Practical Budgeted Submodular Maximization

Abstract

We consider the problem of maximizing a non-negative monotone submodular function subject to a knapsack constraint, which is also known as the Budgeted Submodular Maximization (BSM) problem. Sviridenko (Operat Res Lett 32:41–43, 2004) showed that by guessing 3 appropriate elements of an optimal solution, and then executing a greedy algorithm, one can obtain the optimal approximation ratio of $\alpha =1-{1}/{e} \approx 0.632$ for BSM. However, the need to guess (by enumeration) 3 elements makes the algorithm of Sviridenko (Operat Res Lett 32:41–43, 2004) impractical as it leads to a time complexity of roughly $O(n^5)$ (this time complexity can be slightly improved using the thresholding technique of Badanidiyuru & Vondrák (in: SODA, 1497–1514, 2014) but only to roughly $O(n^4)$). Our main results in this paper show that fewer guesses suffice. Specifically, by making only 2 guesses (and using the thresholding technique of Badanidiyuru & Vondrák (in: SODA, 1497–1514, 2014), we get the same optimal approximation ratio of $\alpha$ with an improved time complexity of roughly $O(n^3)$. Furthermore, by making only a single guess, we get an almost as good approximation ratio of $0.6174 > 0.9767\alpha$ in roughly $O(n^2)$ time. Prior to our work, the only approximation algorithms that were known to obtain an approximation ratio close to $\alpha$ for BSM were the algorithm of Sviridenko (Operat Res Lett 32:41–43, 2004) and an algorithm of Ene & Nguyen (in: ICALP, 53:1–53:12, 2019) that achieves $(\alpha -\varepsilon )$-approximation. However, the algorithm of Ene & Nguyen (in: ICALP, 53:1–53:12, 2019) requires ${(1/\varepsilon )}^{O(1/\varepsilon ^4)}n\log ^2 n$ time, and hence, is of theoretical interest only since ${(1/\varepsilon )}^{O(1/\varepsilon ^4)}$ is huge even for moderate values of $\varepsilon$. In contrast, all the algorithms we analyze are simple and parallelizable, which makes them good candidates for practical use. Recently, Tang et al. (in: Proc ACM Meas Anal Comput Syst, 5(1): 08:1–08:22, 2021) studied a simple greedy algorithm that already has a long research history, and proved that it admits an approximation ratio of at least 0.405 (without any guesses). The last part of this paper improves over the result of Tang et al. (in: Proc ACM Meas Anal Comput Syst, 5(1): 08:1–08:22, 2021) and shows that the approximation ratio of this algorithm is within the range [0.427, 0.462].

Appendices

Appendix A: Missing Proofs of Sect. 4

In this section we give the proofs that have been omitted from Section 4.

Lemma 14

For every $y \in [0, 1/2]$, there is a unique value $z(y) \in [y, 1/2]$ satisfying the equation

$$\begin{aligned} \frac{y}{z(y)} - 1 = \ln \left( \frac{z(y)}{1 - y}\right) . \end{aligned}$$

Moreover, $z'(y) = \frac{z(y) (1 - y - z(y))}{(1 - y)(z(y) + y)}$, z(y) is a non-decreasing function ofy, and  $z(y) \ge z(0) = 1/e$.

Proof

To see that the first part of the lemma holds for $y \in (0, 1/2]$, note that as z(y) increases from y to 1/2, the left the hand side of the inequality defining z(y) continuously decreases from 0 to $2y - 1$, and the right hand side of this inequality continuously increases to $\ln \left( \frac{1/2}{1 - y}\right)$, which is less than 0 and at least $- \ln (2 - 2y) \ge -[e^{\ln (2 - 2y)} - 1] = -(1 - 2y) = 2y - 1$. To see that the first part of the lemma holds for $y = 0$ as well, note that by definition z(0) satisfies $-1 = \ln (z(0))$, and $1/e \in [0, 1/2]$ is the only number satisfying this equality.

Up to this point we have proved that the function z(y) is well defined. Our next objective is to show that it is also differentiable. To do that, consider the function $G(y, z) = y / z - 1 - \ln \left( \frac{z}{1 - y}\right)$ and a point (y, z) obeying $y \in [0, 1/2]$ and $z = z(y)$. Clearly, $G(y, z) = 0$, and the derivative dG(y, z)/dz obeys at this point

$$\begin{aligned} \frac{dG(y, z)}{dz} = -\frac{y}{z^2} - \frac{1 / (1 - y)}{z/(1 - y)} = -\frac{y}{z^2} - \frac{1}{z} < 0 . \end{aligned}$$

Hence, by the implicit function theorem, z(y) is a continuous differentiable function of y for this range of y.

Given the knowledge that the derivative of z(y) exists within the range $y \in [0, 1/2]$, we can now calculate it by taking the derivative with respect to y of both sides of the inequality defining z(y). Doing so yields

$$\begin{aligned} \frac{z(y) - y \cdot z'(y)}{z^2(y)} = \frac{[z'(y) \cdot (1 - y) + z(y)] / (1 - y)^2}{z(y) / (1 - y)} , \end{aligned}$$

and solving for $z'(y)$ gives

$$\begin{aligned} z'(y) = \frac{z(y) (1-y-z(y))}{(1-y)(z(y) + y)} \ge 0 , \end{aligned}$$

where the inequality holds since the first part of the lemma shows that $z(y) \in (0, 1/2] \subseteq (0,1-y]$. Note that, since the derivative $z'(y)$ is non-negative, we get that z(y) is a non-decreasing function, as promised. $\square$

Lemma 15

$\min \{p(x):x \in [0,1/3]\}=\frac{3-\ln 4}{4-\ln 4}$.

