Information Processing Letters 110 (2010) 345–350

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Information Processing Letters

www.elsevier.com/locate/ipl

An improved algorithm for sorting by block-interchanges basedon permutation groups

Yen-Lin Huang a, Cheng-Chen Huang b,c, Chuan Yi Tang a, Chin Lung Lu b,c,∗a Department of Computer Science, National Tsing Hua University, Hsinchu 300, Taiwanb Institute of Bioinformatics and System Biology, National Chiao Tung University, Hsinchu 300, Taiwanc Department of Biological Science and Technology, National Chiao Tung University, Hsinchu 300, Taiwan

a r t i c l e i n f o a b s t r a c t

Article history:Received 13 July 2009Received in revised form 28 February 2010Accepted 3 March 2010Available online 4 March 2010Communicated by F.Y.L. Chin

Keywords:AlgorithmData structureGenome rearrangementPermutation groupBlock-interchangeGeneralized transpositionPermutation tree

Given a chromosome represented by a permutation of genes, a block-interchange isproposed as a generalized transposition that affects the chromosome by swapping twonon-intersecting segments of genes. The problem of sorting by block-interchanges is tofind a minimum series of block-interchanges for sorting one chromosome into another. Inthis paper, we present an O(n + δ log δ) time algorithm for solving the problem of sortingby block-interchanges, which improves a previous algorithm of O(δn) time proposed byLin et al. (2005) [14], where n is the number of genes and δ is the minimum number ofblock-interchanges required to sort a chromosome.

© 2010 Elsevier B.V. All rights reserved.

1. Introduction

Genome rearrangement studies based on genome-wideanalysis of gene orders play an important role in the phylo-genetic tree reconstruction [20,11,12,19,3]. In these studies,the homologous genes of two genomes are usually iden-tified with the integers 1,2, . . . ,n, with a plus or minussign to indicate their direction. Then a chromosome canbe represented by a signed permutation of {1,2, . . . ,n}.To evaluate the evolutionary distance between two relatedgenomes in gene order, various rearrangement events act-ing on genes within or among chromosomes have beenproposed, such as reversals [12,13,1,21], transpositions [2,7,8], block-interchanges [6,14,8], translocations [11,10,4,18],

* Corresponding author at: Institute of Bioinformatics and System Bi-ology, National Chiao Tung University, Hsinchu 300, Taiwan. Tel.: +886 35712121x56949; fax: +886 3 5729288.

E-mail addresses: slippers.bi92g@nctu.edu.tw (Y.-L. Huang),sosorovo@yahoo.com.tw (C.-C. Huang), cytang@cs.nthu.edu.tw (C.Y. Tang),cllu@mail.nctu.edu.tw (C.L. Lu).

0020-0190/$ – see front matter © 2010 Elsevier B.V. All rights reserved.doi:10.1016/j.ipl.2010.03.003

fusions [11,17,15] and fissions [11,17,15]. Reversals, alsocalled inversions, affect a block of consecutive integerson a chromosome by reversing the order of the integersand flipping their signs; transpositions affect two adjacentblocks of consecutive integers on a chromosome by ex-changing their positions; block-interchanges are generalizedtranspositions by allowing the exchanged blocks not be-ing adjacent on a chromosome. In genomes with multiplechromosomes, translocations exchange the end segmentsbetween two chromosomes; fusions join two chromosomesinto a bigger one; fissions break a chromosome into twosmaller ones.

Recently, the study on the genome rearrangement us-ing block-interchanges has increasingly drawn great atten-tion, since the block-interchange event is a generalizationof transposition and, currently, its computational modelsmeasuring the genetic distance are more tractable thanthose modeled by transposition. Christie [6] first intro-duced the concept of block-interchange and also proposedan O(n2) time algorithm using the breakpoint graph ap-proach, where n is the number of genes, to solve the so-

