Computing the Longest Common Subsequence of Multiple RLE Strings

Transcription

1 The 29th Workshop on Combinatorial Mathematics an Computation Theory Computing the Longest Common Subsequence of Multiple RLE Strings Ling-Chih Yao an Kuan-Yu Chen Grauate Institute of Networking an Multimeia Department of Computer Science an Information Engineering National Taiwan University, Taipei, Taiwan 06 Abstract In this paper, we stuy the problem of computing the longest common subsequence (LCS) of multiple compresse strings. The input strings are all represente by run-length encoe format. We present two algorithms for the problem. Given strings, which are compresse into of n,..., n runs respectively, our algorithm computes the LCS of the strings in O(nm) time, where n = i= n i an m = i= n i. The secon algorithm achieves O(nρ Σ ) time, where ρ enotes the number of ark blocks plus one an Σ enotes the alphabet size. Introuction Longest common subsequence (LCS) serves as a well-known measure of the similarity between strings, while run-length encoing (RLE) is a classic metho to compress strings. In this paper, we stuy a problem which lies in the intersection of the two notions, i.e. the problem of computing the similarity between compresse strings. More specifically, we consier the problem of computing longest common subsequence (LCS) of run-length encoing (RLE) strings. Several works on computing the LCS of RLE strings have been propose. Bunke an Csirik [4] propose an algorithm running in O(Nl + km) time for strings of lengths M an N with l an k runs, respectively. The algorithm ivies the ynamic programming table into blocks an computes only the entries on the bounaries of the blocks. Apostolico et al. [3] gave an algorithm running in O(kl(k+l)) time which relies on the notion of force an corner paths. They next improve upon the time of the algorithm to O(kl log kl) time This is the footnote of this paper. by storing all the possible force paths in balance binary search trees. Ann an Yang et al. [2] propose an algorithm of O(kl + min{p, p 2 }) time, where blocks are ivie into ark an light blocks an p an p 2 enote the bottom an right bounaries of ark blocks. Computing the similarity between RLE strings has been stuie in more general istance settings. For example, computing the Levenshtein istance between RLE strings was explore in [5, 9]. Computing the optimal alignment between RLE strings uner unrestricte scoring matrices was investigate in [6, 8]. In this paper, we stuy the problem of computing the LCS between RLE strings. In particular, we consier a natural generalization of the problem, i.e. computing the LCS of multiple strings. In other wors, our algorithms are capable of computing the LCS between not only two RLE strings but multiple RLE strings. The paper is organize as follows. We first escribe an efficient algorithm which can irectly cope with compresse strings, without resorting to any ecompression. Given strings, which are compresse into n,..., n runs respectively, our first algorithm achieves the time complexity of O(nm) time, where n = i= n i an m = i= n i. We next present an improve algorithm which achieves O(nρ Σ ) time, where ρ enotes the number of ark blocks plus one an Σ enotes the alphabet size. 2 Preliminaries A string S is run-length encoe (RLE) if it is escribe as an orere sequence of pairs (σ, i), where σ is an alphabet symbol an i is a positive integer. Each pair correspons to a run of S which consists of i consecutive occurrences of σ. For example, aaabbbbcc can be encoe into 228

