Pasaniuc Presentation

8/2/2019 Pasaniuc Presentation

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Sorting by reversals

Bogdan Pasaniuc

Dept. of Computer Science & Engineering


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Overview

Biological background

Definitions

Unsigned Permutations

Approximation Algorithm

Sorting Signed Permutations

Simplified Algorithm


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What is the evolutionary path ?

What is the ancestor chromosome?

Chromosomes lists of genes permutation

Unknown ancestor

Human (X chrom.)

Mouse (X chrom.)


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Mutation at chromosome level Inversion (1 2 3 4 5 6 7) (1 4 3 2 5 6 7)

Transposition (1 2 3 4 5 6 7) (1 5 6 2 3 4 7) Translocation (1 2 3 4 5 6 7) (1 2 3 4 5 2 3 4 6 7)

Inversions Known as reversals

The most common Most often reflect the differences between and within species

What is the minimum number of reversals required to

transform one perm. into another? Reversal distance good approx. for evolutionary

distance


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1 32

4

10

56

8

9

7

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Reversals

Genes (blocks)


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Reversals

1 32

4

10

56

8

9

7

1, 2, 3, 8, 7, 6, 5, 4, 9, 10


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Reversals

1 32

4

10

56

8

9

7Breakpoints

1, 2, 3, 8, 7, 6, 5, 4, 9, 10


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Breakpointa pair of adjacent positions

(i,i+1) s. t. | i - i+1| 1 The values ii+1are not consecutive If | i - i+1| = 1 then the values ii+1are adjacent

Introduce 0= 0 , n+1 = n+1

(0,1) breakpoint if 1 1

(n,n+1) breakpoint if n n

A reversal affects the breakpoints only atits endpoints

Any reversal can remove or induce at most2 bkpts.


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StripA maximal run of increasing (decreasing)elements.

Identity permutation has no breakpoints and anyother permutation has at least one breakpoint

Greedy at each step remove the maximumnumber of breakpoints.

() = number of breakpoints in While(() > 0)

Choose a reversal that removes the maximum number

of breakpoints. (if there is a tie favor the reversal thatleaves a decreasing strip)

Greedy ends in at most () steps.


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Quality of approximation

Lemma1:Every permutation with a decreasing striphas a reversal that removes one breakpoint.

Proof:

consider the decreasing strip with i being the smallest

i -1 must be in an increasing strip that lies to the left or right

Breakpoint that will be removed


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Lemma2: has a decreasing strip. If every reversalthat removes one bkpt leaves a permutation with nodecreasing strips has a reversal that removes

two bkpts.Proof:

consider the decreasing strip with i being the smallest

increasing strip must be to the left. i

consider the decreasing strip with jbeing the largest

decreasing strip containing j+1must be to the right.j


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Fact 1: i andj must overlap

j must lie in i if it doesnt then oi has the

decreasing strip that contains j i must lie in jif it doesnt then oj has the

decreasing strip that contains i


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Fact 2. i =jIf i -j 0 then

- if i -j contains an increasing stripoj has a decreasing

strip- if i -j contains an decreasing stripoi has a decreasing

strip

Then =i = removes 2 breakpoints.


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Lemma 3:Greedy solves a permutation with adecreasing strip in at most() 1 reversals

Obs:

if

i has no decreasing strip

at step i-1 the reversalremoved 2 bkpts.

we can use one reversal to create a decr. strip existsa reversal that removes at least one bkpt

Theorem1: Greedy sorts every permutation inat most() reversals.

If has a decreasing strip at most () -1reversals

If has no decreasing strip

every reversal inducesa decreasing strip after one step we can apply

lemma3 at most () reversals


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Corollary:Greedy is a 2-approximation algorithm

Every reversal removes at most 2 bkpts OPT() () /2 Greedy() /2

Greedy() 2* OPT() .

Runtime#of steps O(n).

At each step we need to analyze reversalsO(n2).

Total runtime = O(n

3

). analyze only reversals that remove bkpts O(n2).