346 Y.-L. Huang et al. / Information Processing Letters 110 (2010) 345–350

called problem of sorting by block-interchanges that is to finda minimum series of block-interchanges for sorting onechromosome into another. However, the chromosomes heconsidered were restricted to linear chromosomes. Later,by making use of permutation groups in algebra, Lin et al.[14] designed a simpler algorithm for solving the sameproblem on both linear and circular chromosomes withtime complexity of O(δn), where δ is the minimum num-ber of block-interchanges required to sort a chromosomeand can be calculated in O(n) time in advance. Recently,Feng and Zhu [8] proposed a new data structure calledpermutation tree to improve the Christie’s algorithm by re-ducing the time complexity from O(n2) to O(n log n). Inthis paper, we derive a new property that helps us to im-prove Lin et al.’s algorithm by using the data structure ofpermutation tree such that the overall time complexity canbe further reduced down to O(n+δ log δ). Notice that Bónaand Flynn [5] have shown recently that δ is close to n

2 onaverage.

The rest of this paper is organized as follows. Some ba-sic concept and properties of the permutation groups inalgebra are introduced in Section 2. In Section 3, we intro-duce the idea of algorithm proposed by Lin et al. [14] andpresent an improved algorithm. Finally, we make a conclu-sion in Section 4.

2. Preliminaries

In group theory, a permutation is defined to be a one-to-one mapping from a set E into itself. For example, wemay define a permutation α of the set {1,2,3,4,5} byspecifying α(1) = 4, α(2) = 1, α(3) = 2, α(4) = 5 andα(5) = 3. In the study of genome rearrangement, it is con-venient to express the permutation α in cycle form asα = (1,4,5,3,2), in which for each x ∈ E , α(x) is placeddirectly right to x. A cycle of length k, say (a1,a2, . . . ,ak),is simply called k-cycle and it can also be rewritten as(ai,ai+1, . . . ,ak,a1, . . . ,ai−1) (i.e., indices are cyclic), where2 � i � k. Any two cycles are said to be disjoint if theyhave no element in common. In fact, any permutation, sayα, can be written in a unique way as the product of dis-joint cycles, which is called the cycle decomposition of α,if we ignore the order of the cycles in the product. Usu-ally, a cycle of length one in α is not explicitly written andits element, say x, is said to be fixed by α since α(x) = x.Especially, the permutation whose elements are all fixedis called an identity permutation and is denoted by 1, i.e.,1 = (1)(2) · · · (n) if E = {1,2, . . . ,n}.

Given two permutations α and β of E , the composition(or product) of α and β , denoted by αβ , is defined to bea permutation of E with αβ(x) = α(β(x)) for all x ∈ E . Forinstance, if we let E = {1,2,3,4,5,6}, α = (2,3) and β =(2,1,5,3,6,4), then αβ = (2,1,5)(3,6,4). If α and β aredisjoint cycles, then αβ = βα. The inverse of α is definedto be a permutation, denoted by α−1, such that αα−1 =α−1α = 1. If a permutation is expressed by the productof disjoint cycles, then its inverse can be obtained by justreversing the order of the elements in each cycle. For ex-ample, if α = (2,1,5)(3,6,4), then α−1 = (5,1,2)(4,6,3).Clearly, α−1 = α if α is a 2-cycle. Furthermore, it is wellknown that every permutation can be written as a prod-

Fig. 1. The illustration of a permutation α = (1,4,5,3,2), where α(1) = 4,α(2) = 1, α(3) = 2, α(4) = 5 and α(5) = 3.

uct of 2-cycles. For example, (1,2,3,4) = (1,4)(1,3)(1,2).However, there are many ways of expressing a permuta-tion as a product of 2-cycles [9]. Given a permutation α,let f (α) denote the number of the disjoint cycles in thecycle decomposition of α. Notice that f (α) counts alsothe non-expressed cycles of length one. For example, ifα = (1,5)(2,4) is a permutation of E = {1,2, . . . ,5}, thenf (α) = 3, instead of f (α) = 2, since α = (1,5)(2,4)(3).

