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On the Minimum Common Integer Partition Problem Author: Xin Chen, Lan Liu, Zheng Liu, Tao Jiang Presenter: Lan Liu
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Outline Introduction Problem definitions Biological applications Approximation of 2-MCIP Approximation of k-MCIP Conclusion and future work
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Problem Definitions P(n): given an integer n, a partition is a set of integers, say {n 1,n 2,…, n r }, s.t. i=1 r n i =n. Example: given n=4, {2,2} is a P(4); given n=3, {3} is a P(3). Observation: S = IP(S) Example: given S= {3, 3, 4}, {2,2,3,3} is an IP({3,3,4}). IP(S): given a multiset S= {x 1, , x m }, an integer partition is a disjoint union
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Examples CIP(S 1, S 2, …, S k ): given multisets S 1, S 2, …, S k, a common integer partition of all multisets. Example: given S= {3, 3, 4}, T={2,2,6}, {2,2,3,3} is a CIP(S,T); {1,1,2,2,4} is also a CIP(S,T). # P(100)=190569292 Observation: (1) 9 CIP(S 1,…, S k ) $ S 1 =…= S k (2) |CIP(S 1,…, S k )| ¸ |S i |, i 2 {1,..,k} MCIP(S 1, S 2, , S k ): a common integer partition with the minimum cardinality. Example: {2,2,3,3} is a MCIP(S,T).
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MCIP is NP-hard Subset sum · P MCIP Subset sum problem Given a set of integer x 1, x 2, …, x n, s.t. X= i x i, ask if there is a subset with the sum X/2. Reduction to MCIP problem - Let S={X/2, X/2}, T={x 1, x 2, …, x n }, find MCIP(S,T). - If {x 1, x 2, …, x n } is a MCIP(S,T), the answer is “ yes ” to Subset sum problem; otherwise, the answer is “ no ”.
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Biological Applications(1) The distance between two strings a b c d e f g h i j k h h i j k h e f g a b c d Genetic distance between two genomes a b c d e f g h i j k h h i j k h e f g a b c d Minimum Common Substring Partition
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Biological Applications(2) MCIP is a special case of Minimum Common Substring Partition(MCSP) MCIP(S',T') S'= {x 1, x 2, , x m } T'= {y 1, y 2, , y n } MCSP(S,T) S= T=
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Outline Introduction Approximation of 2-MCIP Positive results Negative results Approximation of k-MCIP Conclusion and future work
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Some basic facts |MCIP(S 1,S 2,…,S k )| ¸ max(|S 1 |,|S 2 |,…,|S k |) |MCIP(S,T)| · m+n-1. |S|=m,|T|=n
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Algorithm Analysis |MCIP(S,T)| · m+n-1 |MCIP(S,T)| ¸ max(m,n) Approximation ratio is 2 An example: S= {3, 3, 4},T={2,2,6}
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Definitions for MRSP(1) Related multisets: if S= T and S,T ;, S and T are a pair of related multisets. Example: Basic related multisets: if there are no S' ½ S and T' ½ T, s.t. S' and T' are related. Example:
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Definitions for MRSP(2) Maximum Related Multiset Partition problem(MRSP) Given S and T, partition them into related submultisets with the maximum cardinality. Observation: If S, T are a pair of basic related multisets, |MRSP|=1.
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MRSP $ 2-MCIP CIP ! RSP For each component, #edges ¸ #vertices –1 Each component is related. |CIP| ¸ m+n-|RSP| |MCIP| ¸ m+n-|RSP| ¸ m+n-|MRSP|
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MRSP $ 2-MCIP CIP Ã RSP For each related submultisets (S', T'), we run Greedy_CIP(S', T'), |CIP (S', T')| · |S'|+ |T'|-1 |CIP| · m+ n- |RSP| |MCIP| · |CIP| · m+ n-| MRSP|
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MRSP $ 2-MCIP |MRSP| = m+n –|MCIP| If S, T are a pair of basic related multisets, |MCIP|= m+n-1, because |MRSP|=1. When m+n ¸ 5, |MCIP| =m+n-1 ¸ 4/5(m+n). A new way to solve MCIP Step1. find MRSP; Step2. for each basic related submultiset, run Greedy_CIP(S', T').
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Approximate 2-MCIP Algorithm intuition: Step 1. find related submulitsets Step 2. set packing Step 3. Greedy-CIP mimic MRSP
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Set Packing Problem(1) Set Packing Given a set of subsets S, find the largest number of mutually disjoint subsets from S?
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Set Packing Problem(2) Bad news - It is NP-hard to find related submultisets of large size. - Set packing itself is NP-hard. Good news We can find the small related submultisets and approximate set packing efficiently.
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Approximate 2-MCIP Main idea: use different strategies for the submultisets with different sizes. The approximation ratio is 5/4. If there are no basic related submultisets with size smaller than 5, 4/5 (m+n) · |MCIP| · m+n-1.
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Outline Introduction Approximation of 2-MCIP –Positive results –Negative results Approximation of k-MCIP Conclusion and future work
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General framework Linear Reduction · L OPT P2 (f(x)) · OPT P1 (x) | OPT P1 (x)- g(x,y)| · |OPT P2 (f(x))-y| If P1 cannot be approximated within some constant ratio c, P2 cannot be approximated by some constant ratio c'.
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Maximum 3DM-3 Problem Definition Given a set D µ X £ Y £ Z, where X, Y and Z are disjoint sets, and each element occurs in at most three triples, find a matching with the maximum cardinality. Known fact Maximum 3DM-3 cannot be approximated within some constant ratio. [Kann91]
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L-reduction(1) f: S={4 i | i 2 X [ Y [ Z } T={4 i1 +4 i2 +4 i3 | (i 1,i 2,i 3 ) 2 D} OPT MCIP · 70*OPT 3DM
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g: - CIP ! RSP |OPT RSP –SOL RSP | · |OPT MCIP – SOL MCIP | - RSP ! 3DM OPT 3DM ¼ OPT RSP Each related submultiset includes at least one triple L-reduction(2) |OPT 3DM –SOL 3DM | · |OPT RSP – SOL RSP |
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L-reduction(3) There is a constant c s.t. Maximum 3DM-3 cannot be approximated within c. There is a L-reduction s.t. OPT MCIP · 70*OPT 3DM |OPT 3DM –SOL 3DM | · |OPT MCIP – SOL MCIP | There is a constant c' s.t. 2-MCIP cannot be approximated within c'. c'<5/4
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Outline Introduction Approximation of 2-MCIP Approximation of k-MCIP Conclusion and future work
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Approximate k-MCIP Run Greedy_CIP(S,T) sequentially on S 1,S 2, …, S k. |MCIP(S 1,S 2,…,S k )| · |S 1 |+|S 2 |+…+|S k | |MCIP(S 1,S 2,…,S k )| ¸ max(|S 1 |,|S 2 |,…,|S k |) Approximation ratio is k We can get a {3k(k-1)}/(3k-2)- approximation by removing the common elements.
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Outline Introduction Approximation of 2-MCIP Approximation of k-MCIP Conclusion and future work
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