DeepAI

# 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.

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## 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.,

 max{f(S)∣∣S⊆E,|S|≤k}.

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.

###### Lemma 2 ([1]).

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

 w(OPT)≤92w(S)+f(OPT)2.
###### Proof.

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

 ∑e∈OPTw(e)≤∑e∈S[(4w(e)+w02k)+w(e)2]=92w(S)+w02≤92w(S)+f(OPT)2.

The proof is complete. ∎

###### Lemma 3 ([1]).

For set obtained in the first phase,

 f(S)≥f(OPT)13.
###### Proof.

According to Lemma in [1], we know that

 f(OPT)≤2f(S)+w(OPT). (1)

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

 w(S)≤f(S). (2)

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

###### Theorem 4 ([2]).

For set returned by Algorithm 1, we have

 f(U)≥(1−1/e−ε)⋅f(OPT).
###### Proof.

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

## Acknowledgement

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