3. Sorting a permutation by block-interchanges

Recall that the problem of sorting by block-interchangesis to find a minimum series of block-interchanges forsorting a source chromosome into a target chromosome.By convention, the target chromosome is usually de-noted by I = (1,2, . . . ,n). Meidanis and Dias [16,17]first noted that each cycle of a permutation may rep-resent a circular chromosome of a genome with eachelement of the cycle corresponding to a gene and theorder of the cycle corresponding to the gene order ofthe chromosome. Fig. 1, for example, shows a circu-lar chromosome, which is represented by (1,4,5,3,2).Moreover, they observed that rearrangements, such asfusions and fissions (respectively, transpositions), corre-spond to the composition of a 2-cycle (respectively, 3-cycle) and the permutation representing a chromosome.For instance, let α be any permutation whose cycle de-composition is α1α2 · · ·αr . If ρ = (x, y) is a 2-cycle andx and y are in the different cycles of α, say αp = (a1 ≡ x,a2, . . . ,ai) and αq = (b1 ≡ y,b2, . . . ,b j) where 1 � p,q � r,then in the composition ρα, αp and αq are joined intoa cycle (a1,a2, . . . ,ai,b1,b2, . . . ,b j), i.e., ρ is a fusionevent affecting on α (and also called a join operation ofα here). If ρ = (x, y) is a 2-cycle and x and y are inthe same cycle of α, say αp = (a1 ≡ x,a2, . . . ,ai ≡ y,

ai+1, . . . ,a j) where 1 � p � r, then in the composi-tion ρα, this cycle αp is broken into two disjoint cy-cles (a1,a2, . . . ,ai−1) and (ai,ai+1, . . . ,a j), i.e., ρ is afission event affecting on α (and also called a split op-eration of α here). If ρ = (x, y, z) is a 3-cycle and x, yand z are in the same cycle of α, say αp = (a1 ≡ x,a2, . . . ,ai,b1 ≡ y,b2, . . . ,b j, c1 ≡ z, c2, . . . , ck) where 1 �p � r, then in the composition ρα, the cycle αp becomes(a1,a2, . . . ,ai, c1, c2, . . . , ck,b1,b2, . . . ,b j), i.e., ρ is a trans-position event affecting on α. In [14], Lin et al. further ob-served that a block-interchange affecting on α correspondsto two 2-cycles, say ρ1 and ρ2, such that ρ1 is a split oper-ation of α and ρ2 is a join operation of ρ1α. More clearly,let αp = (a1,a2, . . . ,ak) be a cycle of α, ρ1 = (a1,ai) andρ2 = (ah,a j), where 1 < i � k, 1 � h � i − 1 and i � j � k.

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Algorithm SortBIInput: A permutation α = (a1,a2, . . . ,an) of E = {1,2, . . . ,n};Output: A minimum series of block-interchange operations σ1, σ2, . . . , σδ

for sorting α into I;

1: Let δ = n− f (Iα−1)2 ;

2: for i = 1 to δ do2.1: Arbitrarily choose two adjacent elements x and y in Iα−1;

/* Note that (x, y) is a split operation of α. */2.2: Circularly shift (a1,a2, . . . ,an) such that a1 = x and assume

y = ak ;2.3: for j = 1 to n do

index(a j) = j;end for

2.4: Find two adjacent elements u and v in Iα−1(x, y) such thatindex(u) � k − 1 and index(v) � k;/* Note that (u, v) is a join operation of (x, y)α. */

2.5: σi = (u, v)(x, y);2.6: Compute (u, v)(x, y)α and denote it by α again;2.7: Compute Iα−1(x, y)(u, v) and denote it by Iα−1 again;

end for3: Output σ1, σ2, . . . , σδ ;

Then ρ2ρ1α is the permutation obtained from α by ex-changing the blocks [ah,ai−1] and [a j,ak] of αp . Thatis, ρ2ρ1 is a block-interchange event that affects on αby swapping [ah,ai−1] and [a j,ak], two non-intersectingblocks in α.

As discussed above, any series of block-interchanges re-quired to sort one chromosome α into another I can beexpressed by a product of 2-cycles, say ρkρk−1 . . . ρ1, suchthat ρkρk−1 . . . ρ1α = I . Hence, ρkρk−1 . . . ρ1 = Iα−1, sug-gesting that Iα−1 contains all information, which can beutilized to derive ρ1,ρ2, . . . , ρk for sorting α into I .

