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A Simpler 1.5-Approximation Algorithm for Sorting by Transpositions Tzvika Hartman Weizmann Institute

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Genome Rearrangements During evolution, genomes undergo large- scale mutations which change gene order (reversals, transpositions, translocations). Given 2 genomes, GR algs infer the most economical sequence of rearrangement events which transform one genome into the other.

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Genome Rearrangements Model Chromosomes are viewed as ordered lists of genes. Unichromosomal genome, every gene appears once. Genomes are represented by unsigned permutations fo genes. Circular genomes (e.g., bacteria & mitochondria) are represented by circular perms.

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Sorting by Transpositions A transposition exchanges between 2 consecutive segments of a perm. Example : 1 2 3 4 5 6 7 8 9 1 2 6 7 3 4 5 8 9 Sorting by transpositions : finding a shortest sequence of transpositions which sorts the perm.

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Previous work 1.5-approximation algs for sorting by transpositions [BafnaPevzner98, Christie99]. An alg that sorts every perm of size n in at most 2n/3 transpositions [Erikkson et al 01]. Complexity of the problem is still open.

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Main Results 1. The problem of sorting circular permutations by transpositions is equivalent to sorting linear perms by transpositions. 2. A new and simple 1.5-approximation alg for sorting by transpositions, which runs in quadratic time.

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Linear & Circular Perms A B A C t BADCDBCA t B C Linear transposition : Circular transposition : Circular transpositions can be represented by exchanging any 2 of the 3 segments. A transposition “cuts” the perm at 3 points.

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Linear & Circular Equivalence Thm : Sorting linear perms by transpositions is computationally equivalent to sorting circular perms. Pf sketch: Circularize linear perm by adding an n+1 element and closing the circle. П n+1 ПnПn П1П1 П 1... П n..... Every linear transposition is equivalent to a circular transposition that exchanges the 2 segments that do not include n+1.

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Breakpoint Graph [BafnaPevzner98] Perm : ( 1 6 5 4 7 3 2 ) Replace each element j by 2j-1,2j: = (1 2 11 12 9 10 7 8 13 14 5 6 3 4) Circular Breakpoint graph G( ): 1 10 2 8 7 9 14 5 6 3 4 11 13 12 Vertex for every element. Black edges ( 2i, 2i+1 ) Grey edges (2i, 2i+1)

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Breakpoint Graph (Cont.) Unique decomposition into cycles. c odd ( ) : # of odd cycles in G( ). Define Δc odd ( ,t) = c odd (t · ) – c odd ( ) Lemma [BP98]: t and , Δc odd ( ,t) {0, 2, -2}. 1 10 2 8 7 9 14 5 6 3 4 11 13 12

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Effect on Graph : Example Perm: (1 3 2). After extension: (1 2 5 6 3 4). Breakpoint graph: 1 4 36 5 2 1 4 36 5 2 # of cycles increased by 2

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Effect on Graph : Example Perm : (6 5 4 3 2 1). After extension : (11 12 9 10 7 8 5 6 3 4 1 2). Breakpoint graph : 11 2 1 4 3 6 5 8 7 10 9 12 11 2 1 4 3 6 5 8 7 10 9 12 # of cycles remains 2

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Breakpoint Graph (Cont.) Max # of odd cycles, n, is in the id perm, thus: Lower bound [BP98]: For all , d( ) [n-c odd ( )]/2. Goal : increase # of odd cycles in G. t is a k-transposition if Δc odd ( ,t) = k. A cycle that admits a 2-transposition is oriented.

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Simple Permutations A perm is simple if its breakpoint graph contains only short ( 3) cycles. The theory is much simpler for simple perms. Thm : Every perm can be transformed into a simple one, while maintaining the lower bound. Moreover, the sorting sequence can be mimicked. Corr : We can focus only on simple perms.

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3 - Cycles 2 possible configurations of 3-cycles: Non-oriented 3-cycleOriented 3-cycle

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(0,2,2)-Sequence of Transpositions A (0,2,2)-sequence is a sequence of 3 transpositions: the 1 st is a 0-transposition and the next two are 2-transpositions. A series of (0,2,2)-sequences preserves a 1.5 approximation ratio. Throughout the alg, we show that there is always a 2-transposition or a (0,2,2)- sequence.

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Interleaving Cycles 2 cycles interleave if their black edges appear alternatively along the circle. Lemma : If G contains 2 interleaving 3-cycles, then a (0,2,2)-sequence.

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Shattered Cycles Lemma : If G contains a shattered cycle, then a (0,2,2)-sequence. 2 pairs of black edges intersect if they appear alternatively along the circle. Cycle A is shattered by cycles B and C if every pair of black edges in A intersects with a pair in B or with a pair in C.

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Shattered Cycles (Cont.) Lemma : If G contains no 2-cycles, no oriented cycles and no interleaving cycles, then a shattered cycle.

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The Algorithm While G contains a 2-cycle, apply a 2-transposition [Christie99]. If G contains an oriented 3-cycle, apply a 2- transposition on it. If G contains a pair of interleaving 3-cycles, apply a (0,2,2)-sequence. If G contains a shattered unoriented 3-cycle, apply a (0,2,2)-sequence. Repeat until perm is sorted.

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Conclusions We introduced 2 new ideas which simplify the theory and the alg: 1. Working with circular perms simplifies the case analysis. 2. Simple perms avoid the complication of dealing with long cycles (similarly to the HP theory for sorting by reversals).

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Open Problems Complexity of sorting by transpositions. Models which allow several rearrangement operations, such as trans-reversals, reversals and translocations (both signed & unsigned).

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Acknowledgements Ron Shamir.

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Thank you !

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