To make our calculations easier, it is useful to define $y=\frac{x}{1-x}$, which implies $x=\frac{y}{1+y}$ and $1-x=\frac{1}{1+y}$, and therefore, also

$$\begin{aligned} p(x)=x+(1 -x) \cdot \left( 1-z\left( \frac{x}{1 - x}\right) \right) =\frac{y}{1+y}+\frac{1-z(y)}{1+y}=1-\frac{z(y)}{1+y} . \end{aligned}$$

We denote the rightmost side of the last equality by q(y). Since y goes exactly over all the values of the range [0, 1/2] when x grows from 0 to 1/3, $\min \{p(x):x \in [0,1/3]\}=\min \{q(y):y \in [0,1/2]\}$. Thus, to prove the lemma it suffices to show that $\min \{q(y):y \in [0,1/2]\} = \frac{3-\ln 4}{4-\ln 4}$, which we do in the rest of this proof.

Let us now use the shorthand $z = z(y)$. Then,

$$\begin{aligned} q'(y) ={}&-\frac{z'(y)(1+y)-z}{(1+y)^2} = \frac{z - \frac{z(1 + y) (1 -y - z)}{(1 - y)(z + y)}}{(1+y)^2} \\ ={}&\frac{z}{(1 + y)^2} \cdot \left( 1 - \frac{(1 + y) (1 - y - z)}{(1 - y)(z + y)}\right) \\ ={}&\frac{z}{(1 + y)^2} \cdot \left( \frac{2(z + y) - (1 + y)}{(1 -y)(z + y)}\right) \\ ={}&\frac{z}{(1 + y)^2(1 - y)(z + y)} \cdot (2z + y- 1) , \end{aligned}$$

where the second equality follow from Lemma 4. Since Lemma 4 guarantees that $z \ge z(0) = 1/e$, the sign of $q'(y)$ (within the range [0, 1/2]) is equal to the sign of $2z + y - 1$, and the last expression is a (strictly) increasing function of y since $z = z(y)$ is a non-decreasing function of y by Lemma 4. Hence, q(y) is a convex function within this range which takes its minimum value at the point $y_0$ in which $q'(y_0) = 0$ (assuming there is such a point). In other words, to complete the proof of the lemma it remains to show that there is a point $y_0 \in [0, 1/2]$ obeying $q'(y_0) = 0$ and $q(y_0) = \frac{3-\ln 4}{4-\ln 4}$.

We show that $y_0 = \frac{1-\ln 2}{3-\ln 2} \in [0, 1/3]$ has the above mentioned properties. According to Lemma 4, $z(y_0)$ is the sole value in $[y_0, 1/2]$ obeying the inequality

$$\begin{aligned} \frac{y_0}{z(y_0)} - 1 = \ln \left( \frac{z(y_0)}{1 - y_0}\right) , \end{aligned}$$

and one can verify that $z(y_0) = (1 - y_0)/2 \in [y_0, 1/2]$ obeys this inequality. Hence,

$$\begin{aligned} q(y_0) ={}&1-\frac{z(y_0)}{1+y_0} = 1-\frac{(1 - y_0)/2}{1+y_0} = 1 - \frac{1 - (1 - \ln 2) / (3 - \ln 2)}{2(1 + (1 - \ln 2) / (3 - \ln 2))}\\ ={}&1 - \frac{(3 - \ln 2) - (1 - \ln 2)}{2((3 - \ln 2) + (1 - \ln 2))} = 1 - \frac{1}{4 - 2\ln 2} = \frac{3 - \ln 4}{4 - \ln 4} \end{aligned}$$

and

$$\begin{aligned} q'(y_0) ={}&\frac{z(y_0)}{(1 + y_0)^2(1 - y_0)(z(y_0) + y_0)} \cdot (2z(y_0) + y_0 - 1)\\ ={}&\frac{z(y_0)}{(1 + y_0)^2(1 - y_0)(z(y_0) + y_0)} \cdot ((1 - y_0) + y_0 - 1) = 0 . \end{aligned}$$

$\square$

Appendix B: Code

This appendix includes the code used to get a lower bound on the expression

$$\begin{aligned} \min \left\{ \rho , \min _{(\tilde{c}(r), \tilde{c}(r')) \in \mathscr {C}(\delta )} m(\rho , \tilde{c}(r), \tilde{c}(r'), \delta ^{-1})\right\} \end{aligned}$$

mentioned in Proposition 1. This code consists of two parts. The first part, given in Sect. 1, describes a structure used to represent non-negative rational numbers. The second part, given in Sect. 1, describes the main program, which uses the structure defined by the first part.

1.1 B.1 Structure Representing a Non-negative Rational Number

In this section we describe an immutable structure used to represent a non-negative rational number. The public member functions of this structure support various operations on such numbers. Some of these operations return the output one would expect, while others return a lower bound on this output. Member functions of the last kind are denoted by the prefix LB.

1.2 B.2 Main Program

In this section we give the main program used to evaluate the expression given in Proposition 1. This program uses the structure defined in Section 1.