A 2-Approximation Algorithm for Sorting by Prefix Reversals

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1 A 2-Approximation Algorithm for Sorting by Prefix Reversals c Springer-Verlag Johannes Fischer and Simon W. Ginzinger LFE Bioinformatik und Praktische Informatik Ludwig-Maximilians-Universität München Amalienstr. 17 D München {Johannes.Fischer Simon.Ginzinger}@bio.ifi.lmu.de Abstract. Sorting by Prefix Reversals, also known as Pancake Flipping, is the problem of transforming a given permutation into the identity permutation, where the only allowed operations are reversals of a prefix of the permutation. The problem complexity is still unknown, and no algorithm with an approximation ratio better than 3 is known. We present the first polynomial-time 2-approximation algorithm to solve this problem. Empirical tests suggest that the average performance is in fact better than 2. 1 Introduction The problem of sorting a permutation by prefix reversals, also known as pancake flipping, was first considered in a computational context by Gates and Papadimitriou [7]. The problem may informally be described as follows: given a permutation of the first n integers, transform it into the sorted sequence 1,..., n by using as few prefix reversals as possible, an operation that flips the first x elements. In [7], and likewise in a subsequent article by Heydari and Sudborough [11], bounds were given that only depend on the size n of the permutation, disregarding special properties of the given permutation. This reflects the concept of the diameter f(n) of the pancake network of size n, i.e. the maximal number of prefix reversals that is needed to sort an (arbitrary) permutation of 1,..., n. In summary, we now know [7, 11] that (15/14)n f(n) (5n + 5)/3. In this article, we tackle the problem of finding a minimal sequence of prefix reversals that sorts a given permutation π (MIN-SBPR). It should be clear that for n fixed, some permutations of length n are easier to sort than others. This intuition can be formalized by the concept of breakpoints and breakpoint graphs, first introduced by Bafna and Pevzner [1]. They considered a different version of the problem, the task of sorting a permutation by (arbitrary) reversals (MIN-SBR). The problem was shown to be NP-complete by Caprara [4], but Hannenhalli and Pevzner [9] gave a polynomial time algorithm for the slightly restricted case of signed permutations which is highly relevant in computational biology. Despite the NP-completeness of MIN-SBR, substantial progress has been made in finding approximation algorithms, starting with Kececioglu and Sankoff s algorithm [14] which has a performance guarantee of 2. The currently best known algorithm is a approximation due to Berman et al. [3]. A related problem is the one of sorting by transpositions (MIN-SBT), where one seeks to find the minimum number of transpositions needed to sort a permutation. In contrast to MIN-SBR, it is still unknown whether MIN-SBT is in P or NP-hard. However, several 1.5-approximation algorithms have been devised (e.g. [2], the most recent being [10]). Similar to the case of reversals, the problem of prefix transpositions has been considered. Here, the currently best result is a 2-approximation by Dias and Meidanis [6]. However, in the case of prefix reversals, little progress has been made. Although Heydari [12] has proven the NP-completeness of a modified version of MIN-SBPR, it remains unknown

2 Fig. 1. The breakpoint graph of the permutation (8 2, 1 9 5, 6 3, 4 7 ). whether or not the original problem is in P. Similarly, although it is easy to come up with a 3-approximation for MIN-SBPR 1, no approximation algorithms with a performance guarantee less than 3 have been found. We give the first 2-approximation algorithm for MIN-SBPR. The remainder of this article is organized as follows. Sect. 2 gives a formal definition of the problem and introduces the notion of breakpoint graphs and related concepts. It also states the lower bound on which our algorithm is based. Sect. 3 develops the 2-approximation algorithm. Sect. 4 shows by empirical tests that the actual performance of our approximation is much better than 2. Finally, Sect. 5 concludes and gives an outlook to future work. 2 Preliminaries For a permutation π = (π 1,..., π n ) of {1,..., n}, a prefix reversal φ(r) is defined as the operation that flips the first r elements of π for 2 r n, i.e. π φ(r) = (π r,..., π 1, π r+1,..., π n ). The prefix reversal distance d(π) is the minimal number of prefix reversals that is needed to transform nto the identity permutation ι := (1,..., n). Determining d for a given permutation is known as the problem of Sorting by Prefix Reversals (MIN-SBPR) and is the issue of this article. We now extend π by setting π 0 := 0 and π n+1 := n + 1, yielding π = (0, π 1,..., π n, n + 1). For convenience, we will also write π for