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problem, we are given as input two sequences, X (*1, *2, ...,Im) nd Y = (y1,y2;-.., Yn) and would like to find a longest subsequence common to both. Towards this end, we define c[i, j] := length of LCS of X; and Yj, where X; := (x1,12, ...r;) nd Y; := (y1, 42, ..Y;) and obtain the following recurrence. if i = 0 or j = 0 if i, j > 0 and r; = Yj (max(c[i, j – 1], c[i – 1, j]) if i, j > 0 and r; # yj cļi, j] = { c[i – 1,j – 1] +1

problem, we are given as input two sequences, X (*1, *2, ...,Im) nd Y = (y1,y2;-.., Yn) and would like to find a longest subsequence common to both. Towards this end, we define c[i, j] := length of LCS of X; and Yj, where X; := (x1,12, ...r;) nd Y; := (y1, 42, ..Y;) and obtain the following recurrence. if i = 0 or j = 0 if i, j > 0 and r; = Yj (max(c[i, j – 1], c[i – 1, j]) if i, j > 0 and r; # yj cļi, j] = { c[i – 1,j – 1] +1

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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In the LCS problem, we are given as input two sequences, X = (r1, x2,..., Tm) and Y = (y1, Y2,..., Yn) and would like to find a longest subsequence common to both. Towards this end, we define c[i, j] := length of LCS of X; and Y;, where X; := (x1, r2, .r;) and Y; := (y1, y2,.y;) and obtain the following recurrence. if i = 0 or j = 0 if i, j > 0 and r; = Yj c[i, j] c[i – 1, j – 1] +1 max(c[i, j – 1], c[i – 1,j]) if i, j > 0 and x; # yj Fill out the following LCS DP table of entries c[i, j]. j012345 ACBAB i Yj 0 xi 1 A 2 В 3 C 4 В
Transcribed Image Text:In the LCS problem, we are given as input two sequences, X = (r1, x2,..., Tm) and Y = (y1, Y2,..., Yn) and would like to find a longest subsequence common to both. Towards this end, we define c[i, j] := length of LCS of X; and Y;, where X; := (x1, r2, .r;) and Y; := (y1, y2,.y;) and obtain the following recurrence. if i = 0 or j = 0 if i, j > 0 and r; = Yj c[i, j] c[i – 1, j – 1] +1 max(c[i, j – 1], c[i – 1,j]) if i, j > 0 and x; # yj Fill out the following LCS DP table of entries c[i, j]. j012345 ACBAB i Yj 0 xi 1 A 2 В 3 C 4 В
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