2 The 29th Workshop on Combinatorial Mathematics an Computation Theory a 3 b 4 c Notation Consier strings S i for i =, 2,..., over a finite alphabet Σ. Suppose that S i can be encoe as an RLE string of n i runs, enote by S i ()S i (2)... S i (n i ), where S i (j) is the j-th run of string S i. Note that we assume that the RLE string of S i is optimally encoe, i.e. each run S i (j) is a maximal run of ientical characters. We let S i (j) enote the run length an symbol(s i (j)) enote the encoe symbol of run S i (j). We let S (r) i enote the first r runs of S i, i.e. S (r) i = S i ()S i (2)... S i (r). We use LCS(S,,..., S ) to enote the length of LCS of strings S to S. Let M be a matrix of blocks, such that M[r, r 2,..., r ] contains the the value of LCS(S (r), S (r2) 2..., S (r ) ), where r i =, 2,..., n i. We say that block (r, r 2,..., r ) is ark if all the runs S (r ), (r 2 ),..., an S (r ) encoe an ientical symbol. Otherwise, we say that block (r, r 2,..., r ) is light. 2.2 Properties for LCS of two RLE strings Before escribing the Lemmas of multiple strings, we first review the known properties of LCS between two RLE strings. Lemma. (see [7]) Let S an be two strings an e r be a run encoing symbol e. Let S e r an e r enote the strings obtaine by appening the run e r to S an. We have that LCS(S e r, e r ) = LCS(S, ) + r. Lemma 2. (see [7]) Let S an be two strings an e r an e r 2 2 be two runs encoing istinct symbols e an e 2. Let S e r an e r 2 2 enote the strings obtaine by appening the run e r to S an the run e r 2 2 to. We have that LCS(S e r, S2er 2 2 ) = max{lcs(s e r, S2), LCS(S, S2er 2 2 )}. 3 LCS of multiple RLE strings We now escribe several properties of the LCS of multiple RLE strings. Lemma 3 is eicate to the case that all the input strings en with the same symbol, an Lemma 4 is eicate to the case that the input strings not en with the same symbol. Lemma 5 is eicate to the case when upating the value of a light block (r, r 2,...r ). Lemma 3. Let S,,..., S be strings an e r be a run. Let S e r, e r,..., S e r enote the strings obtaine by appening run e r to S,,..., S, respectively. We have that LCS(S e r, e r,..., S e r ) = LCS(S,,..., S ) + r. Proof. If LCS(S e, e,..., S e) = LCS(S,,..., S ) + hols, then the lemma can be obtaine by recursively applying the above result r times. Therefore, in the following we aim to show that LCS(S e, e,..., S e) = LCS(S,,..., S ) +. Note that the LCS of S e, e,..., an S e either en with symbol e or not. Case : if the LCS of S e, e,..., an S e en with symbol e, then wherever the ening symbol e appears in S i e for i =,...,, we can always replace it by the last e of S i e. Hence, we obtain that LCS(S e, e,..., S e) = LCS(S,,..., S ) +. Case 2: if the LCS of S e, e,..., an S e oes not en with symbol e, then we have that LCS(S e, e,..., S e) = LCS(S,,..., S ). This leas to a contraiction, for we can obtain a longer common subsequence by appening the last common e to the LCS of S,,..., an S. Lemma 4. Let S,,..., S be strings an let e, e 2,..., e be symbols which are not all the same, i.e. e = e 2 =... = e cannot hol. We have that for any σ {e, e 2,..., e }, LCS(S e r, e r 2 2,..., S e r { LCS(τ(Se r, σ), τ(e r 2 2, σ),..., τ(s e r = max, σ)), LCS( τ(s e r, σ), τ(e r 2 2, σ),..., τ(s e r, σ)), ) { where τ(s ie r i Sie r i, σ) = i i, if ei = σ; S i, otherwise, { τ(s ie r i Si, i, σ) = if e i = σ; S ie r i otherwise. i, an Proof. Given symbol σ {e, e 2,..., e }, we know that the LCS of strings S,,..., S either ens with σ or not. If the LCS ens with symbol σ, the last runs of the strings not encoing symbol σ nee not to be consiere, whereas the the last runs of the strings encoing symbol σ nee to be consiere. The function τ helps us to o this. On the other han, if the LCS oes not en with 229