Based on the above idea, Lin et al. [14] used per-mutation groups to design an O(δn) time algorithm tosort a permutation by block-interchanges, where δ is theminimum number of block-interchanges required to sortthe permutation. In [14], Lin et al. first showed that δ =n− f (Iα−1)

2 , which can be calculated in O(n) time in ad-vance. They then used a greedy method to derive an opti-mal series of δ block-interchanges from Iα−1. Recall that ablock-interchange can be modeled by a split operation fol-lowed by a join operation. Indeed, in each iteration of thegreedy method, they showed that the block-interchangecan be obtained by first arbitrarily choosing two adjacentelements from Iα−1 that serve as a split operation of αand then finding the corresponding join operation in O(n)

time, where two different elements x and y are said tobe adjacent in a permutation β if β(x) = y or β(y) = x.The details of this algorithm are described in AlgorithmSortBI in this paper. It is not hard to see, the bottleneck ofcomputation in the above algorithm is to find the join op-erations. Here, we observe the following property that canhelp us find the join operations in a more efficient way, sothat the overall time complexity of Algorithm SortBI canbe further improved.

Lemma 1. Let α′ be a permutation of E = {1,2, . . . ,n} and z ∈E. Suppose that z and I(z) are in different cycles of α′ . Thenα′(z) and I(z) are adjacent in Iα′−1 , and (α′(z), I(z)) is a joinoperation of α′ .

Proof. It is not hard to see that Iα′−1(α′(z)) = I(z), whichmeans that α′(z) and I(z) are adjacent in Iα′−1. Moreover,

Algorithm FastSortBIInput: A permutation α = (a1,a2, . . . ,an) of E = {1,2, . . . ,n};Output: A minimum series of block-interchange operations σ1, σ2, . . . , σδ

for sorting α into I;

1: Let δ = n− f (Iα−1)2 ;

2: for i = 1 to δ do2.1: Arbitrarily choose two adjacent elements x and y in Iα−1;

/* Note that (x, y) is a split operation of α. */2.2: α′ = (x, y)α;

/* Note that α′ has two cycles, say C1 and C2. */2.3: Find the maximum elements m1 in C1 and m2 in C2;

Let z = min(m1,m2);Let u = α′(z) and v = I(z);/* Note that (u, v) are adjacent in Iα′−1 and (u, v) is a joinoperation of α′ . */

2.4: σi = (u, v)(x, y);2.5: Compute (u, v)α′ and denote it by α again;2.6: Compute Iα−1(x, y)(u, v) and denote it by Iα−1 again;

end for3: Output σ1, σ2, . . . , σδ ;

since z and α′(z) are in the same cycle of α′ , α′(z) andI(z) are in different cycles of α′ . Therefore, (α′(z), I(z)) isa join operation of α′ . �Lemma 2. Let α′ be a permutation of E = {1,2, . . . ,n}, C bea cycle of α′ , and z be the maximum element in C . Supposethat z �= n. Then α′(z) and I(z) are adjacent in Iα′−1 , and(α′(z), I(z)) is a join operation of α′ .

Proof. Since z �= n, I(z) = z + 1. Moreover, since z is themaximum element in C , I(z) is not in C . That is, z andI(z) are in different cycles of α′ . Clearly, this lemma holdsaccording to Lemma 1. �

By Lemma 2, we have the following observation imme-diately.

Observation 1. Let α′ be a permutation of E = {1,2, . . . ,n}with two cycles C1 and C2. Let m1 and m2 be the max-imum elements in C1 and C2, respectively. Suppose thatm1 < m2. Then m2 = n, α′(m1) and I(m1) are adjacent inIα′−1, and (α′(m1), I(m1)) is a join operation of α′ .

Based on the above observation, step 2.4 of AlgorithmSortBI can be done through finding the maximum ele-ments of the two cycles in the permutation. The detailsare described in Algorithm FastSortBI. The maximum ele-ment in a cycle of n different elements can be found inO(n) time. Actually, we observe that this job can be donemore efficiently by using a data structure called permuta-tion tree, which was first proposed by Feng and Zhu [8].

