Finding All Approximate Gapped Palindromes

Transcription

1 Fining All Approximate Gappe Palinromes Ping-Hui Hsu 1, Kuan-Yu Chen 1, an Kun-Mao Chao 1,2,3 1 Department of Computer Science an Information Engineering 2 Grauate Institute of Biomeical Electronics an Bioinformatics 3 Grauate Institute of Networking an Multimeia National Taiwan University, Taipei, Taiwan 106 kmchao@csie.ntu.eu.tw Abstract. We stuy the problem of fining all maximal approximate gappe palinromes in a string. More specifically, given a string S of length n, a parameter q 0 an a threshol k>0, the problem is to ientify all substrings in S of the form uvw such that (1) the Levenshtein istance between u an w r is at most k, wherew r is the reverse of w an (2) v is a string of length q. The best previous work requires O(k 2 n) time. In this paper, we propose an O(kn)-time algorithm for this problem by utilizing an incremental string comparison technique. It turns out that the core technique actually solves a more general incremental string comparison problem that allows the insertion, eletion, an substitution of multiple symbols. Keywors: palinrome, incremental string comparison, string matching. 1 Introuction A wor is calle a palinrome if it reas the same both forwar an backwar. In other wors, a palinrome is a wor of the form uau r,whereu is a string, a is a symbol (or an empty wor), an u r is the reverse of u. A palinrome in astringismaximal if it can not be extene outwar while preserving a palinromic structure. The recognition of palinromes in a string has always been an intriguing question both in theory an practice. For example, fining all maximal palinromes in a string was stuie an well solve in linear time with the help of suffix tree [15] an constant-time LCA (least common ancestor) queries [12]. In [4,5], the authors stuie the problems of palinromes in Sturmian wors or in ternary square-free wors. Moreover, the problem of ientifying palinromes in compresse texts were investigate in [10]. In DNA an RNA sequences, there is a similar structure calle quasipalinrome which plays an important role in genome research [3,8,16]. A quasipalinrome can be seen as a pair of reverse complementary repeats in a string that are separate by a number of characters. The complementarity relation on This research was supporte in part by NSC grants NSC E MY3 an NSC E MY3 from the National Science Council, Taiwan. Y. Dong, D.-Z. Du, an O. Ibarra (Es.): ISAAC 2009, LNCS 5878, pp , c Springer-Verlag Berlin Heielberg 2009

2 Fining All Approximate Gappe Palinromes 1085 nucleoties means that A is complementary to T (U) anc is complementary to G. Several functions of the quasi-palinrome in genomes have been iscovere. For example, the quasi-palinromic structure is a sign of replication origins in the nucleotie sequence [3]. By estimating the appearance of replication origins in avance, the biologists can avoi much labor-intensive work. A stuy of quasi-palinromes also shows that they may control male germ-line gene expression [16]. Thus, the recognition of quasi-palinromic structure raws much attention an raises interesting computational problems. For example, the gappe palinrome problem is to recognize a wor structure of the form uvu r,where strings u an u r are calle arms an string v is calle gap. In [8], the computation of gappe palinromes in a string, with some length constraints on the arms an the gap, are one in linear time. Since mutations may occur in DNA an RNA sequences uring evolution, looking for the approximate quasi-palinromes (or gappe palinromes) is in a sense biologically more meaningful. A tool calle Inverte Repeats Finer ( ientifies approximate gappe palinromes heuristically. The tool uses a set of statistically base criteria to etect caniates of approximate gappe palinromes. By applying some alignment techniques, it then conforms whether these caniates are palinromes. In this paper, we stuy the problem of fining all approximate gappe palinromes in a string. More specifically, we allow the Levenshtein istance between two arms to be at most k an the length of gap to be a fixe q. It shoul be note that we ientify all such palinromes while a previous work [1] can only ientify a set of them. We show that solving our problem is essentially solving the incremental string comparison problem. The research of incremental string comparison was initiate by Lanau et al.. The incremental approach propose in [9] allows one to appen a single symbol to one string at a time. Throughout the paper we aim to solve a more general incremental problem in which we allow multiple symbols to be appene to an elete from the strings. As a result, our algorithm fins all maximal approximate gappe palinromes in O(kn) time while the best previous work requires O(k 2 n) time [11], where n is the string size. 2 Preliminaries The Levenshtein istance (eit istance) between two strings is the minimum number of eiting steps that convert one string into another. Given two string A[1..m] anb[1..n], one can calculate their eit istance by ynamic programming. We refer to matrix D[0..m, 0..n] astheeit-istance table of strings A an B. Initially, D[i, 0] = i for 0 i m an D[0,j]=j for 1 j n. Then the cell D[i, j], where i, j > 0, stores the eit istance of strings A[1..i] an B[1..j]. We also say that the cells D[i, j] wherej i = are on iagonal of D. Below we introuce some important properties of the eit-istance table that are constantly use in later iscussion.

