a simpler 1.5-approximation algorithm for sorting by transpositions tzvika hartman weizmann...
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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
BA DC DBCAt
BC
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.
Пn+1Пn П1
П1 . . . Пn .
.. .
.
• 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
914
5
6
34
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. codd() : # of odd cycles in G().
Define Δcodd(,t) = codd(t · ) – codd()
Lemma [BP98]: t and , Δcodd(,t) {0, 2, -2}.
1
10
2
8 7
914
5
6
34
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
3 6
5
21
4
3 6
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
36
58
7
10
912
11
2
1
4
36
58
7
10
912
• # 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-codd()]/2.
Goal : increase # of odd cycles in G. t is a k-transposition if Δcodd(,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-cycle Oriented 3-cycle
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(0,2,2)-Sequence of Transpositions
A (0,2,2)-sequence is a sequence of 3 transpositions: the 1st 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!