Longest Common Subsequence Using Dynamic Programming

Submitted By:Swati NautiyalRoll No.162420ME-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 SubsequenceA substring of a string S is another string S ′ that occurs in S and all the letters are contiguous in SE.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 contigu-ous in SE.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 MethodGiven two strings X of length m and Y of length n, find a longest sub-sequence 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 MethodGiven two strings X of length m and Y of length n, find a longest sub-sequence 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 substructureWe 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 subprob-lems. 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 LCSBEST CASEX={A,A,A,A}Y={A,A,A,A}

C(4,4)

0 ; i=0 or j=0c[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 = 4T.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 CASEX={A,A,A}Y={B,B,B}

0 ; i=0 or j=0c[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 pro-gramming.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 elementC(2,2

)

C(2,1)

C(1,2)

We can compute either row wise or column wise

AlgorithmAlgorithm LCS(X,Y ):Input: Strings X and Y with m and n elements, respectivelyOutput: 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 subsequencesBCBABCABBDAB

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 matchingDNA comprises of {A,C,G,T}.

DNA1= AGCCTCAGTDNA2=ATCCTDNA3=AGTAGC

DNA 1 and DNA 3 are more similar.Edit DistanceThe Edit Distance is defined as the minimum number of edits needed 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_problem•http://www.slideshare.net/ShahariarRabby1/longest-common-subsequence-lcs?qid=49affff4-9c19-4957-bdb9-7877801a569e&v=&b=&

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