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Longest Common Sequence (LCS)

A subsequence of a given sequence is just the given sequence with some elements left out.

Given two sequences X and Y, we say that the sequence Z is a common sequence of X and Y if Z is a subsequence of both X and Y.

In the longest common subsequence problem, we are given two sequences X = (x1 x2....xm) and Y = (y1 y2 yn) and wish to find a maximum length common subsequence of X and Y. LCS Problem can be solved using dynamic programming.

Characteristics of Longest Common Sequence

A brute-force approach we find all the subsequences of X and check each subsequence to see if it is also a subsequence of Y, this approach requires exponential time making it impractical for the long sequence.

Given a sequence X = (x1 x2.....xm) we define the ith prefix of X for i=0, 1, and 2...m as Xi= (x1 x2.....xi). For example: if X = (A, B, C, B, C, A, B, C) then X4= (A, B, C, B)

Optimal Substructure of an LCS: Let X = (x1 x2....xm) and Y = (y1 y2.....) yn) be the sequences and let Z = (z1 z2......zk) be any LCS of X and Y.

  • If xm = yn, then zk=x_m=yn and Zk-1 is an LCS of Xm-1and Yn-1
  • If xm ≠ yn, then zk≠ xm implies that Z is an LCS of Xm-1and Y.
  • If xm ≠ yn, then zk≠yn implies that Z is an LCS of X and Yn-1

Step 2: Recursive Solution: LCS has overlapping subproblems property because to find LCS of X and Y, we may need to find the LCS of Xm-1 and Yn-1. If xm ≠ yn, then we must solve two subproblems finding an LCS of X and Yn-1.Whenever of these LCS's longer is an LCS of x and y. But each of these subproblems has the subproblems of finding the LCS of Xm-1 and Yn-1.

Let c [i,j] be the length of LCS of the sequence Xiand Yj.If either i=0 and j =0, one of the sequences has length 0, so the LCS has length 0. The optimal substructure of the LCS problem given the recurrence formula

DAA Characteristics of Longest Common Sequence

Step3: Computing the length of an LCS: let two sequences X = (x1 x2.....xm) and Y = (y1 y2..... yn) as inputs. It stores the c [i,j] values in the table c [0......m,0..........n].Table b [1..........m, 1..........n] is maintained which help us to construct an optimal solution. c [m, n] contains the length of an LCS of X,Y.

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