1 / 12

CSC 213

CSC 213. Lecture 19: Dynamic Programming and LCS. Subsequences ( § 11.5.1). 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

tuyen
Download Presentation

CSC 213

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CSC 213 Lecture 19:Dynamic Programmingand LCS

  2. Subsequences (§ 11.5.1) • A subsequence of a string x0x1x2…xn-1 is a string of the form xi1xi2…xik, where ij < ij+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

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

  4. Longest Common Subsequence (LCS) Problem • Given two strings X and Y, find longest subsequence in both X and Y • Applications in DNA testing (S={A,C,G,T}) • Example the LCS for: ABCDEFGandXZACKDFWGH isACDFG

  5. 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(2n) • For a class of problems, however, the solution is: Dynamic Programming

  6. 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

  7. 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 2n subsequences • While waiting, you can not only get coffee, but could first fly to Columbia and pick the beans!

  8. 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

  9. Dynamic-Programming Solution • Use an array Lto 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!

  10. Dynamic-Programming Solution • Solve for remaining L[i,j] as follows: • If xi=yj, then L[i,j] = L[i -1, j -1] +1 • E.g., one more than previous solution • If xi≠yj, 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:

  11. LCS Algorithm AlgorithmLCS(String X, String Y): fori 1 ton-1 L[i,-1]  0 for j 0 tom-1 L[-1, j]0 for i0 ton-1 forj 0 to m-1 ifxi= yjthen L[i, j] L[i-1, j-1] + 1 else L[i, j] max(L[i-1, j], L[i, j-1]) returnL

  12. Visualizing the LCS Algorithm

More Related