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Multi-Pass Streaming Algorithms for Monotone Submodular Function Maximization

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Abstract

We consider maximizing a monotone submodular function under a cardinality constraint or a knapsack constraint in the streaming setting. In particular, the elements arrive sequentially and at any point of time, the algorithm has access to only a small fraction of the data stored in primary memory. We propose the following streaming algorithms taking O(ε− 1) passes: (1) a (1 − e− 1 − ε)-approximation algorithm for the cardinality-constrained problem, (2) a (0.5 − ε)-approximation algorithm for the knapsack-constrained problem. Both of our algorithms run deterministically in O∗(n) time, using O∗(K) space, where n is the size of the ground set and K is the size of the knapsack. Here the term O∗ hides a polynomial of \(\log K\) and ε− 1. Our streaming algorithms can also be used as fast approximation algorithms. In particular, for the cardinality-constrained problem, our algorithm takes \(O(n\varepsilon ^{-1} \log (\varepsilon ^{-1}\log K) )\) time, improving on the algorithm of Badanidiyuru and Vondrák that takes \(O(n \varepsilon ^{-1} \log (\varepsilon ^{-1} K) )\) time.

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Notes

  1. Independently of our work, Norouzi-Fard et al. [43] proposed another algorithm with the same performance as the second one in Theorem 1

  2. Recently, it is shown in [44] that it suffices to enumerate all the sets of at most two items.

  3. This theorem is essentially a rephrasing of Theorem 5.

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Acknowledgements

The authors thank the referees for their valuable comments on this manuscript.

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Authors and Affiliations

  1. CNRS, École Normale Supérieure, Paris, France

    Chien-Chung Huang

  2. Keio University, Yokohama, Japan

    Naonori Kakimura

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The first author was supported by French National Research Agency (ANR), with grant ANR-19-CE48-0016 and ANR-18-CE40-0025-01.The second author was supported by JSPS KAKENHI Grant Numbers JP17K00028, JP18H05291, 20K21834, and 20H05795.

Appendix A: Proof of Lemma 7

Appendix A: Proof of Lemma 7

We discuss how to obtain a (0.5 − O(ε))-approximation when c1 + c2 is almost 1.

Lemma 22

Suppose that \(f(o_{1}+o_{2})\geq v^{\prime }\). We can find a set S using two passes and O(ε− 1K) space such that |S| = 2 and \(f(S)\geq \left (\frac {2}{3}-\varepsilon \right )v^{\prime }\).

We begin by reviewing the algorithmFootnote 3 in [27].

Theorem 8

Let \(E_{\mathrm {R}} \subseteq E\) be a subset of the ground set (and we call ER red items). Let \(X\subseteq E\) such that v ≤ f(X) ≤ (1 + ε)v. Assume that there exists x ∈ X ∩ ER such that \(\underline {\tau } v \leq f(x) \leq \overline {\tau }v\). Then we can find a set \(Y \subseteq E_{\mathrm {R}}\) of red items, in one pass and O(n) time, with \(|Y| = O(\log _{1+\varepsilon } \frac {\overline {\tau }}{ \underline {\tau }})\) such that some item e∗ in Y satisfies f(X − x + e∗) ≥ (2/3 − O(ε))v.

Proof of Lemma 22

For each \(t=1,2,\dots , K/2\), define Et = {e ∈ E∣t ≤ c(e) ≤ K − t} as the red items. The critical thing to observe is that, if t ≤ c(o2), we see o1 ∈ Et.

The above observation suggests the following implementation. In the first pass, for each set Et, apply Theorem 8 to collect a set \(X_{t} \subseteq E_{t}\) (apparently we can set \(\overline {\tau }= 2/3\) and \(\underline {\tau }=1/3\)). Since \(|X_{t}|=O(\log _{1+\varepsilon }2)=O(\varepsilon ^{-1})\), it takes O(ε− 1K) space and O(n) time in total. Then it follows from Theorem 8 that, for each t with t ≤ c(o2), there exists e∗∈ Xt such that \(f(o_{2}+e^{\ast })\geq (2/3- O(\varepsilon )) v^{\prime }\) and c(e∗) ≤ K − c(o2). In the second pass, for each item e in E, check whether there exists \(e^{\prime }\) in Xc(e) such that \(c(e+e^{\prime })\leq K\) and \(f(e + e^{\prime }) \geq (2/3-O(\varepsilon ))v^{\prime }\). It follows that there exists at least one pair of e and \(e^{\prime }\) satisfying the condition. The second pass also takes O(ε− 1K) space as we keep Xt’s. Since |Xt| = O(ε− 1), the second phase takes O(ε− 1n) time. □

Suppose that v ≤ f(OPT) ≤ (1 + ε)v. If f(o1 + o2) ≥ 0.75v, then we are done using Lemma 22. So assume otherwise, meaning that f(OPT − o1 − o2) ≥ 0.25v. Notice that we can also assume that f(OPT − o1) ≥ 0.5v. Now consider two possibilities.