3 1086 P.-H. Hsu, K.-Y. Chen, an K.-M. Chao Lemma 1 ([13]). Given an eit-istance table D, we have D[i, j] D[i 1,j 1] {0, 1}. In other wors, Lemma 1 implies that D[i, j] D[i 1,j 1], i.e. the values of the cells on the same iagonal are monotonically increasing. To compute table D, as notice by Ukkonen [13], it suffices to fin the last cell that has value h, for all h, on each iagonal of table D. Inotherwors,allvaluesinD are etermine when those last cells that have value h, for all h, are etermine. To specify the last cells with value h, thetermsl D (h) anh-cell are efine as follows. Definition 1. Given an eit-istance table D, we efine L D (h) to be the last cell on iagonal that has value h. Inotherwors,L D (h) is the cell D[i, j] where i =max{i : D[i,i + ] =h} an j = i +. Furthermore, we refer to such cell L D (h) as the h-cell on iagonal of D. As mentione in [9,13], one can compute the inex (i, j) ofl D (h) for all h>0 by the following recurrence: i 1 i = Slie max i 2 +1,,j = i +, i 3 +1 where Slie (i) =max{i : A[i..i ]=B[i +, i + ]} an i 1,i 2,ani 3 are the row inices of L D 1 (h 1),LD (h 1) an LD +1 (h 1). Notice that, for brevity we o not explicitly specify the bouns of the value of h. However, all the h-cells are assume to be vali while mentione. The following lemmas emonstrate the relationship between the values of ajacent cells of D. Lemma 2 ([14]). Given an eit-istance table D, we have D[i, j] D[i 1,j],D[i, j] D[i, j 1] { 1, 0, 1}. Lemma 3. Given an eit-istance table D, we have D[i +1,j],D[i, j +1] {h, h +1} for any h-cell D[i, j] in D. Proof. Lemma 3 follows Lemmas 1, 2 immeiately. 3 Incremental String Comparison Recall that given a string S of size n, aparameterq 0 an a threshol k>0, our problem is to fin all maximal approximate gappe palinromes in S that are of gap size q an the eit istance between its two arms is at most k. Here we sketch a basic algorithm for fining all approximate gappe palinromes. Note that the maximal property implies that each pair of prefixes foun in the basic algorithm cannot be further extene. At iteration i, tofinallpairwise prefixes one can compute the eit-istance table of strings S[1..i] r an S[i + q +1..n]. At iteration i + 1, we nee to compute the eit-istance table of Cat(S[i +1],S[1..i]) r )andel(1,s[i + q +1..n]). Here Cat(Ŝ,S) enotes the string obtaine by concatenating string Ŝ an string S, andel(c, S) enotes