3 The 29th Workshop on Combinatorial Mathematics an Computation Theory symbol σ, the function τ removes the last runs of the strings encoing symbol σ, while preserving the last runs of the strings not encoing symbol σ. for r 3, r 2 3, r 3 3 are alreay compute. We now show how to compute the value of LCS(S (4), S(4) 2, S(4) 3 ). We recursively apply Lemma 3 an Lemma 4 as follows: Lemma 5. Suppose that runs S (r ), (r 2),..., S (r ) o not encoe the same symbol, i.e. symbol(s (r ))=symbol((r 2))=...=symbol(S (r )) cannot hol. We have that LCS(S (r ), S (r 2) 2,..., S (r ) ) LCS(S (r ), S (r 2) 2,..., S (r ) ), LCS(S (r ), S (r 2 ) 2,..., S (r ) ), = max... LCS(S (r ), S (r 2) 2,..., S (r ) ). Proof. Suppose that the LCS of S (r), S (r2) 2,..., an S (r ) en with symbol σ Σ. Since symbol(s (r )) = symbol( (r 2 )) =... = symbol(s (r )) cannot hol, at least one of the runs S (r ), (r 2 ),..., S (r ) oes not encoe σ. In other wors, there is at least one string whose last run can be ignore. 4 The first algorithm Before escribing the etails of our algorithm, we first take an example to illustrate our basic iea. 4. The basic iea Given three strings S,, an, of n, n 2, an n 3 runs respectively, we aim to compute LCS(S (n), S (n2) 2, S (n3) 3 ). In the following example, n = 4, n 2 = 4, an n 3 = 4. S a 6 c 6 b 3 a 8 a 8 c 2 b 3 a 6 a 4 c 0 b 2 a 4. By Lemma 3, LCS(S (4), S(4) 2, S(4) 3 ) By Lemma 4, max{, S(3) 2, S(3) 3 ), LCS(S(3) 2 a2, S (2) 3 )}. 3. By Lemma 4, 2 a2, S (2) max{, S(3) 2, S(2) 3 ), LCS(S(3) 2 a2, S () 3 )}. 4. By Lemma 3, 2 a2, S () a2, S (3) 2, a2 ) By Lemma 4, a2, S (3) max{, S(3) 2, ɛ), LCS(S(3) a2, S (2) 2, a2 )}. 6. By Lemma 4, a2, S (2) max{, ɛ), LCS(S(3) a2, S () 2, a2 )}., S(2) 2 7. By Lemma 3, a2, S (), a6, ɛ) + 2. Combining the above equations, we can erive that: LCS(S (3), S(3) 2, S(3) 3 ) + 4, LCS(S (4), S(4) 2, S(4) max, S(3) 2, ɛ) + 6,, a6, ɛ) + 8 Since by assumption the value of, S(3) 2, S(3) 8 is alreay compute, an obviously, S(3) 2, ɛ) = 0 an, a6, ɛ) = 0, we obtain that LCS(S,, ) = max{8 + 4, 0 + 6, 0 + 8} = 2. Clearly, the time for the above proceure epens on the epth of the recursions. Figures (a) to () show that the epth of the recursions is no more than the total number of runs in S,, an. For clarity, we remove all the non- a runs of strings S,, an in our illustrations. Our algorithm runs in a ynamic fashion. Suppose that all the values of LCS(S (r), S (r2) 2, S (r3) 3 ) 230

4 The 29th Workshop on Combinatorial Mathematics an Computation Theory LCS(S (4) a 6 a S 8, S(4) 2, S(4) 3 ) a2, S (3) 2, a2 ) a 6 a 8 S By Lemma 3, LCS(S (4), S(4) 2, S(4) 3 )+4. (a) Initially, the vertical line begins at the right bounaries runs S (4), (4), an (4). When applying Lemma 3, the vertical line is move to the left by four characters, stopping at the location shown in Figure (b). a 6 a S 8 3 ) By Lemma 4 (twice), max { 2 a2, S () 3 ), LCS(S(3), S(3) 2, S(3) 3 )}. By Lemma 4 (twice), a2, S (3) max{ a2, S () 2, a2 ),, S(3) 2, ɛ)}. By Lemma 3, a2, S (), a6, ɛ)+2. (c) When applying Lemma 4 an then Lemma 3, the vertical line is move to the left by two characters, stopping at the location shown in Figure ()., a6, ɛ) a 6 a 8 S By Lemma 3, 2 a2, S () a2, S (3) 2, a2 )+2. (b) When applying Lemma 4 an then Lemma 3, the vertical line is move to the left by two characters, stopping at the location shown in Figure (c). () The vertical line reaches the left en of string an the recursion is terminate. 23

