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A Simpler 1.5-Approximation
Algorithm for Sorting by
Transpositions
Tzvika Hartman
Weizmann Institute
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.
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.
Sorting by Transpositions
 A transposition exchanges between 2
consecutive segments of a perm.
 Example :
123456789
12673458
9
Sorting by transpositions
: finding a
shortest sequence of transpositions which
sorts the perm.
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.
Main Results
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.
1.
Linear & Circular Perms
A transposition “cuts” the perm at 3 points.
Linear transposition :
A
B
C
t
D
A
A
C
B
A
t
Circular transposition :
B
C
C
• Circular transpositions can be represented by
exchanging any 2 of the 3 segments.
B
D
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 Пn+1 П1
П1 . . . Пn
.
.
.
.
.
• Every linear transposition is equivalent to a
circular transposition that exchanges the 2
segments that do not include n+1.
Breakpoint Graph [BafnaPevzner98]
Perm : ( 1
6
5
4
7
Replace each element j by 2j-1,2j:
3
2 )
 = (1 2 11 12 9 10 7 8 13 14 5 6 3 4)
Circular Breakpoint graph G():
4 1
3
Vertex for every element.
Black edges (2i, 2i+1)
2
6
11
5
12
Grey edges (2i, 2i+1)
14
9
13
8
10
7
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}.
4
3
1
2
6
11
5
12
14
9
13
8
10
7
Effect on Graph : Example
 Perm: (1 3 2).
 After extension: (1 2 5 6 3 4).
 Breakpoint graph:
1
2
4
1
5
3
2
4
6
• # of cycles increased by 2
5
3
6
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
11
12
1
10
7
4
3
6
9
2
9
2
12
8
5
• # of cycles remains 2
1
10
7
4
3
6
8
5
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.
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.
3 - Cycles
 2 possible configurations of 3-cycles:
Non-oriented 3-cycle
Oriented 3-cycle
(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.
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.
Shattered Cycles
 2 pairs of black edges intersect if they appear
alternatively along the circle.
 Cycle A is shattered by cycles
and C if every pair of black
edges in A intersects with a pair
in B or with a pair in C.
B
 Lemma : If G contains a shattered cycle, then
 a (0,2,2)-sequence.
Shattered Cycles (Cont.)
 Lemma : If G contains no 2-cycles, no
oriented cycles and no interleaving cycles,
then  a shattered cycle.
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.
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).
Open Problems
 Complexity of sorting by transpositions.
 Models which allow several rearrangement
operations, such as trans-reversals, reversals
and translocations (both signed & unsigned).
Acknowledgements
 Ron Shamir.
Thank you !