4 Fining All Approximate Gappe Palinromes 1087 proceure Basic Algorithm 1 for i =1to n q 1 o 2 fin all possible pairs of prefixes of (S[1..i]) r an S[i + q +1..n] such that 3 the eit istance of pairwise prefixes is at most k. 4 en for Fig. 1. A basic algorithm for fining all approximate gappe palinromes the string obtaine by eleting the first c symbols from S. Thus the challenge lies in, given two strings A an B how to compare Cat(a, A) andel(1,b), provie the comparison result of A an B, wherea is an aitional symbol. That is we aim to solve an incremental string comparison problem. In what follows we formulate this problem in a more general form. Problem 1. The (+t 1, t 2 )-Incremental-String-Comparison Problem (abbr. (+t 1, t 2 )-ISC). Given the eit-istance table of strings A an B, how to efficiently compute the eit-istance table of strings Cat(Â, A) an Del(t 2,B), where  is an arbitrarily appene string with size t 1 (t 1 =  ) an t 2 is the number of elete symbols from B? The (+1,0)-ISC problem has been stuie by Lanau et al.. In [9], they show that there exist nice properties between the given eit-istance table an the esire eit-istance table. Base on the properties observe, they further evise a clever incremental algorithm to compute the h-cells for all h 0 on each iagonal of the esire table. In this paper we exten their ieas an show that the same properties hol even for the more general (+t 1, t 2 )-ISC problem, where t 1 0 an 0 t 2 n. Let D enote the eit-istance table of strings A[1..m] anb[1..n], an let D enote the eit-istance table of Cat(Â, A) andel(t 2,B), where  = t 1.The (+t 1, t 2 )-ISC problem is to compute table D from table D. For convenience of our iscussion, we label the inices of table D by an offset. We assume that the first row of D is of inex t 1 an the first column of D is of inex t 2, i.e. D [ t 1..m, t 2..n]. Notice that D[i, j] stores the eit istance of A[1..i]anB[1..j], an D [i, j] stores the eit istance of Cat(Â, A[1..i]) an Del(t 2,B[1..j]). We efine the ifference table C of D an D as follows. Definition 2. The ifference table of D an D is table C[0..m, t 2..n] whose entry C[i, j] =D [i, j] D[i, j] for 0 i m an t 2 j n. In other wors, table C can be obtaine by (1) putting table D on the top of table D such that the cells of the same inices between the two tables are overlappe, an (2) storing the ifference of values of two overlappe cells in a cell of C. Figure 2 shows an example where t 1 =1ant 2 =2. Kim an Park [7] showe that for the case where either t 1 =1ort 2 =1,the value range of the cells in C is constraine. Besies, there exist some monotonic properties in table C. We shall emonstrate that these properties hol for the

5 1088 P.-H. Hsu, K.-Y. Chen, an K.-M. Chao Column n Table D [-1..m, 2..n] Row -1 Row 0 Row 1 Row 2 Row 3 Table C[0..m, 2..n] Row m Table D[0..m,0..n] Diagonal 1 Diagonal 0 Diagonal -1 Fig. 2. Table D [ 1..m, 2..n] (gray cells) is place on the top of table D[0..m, 0..n] (white cells) an the overlappe region forms the ifference table C[0..m, 2..n] (ark cells) more general cases where t 1 0an0 t 2 n. Due to space limitations, the proofs of the following lemmas are omitte. Lemma 4. For 1 i m an t 2 +1 j n, min{c[i 1,j],C[i 1,j 1],C[i, j 1]} C[i, j] max{c[i 1,j],C[i 1,j 1],C[i, j 1]}. Lemma 5. For any column j of C, we have C[0,j] C[1,j]... C[m, j]. Lemma 6. For any row i of C, we have C[i, t 2 ] C[i, t 2 +1]... C[i, n]. We introuce the following notations for comparing the orer of inices of cells in D an D. Definition 3. Given cell D[i, j] an cell D [i,j ], we efine D[i, j] D [i,j ] iff i i. Similarly, we efine D[i, j] D [i,j ] iff i>i. Moreover, we efine D[i, j] D [i,j ] iff i = i an j = j.wesaycelld[i, j] coincies with cell D [i,j ] iff D[i, j] D [i,j ] Suppose that table D is place on the top of D as before. For those h-cells L D (h) lying in the overlappe region, we efine function f, analogous to the notion of key values mentionein[9]. Definition 4. For an h-cell L D (h), weletf(ld (h)) = g if LD (h) LD (g) for some integer g. Otherwise, we let f(l D (h)) = g 1 2 such that g =min{g : L D (g ) L D (h)}. Now we prove the central theorem of our algorithm. Again, we only consier the h-cells L D (h) lying in the overlappe region. Theorem 1. For h-cells L D (h) an LD +1 (h), we have f(ld +1 (h)) f(l D (h)).

