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Longest Common Subsequence

The document discusses the longest common subsequence (LCS) problem and how to solve it using dynamic programming. It begins by defining LCS as the longest sequence of characters that appear left-to-right in two given strings. It then describes solving LCS using a brute force method with exponential time complexity and using dynamic programming with polynomial time complexity. Finally, it provides an example of finding the LCS of two strings and discusses applications and references.

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Longest Common Subsequence Using Dynamic Programming Submitted By: Swati Nautiyal Roll No.162420 ME-CSE(R) Submitted To: Mrs. Shano Solanki (Assistant Professor CSE)
Contents •Difference in substring and subsequence •Longest common subsequence •LCS with brute force method •LCS with dynamic programming •Recursion tree of LCS •LCS example •Analysis of LCS •Applications •References
Substring and Subsequence A substring of a string S is another string S ′ that occurs in S and all the letters are contiguous in S E.g. Amanpreet substring1 : Aman substring2 : preet A subsequence of a string S is another string S ′ that occurs in S and all the letters need not to be contiguou s in S E.g Amanpreet subsequence1 : Ant subsequence2 : mnet
Longest Common Subsequence The Longest Common Subsequence (LCS) problem is as follows. We are given two strings: string A of length x and string B of length y. We have to find the longest common subsequence: the longest sequence of characters that appear left-to-right in both strings. Example, A= KASHMIR B= CHANDIGARH
Longest Common Subsequence The Longest Common Subsequence (LCS) problem is as follows. We are given two strings: string A of length x and string B of length y. We have to find the longest common subsequence: the longest sequence of characters that appear left-to-right in both strings. Example, A= KASHMIR B= CHANDIGARH LCS has 3 length and string is HIR
Brute Force Method Given two strings X of length m and Y of length n, find a longest subse quence common to both X and Y STEP1 : Find all subsequences of ‘X’. STEP2: For each subsequence, find whether it is a subsequence of ‘Y’. STEP3: Find the longest common subsequence from available subsequences
Brute Force Method Given two strings X of length m and Y of length n, find a longest subse quence common to both X and Y STEP1 : Find all subsequences of ‘X’. 2m STEP2: For each subsequence, find whether it is a subsequence of ‘Y’. n*2m STEP3: Find the longest common subsequence from available subsequences. T.C= O(n2m) To improve time complexity, we use dynamic programming
Dynamic Programing Optimal substructure We have two strings X= { x1,x2,x3,……,xn} Y= {y1,y2,y3,…….,ym} •First compare xn and ym. If they matched, find the subsequence in the remaining string and then append the xn with it. •If xn ≠ ym, • Remove xn from X and find LCS from x1 to xn-1 and y1 to ym • Remove ym from Y and find LCS from x1 to xn and y1 to ym-1 In each step, we reduce the size of the problem into the subproblems . It is optimal substructure.
