# Improved Algorithms for Inferring the Minimum Mosaic of a Set of Recombinants Yufeng Wu and Dan Gusfield UC Davis CPM 2007.

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Improved Algorithms for Inferring the Minimum Mosaic of a Set of Recombinants Yufeng Wu and Dan Gusfield UC Davis CPM 2007

2 Recombination Recombination: one of the principle genetic forces shaping sequence variations within species. Two equal length sequences generate a new equal length sequence. 110001111111001 000110000001111 Prefix Suffix 110000000001111 Breakpoint

Founders and Mosaic Current sequences are descendents of a small number of founders. –A current sequence is composed of blocks from the founders, due to recombination. –No mutations since formation of founders. 000000 111111 000000 111111 001111 000000 111111 001111 111100 Breakpoint Founders Sampled sequences in current population 000000 001111 111100 011100 Mosaic 000000 001111 111100 011100

4 The Minimum Mosaic Problem Given a set of aligned binary sequences in the current population and assume the number of founders is known to be K f, find set of founders and the mosaic with the minimum number of breakpoints. 1101101 1010001 0111111 0110100 1100011 Assume K f =3 1101101 1010001 0111111 0110100 1100011 1101111 1010001 0110100 Three Founders Four breakpoints: minimum for all possible three founders

5 Status of the Minimum Mosaic Problem First studied by E. Ukkonen (WABI 2002). –Dynamic programming method. Not practical when the number of rows is more than 20 and K f >2. No polynomial-time algorithm was known even when K f is small. No NP-completeness result is known. Our results: –A simple polynomial-time algorithm for K f = 2 case. –Exact and practical method for data of medium range for K f  3.

The Two-Founder Case 110111101 100100101 010111111 010101100 110000111 Key: at columns 1 and 2, the founders are either or. There are two rows with 00/11, and three rows with 01/10. So, at least two breakpoints between columns 1 and 2 with founders as. 1111101 1010001 0111111 0110100 1100011 Founders Remove uniform columns 0?1?0?1? 1111101 1010001 0111111 0110100 1100011 0? 1? 01100110  2 breakpoints between c1 and c2 00 11  2 breakpoints between c2 and c3 Study pairs of neighboring columns

The Two-Founder Case (Cont.) No matter which founder states are chosen for previous column, we can always choose the needed founders for current column. 2 22122 # breakpoints between two columns Local founders c1c2 c3c4 c5 c6 c7 At least 2 + 2 + 2 +1 +2 +2 = 11 breakpoints needed. On the other hand, we can construct two founders that use the same local optimal founders, and thus 11 breakpoints is global optimum. Founders 0101 1010 1010 1010 1010

8 Three or More Founders: Assuming Known Founders 1101101 1010001 0111111 0110100 1100011 Three Founders 1101101 1101111 1010001 0110100 With known founders, can minimize breakpoints for each sequence, and thus also minimize the total number of breakpoints. For each input sequence, starting from the left, insert a breakpoint at the end of longest segments matching one founder. Founder mapping: at each position c in any input sequence s, which founder s[c] takes its value from. Breakpoint! Input Sequences 1101101 Founder 1Founder 2 Founder Mapping

Enumerating Founders for Founder- Unknown Case In reality, founders are not known. A straightforward way is to simply enumerate all possible sets of founders, and then run the previous method to find the minimum mosaic. 100100 001001 011011 101101 110110 010010 At each column, there are 2 kf –2 founder settings. Let m be the number of columns, fully enumerate all possible sets of founders takes  (2 m*kf ) time. Infeasible when m or K f is large. Need more ideas to develop a practical method. First, we do the enumeration in the form of search paths in a search tree.

Search Paths and Search Tree It works but exponential blowup of the search paths! Obvious idea to reduce search space: branch and bound (compute a lower bound and …). But we found a different idea is more useful. 001001 0 Founder setting at column one Num of tot. breakpoints up to current column 011011 0 c1 c3 001001 2 010010 1 c2 001001 001001 100100 0 001001 011011 1 001001 101101 0 001001 110110 2 001001 010010 5 On-line computation: Compute partial solution up to the current column for speedup. 010010 001001 Founder settings up to column 3 The founder-known method can be run with partially-known founders! Assume three founders

Dropping Search Paths that are Beaten by Another Search Path 001001 0 011011 6 P1 and P2 are two search paths up to column 2. Can we say P1 is better than P2? Not really, because maybe P2 can lead to fewer breakpoints later on. But, suppose the number of input sequences is 5. We can then say P1 beats P2 (and so drop P2). Why? P1 P2 <=39 <= 5 bkpts>= 0 bkpts An optimal search path following P2 40 Assume three founders 011011 101101 Founder Config.

A More Powerful Beaten Rule 001001 0 011011 4 P1 P2 Still five input rows. Now can not say P1 beats P2. But remember we have founder matching… 5 4 3 2 1 MatchRows 5 4 3 2 1 MatchRows So P1 beats P2 since at most 3 rows need extra breakpoints to get onto a path from P2, and P2 uses 4 more breakpoints than P1. These two rows have the same founder mappings. P1Row2 P2Row2 No extra breakpoints at rows 2 and 4 If no bkpt at P2, no bkpt at p1 too. 011011 001001

How Practical Is Our Method? Source of data and image: UNC Chapel Hill Five founders 20 rows, 36 columns UNC’s heuristic solution: 54 breakpoints Enumerating 2 180 founder states is impossible! Our method takes 5 minutes to find the optimal solutions: 53 breakpoints. It is also practical for 50x50 matrix with four founders.

14 Open Problems and Software Is the minimum mosaic problem NP- complete? Is there a polynomial-time algorithm for the minimum mosaic problem for small (say three to ten) number of founders? Software available at: http://wwwcsif.cs.ucdavis.edu/~wuyu Thank you.

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