6 Fining All Approximate Gappe Palinromes 1089 Fig. 3. (a) Case 1: D [i, j+1] = h (the gray cell). In this case, we show D[i h +1,j h +1] = min{d[i h +1,j h ]+1,D[i h,j h ]+δ(i h +1,j h +1),D[i h,j h +1]+1} >g. Note that each circle represents two overlappe cells when table D is place on the top of table D.The top-right value in a circle inicates the value of the cell from D an the bottom-left value in a circle inicates the value of the cell from D. (b)case2:d[i, i++1] = h+1 (gray cell). In this case, we show that D [i g,j g] h by first proving that D[i g+1,j g]=g an D [i g +1,j g] h. Proof. Let (i, j) be the position of L D (h) an D[i, j] = g. To prove f(l D +1 (h)) f(ld (h)), it must be shown that f(ld +1 (h)) g. By Lemma 1, D [i, j +1] {h, h +1}. Depening on the value of cell D [i, j + 1], we ivie the proof into two cases. Case 1: D [i, j +1]=h. (See Figure 3(a) for an illustration.) Observe that, by efinition, f(l D +1 (h)) g iff LD +1 (h) LD +1 (g). Let (i h,j h ) be the position of L D +1 (h). We have LD +1 (h) LD +1 (g) iff D[i h +1,j h +1] >g. By the recursive relation, we have D[i h +1,j h +1] = min{d[i h +1,j h ]+1,D[i h,j h ]+δ(i h +1,j h +1),D[i h,j h +1]+1}. In the following, we show that the three parameters of function min are greater than g, implying D[i h +1,j h +1]>g. By Lemma 1, D[i h +1,j h ] D[i, j] =g since they both are cells on iagonal an i h +1 is greater than i. Thus,D[i h +1,j h ]+1 > g.wenextshow that the secon parameter, D[i h,j h ]+δ(i h +1,j h + 1), is greater than g. By Lemma 6, C[i, j] C[i, j +1]. That is, D [i, j] D[i, j] D [i, j +1] D[i, j +1]. Because D [i, j] =h, D[i, j] =g an D [i, j +1]=h, wehaved[i, j +1] g. Accoring to Lemma 1, D[i h,j h ] D[i, j +1] g (because they are cells on iagonal +1 an i h i). Besies, since L D +1 (h) = cell D [i h,j h ], which implies A[i h +1] B[i j +1],we haveδ(i h +1,j h +1)=1.Henceweobtain D[i h,j h ]+δ(i h +1,j h +1)>g.

