CSE 421 Algorithms Richard Anderson Lecture 19 Longest Common Subsequence.

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CSE 421 Algorithms Richard Anderson Lecture 19 Longest Common Subsequence

C=c 1 …c g is a subsequence of A=a 1 …a m if C can be obtained by removing elements from A (but retaining order) LCS(A, B): A maximum length sequence that is a subsequence of both A and B ocurranec occurrence attacggct tacgacca Instructor Example

Determine the LCS of the following strings Student Submission BARTHOLEMEWSIMPSON KRUSTYTHECLOWN

String Alignment Problem Align sequences with gaps Charge  x if character x is unmatched Charge  xy if character x is matched to character y CAT TGA AT CAGAT AGGA

LCS Optimization A = a 1 a 2 …a m B = b 1 b 2 …b n Opt[j, k] is the length of LCS(a 1 a 2 …a j, b 1 b 2 …b k )

Optimization recurrence If a j = b k, Opt[j,k] = 1 + Opt[j-1, k-1] If a j != b k, Opt[j,k] = max(Opt[j-1,k], Opt[j,k-1])

Give the Optimization Recurrence for the String Alignment Problem Charge  x if character x is unmatched Charge  xy if character x is matched to character y Student Submission

Dynamic Programming Computation

Write the code to compute Opt[j,k] Student Submission

Storing the path information A[1..m], B[1..n] for i := 1 to m Opt[i, 0] := 0; for j := 1 to n Opt[0,j] := 0; Opt[0,0] := 0; for i := 1 to m for j := 1 to n if A[i] = B[j] { Opt[i,j] := 1 + Opt[i-1,j-1]; Best[i,j] := Diag; } else if Opt[i-1, j] >= Opt[i, j-1] { Opt[i, j] := Opt[i-1, j], Best[i,j] := Left; } else { Opt[i, j] := Opt[i, j-1], Best[i,j] := Down; } a 1 …a m b 1 …b n

How good is this algorithm? Is it feasible to compute the LCS of two strings of length 100,000 on a standard desktop PC? Why or why not. Student Submission

Observations about the Algorithm The computation can be done in O(m+n) space if we only need one column of the Opt values or Best Values The algorithm can be run from either end of the strings

Divide and Conquer Algorithm Where does the best path cross the middle column? For a fixed i, and for each j, compute the LCS that has a i matched with b j

Constrained LCS LCS i,j (A,B): The LCS such that –a 1,…,a i paired with elements of b 1,…,b j –a i+1,…a m paired with elements of b j+1,…,b n LCS 4,3 (abbacbb, cbbaa)

A = RRSSRTTRTS B=RTSRRSTST Compute LCS 5,1 (A,B), LCS 5,2 (A,B),…,LCS 5,9 (A,B) Student Submission

A = RRSSRTTRTS B=RTSRRSTST Compute LCS 5,1 (A,B), LCS 5,2 (A,B),…,LCS 5,9 (A,B) Instructor Example jleftright

Computing the middle column From the left, compute LCS(a 1 …a m/2,b 1 …b j ) From the right, compute LCS(a m/2+1 …a m,b j+1 …b n ) Add values for corresponding j’s Note – this is space efficient

Divide and Conquer A = a 1,…,a m B = b 1,…,b n Find j such that –LCS(a 1 …a m/2, b 1 …b j ) and –LCS(a m/2+1 …a m,b j+1 …b n ) yield optimal solution Recurse

Algorithm Analysis T(m,n) = T(m/2, j) + T(m/2, n-j) + cnm

Prove by induction that T(m,n) <= 2cmn Instructor Example