Given a permutation α = (a1,a2, . . . ,an) of E = {1,2,

. . . ,n}, a permutation tree corresponding to α is defined tobe a balanced binary tree with n leaves, whose nodes arelabeled according to the following two rules. (1) The leavesare labeled by a1,a2, . . . ,an , respectively, and in this or-der. (2) Each of other nodes is labeled by the maximumof its children. By definition, the root r of a permutationtree has a label that is the maximum element in α. Asan example, Fig. 2 is a permutation tree correspondingto α = (2,5,3,1,4,6,7,8). Basically, there are followingthree operations that can be applied to a permutation tree.

348 Y.-L. Huang et al. / Information Processing Letters 110 (2010) 345–350

Fig. 2. A permutation tree corresponding to α = (2,5,3,1,4,6,7,8).

They are: (1) BUILD, which is to build a permutation treecorresponding to a permutation, (2) JOIN, which is to jointwo permutation trees into a bigger one, and (3) SPLIT,which is to split a permutation tree into two smaller ones.The explicit definitions of the three operations are given inDefinitions 1, 2 and 3, respectively. Feng and Zhu [8] haveshown that the costs of the BUILD, JOIN and SPLIT oper-ations are O(n), O(log n) and O(log n), respectively.

Definition 1. (See Feng and Zhu [8].) BUILD(α) creates apermutation tree, which corresponds to a given permuta-tion α.

Definition 2. (See Feng and Zhu [8].) SPLIT(T ,m) separatesa permutation tree T corresponding to α = (a1, . . . ,am−1,

am, . . . ,an) into two trees such that one corresponds toαl = (a1,a2, . . . ,am−1), and the other to αr = (am,am+1,

. . . ,an).

Definition 3. (See Feng and Zhu [8].) JOIN(Tl, Tr) mergestwo permutation trees Tl corresponding to αl = (a1,a2, . . . ,

am−1) and Tr corresponding to αr = (am,am+1, . . . ,an) intoa bigger permutation tree, which corresponds to (a1, . . . ,

am−1,am, . . . ,an).

Below, we describe how to apply the data structure ofpermutation tree to Algorithm FastSortBI. Initially, we builda permutation tree corresponding to the source genome α.In step 2.2 of FastSortBI algorithm, (x, y) splits α into α′and hence the permutation tree corresponding to α is sep-arated into two smaller trees corresponding to two cyclesof α′ , say C1 and C2, respectively. Then, it takes con-stant time to obtain the maximum elements of C1 and C2from the roots of their corresponding permutation trees. Instep 2.5, (u, v) joins α′ into α and the permutation treescorresponding to C1 and C2, respectively, are then mergedinto a bigger tree corresponding to α.

It should be noted that because a permutation is repre-sented in cycle form, there is no difference for this permu-tation to take which of its elements as the first one in a cy-cle of the permutation. For example, (2,5,3,1,4,6,7,8) =(3,1,4,6,7,8,2,5). However, for a permutation tree, itsleaves are labeled according to a linear order of the ele-ments in the corresponding permutation. Hence in step 2.2of FastSortBI algorithm, although a 2-cycle splits α intoα′ = C1C2 with two cycles, we may not obtain the twosmaller permutation trees corresponding to C1 and C2,respectively, by using one SPLIT operation. For example,(3,6) is a split operation to α = (2,5,3,1,4,6,7,8) and(3,6)α = (3,1,4)(6,7,8,2,5). However, we cannot use a

SPLIT operation to transform the permutation tree cor-responding to (2,5,3,1,4,6,7,8) into two permutationtrees corresponding to (3,1,4) and (6,7,8,2,5), respec-tively. In this case, two SPLIT and one JOIN operationsare needed for its correct transformation (see Lemma 3 fordetails). A similar argument is applicable to the join oper-ation of step 2.5.