7 1090 P.-H. Hsu, K.-Y. Chen, an K.-M. Chao Finally, we show that the thir parameter, D[i h,j h + 1] + 1, is greater than g. Accoring to Lemma 6, C[i h,j h ] C[i h,j h + 1]. That is, D [i h,j h ] D[i h,j h ] D [i h,j h +1] D[i h,j h + 1]. Together with Lemma 3, we have (1) D [i h,j h ]=h, (2) D[i h,j h ] g an (3) D [i h,j h +1] {h, h +1}. Thus, we conclue that D[i h,j h +1] g, implying D[i h,j h +1]+1>g. Case 2: D [i, j +1]=h + 1. (See Figure 3(b) for an illustration.) If L D +1 (g) oes not exist, i.e. the smallest value of the cells on iagonal +1 of D is at least g +1, we have f(l D +1 (h)) g immeiately. Thus, we assume LD +1 (g) exists. By efinition, f(l D +1 (h)) g iff LD +1 (h) LD +1 (g). Let (i g,j g )bethe position of L D +1 (g). We have LD +1 (h) LD +1 (g) iff h D [i g,j g ]. Hence, in the following we show that h D [i g,j g ]. First we show that i g <i. By Lemma 6, D [i, j] D[i, j] D [i, j+1] D[i, j+ 1]. Besies, we have (1) D [i, j] =h, (2)D[i, j] =g an (3) D [i, j +1] = h+1. It follows that D[i, j +1] g +1, implying i>i g. Combining i>i g with Lemma 1, we further evaluate D [i g +1,j g ]and[i g +1,j g ] as follows. Since D [i g +1,j g ] D [i, j] by Lemma 1 an D [i, j] =h, weerived [i g + 1,j g ] h. Since D[i g +1,j g ] D[i, j] by Lemma 1 an D[i, j] =g, weerived[i g + 1,j g ] g. Moreover,sinceD[i g +1,j g ] {g, g +1} by Lemma 3, it follows that D[i g +1,j g ]=g. By Lemma 5, D [i g +1,j g ] D[i g +1,j g ] D [i g,j g ] D[i g,j g ]. Since D [i g + 1,j g ] h, D[i g +1,j g ]=g, and[i g,j g ]=g, we conclue that h D [i g,j g ]. 4 The Algorithm In this section, inspire by [9], we escribe an algorithm for the (+t 1, t 2 )-ISC problem as follows. Given an eit-istance table D, forafixeh we refer to the list of all h-cells of D as the h-wave of D. A wave is implemente as a oubly-linke list, in which each noe stores the table inex of an h-cell in D. Thus, we can access the h-cells on iagonals 1an + 1 in constant time when provie the pointer to the h-cell on iagonal. Besies, we also construct links across the waves to connect those cells on the same iagonal. Therefore, we can also access the (h-1)-cell an (h+1)-cell on iagonal in constant time when given the pointer to the h-cell on iagonal. As mentione before, let D enote the eit-istance table of strings A an B, anletd enote the eit-istance table of Cat(Â, A) andel(t 2,B), where  = t 1. Note that we label the inices of D by an offset, i.e. the first row is labele t 1 an the first column is labele t 2. The algorithm takes the waves of D as input, an aims to compute the 0-wave, 1-wave,..., k-wave of D.Below we show how the scheme works.

8 Fining All Approximate Gappe Palinromes 1091 We partition a wave of D into a series of blocks accoring to their f s values, i.e. the cells having the ientical f s value are in the same block. By Theorem 1, each block contains either a single cell or several consecutive cells of the wave. For those cells in the same block, if their f s values are integer g, weknoweach of those cells coincies with a certain g-cell of D. Observe that if two noes of the oubly-linke lists represent a cell x of D an x s coincie cell in D respectively, these two noes store the same table inex. That is, we can imitate one noe by the other noe if they represent the coincie cells. Thus, in the implementation we can construct this block by cutting a piece from the g- wave. Ifthef s value of a cell is not an integer, i.e. g 1 2, we compute this cell by the recurrence in Section 2. Observe that the waves of D may inclue cells lying outsie the overlapping region of D an D. We can not obtain those cells (which are outsie the overlapping region) of D by cutting the waves of D; however, they can be compute by the recurrence in Section 2. Definition 5. For h 0, the partition of the h-wave can be parameterize as: s(h): the number of blocks; b i (h): thef s value of cells in the i th (for 1 i s(h)) block; l i (h): the left bounary of the i th (for 1 i s(h)) block (formally, l i (h) is the smallest such that f(l D (h)) = bi (h)); r i (h): the right bounary of the i th (for 1 i s(h)) block (formally, r i (h) is the largest such that f(l D (h)) = bi (h)). The number of blocks, s(h), epens on the possible f s values of cells. Lemma 7 shows that the value of s(h) is boune by O(t 1 + t 2 ). Besies, with Lemma 8 an Corollary 1 we can obtain the f s values of h-cells by checking the f s values of (h-1)-cells. As space is limite, the proofs of these lemmas are omitte. Lemma 7. For h 0, we have s(h) =O(t 1 + t 2 ). Lemma 8. For iagonals an, <,iff(l D (h 1)) = f(ld (h 1)) = g, we have f(l D (h)) = g +1 for < <. Corollary 1. For bounaries l i (h 1) an r i (h 1) of the i th block of the (h 1)-wave, we have f(l D (h)) = bi (h 1) + 1 for l i (h 1) <<r i (h 1). By Corollary 1 we know how to obtain the f s values of h-cells lying in between the bounaries of each block of the (h-1)-wave. These h-cells are calle trivial since we can immeiately retrieve their f s values by examining the partition. As for the remaining h-cells, we compute them by the recurrence in Section 2 an then obtain their f s values. As for the remaining h-cells, theirf s values can be obtaine once we have their table inices, which can be compute by the recurrence in Section 2. Moreover, those h-cells are calle non-trivial. Now we have the f s value (if exists) of each h-cell. The partition of the h- wave is easily etermine by checking the f s values of all the non-trivial h-cells.