2

n


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Signed permutations:

reversals change the sign:(1,2,3,4,5,6,7,8,9,10)

(1,2,3,-8,-7,-6,-5,-4,9,10)

Problem:

Given a signed perm., find the minimum lengthseries of reversals that transforms it into the

identity perm.

polynomial algorithm (Hannenhalli&Pevzner 95)

relies on several intermediary constructions

these constructions have been simplified

first completely elementary treatment of the problem(Bergeron 05)


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Oriented pair a pair of consecutive integers withdifferent signs

(0,3,1,6,5,-2,4,7) o.p. (3,-2) and (1,-2).

o.p. reversals that create consecutive integers

(3,-2) : (0,3,1,6,5,-2,4,7)

(0,3,2,-5,-6,-1,4,7)(1,-2) : (0,3,1,6,5,-2,4,7) (0,3,-5,-6,-1,-2,4,7)

Oriented reversal:reversal that creates consecutive

integers Score of a reversal:# of oriented pairs it creates.


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Algorithm1:As long as has an oriented pair, choose theoriented reversal that has the maximal score.

output will be a permutation with positive elements.

0 and n+1 are positive;

if there is a negative element there exists an o.p.

Claim1: If Alg1 applies k reversals to , yielding thend() = d() + k.


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Sorting positive perms.:

- signed perm. with positive elements

- circular order: 0 successor of n+1.

- reduced if it does not contain consecutive elements.

framed interval in : i j+1j+2j+k-1i+k

s.t. i < j+1j+2 j+k-1 < i+k

(0 2 5 4 3 6 1 7 )

hurdle a framed int. that contains no shorter framed int.

(0 2 5 4 3 6 1 7 )

(0 2 5 4 3 6 1 7 )

(0 2 5 4 3 6 1 7 )


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Idea: create oriented pairs and then apply Algorithm1

Operations on Hurdles:

Hurdle Cutting: i j+1j+2i+1j+k-1i+k

(0 1 4 3 2 5) (0 -3 -4 -1 2 5)

Hurdle Merging: i i+k i ii+k

(0 2 5 4 3 6 1 7)

Simple hurdle if cutting it decreases the # of hurdles

Super hurdles if cutting it increases the # of hurdles

(0 2 5 4 3 -6 1 7 )


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Algorithm2:

has 2k hurdles merge any two non-consecutivehurdles

has 2k+1 hurdles cut one simple hurdle (if it has nonemerge any two non-consecutive)

Claim2: Alg1 + Alg2 optimally sort any signed perm.


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Proof of claims:

breakpoint graph

1. each positive el x 2x-1,2x and each negative (-x) 2x,2x-1

(0 -1 3 5 4 6 -2 7)

(0 2 1 5 6 9 10 7 8 11 12 4 3 13 )

arcs


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Arcs oriented if they span an odd # of elements

Arc overlap graph:

Vertices -> arcs from breakpoint graph

Edges arcs overlap


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Every oriented vertex corresponds to an orientedpair.

Fact2: Score of an oriented reversal (orientedvertex v) is T+U-O+1.

T= #oriented vertices.

U= #unoriented vertices adjacent to v O= #oriented vertices adjacent to v

Oriented component if it contains an oriented v

Safe reversal does not create new unorientedcomponents.


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Theorem (Hannenhalli&Pevzner). Anysequence of oriented safe reversals is optimal.

Theorem. An oriented reversal of maximal scoreis safe.

claim1 holds.

Claim2 is proven in a similar manner.


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J. Kececioglu and D. Sankoff. Exact and

approximation algorithms for sorting by reversals, withapplication to genome rearrangement. 1995.

A. Bergeron. A very elementary presentation of the

Hannenhalli-Pevzner Theory. 2005

A. Caprara. Sorting by reversals is difficult. 1997

S. Hannenhalli and Pavel Pevzner.Transforming

cabbage into turnip: polynomial algorithm for sorting

signed permutations by reversals. 1999

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