5 The 29th Workshop on Combinatorial Mathematics an Computation Theory 4.2 The algorithm Our algorithm for computing the LCS of multiple RLE strings is given in Algorithm. Recall that block (r, r 2,..., r ) is a ark block when S (r ), (r 2 ),..., an S (r ) all encoe the same symbol, an we call the common encoe symbol the ark symbol of block (r, r 2,..., r ). Algorithm. lcs(s,,..., S ) Matrix M is initialize as a zero matrix; 2 for r = to n 3 for r 2 = to n for r = to n 6 if (block (r, r 2,..., r ) is light) 7 M[r, r 2,..., r ] = max{ M[r, r 2,..., r ], M[r, r 2,..., r ],... M[r, r 2,..., r ]}; 8 else 9 Let σ be the ark symbol of block (r, r 2,..., r ); 0 for i = to o D[i] = S i (r i ) ; matche = min{d[i] i =,..., }; 2 M[r, r 2,..., r ] = M[r, r 2,..., r ] + matche; 3 DarkP ath(r,..., r, M, matche, D, σ); in substring S i (r i )... S i(r i ) of string S i. In line 6, variable matche inicates the number of matche ark symbols σ which substrings S (r )... S (r ) to S (r )... S (r ) can contribute to the LCS. In line 7, the value of M[r, r 2,..., r ] is upate base on Lemma 3 an Lemma 4. The while-loop continues until there exists some string whose runs encoing ark symbol σ are completely exhauste. Since there are i= n i runs in total, the while-loop is execute no more than i= n i times. For each iteration of the while-loop, it takes O() time. Hence, the proceure of DarkPath spens O( i= n i) time. Theorem. Given strings S,..., S, compresse into n,..., n runs respectively, Algorithm computes the LCS of S,..., S in O(nm) time, where n = i= n i an m = i= n i. Proof. Each light block takes O() time to upate, while each ark block nees O(n) time to complete proceure DarkPath. As there are m blocks in total, the total time require is O(nm). In line 7, value of M[r, r 2,..., r ] can be upate irectly from Lemma 5. In line 0, D[i] is initialize to the number of ark symbols σ in run S i (r i ). In line, variable matche inicates the number of matche ark symbols σ which runs S (r ) to S (r ) can contribute to the LCS. DarkPath(r,..., r, M, matche, D, σ) r = r,..., r = r ; 2 while (every string contains at least one run encoing ark symbol σ which has not been exhauste) 3 for i = to 4 if (D[i] == matche) 5 upate r i to inicate the former run of S i (r i ) encoing ark symbol σ an upate D[i] to D[i] + S i (r i ) ; 6 matche = min{d[i] i =,..., }; 7 M[r,..., r ] = max{ M[r,..., r ] + matche, M[r,..., r ]}; In lines 3 5, proceure DarkPath ientifies those strings S i whose run S i (r i ) is exhauste. The proceure then upates r i to inicate the former run of S i (r i ) encoing ark symbol σ an upates D[i] to D[i] + S i (r i ). Note that in line 5, D[i] is upate to the number of ark symbols σ 5 A faster algorithm In this section, we improve upon the time of the first algorithm by exploiting the sparsity inherent in the LCS computation. Algorithm 2 will only store values of ark blocks. In aition, we propose Lemma 6 when we nee a value of an light block by comparing the values of ark blocks. Lemma 6. Suppose that runs S (r ), (r 2),..., S (r ) o not encoe the same symbol, i.e. symbol(s (r ))=symbol((r 2))=...=symbol(S (r )) cannot hol. We have that LCS(S (r ), S (r 2) 2,..., S (r ) ) = max{lcs( τ(s (r ), α), τ(s (r 2) 2, α),..., τ(s (r ), α)) α Σ}, where τ(s (r i) i, α) = S (r i) i if symbol(s i(r i)) = α, an τ(s (r i) i, α) = S (r i ) i otherwise. Here, S i(r i) is the former run of S i(r i) which encoes symbol α. Proof. Suppose that the LCS of S (r), S (r2) 2,..., an S (r ) ens with symbol α Σ. Then, the last runs of the strings encoing α nee to be consiere, whereas the last runs of the strings not encoing α nee not to be consiere. The function τ helps us to o this. 232

