \log {n\delta ^{-1}} )$. Hence the total runtime of Frank–Wolfe is $O(\varepsilon ^{-1} n^2\delta ^{-3} \alpha ^{-1} T^2 \log {n\delta ^{-1}})$.

As specified in the main text, relating an approximate solution of (17) to a solution of (14) is nontrivial. Assume $\rho ^*$ has distinct elements separated by $\eta$, and chose $\varepsilon$ to be less than $\eta ^2 \alpha \delta / 8$. If $\rho$ is $\varepsilon$-suboptimal, then by $\alpha \delta$-strong convexity we must have $||\rho - \rho ^* ||_2 < \eta /2$, and therefore $||\rho - \rho ^* ||_\infty < \eta /2$. Since the smallest consecutive gap between elements of $\rho ^*$ is $\eta$, this implies that $\rho$ and $\rho ^*$ have the same ordering, and therefore admit the same solution x after thresholding. Accounting for this choice in $\varepsilon$, we have an exact solution to (14) in total runtime of $O(\eta ^{-2} n^2\delta ^{-4} \alpha ^{-2} T^2 \log {n\delta ^{-1}})$.