Lemma 3. Let a permutation α = (a1,a2, . . . ,an) and ρ1 bea split operation that separates α into two cycles, say C1 andC2 , that is ρ1α = C1C2 . Then, by using at most two SPLIT andone JOIN operations, we transform a permutation tree T corre-sponding to α into two smaller permutation trees correspondingto C1 and C2 , respectively.

Proof. Let ρ1 = (ai,a j), where 1 � i < j � n. If ai = a1,we can use SPLIT(T , j) to split T into two permuta-tion trees that correspond to C1 = (a1,a2, . . . ,a j−1) andC2 = (a j,a j+1, . . . ,an), respectively. On the other hand,if ai �= a1, SPLIT(T , i) leads to two permutation trees,Tl and Tr . Then, JOIN(Tr, Tl) merges the two permuta-tion trees into a new one corresponding to the permuta-tion (ai, . . . ,a j, . . . ,an,a1, . . . ,ai−1). We reassign the newtree as T , which still corresponds to α, but the first ele-ment in the cycle of α is changed from a1 to ai . Finally,SPLIT(T , j − i + 1) operates on T to produce two per-mutation trees that correspond to C1 = (ai, . . . ,a j−1) andC2 = (a j, . . . ,an,a1, . . . ,ai−1), respectively. �Lemma 4. Let α′ = C1C2 be a permutation with two cycles C1and C2 , and ρ2 be a join operation of α′ , that is α = ρ2α

′ . Then,by using at most two SPLIT and three JOIN operations, wetransform the two permutation trees corresponding to C1 andC2 , respectively, into a permutation tree corresponding to α.

Proof. Let C1 = (a1,a2, . . . ,am) and C2 = (b1,b2, . . . ,bp).Let T1 and T2 be permutation trees corresponding to C1and C2, respectively, and ρ2 = (ai,b j), where i � m andj � p. If ai �= a1, then, as was discussed in Lemma 3,we can use one SPLIT followed by one JOIN to ob-tain a new permutation tree T1 that corresponds to(ai, . . . ,am,a1, . . . ,ai−1). Similarly, if b j �= b1, one SPLITand one JOIN can be used to obtain T2 corresponding to(b j, . . . ,bp,b1, . . . ,b j−1). Finally, JOIN(T1, T2) merges thetwo trees T1 and T2 into a bigger permutation tree corre-sponding to α. �Theorem 1. The problem of sorting by block-interchanges canbe solved by Algorithm FastSortBI in O(n + δ log n) time.

Proof. As discussed previously, Algorithm FastSortBI sortsα into I using a minimum number of block-interchangeoperations. Now, we analyze its time complexity as fol-lows. In step 1, we build a permutation tree correspond-ing to α in O(n) time. As to step 2, there are δ iter-ations. In step 2.2, by Lemma 3, we obtain the permu-tation trees corresponding to C1 and C2, respectively, byat most two SPLIT and one JOIN operations, which to-tally take only O(log n) time. In step 2.3, the maximumelements of the two cycles are obtained in constant time

Y.-L. Huang et al. / Information Processing Letters 110 (2010) 345–350 349

from the roots of their corresponding permutation trees.In step 2.5, by Lemma 4, it takes O(log n) time to obtaina new permutation tree corresponding to α. As a result,step 2 costs O(log n) time in each of δ iterations. There-fore, the total time complexity of Algorithm FastSortBI isO(n + δ logn). �

In the following, we furthermore improve the cost ofstep 2 in Algorithm FastSortBI from O(log n) to O(log δ).Hence, the overall time complexity of Algorithm FastSortBIcan be further reduced down to O(n + δ log δ).

Definition 4. Let α be the input permutation of AlgorithmFastSortBI. Then, let Φ = {I−1(x) | Iα−1(x) �= x, that is, x isa non-fixed element of Iα−1}.

Observation 2. Once an element is fixed in Iα−1 or Iα′−1

in the process of Algorithm FastSortBI, it is always fixedin Iα−1 and Iα′−1 during the subsequence process of thealgorithm.

Theorem 2. The problem of sorting by block-interchanges canbe solved by Algorithm FastSortBI in O(n + δ log δ) time.