the extended permutation π. We say that there is a breakpoint between and +1 if +1 1 for 1 i n. 2 That is, there is a breakpoint between 2 elements that are adjacent in π, but not adjacent in the identity permutation ι := (1,..., n + 1). As an example, the breakpoints in the following permutation are marked with a horizontal bar: (8 2, 1 9 5, 6 3, 4 7 ). We define b(π) to be the number of breakpoints in π. An immediate consequence is Lemma 1. For a permutation π = (π 1,..., π n ) of {1,..., n} with b(π) breakpoints, we have d(π) b(π). Proof. Let φ(r 1 ),..., φ(r d(π) ) be an optimal sequence of prefix reversals that sorts π, i.e. π φ(r 1 ) φ(r d(π) ) = ι. Note that a reversal φ(r) can eliminate at most one breakpoint, namely the one between π r and π r+1 (if any). Since ι is the only permutation having 0 breakpoints, the claim follows. We note that this bound is not sharp for all cases. For example, the permutation (3, 4 1, 2 ) has two breakpoints, but needs at least three prefix reversals to be transformed into the identity permutation. (The prefix reversals are φ(2), φ(4) and φ(2).) The breakpoint graph G π = (V, E) of a permutation s defined as follows: the set of vertices in G s V := {π 0,..., π n+1 }, and the set of (directed) edges E is the union of so-called red edges R and blue edges B to be defined next. An edge e is in R if and only if e = (, +1 ) 1 The obvious 3-approximation flips the strip with the highest unsorted element to the beginning, brings it in the correct direction and then flips it to its proper place at the end where it remains untouched. See Sec. 2 for the definition of strips. 2 Because of the inherent asymmetry of prefix reversals, we never say that there is a breakpoint between π 0 and π 1. However, in the breakpoint graph (to be defined after Lemma 1) we do have a red edge between π 0 and π 1 if π 0 π 1 1.

3 (a) Type 1 (b) Type 2 (c) Type 3 (d) Type 4 Fig. 2. Different types of blue edges. and there is a breakpoint between and +1, or e = (π 0, π 1 ) and π 0 π 1 1. Further, an edge e is in B if and only if e = (, π j ) for some 0 i < j n + 1, π j = ± 1 and e for some red edge e. Note that the number of blue edges equals the number of red edges. See Fig. 1 for an example of a breakpoint graph, where the blue edges are drawn as dashed lines to distinguish them from the red edges. We stick to the convention that the breakpoint graph is drawn from left to right, and that red edges are drawn as straight lines, whereas blue edges are drawn as arcs. Whenever we talk of the first or leftmost blue edge we mean the blue edge argmin (πi,π j ) Bi. Likewise, the last or rightmost blue edge is defined as argmax (πi,π j ) Bj. For example, in Fig. 1 the first blue edge connects 0 and 1 and the last blue edge connects 9 and 10. We say that,..., π j form a strip for 1 i j n if π k = π k+1 ± 1 for all i k < j, 1 ± 1 and π j π j+1 ± 1. A strip is called a singleton if it consists of only one element, with the exception that π 1 = 1 and π n = n are never considered as singletons. A strip of length 2 is called ascending if π k = π k+1 1 for all i k < j and descending otherwise. Now consider a blue edge between 2 elements and π j := ± 1. Because of the definition of blue edges there is at least one adjacent red edge on each side of the blue edge; we can thus classify such a blue arc into one of the four different types shown in Fig. 2, depending on the directions of the adjacent red edges. Note that if s or π j s strip is a singleton, the blue arc may be classified into at least 2 different types. As an example, consider the blue arc (6, 7) in Fig. 1 which is of type 2 and 3. Note that we will not use the breakpoint graph for theoretical considerations such as relating the maximal number of alternating cycles in it to the reversal distance as in [9]; it will rather be used for expository purposes. In the rest of this paper we will only consider permutations where there is a breakpoint between π n and π n+1. This is simply because if there are x ordered elements at the end of π, say π = (π 1,..., π n x, n x + 1,..., n 1, n), we can reduce the problem of sorting π to sorting the permutation π := (π 1,..., π n x ). We will further restrict ourselves to prefix reversals that act on a breakpoint, which means that φ(r) is applied to π only if there is a breakpoint between π r and π r+1. We will see later that this restriction is sufficient for a 2-approximation. 3 3 The 2-Approximation Algorithm We are now ready to present the details of our 2-approximation. The next lemma gives a nice property of the breakpoint graph that allows us to eliminate a breakpoint by using at most two prefix reversals. 3 In analogy to Hannenhalli and Pevzner s Theorem 3.1 in [8] we can in fact show that there is an optimal sorting of prefix reversals that does not cut a strip of length 3. However, for strips of length 2 this is not the case.