9 1092 P.-H. Hsu, K.-Y. Chen, an K.-M. Chao We then cut the corresponing pieces of waves of D to assemble the h-wave of D. Combining those pieces with the non-trivial h-cells compute, we erive the whole h-wave. Theorem 2. There exists an algorithm that computes the 0-wave,..., an k-wave of D an their partitions in O((t 1 + t 2 ) k) time. Proof. We analyze the time of computing the non-trivial cells an the time of oing the cutting operations. For an h-cell, the(h-1)-cell on the same iagonal may (1) lie in the overlappe region, (2) lie outsie the overlappe region, or (3) not exist. Therefore, the nontrivial h-cells can be ivie into three groups. In the following we show that there are in total O((t 1 + t 2 ) k) non-trivial h-cells for h =0, 1,..., an k. The first group are those cells on the iagonals l 1 (h 1), r 1 (h 1), l 2 (h 1), r 2 (h 1),...,l s(h 1) (h 1), an r s(h 1) (h 1). Thus, by Lemma 7 we know that there are O(t 1 + t 2 ) cells for each fixe h, implying that there are in total O((t 1 + t 2 ) k) cells for h =0, 1,..., an k. To calculate the size of the secon group, we count the number of (h-1)-cell lying outsie the overlappe region. We have that the number of 0-cells,1-cells,..., an (k-1)-cells lying outsie the overlappe region is O(t 1 k). The reasons are that (1) all the 0-cells,1- cells,..., an (k-1)-cells lie on O(k) ifferent iagonals an (2) each iagonal of D contains at most t 1 cells lying outsie the overlappe region. The thir group only contains the leftmost an rightmost cells of each wave of D, leaing to a total of O(k) cells for h =1, 2,...,ank. With the help of suffix tree (constructe in avance) an constant-time LCA queries, each non-trivial cell can be compute in constant time. Thus, the computation time of those nontrivial h-cells is O((t 1 + t 2 ) k) time. Due to the oubly-linke list structure, each cutting operation can be one in constant time. Besies, for each wave of D there are O(t 1 + t 2 )blocksby Lemma 7. Therefore, we can o all the cutting operations in O((t 1 + t 2 ) k) time. Theorem 3. Given a string S of size n an a threshol k, wherek specifies the eit istance between two arms, the problem of fining all maximal approximate gappe palinromes (with fixe gap length) can be solve in O(kn) time. Proof. We have shown that to solve the problem is essentially to solve the (+1,- 1)-ISC problem. We first construct the generalize suffix tree for strings S an S r (the reverse of S) in linear time. Thus, the Slie function in Section 2 can be one in constant time. It immeiately implies that uring the process of solving the (+1, 1)-ISC problem, we can compute the h-cell in constant time. Thus, by Theorem 2 each iteration (except the first one) in the basic algorithm of Figure 1 spens O(k) time. By using the approach of [14], the first iteration can be one in O(k 2 ) time. Hence the total time is O(kn).

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