1 A Linear Space Algorithm for Computing Maximal Common Subsequences Author: D.S. Hirschberg Publisher: Communications of the ACM 1975 Presenter: Han-Chen.

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A Linear Space Algorithm for Computing Maximal Common Subsequences

Author:D.S. HirschbergPublisher:Communications of the ACM 1975Presenter:Han-Chen ChenDate:2010/04/07

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Outline

Introduction Algorithm A Algorithm B Algorithm C

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Introduction

LCS (Longest Common Subsequence) of two strings has been solved in quadratic time and space.

We present an algorithm which will solve this problem in quadratic time and in linear space.

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Algorithm A

Input string A1m and B1n output matrix L

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Analysis of Algorithm A

Time Complexity :

execute m*n times → O(mn) Space Complexity :

input arrays m + n

output array (m+1)*(n+1)

space require → O(mn)

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Algorithm B (I)

Space require : O(m+n) It can output the max common length but ca

nnot record the max common subsequence.

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Algorithm B (II)

Input string A1m and B1n output matrix LL

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Analysis of Algorithm B

Time Complexity :

execute m*n times → O(mn) Space Complexity :

input arrays m + n

output array n+1

space require → O(m+n)

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Algorithm C Divide and conquer

i=m/2

String B

Strin

g

A

Find j

1

1 n

m

ALG B

ALG B

A1i

B1j

Bj+1,n

Ai+1,m

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Algorithm C L(i,j) j=0 … n the maximum lengths of common subsequence A1i and

B1j

L*(i,j) j=0 … n the maximum lengths of common subsequence Am,i+1 and Bn,j+1

Define M(i) = max{ L(i,j) + L*(i,j) } 0 j n≦ ≦ Theorem M(i) = L(m,n) Proof: for all L(i,j) + L*(i,j) L(m,n)≦

S(i,j) : any maximal common subsequence of A1i and B1j

S*(i,j) : any maximal common subsequence of Ai+1,m and Bj+1,n

Then C= S(i,j) || S*(i,j) is a common subsequence of A1m and B1n of length M(i). Thus L(m,n) L(i,j) + L*(i,j) ≧

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Algorithm C exist some L(i,j) + L*(i,j) L(m,n)≧

S(m,n) : any maximal common subsequence of A1m and B1n

S(m,n) is a subsequence of A1m so S(m,n) = S1 || S2 that

S1 is a subsequence of A1i , S2 is a subsequence of Ai+1,m

Also S(m,n) is a subsequence of B1n so there exists j such that S1 is a

subsequence of B1j and S2 is a subsequence of Bj+1,n

By definition of L and L* , |S1| L(i,j) and |S≦ 2| L*(i,j)≦Thus L(m,n) = |S(m,n)| = |S1| + |S2| L(i,j) + L*(i,j)≦

So M(i) = max{ L(i,j) + L*(i,j) } = L(m,n)

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Algorithm C

m,i+1

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Algorithm C

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Analysis of Algorithm C (I)

Time analysis: O(mn) + O(1/2mn) + O(1/4mn) + …

= O(mn(1+1/2+1/4+…)) = O(mn)

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Analysis of Algorithm C (II)

Space analysis:

we calls ALG B use temporary storage which is m and n.

Exclusive of recursive calls to ALG C, ALG C uses a constant amount of memory space. There are 2m-1 calls to ALG C, so ALG C require memory space O(m+n).

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Proof 2m-1 calls to ALG C

Let m 2≦ r m=1 there are 2*1 – 1 = 1 call to ALG C Assume m 2≦ r = M there are 2m-1 calls to

ALG C For m’ = 2r+1 = 2M. First call ALG C to

partition 2 part, each calls call 2m-1 times ALG C. So there are 1 + (2m-1) + (2m-1) = 4m - 1 = 2m’ – 1 calls.