6 The 29th Workshop on Combinatorial Mathematics an Computation Theory Algorithm 2. lcs(s,..., S ) Matrix M is initialize as a zero matrix; 2 for r = to n 3 σ = symbol(s (r )) 4 for r 2 = [σ].begin to [σ].en for r = S [σ].begin to S [σ].en 7 for i = to o D[i] = S i (r i ) ; 8 matche = min{d[i] i =,..., }; 9 if (M[r,..., r ] == 0) 0 prematche = max{ M[pre(S, r, α),..., pre(s, r, α)] α Σ }; else prematche = M[r,..., r ]; 2 M[r,..., r ] = prematche + matche; 3 DarkBlockOnly(r,..., r, M, matche, D, σ); 4 if (M[n,..., n ] == 0) 5 M[n,..., n ] = max{ M[pre(S, n, α),..., pre(s, n, α)] α Σ } In line 3, symbol(s (r )) returns the encoe symbol σ of run S (r ). In lines 4 6, array S i [σ] recors the inices of runs encoing σ in S i. For example, for string S i = a 5 b 4 c 7 b 3 a 4, S i [ a ] = {, 5}. Preprocessing S i [σ] for σ Σ can be one in O( i= n i) time. Then after line 6, block (r, r 2,..., r ) is a ark block encoing symbol σ, i.e. σ is the ark symbol of block (r, r 2,..., r ). In line 7, D[i] is initialize as the number of ark symbols σ in run S i (r i ). In line 8, variable matche inicates the number of matche ark symbols σ which runs S (r ) to S (r ) can contribute to the LCS. In lines 9, variable prematche inicates value of the LCS of S (r ) to S (r ). If value of M[r,..., r ] is zero, i.e. block (r,..., r ) is light, we compute its value irectly from Lemma 6. Function pre(s i, r i, α) returns the inex of the former run of S i (r i ) which encoes symbol α (inclusive of S i (r i )). For example, for string S = a 5 b 4 c 7 b 3 a 4 c 3, pre(s, 6, a) = 5, pre(s, 3, c) = 3 an pre(s,, b) = 0. Preprocessing pre(s i, r i, α) for α Σ can be one in O( Σ i= n i) time. In lines 4 5, if value of M[n,..., n ] is zero, i.e. block (n,..., n ) is light, we compute its value irectly from Lemma 6 for the LCS of S to S. DarkBolckOnly(r,..., r, M, matche, D, σ) r = r,..., r = r ; 2 while (every string contains at least one run encoing ark symbol σ which has not been exhauste) 3 for i = to 4 if (D[i] == matche) 5 upate r i to inicate the former run of S i (r i ) encoing ark symbol σ an upate D[i] to D[i] + S i (r i ) ; 6 matche = min{d[i] i =,..., }; 7 if (M[r,..., r ] == 0) 8 prematche = max{ M[pre(S, r, α),..., pre(s, r, α)] α Σ } 9 else prematche = M[r,..., r ]; 0 M[r,..., r ] = max{ prematche + matche, M[r,..., r ]} In lines 3 5, proceure DarkBolckOnly ientifies those strings S i whose run S i (r i ) is exhauste. The proceure then upates r i to inicate the former run of S i (r i ) encoing ark symbol σ an upates D[i] to D[i] + S i (r i ). Note that in line 5, D[i] is upate to the number of ark symbols σ in substring S i (r i )... S i(r i ) of string S i. In line 6, variable matche inicates the number of matche ark symbols σ which substrings S (r )... S (r ) to S (r )... S (r ) can contribute to the LCS. In lines 7 9, variable prematche inicates value of the LCS of S (r ) to S (r ). If value of M[r,..., r ] is zero, i.e. block (r,..., r ) is light, we compute its value irectly from Lemma 6. It can be one in O( Σ ) time. In line 0, the value of