4 (a) The blue edge from π 1 = 8 to π 9 = 7 satisfies condition 1 of Lemma 2. We apply the prefix reversal φ(8); see the next picture for the result (b) Now the blue edge from π 2 = 3 to π 7 = 2 satisfies condition 2. We thus apply the reversals φ(7) (c)... and φ(5) to create an adjacency between 2 and (d) Here, the highlighted blue edge is of type 3. The reversals φ(3) (e)... and φ(7) (f)... create the desired adjacency. Fig. 3. An example. We want to sort π = (8 2, 1 9 5, 6 3, 4 7 ) (the six breakpoints are marked with a bar). The adjacencies that will be created next are highlighted. The example will be continued in figure 7, when the necessary tools have been introduced. Lemma 2. Let π be a permutation of {1,..., n} and G π be its associated breakpoint graph. Then there exists a sequence of at most two prefix reversals φ(r) and φ(s) that eliminates a breakpoint if at least one of the three conditions holds: 1. G π contains a blue arc (, π j ) of type 1 with i = G π contains a blue arc (, π j ) of type 2, where i G π contains a blue arc (, π j ) of type 3. Proof. Because (, π j ) is blue we must have π j = ±1. We show that in all cases we can create an adjacency between and π j without introducing any new breakpoints, thereby eliminating a breakpoint. The reader is encouraged to follow the examples shown in Fig. 3. In case 1, the single prefix reversal φ(j 1) creates the desired adjacency between π 1 and π j without introducing a new breakpoint. See Fig. 3 (a) for an example. In case 2, φ(j) and φ(j i) suffice: we have (π 1,..., π n ) φ(j) = (π j,...,,..., π 1, π j+1,..., π n ) =: π ; so G π contains a blue arc (π 1, π j i+1 ) of type 1, and we are thus in the situation of case 1. The condition i 0 is necessary because with i = 0 we have π 0 = 0, so π j = 1 and we cannot create an adjacency by our definition of breakpoints. Note also that an arc of type 2 cannot be the last blue edge in G π ; this implies in particular j n + 1, so φ(j) is a valid reversal. See Fig. 3 (b) (c) for an example. In case 3, the adjacency is created by the prefix reversals φ(i) and φ(j 1): again, with (π 1,..., π n ) φ(i) = (,..., π 1, +1, π j,..., π n ) =: π we see that G π contains a blue arc (π 1, π j ) of type 1. See Fig. 3 (d) (f) for an example. For the rest of this section, we will say that a blue edge of type 1,2 or 3 is a good edge if it satisfies one of the respective conditions given in Lemma 2. The next two lemmas are the key to

5 1 + (a) If there is a blue edge going from to the right, then it is either type (b)... or type 2, depending on the direction of the incident red edge in ± p p (c) The blue edge between and +1 must point to the left, as well as the red edge incident to +1. (d) The highlighted edges correspond to exactly the same situation which we have already considered. Fig. 4. Illustrations to the proof of Lemma 3. our 2-approximation. They characterize those permutations that do not contain a good edge. In fact, we will show that these permutations all resemble a certain prototype permutation (given in Eq. ( ) on p. 6) which can then be solved by the generic sorting sequence given in Lemma 6. Lemma 3. Let π be a permutation of {1,..., n} that does not contain a good blue edge and let G π be its associated breakpoint graph. Then π does not contain any singletons. Proof. Assume π contains a singleton, say at position 1 i n. Then there are two blue edges beginning or ending at. Further, there is a red edge ending at and a red edge beginning at. If one of the two blue edges had as its left endpoint, then this edge would be of type 2 or 3 (or both), so at least one of the conditions 2 or 3 in Lemma 2 would hold, a contradiction (see Fig. 4 (a) (b)). So both blue edges have as their right endpoint, in particular the edge starting at + 1. There cannot be a red edge on the right side of + 1, because otherwise the blue edge ( + 1, ) would be of type 3 and thus be good. So the only red edge incident to + 1 ends there (Fig. 4 (c)). This means that there is an ascending strip from + 1 to + p