Claim

If \(c_{1} \geq 1- \sqrt {\varepsilon }\), then we can find a set S in O(ε− 1) passes and O(K) space such that c(S) ≤ K and f(S) ≥ (0.5 − O(ε))v.

Proof

Since \(c_{1} \geq 1- \sqrt {\varepsilon }\), we have \(c(\text {OPT} - o_{1}) \leq \sqrt {\varepsilon } K\). Consider the problem (1) to approximate OPT − o1. Then the largest item in OPT − o1 is c(o2) which is at most \(\sqrt {\varepsilon }K\). By Corollary 1, Simple\((\mathcal {I};0.5v, \sqrt {\varepsilon }K)\) can obtain a set S satisfying that

$$ f(S)\geq 0.5 \left( 1-e^{-\frac{1 - \sqrt{\varepsilon} }{\sqrt{\varepsilon}}} - O(\varepsilon)\right)v\geq (0.5 - O(\varepsilon))v, $$

where the last inequality follows because \(e^{-\frac {1 - \sqrt {\varepsilon } }{\sqrt {\varepsilon }}} \leq \varepsilon \) when ε ≤ 1. □

Claim

If \(c_{1} < 1- \sqrt {\varepsilon }\), then we can find a set S in O(ε− 1) passes and O(K) space such that c(S) ≤ K and f(S) ≥ (0.5 − O(ε))v.

Proof

Consider the problem:

$$ \text{maximize\ \ }f(S) \quad \text{subject to \ } c(S)\leq \sqrt{\varepsilon} K,\quad S\subseteq E,\\ $$

to approximate OPT − o1 − o1. Let \(\mathcal {I}^{\prime }\) be the corresponding instance. Since f(OPT − o1 − o2) ≥ 0.25v and c(OPT − o1 − o2) ≤ εK, Corollary 1 implies that Simple\((\mathcal {I}^{\prime };\) 0.25v,εK) can obtain a set Y satisfying that

$$ f(Y)\geq 0.25 \left( 1-e^{-\frac{\sqrt{\varepsilon} - \varepsilon}{\varepsilon}} - O(\varepsilon)\right)v\geq (0.25 - O(\varepsilon))v, $$

since the largest item in OPT − o1 − o2 has size at most ε. After taking the set Y, we still have space for packing either o1 or o2, since \(c(Y)\leq \sqrt {\varepsilon } K <K-c(o_{1})\).

Define g := f(⋅∣Y ). If some element e satisfies c(Y ) + c(e) ≤ K and f(Y + e) ≥ 0.5v, then we are done. Thus we may assume that no such element exists, implying that f(Y + oℓ) < 0.5v for ℓ = 1, 2. Hence it holds that

$$ g(\text{OPT} -o_{1})\geq g(\text{OPT})-g(o_{1})\geq \left( f(\text{OPT})-f(Y) \right) - \left( f(Y+o_{1})-f(Y)\right)\geq 0.5 v, $$

This implies that

$$ \begin{array}{@{}rcl@{}} g(\text{OPT}-o_{1}-o_{2}) & \geq& g(\text{OPT} - o_{1}) - g(o_{2}) \\ & \geq& 0.5 v - (f(Y+o_{1})-f(Y)) \geq f(Y)\geq (0.25 - O(\varepsilon))v. \end{array} $$

Consider the problem:

$$ \text{maximize\ \ }g(S) \quad \text{subject to \ } c(S)\leq K - c(Y),\quad S\subseteq E,\\ $$

to approximate OPT − o1 − o1. Denote by \(\mathcal {I}^{\prime \prime }\) the corresponding instance. Since \(K-c(Y)\geq (1-\sqrt {\varepsilon })K\) and g(OPT − o1 − o2) ≥ (0.25 − O(ε))v, Corollary 1 implies that Simple\((\mathcal {I}^{\prime \prime }; (0.25 - O(\varepsilon ))v, \varepsilon K)\) can obtain a set S satisfying that

$$ f(S)\geq (0.25- O(\varepsilon)) \left( 1-e^{-\frac{1-\sqrt{\varepsilon} - \varepsilon}{\varepsilon}} - O(\varepsilon)\right)v\geq (0.25 - O(\varepsilon))v. $$

Therefore, Y ∪ S satisfies that c(Y ∪ S) ≤ K and

$$ f(Y\cup S) = f(Y) + g(S)\geq (0.5 - O(\varepsilon))v. $$

□

For a given v, the above can be done in O(ε− 1K) space using O(ε− 1) passes. The total running time is O(nε− 1). This completes the proof of Lemma 7.

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Huang, CC., Kakimura, N. Multi-Pass Streaming Algorithms for Monotone Submodular Function Maximization. Theory Comput Syst 66, 354–394 (2022). https://doi.org/10.1007/s00224-021-10065-6

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