Cont. Recursive Equation X= { x1,x2,x3,……,xn} Y= {y1,y2,y3,…….,ym} C[I,j] is length of LCS in X and Y {0 ; i=0 or j=0 c[I,j] = 1+c[i-1,j-1] ; I,j>0 and xi=yi max(c[i-1,j],c[I,j-1]) ; I,j>0 and xi≠yi
Recursion Tree Of LCS BEST CASE X={A,A,A,A} Y={A,A,A,A} C(4,4) 0 ; i=0 or j=0 c[I,j] = 1+c[i-1,j-1] ; I,j>0 and xi=yi max(c[i-1,j],c[I,j-1]); I,j>0 and xi≠yi { 1+C(3,3) 1+C(2,2) 1+C(1,1) 1+C(0,0) 0 =1 =2 =3 =4 =4 LCS = 4 T.C. = O(n)
Cont. C(3,3) C(3,2) C(3,1) C(3,0) C(2,1) C(2,0) C(1,1) C(1,0) C(0,1) C(2,2) C(2,1) C(2,0) C(1,1) C(1,0) C(0,1) C(1,2) C(1,1) C(1,0) C(0,1) C(0,2) C(2,3) C(2,2) C(2,1) C(2,0) C(1,1) C(1,0) C(0,1) C(1,2) C(1,3) WORST CASE X={A,A,A} Y={B,B,B} 0 ; i=0 or j=0 c[I,j] = 1+c[i-1,j-1] ; I,j>0 and xi=yi max(c[i-1,j],c[I,j-1]); I,j>0 and xi≠yi { As here, the overlapping problem exits, we can apply the dynamic programming. There are 3*3 unique subproblems. So we compute them once and save in table for further refrence so n*n memory space required. No of nodes O(2ⁿ+ⁿ)
Cont. 00 01 02 0m 10 11 12 1m 20 21 22 2m n0 n1 n2 nm 0 1 - - m 0 1 2 - n (m+1)*(n+1) =O(m*n) Every element is depend on diagonal, left or above element C(2,2) C(2,1) C(1,2) We can compute either row wise or column wise
Algorithm Algorithm LCS(X,Y ): Input: Strings X and Y with m and n elements, respectively Output: For i = 0,…,m; j = 0,...,n, the length C[i, j] of a longest string that is a subsequence of both the strings. for i =0 to m c[i,0] = 0 for j =0 to n c[0,j] = 0 for i =0 to m for j =0 to n do if xi = yj then c[i, j] = c[i-1, j-1] + 1 else L[i, j] = max { c[i-1, j] , c[i, j-1]} return c
Example X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 A B C B D A B for i =1 to m c[i,0] = 0 for j =0 to n c[0,j] = 0
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 B 0 C 0 B 0 D 0 A 0 B 0 for i =1 to m c[i,0] = 0 for j =0 to n c[0,j] = 0 if xi = yj then c[i, j] = c[i-1, j-1]+1 else L[i, j]=max{c[i-1,j], c[i, j-1]
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 B 0 C 0 B 0 D 0 A 0 B 0 if xi = yj then c[i, j] = c[i-1, j-1]+1 else L[i, j]=max{c[i-1,j], c[i, j-1]
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 0+1 B 0 C 0 B 0 D 0 A 0 B 0 if xi = yj then c[i, j] = c[i-1, j-1]+1 else L[i, j]=max{c[i-1,j], c[i, j-1]
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4 if xi = yj then c[i, j] = c[i-1, j-1]+1 else L[i, j]=max{c[i-1,j], c[i, j-1]
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4 Subsequence=BCBA
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4 Subsequence = BCAB
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4
Cont. X={B,D,C,A,B,A} Y={A,B,C,B,D,A,B} 0 B D C A B A 0 0 0 0 0 0 0 0 A 0 0 0 0 1 1 1 B 0 1 1 1 1 2 2 C 0 1 1 2 2 2 2 B 0 1 1 2 2 3 3 D 0 1 2 2 2 3 3 A 0 1 2 2 3 3 4 B 0 1 2 2 3 4 4 Subsequence = BDAB
Cont We get three longest common subsequences BCBA BCAB BDAB Length of longest common subsequence is 4
Analysis Of LCS We have two nested loops – The outer one iterates n times – The inner one iterates m times – A constant amount of work is done inside each iteration of the inner loop – Thus, the total running time is O(nm) Space complexity is also O(nm) for n*m table
Application DNA matching DNA comprises of {A,C,G,T}. DNA1= AGCCTCAGT DNA2=ATCCT DNA3=AGTAGC DNA 1 and DNA 3 are more similar. Edit Distance The Edit Distance is defined as the minimum number of edits neede d to transform one string into the other.
REFERENCES •Textbook Introduction to Algorithm by Coreman •http://www.perlmonks.org/?node_id=652798 •https://www.ics.uci.edu/~eppstein/161/960229.html •https://en.wikipedia.org/wiki/Longest_common_subsequence_probl em •http://www.slideshare.net/ShahariarRabby1/longest-common-subse quence-lcs?qid=49affff4-9c19-4957-bdb9-7877801a569e&v=&b=&fr om_search=1
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