Longest Increasing Subsequences in Windows Based o…….ppt

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Longest Increasing Subsequences in Windows Based o…….ppt

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* * Future Works… Questions? * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Longest Increasing Subsequences in WindowsBased on Canonical Antichain Partition Erdong Chen (Joint work with Linji Yang Hao Yuan) Shanghai Jiao Tong Univ. Outline Problem Definition Canonical Antichain Partition Sweep Algorithm Complexity Analysis Conclusion Longest Increasing Subsequence (LIS) 6 9 8 2 3 5 1 4 7 2 3 4 7 2 3 5 7 Input sequence: All Longest Increasing Subsequences: LIS in a Window 6 9 8 2 3 5 1 4 7 6 9 8 The length of LIS within the window is 2 6 9 Sequence a Window Longest Increasing Sequence in A Set of Variable-size Windows (LISSET) 6 9 8 2 3 5 1 4 7 8 9 6 5 1 4 3 5 3 2 8 9 5 1 4 7 3 2 8 9 6 9 6 5 3 5 3 2 5 7 3 2 Length of LIS = 2 Length of LIS = 3 Length of LIS = 2 Length of LIS = 4 OUTPUT = 11 + = + + Longest Increasing Sequence in A Set of Variable-size Windows (LISSET) 6 9 8 2 3 5 1 4 7 9 6 5 3 5 3 2 5 7 3 2 Length of LIS = 2 Length of LIS = 3 Length of LIS = 2 Length of LIS = 4 OUTPUT = 11 + = + + Related Works Knuth proposed an O(n log n) algorithm for LIS problem Fredman proved an Ω(n log n) lower bound under the decision tree model An O(n log log n) algorithm is possible by using van Emde Boas tree on a permutation. Related Works (Cont.) Longest Increasing Subsequences in Sliding Windows (LISW problem) (by Michael H. Albert et al), Time Complexity O(OUTPUT + n log log n) We called it Longest Increasing Subsequences in Fixed-size windows LISW Problem 6 9 8 2 3 5 8 9 6 2 8 9 3 2 8 5 3 2 3 5 6 6 9 OUTPUT = 14 + = + + + + + + 5 n = 6 w = 3 Length of LIS = 1 Length of LIS = 2 Length of LIS = 2 Length of LIS = 1 Length of LIS = 2 Length of LIS = 3 Length of LIS = 2 Length of LIS = 1 9 6 9 3 2 5 3 2 5 3 5 6 9 6 n+w-1 = 8 windows Our Contribution A algorithm for the generalized problem LISSET To solve the sub case LISW problem, our algorithm runs in O(OUTPUT) time. The best result am