M[r, r 2,..., r ] is upate base on Lemma 3 an Lemma 4. The while-loop continues until there exists some string whose runs encoing ark symbol σ are completely exhauste. Since there i= n i runs in total, the while-loop is execute no more than i= n i times. For each iteration of the whileloop, it takes O( Σ ) time. Hence, the proceure of DarkBolckOnly spens O( Σ i= n i) time. Theorem 2. Algorithm 2 takes O(nρ Σ ) times, where ρ enotes the number of ark blocks plus one an n = i= n i enotes the total number of runs of strings S,,..., S. Proof. The preprocessing for S i [σ], σ Σ, an pre(s i, r i, α), α Σ, can be one in O(n) an 233

7 The 29th Workshop on Combinatorial Mathematics an Computation Theory O(n Σ ) time, respectively. The light blocks nee not to be upate, while the ark blocks take O(n Σ ) time to run the proceure of Dark- BolckOnly. Hence, the total time require is O(n)+O(n Σ )+ O(n Σ ) (ρ ) time. The time complexity thus follows. Note that Algorithm 2 limits the LCS computation to those ark blocks, whose number is specifie by parameter ρ. Since ρ = O(m), the time of algorithm 2 is never worse than that of algorithm for constant-size alphabets. In other wors, for cases where ark blocks are sparse, algorithm 2 is preferable over algorithm for small alphabets. On the other han, for ranom input strings the probability of the appearance of a ark block is, which implies that ρ = O( m). Hence, we obtain Σ Σ that in average case the time of algorithm 2 is O(nρ Σ ) = O(nm), which is superior to Σ 2 that of algorithm. 6 Concluing Remarks In this paper, we stuy the problem of computing LCS of multiple RLE strings. Two algorithms are presente, both can irectly cope with the RLE strings without resorting to ecompression. The first algorithm works in a ynamic fashion, recursively applying the lemmas escribe in Section 3. The secon algorithm improves upon the time by exploiting the sparsity inherent in the LCS computation. It remains open whether the time can be further reuce by utilizing more efficient ata structures an algorithmic techniques. Common Subsequence of Run-Length Encoe Strings. Information Processing Letters 08(6): , 2008 [3] Alberto Apostolico, Ga M. Lanau, an Steven Skiena: Matching for Run-Length Encoe Strings. Journal of Complexity 5():4 6, 999. [4] Horst Bunke an Janos Csirik: An Improve Algorithm for Computing the Eit Distance of Run-Length Coe Strings. Information Processing Letters 54(2): 93 96, 995. [5] Kuan-Yu Chen, Kun-Mao Chao: A Fully Compresse Algorithm for Computing the Eit Distance of Run-Length Encoe Strings. Algorithmica, to appear. [6] Maxime Crochemore, Ga M. Lanau, Michal Ziv-Ukelson: A Subquaratic Sequence Alignment Algorithm for Unrestricte Scoring Matrices. SIAM Journal on Computing 32(6): , [7] Valerio Freschi, Alessanro Bogliolo: Longest Common Subsequence between Run-Length- Encoe Strings: a New Algorithm with Improve Parallelism. Information Processing Letters 90(4): 67 73, [8] Guan-Shieng Huang, Jia Jie Liu, Yue- Li Wang: Sequence Alignment Algorithms for Run-Length-Encoe Strings. COCOON, , [9] Veli Mäkinen, Esko Ukkonen, Gonzalo Navarro: Approximate Matching of Run- Length Compresse Strings. Algorithmica 35(4): , Acknowlegements The authors woul like to thank Kun-Mao Chao for his constructive comments. References [] Amihoo Amir an Gary Benson: Efficient Two-Dimensional Compresse Matching. DCC : , 992. [2] Hsing-Yen Ann, Chang-Biau Yang, Chiou- Ting Tseng, Chiou-Yi Hor: A Fast an Simple Algorithm for Computing the Longest 234