CSC 213 Lecture 19: Dynamic Programming and LCS

Subsequences (§ ) A subsequence of a string x 0 x 1 x 2 …x n-1 is a string of the form x i 1 x i 2 …x i k, where i j < i j+1 This is not the same things as a substring! Subsequences can skip letters in the string Substrings must use consecutive letters Example string: ABCDEFGHIJK Subsequence (& substring): DEFGH Subsequence (& NOT substring): ACEFHJ Not subsequence or substring: DAGH

Longest Common Subsequence (LCS) Problem Given two strings X and Y, find longest subsequence in both X and Y Applications in DNA testing (  ={A,C,G,T}) Example the LCS for: ABCDEFG and XZACKDFWGH is ACDFG

Longest Common Subsequence (LCS) Problem Given two strings X and Y, find longest subsequence in both X and Y Applications in DNA testing (  ={A,C,G,T}) Example the LCS for: ABCDEFG and XZACKDFWGH is ACDFG

Dynamic Programming Some problems appear hard There does not seem to be a simple solutions Require a brute force approach --- evaluate every solution This means constantly reevaluating a lot of options Ultimately, this takes exponential time -- O(2 n ) For a class of problems, however, the solution is:

Dynamic Programming Works from problems with: Simple subproblems: can be defined using only a few simple variables Subproblem optimality: can define how to solve problem using the subproblem solutions Subproblem overlap: subproblems overlap such that the solution to a first subproblem can (help) solve later subproblems

How Not to Solve LCS in your Lifetime Brute-force solution: List all subsequences of X Check each subsequence to see if it is also a subsequence of Y Return the longest one of these Analysis: If X has length n, it has 2 n subsequences While waiting, you can not only get coffee, but could first fly to Columbia and pick the beans!

How to Solve LCS Quickly If X and Y are 1 character, LCS is 0 or 1 If we then add 1 character to X and Y, LCS increases by at most 1 Note that we do not need to rescan the first character X aa Y ba X abababab Y bdbdadad

Dynamic-Programming Solution Use an array L to hold solution to subproblems L[i,j] stores LCS of X[0..i] and Y[0..j] Define array to include an index of -1 L[-1, * ] computes LCS for X[0..-1] = “” L[ *,-1] computes LCS for Y[0..-1] = “” L[-1, * ] = 0 and L[ *, -1] = 0 since there are no characters to match!

Dynamic-Programming Solution Solve for remaining L[i,j] as follows: If x i = y j, then L[i,j] = L[i -1, j -1 ] +1  E.g., one more than previous solution If x i ≠ y j, then L[i,j] = max( L[i -1, j], L[i, j -1 ] )  E.g. use however good we did before Final result will be stored in L[n,m] Case 1:Case 2:

LCS Algorithm Algorithm LCS(String X, String Y): for i  1 to n-1 L[i,-1]  0 for j  0 to m-1 L[-1, j]  0 for i  0 to n-1 for j  0 to m-1 if x i = y j then L[i, j]  L[i-1, j-1] + 1 else L[i, j]  max(L[i-1, j], L[i, j-1]) return L

Visualizing the LCS Algorithm