for some p 2. Now we either have + p = n, in which case there must be a blue edge from + p = n to π n+1 = n + 1, which would then be of type 3, a contradiction. If, on the other hand, + p n, by the same reasoning as above we know that the blue edge incident to + p must point to the left, and again the red edge incident to + p + 1 must also point to the left (Fig. 4 (d)). We are thus in the same situation as before and must eventually reach n which gives us a blue edge of type 3, a contradiction. This proves that there are no singletons in π. Lemma 4. Let π ι be a permutation of {1,..., n} that does not contain a good blue edge and let G π be its associated breakpoint graph. Then all of G π s blue edges have a unique type, the first is of type 2, the last is of type 1, and all other blue edges are of type 4. Proof. The fact that all blue edges have a unique type follows immediately from Lemma 3. Since condition 3 of Lemma 2 is not satisfied, the first blue edge cannot be of type 3 and must thus be of type 2. (The first blue edge can never be of type 1 or 4.) We now prove that all other blue edges apart from the last are of type 4. In fact, all we need to show is that they are not of type 1, for if there were an arc of type 2 or 3, one of the conditions 2 or 3 in Lemma 2 would hold. Note further that because π contains no singletons, there are either 0 or 2 edges incident to each vertex of G π. So G π forms a unique cycle. Following this cycle from the end of the leftmost

6 f e f e π + i 1 + e (a) Assume edge e is the first of type 1 (all shown edges between the first and the last are of type 4). (b) The next blue edge e must point to the left, as well as the red edge e incident to ± 1. Fig. 5. Illustrations to the proof of Lemma 4. blue edge f, look at the first blue edge e of type 1, see Fig. 5 (a). (We will actually show that this edge must be the last.) Define as the element to the left of the right endpoint of e. By the same reasoning as in Lemma 3, the blue edge e incident to cannot point to the right, and likewise the next red edge e must point to the left (Fig. 5 (b)), because otherwise e would be of type 2. Now either e = f (in which case we are done), or the argument can be continued inductively until we eventually reach edge f. The proof is finished by noting that there must be at least one blue edge of type 1, for otherwise G π would not form a cycle. For the proof of the following lemma we need another definition for blue edges [9]: two blue edges (, π j ) and (π k, π l ) are said to be interleaving if the intervals [i, j] and [k, l] overlap but neither of them contains the other. Lemma 5. Let π ι be a permutation of {1,..., n} that does not contain a good blue edge. Then s of the form π = (p 1,..., 1, p }{{} 2,..., p 1 + 1,......, n,..., p }{{} b(π) 1 + 1), ( ) }{{} l 1 l 2 l b(π) i.e. π consists of b(π) 2 descending strips of length l i 2 for all 1 i b(π). Proof. By Lemma 4, G π consists of one blue edge of type 2 at the beginning, one blue edge of type 1 at the end, and all blue edges in between are of type 4. Because G π forms a unique cycle and blue edges of type 4 must interleave with at least two other blue edges, the only possible arrangement of these edges looks as shown in Fig. 6, with (, π j ) being its first blue edge. π π j n+1 Fig. 6. The only possible breakpoint graph when we cannot apply two reversals that eliminate a breakpoint. First note that i = 0, for otherwise the blue edge (, π j ) would be good. Following the first blue arc from = 0, we see that π j = 1. Therefore the first strip must be p 1,..., 1 with l 1 = p 1 2, for if p 1 = 1, the strip would be a singleton. Continuing this line of arguments we get that π must of the form ( ). To see that b(π) 2, assume that there is only one breakpoint in π. Then there would be no edge of type 4 in G π, so the blue edge of type 1 would start at π 1 and would thus be good. The previous lemma has characterized permutations that do not contain a good blue edge. We now show how to cope with these hard instances.