Proof. In step 2.3, the u and v are adjacent in Iα′−1 byLemma 2. This implies that at that moment, v is a non-fixed element of Iα′−1. According to Observation 2, v is anon-fixed element before running step 2 in Algorithm Fast-SortBI and, therefore, z ∈ Φ since v = I(z) (i.e., z = I−1(v)).This suggests that before running step 2, we can use onlythose elements that are in Φ ∪ {n} to build a permuta-tion tree T for more simply corresponding to α and thoseelements that are not in Φ ∪ {n} can be skipped. The rea-son for including n in the construction of T above is thatmax{m1,m2} = n in step 2.3 and hence n must be in Teven though it may not be in Φ . It is not hard to see thatthere are at most O(δ) elements in Φ ∪ {n}. In this case,however, the x and y in step 2.2 may not be in Φ ∪ {n}and hence they may not be in T , resulting in that T cannotbe separated into two permutation trees using SPLIT(T , x)and/or SPLIT(T , y), as was discussed in Lemma 3. To fixthis problem, we first define Θ(w) = αk(w), where k isthe smallest positive integer such that αk(w) ∈ Φ ∪ {n}.Then we can use SPLIT(T ,Θ(x)) and/or SPLIT(T ,Θ(y))

to correctly divide T into two permutation trees. Again, itis not hard to see that the computation of Θ(w) for allw /∈ Φ ∪ {n} can be done in advance in O(n) time. There-fore, the total time complexity of Algorithm FastSortBI isO(n + δ log δ). �

Let us take α = (7,8,6,5,2,3,1,4) for an example.We note that Iα−1 = (1,4,2,6)(5,7)(3)(8) and n = 8.According to Algorithm FastSortBI, we understand thatthe block-interchange distance between α and I is δ =(8−4)

2 = 2, which means that α can be sorted into I us-ing two block-interchange operations. Below, we show howto find these two block-interchange operations, say σ1 =ρ12ρ11 and σ2 = ρ22ρ21. Initially, we build a permutationtree T for corresponding to α by just using those ele-ments in Φ ∪ {n}. In this case, Φ ∪ {n} = {1,3,4,5,6,8}

and hence we build a permutation tree that correspondsto (8,6,5,3,1,4). By definition, we have Θ(2) = 3 andΘ(7) = 8. Then, we let ρ11 = (4,2) since 4 and 2 are ad-jacent in Iα−1. After applying ρ11 to α, we obtain α′ =ρ11α = (4,2)(7,8,6,5,2,3,1,4) = (4,7,8,6,5)(2,3,1). Inthis case, two SPLIT and one JOIN operations are neededto obtain the two permutation trees. First, SPLIT(T ,4)

divides T into Tl and Tr corresponding to (8,6,5,3,1)

and (4), respectively. Next, JOIN(Tr, Tl) merges Tr andTl into a new T corresponding to (4,8,6,5,3,1). Finally,SPLIT(T ,Θ(2)) = SPLIT(T ,3) further divides T into twopermutation trees that correspond to (4,8,6,5) and (3,1),respectively. Next, in step 2.3, we find z = 3 and henceρ12 = (α′(3), I(3)) = (1,4) is a join operation of α′ . In thiscase, one SPLIT and two JOIN operations are required tomerge the two permutation trees into a bigger one corre-sponding to (1,3,4,8,6,5). After the above iteration, wehave α = (1,2,3,4,7,8,6,5) and Iα−1 = (1,6)(5,7). Sim-ilarly, in the second iteration, we can find that ρ21 = (1,6)

and ρ22 = (α′(6), I(6)) = (5,7).

4. Conclusions

In this paper, we studied the problem of sorting byblock-interchanges that is to find a minimum series ofblock-interchanges to sort the permutation of the givenchromosome. Here, we proposed an improved algorithmof O(n + δ log δ) time based on permutation groups andpermutation trees. It will be interesting to study whetherthe permutation groups and permutation trees can be fur-ther applied to other problems of genome rearrangementinvolved with other rearrangements, such as reversals andtranslocations.

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