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Find the recccurence relation of longest common substring- Algorithm b.docx
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### Algorithm below--- a) IF n - 0 return 0 (b) IF m - 0 return 0 (c) IF w.docx

1. Algorithm below... a) IF n = 0 return 0 (b) IF m = 0 return 0 (c) IF w 1 = v 1 return 1 + LCS(w 2 ..w n , v 2 ..v m ) (d) Return max(LCS(w 2 ..w n , v 1 ..v m ), LCS(w 1 ..w n , v 2 ..v m ) Solution Let the input sequences be X[0..m-1] and Y[0..n-1] of lengths m and n respectively. And let L(X[0..m-1], Y[0..n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0..m-1], Y[0..n-1]). If last characters of both sequences match (or X[m-1] == Y[n-1]) then L(X[0..m-1], Y[0..n-1]) = 1 + L(X[0..m-2], Y[0..n-2]) If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then L(X[0..m-1], Y[0..n-1]) = MAX ( L(X[0..m-2], Y[0..n-1]), L(X[0..m-1], Y[0..n-2]) Examples: 1) Consider the input strings â€œAGGTABâ€ and â€œGXTXAYBâ€. Last characters match for the strings. So length of LCS can be written as: L(â€œAGGTABâ€, â€œGXTXAYBâ€) = 1 + L(â€œAGGTAâ€, â€œGXTXAYâ€) 2) Consider the input strings â€œABCDGHâ€ and â€œAEDFHR. Last characters do not match for the strings. So length of LCS can be written as: L(â€œABCDGHâ€, â€œAEDFHRâ€) = MAX ( L(â€œABCDGâ€, â€œAEDFHRâ€), L(â€œABCDGHâ€, â€œAEDFHâ€) ) /* Returns length of LCS for X[0..m-1], Y[0..n-1] */ int lcs( char *X, char *Y, int m, int n ) {
2. if (m == 0 || n == 0) return 0; if (X[m-1] == Y[n-1]) return 1 + lcs(X, Y, m-1, n-1); else return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n)); }
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