7 (a) Here we are in the situation of Lemma 4. The permutation is sorted by first applying the prefix reversal φ(9) (b)... then φ(5) (c)... again φ(9) (d)... then φ(7) (e)... again φ(9) (f)... and finally φ(6) (g) The sorted permutation. Fig. 7. Continuing the running example: This figure shows the effect of the generic sorting sequence. Lemma 6. Assume s of the form ( ), and let l 1,..., l b(π) be defined as in Lemma 5. Then the sequence of 2b(π) prefix reversals φ(n) φ(n l 1 ) φ(n) φ(n l 2 ) φ(n) φ(n l b(π) ) applied to π yields the identity permutation. Proof. Applying the first two reversals to π yields π := (p 2,..., p 1 + 1, p 3,..., p 2 + 1,......, n,..., p b(π) 1 + 1, 1,..., p 1 ). We now prove by induction on the number of breakpoints in π (equal to b(π)) that applying the next 2b(π) 2 prefix reversals yields the identity. The base is when b(π ) = 2. So π = (n,..., p 1 + 1, 1,..., p 1 ), and π φ(n) φ(n l 2 ) is clearly equal to ι. For the induction step, let π = (p 2,..., p 1 + 1, p 3,..., p 2 + 1,......, n,..., p b(π) 1 + 1, 1,..., p 1 ). Now π φ(n) φ(n l 2 ) = (p 3,..., p 2 + 1,......, n,..., p b(π) 1 + 1, 1,..., p 1, p 1 + 1,..., p 2 ), which is of the same form as π but has one breakpoint less. The claim follows. See Fig. 7 for an example of the generic sorting sequence. We note that in the above lemma, the first two prefix reversals (φ(n) and φ(n l 1 )) do not eliminate a breakpoint, whereas the last two prefix reversals (φ(n) and φ(n l b(π) )) both create an adjacency, and all other pairs of reversals create exactly one adjacency. That is, the last two prefix reversals compensate for the first two that could not create any adjacency. We now come to our main result:

8 Corollary 1. MIN-SBPR is 2-approximable. Proof. While the breakpoint graph of π contains a good blue edge, choose on of these edges and apply at most 2 prefix reversals to eliminate a breakpoint. If this is impossible, by Lemma 5 π must be of the form ( ), which can be transformed into the identity permutation ι by the generic sequence of prefix reversals given in Lemma 6. The number of prefix reversals used is at most 2b(π). The claim follows with the lower bound given in Lemma 1. 4 Empirical Results We implemented the algorithm that drops out from the proof of Cor. 1. The obvious strategy used was that good arcs of type 1 were preferred over good arcs of type 2 or 3 because the former just need one prefix reversal to create an adjacency instead of two. This algorithm was compared to a branch-and-bound method to compute the optimal number of prefix reversals to sort a given permutation. Due to the enormous size of the group of permutations (and the even more numerous number of possible sorting sequences to be inspected by the branch-and-bound algorithm), we chose to select at random 10,000 permutations of length up to 71 and computed the actual approximation ratios of our 2-approximation. The results can be seen in Fig approximation ratio size of permutation Fig. 8. The actual approximation ratios of our 2-approximation algorithm. The size of the permutations is plotted against the number of operations performed by our algorithm, divided by the minimum number of prefix reversals needed to sort these permutations. All numbers are averaged over 10,000 random permutations of the shown size. It is interesting to see that the actual approximation ratio is much better than 2. This suggests that with a deeper analysis of the algorithm the theoretical approximation ratio could even be lowered. For example, we were able to prove that certain blue arcs of type 4 (similar to the ones in Fig. 6) contribute to d(π) by an extra prefix reversal. Nevertheless this is not sufficient to lower the theoretical approximation ratio of the algorithm: To do so, one would

