Log In Sign Up

A Note on Monotone Submodular Maximization with Cardinality Constraint

We show that for the cardinality constrained monotone submodular maximization problem, there exists a (1-1/e-ε)-approximate deterministic algorithm with linear query complexity, which performs O(n/ε) queries in total.


page 1

page 2

page 3


On the Complexity of Dynamic Submodular Maximization

We study dynamic algorithms for the problem of maximizing a monotone sub...

Towards Practical Constrained Monotone Submodular Maximization

We design new algorithms for maximizing a monotone non-negative submodul...

A polynomial lower bound on adaptive complexity of submodular maximization

In large-data applications, it is desirable to design algorithms with a ...

Altitude Terrain Guarding and Guarding Uni-Monotone Polygons

We show that the problem of guarding an x-monotone terrain from an altit...

Efficient Algorithms for Monotone Non-Submodular Maximization with Partition Matroid Constraint

In this work, we study the problem of monotone non-submodular maximizati...

Sensitivity Analysis of Submodular Function Maximization

We study the recently introduced idea of worst-case sensitivity for mono...

Non-monotone Submodular Maximization in Exponentially Fewer Iterations

In this paper we consider parallelization for applications whose objecti...

1 Introduction

A set function defined on ground of size is submodular, if inequality holds for any two subsets . It is monotone non-decreasing if holds for any two sets . In this paper we consider the classic problem of maximizing a monotone submodular function subject to a cardinality constraint , i.e.,

Our result is stated as following.

Theorem 1.

There exists an -approximate algorithm for the cardinality constrained monotone submodular maximization problem, which makes queries in total.

This result mainly relies on the algorithm in [1], together with that in [2].

2 Algorithm with linear query complexity

Similar to [2], Algorithm consists of two phases—a preprocessing procedure and a refined threshold decreasing procedure. We use the algorithm in [1] as the first phase, while the second phase is the same as that in [2]. To reduce the running time of the implementation, we maintain sets , where stores the elements with weight belonging to interval , elements stored in have a weight less than . Instead of considering the element in the candidate solution with minimum weight, each time we select an arbitrary element in , where represents the largest index such that is non-empty. It can be verified that this implementation has a running time of . We do not attempt to obtain the most efficient implementation of the algorithm, instead we focus on the query complexity.

1 Initialization: , ,
2 for  do
3       ,
4        , an arbitrary element in
5        if  then
6              ,
7       else if  then
8              ,
9               ,
12 while  do
13       for  do
14               if  and  then
Algorithm 1 A Linear Query Complexity Algorithm for Cardinality Constraint
Lemma 2 ([1]).

For any set , we use to denote the total weights of elements in . Then


The proof of this lemma follows from the charging scheme in [1]. We simplify the arguments of [1] in the context of cardinality constraint, and slightly modify the analysis to show the correctness of preprocessing phase. Let be the value of before considering element . For each , the element is charged if it is added into . Otherwise holds for , we charge to an uncharged element . We remark that there always exists an uncharged element in , since there are elements in and hence there are at most charged elements in . We redistribute the charge as follows. When element is added into while is removed, the charge on is transferred to .

Note that for any , initially it is charged at most , the amount of charge obtained from transfer is no more than . In addition, the element will not be charged again after it is added into the candidate set. Hence we have

The proof is complete. ∎

Lemma 3 ([1]).

For set obtained in the first phase,


According to Lemma in [1], we know that


In addition, we have the following inequality from Lemma in [1],


Combining (1)-(2) with Lemma 2, the proof is complete. ∎

Theorem 4 ([2]).

For set returned by Algorithm 1, we have


The proof is similar to the proof of Theorem 1 in [2]. ∎


The author would like to thank the anonymous reviewer for making this note possible.