New Efficient Algorithms for LCS and Constrained LCS Problem

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Algorithm 2 AlgCLCSNew1: Construct the set M using Algorithm 1 of [24]. Let M i = (i, j) ∈ M, 1 ≤ j ≤ n.2: globalCLCS.Instance = ɛ3: globalCLCS.Value = ɛ4: H −10 = ɛ5: H 0 0 = ɛ6: for i = 1 to n do7: for k = 0 to p do8:9:Hi k = Hi−1kfor each (i, j) ∈ M i do10: max 1 = BoundedMax(H k−1i−1 , j)11: max 2 = BoundedMax(Hi−1, k j){First, we consider Equations 7 and 8}12: Val 2 .Value = 013: Val 2 .Instance = ɛ14: if ((k = 0) or (max 2 .Value > 0)) then15: Val 2 = max 216: end if{Now, we consider Equations 5 and 6}17: if (X[i] = Z[k]){This means X[i] = Y [j] = Z[k]} then18: Val 1 .Value = 019: Val 1 .Instance = ɛ20: if (k = 1) or ((k > 1) and (max 1 .V alue > 0)) then21: Val 1 = max 122: end if23: end if{Finally, we consider Equation 2}24: maxresult = max{Val 1 , Val 2 }25: T k [i, j].Value = maxresult.Value + 126: T k [i, j].Prev = maxresult.Instance27: if globalLCS.Value < T k [i, j].Value then28: globalLCS.Value = T k [i, j].Value29: globalLCS.Instance = (i, j, k)30: end if31: IncreaseValue(Hi k , j, T [i, j].Value, (i, j, k)).32: end for33: Delete H k−1i−1 .34: end for35: end for36: return globalLCS
– We update H k i , which is initialized to H k i−1 when we start row i. H k i is later usedwhen we consider the matches of row i + 1 of the two matrices T k and T k+1Due to Fact 4, while computing a T k [i][j], (i, j) ∈ M, 1 ≤ i, j ≤ n, 1 ≤ k ≤ p, we canemploy the same technique with Hi k , Hi−1 k and H k−1i−1 used in AlgLCSNew (with H iand H i−1 to compute T [i, j], (i, j) ∈ M, 1 ≤ j ≤ n). On the other hand, courtesyto Fact 3, to realize Equations 5−8, we need only keep track of H k−1i−1 and Hk i−1. Thesteps are formally stated in Algorithm 2. In light of the analysis of AlgLCSNew, it isquite easy to see that the running time is O(pR log log n + n). The space complexity,like AlgLCSNew, is dominated by the preprocessing step of [24] because in the mainalgorithm, we only need to keep track of 3 BoundedHeap data structures and thethird string Z, requiring in total O(n) space.Theorem 3. Problem CLCS can be solved in O(pR log log n + n) time requiringθ(max{R, n}) space.5 ConclusionIn this paper, we have studied the classic and well-studied longest common subsequence(LCS) problem and a recent variant of it namely constrained LCS (CLCS)problem. In LCS, given two strings, we want to find the common subsequence havingthe highest length; in CLCS, in addition to that, the solution to the problemmust also be a supersequence of a third given string. The traditional LCS problemhas extensive applications in diverse areas of computer science including DNA sequenceanalysis, genetics and molecular biology. CLCS, a relatively new variant ofLCS, gets its motivation from computational bio-informatics as well. In this paper,we have presented an efficient algorithm for the traditional LCS problem that runsin O(R log log n + n) time. Then, using this algorithm, we have devised an algorithmfor the CLCS problem having time complexity O(pR log log n + n) in the worst case.Note that, if R = o(n 2 ), our algorithm will perform very well but, if R = O(n 2 ), then,due to the log log n term, our algorithms will behave slightly worse than the existingalgorithms. It would be interesting to see whether our techniques can be employed toother variants of LCS problem to devise similar efficient algorithms.References1. V. Arlazarov, E. Dinic, M. Kronrod, and I. Faradzev. On economic construction of the transitive closureof a directed graph (english translation). Soviet Math. Dokl., 11:1209–1210, 1975.2. A. N. Arslan and Ö. Eğecioğlu. Algorithms for the constrained longest common subsequence problems.In M. Simánek and J. Holub, editors, The Prague Stringology Conference, pages 24–32. Department ofComputer Science and Engineering, Faculty of Electrical Engineering, Czech Technical University, 2004.3. A. N. Arslan and Ö. Eğecioğlu. Algorithms for the constrained longest common subsequence problems.Int. J. Found. Comput. Sci., 16(6):1099–1109, 2005.4. S. Bereg and B. Zhu. RNA multiple structural alignment with longest common subsequences. InL. Wang, editor, Computing and Combinatorics (COCOON), volume 3595 of Lecture Notes in ComputerScience, pages 32–41. Springer, 2005.
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Page 3 and 4: Given two strings X[1..n] and Y [1.
Page 5 and 6: to H i−1 . Assume that we are con
Page 7: V 1 = max {T [l i , l j , k − 1]}

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