9 have to make sure that, among other things, such situations are not created unless they are inevitable. Another point to note on Fig. 8 is that the actual approximation ratio seems to level off at 1.2. One possible explanation for this could be that the number of hard permutations for our method converges against a constant fraction of the size of the group. However, because we could only sample a constant number of permutations for every n (namely 10,000), it could also be that the really hard instances were not covered by our randomly chosen permutations and the true approximation ratio is worse than the graph shows. 5 Conclusions and Outlook We have seen an algorithm to solve MIN-SBPR with an approximation ratio of 2. We note that Cohen and Blum [5] give a similar 2-approximation for the signed version of MIN-SBPR, parts of which could also have been used for our problem. While our result is rather of theoretical interest, empirical tests have shown that on average, our algorithm is within 1.2 of the optimal for permutations of length up to 71. This suggests that there is a hidden parameter in the prefixreversal-distance d(π), in a similar way as hurdles and fortresses account for the reversal-distance in MIN-SBR. Future work will be directed towards raising the lower bound by identifying the parameters influencing the prefix-reversal distance. From a theoretical standpoint, another interesting topic of research is to prove the theoretical computational complexity of both the signed and the unsigned version of the problem. Acknowledgments We wish to thank Volker Heun for fruitful discussions and helpful suggestions on this subject. Further thanks go to an anonymous reviewer who pointed out the connection to the signed version of MIN-SBPR [5]. References 1. V. Bafna, P. A. Pevzner: Genome Rearrangements and Sorting by Reversals. SIAM J. on Computing, 25(2): , V. Bafna, P. A. Pevzner: Sorting by Transpositions SIAM J. on Discrete Mathematics 11(2), , P. Berman, S. Hannenhalli, M. Karpinski: Approximation Algorithm for Sorting by Reversals. Proc. ESA 02, Lecture Notes in Computer Science 2461: , A. Caprara: Sorting Permutations by Reversals and Eulerian Cycle Decompositions. SIAM J. on Discrete Mathematics 12(1): , D. S. Cohen, M. Blum: On the Problem of Sorting Burnt Pancakes. Discrete Applied Mathematics 61: , Z. Dias, J. Meidanis: Sorting by Prefix Transpositions. Proc. SPIRE 02, Lecture Notes in Computer Science 2476: 65 76, W. H. Gates, C. H. Papadimitriou: Bounds for Sorting by Prefix Reversals. Discrete Mathematics 27: 47 57, S. Hannehalli, P. Pevzner: TO CUT... OR NOT TO CUT (Applications of Comparative Physical Maps in Molecular Evolution). Proceedings of the 7th ACM Symposium on Discrete Algorithms (SODA 96), , S. Hannenhalli, P. A. Pevzner. Transforming Cabbage into Turnip: Polynomial Algorithm for Sorting Signed Permutations by Reversals. J. of the ACM 46(1): 1 27, T. Hartman, R. Shamir: A Simpler 1.5-Approximation Algorithms for Sorting by Transpositions, Proc. CPM 03, Lecture Notes in Computer Science 2676, , M. H. Heydari, I. H. Sudborough: On the Diameter of the Pancake Network. J. of Algorithms 25: 67 94, M. H. Heydari. The Pancake Problem. PhD-thesis, University of Wisconsin at Whitewater, 1993.

10 13. J. Kececioglu, D. Sankoff. Efficient Bounds for Oriented Chromosome Inversion Distance. Proc. CPM 94, Lecture Notes in Computer Science 807: , J. Kececioglu, D. Sankoff. Exact and Approximation Algorithms for Sorting by Reversals, with Application to Genome Rearrangements. Algorithmica 13: , 1995.