We study the classic NP-Hard problem of finding the maximum k-set coverage in the data stream model: given a set system of m sets that are subsets of a universe {1,…,n}\documentclass[12pt]{minimal}
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\begin{document}$\{1,\ldots ,n \}$\end{document}, find the k sets that cover the most number of distinct elements. The problem can be approximated up to a factor 1−1/e\documentclass[12pt]{minimal}
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\begin{document}$1-1/e$\end{document} in polynomial time. In the streaming-set model, the sets and their elements are revealed online. The main goal of our work is to design algorithms, with approximation guarantees as close as possible to 1−1/e\documentclass[12pt]{minimal}
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\begin{document}$1-1/e$\end{document}, that use sublinear space o(mn)\documentclass[12pt]{minimal}
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\begin{document}$o(mn)$\end{document}. Our main results are:
Two (1−1/e−𝜖)\documentclass[12pt]{minimal}
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\begin{document}$(1-1/e-\epsilon )$\end{document} approximation algorithms: One uses O(𝜖−1)\documentclass[12pt]{minimal}
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\begin{document}$O(\epsilon ^{-1})$\end{document} passes and Õ(𝜖−2k)\documentclass[12pt]{minimal}
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\begin{document}$\tilde {O}(\epsilon ^{-2} k)$\end{document} space whereas the other uses only a single pass but Õ(𝜖−2m)\documentclass[12pt]{minimal}
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\begin{document}$\tilde {O}(\epsilon ^{-2} m)$\end{document} space. Õ(⋅)\documentclass[12pt]{minimal}
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\begin{document}$\tilde {O}(\cdot )$\end{document} suppresses polylog factors.We show that any approximation factor better than (1−(1−1/k)k)≈1−1/e\documentclass[12pt]{minimal}
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\begin{document}$(1-(1-1/k)^{k})\approx 1-1/e$\end{document} in constant passes requires Ω(m)\documentclass[12pt]{minimal}
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\begin{document}${\Omega }(m)$\end{document} space for constant k even if the algorithm is allowed unbounded processing time. We also demonstrate a single-pass, (1−𝜖)\documentclass[12pt]{minimal}
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\begin{document}$(1-\epsilon )$\end{document} approximation algorithm using Õ𝜖−2m⋅min(k,𝜖−1)\documentclass[12pt]{minimal}
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\begin{document}$\tilde {O}\left (\epsilon ^{-2} m \cdot \min (k,\epsilon ^{-1})\right )$\end{document} space.
