simple and fast linear space computation of longest common subsequences claus rick, 1999

Simple and fast linear space computation of Longest common subsequences

Claus Rick, 1999

What is the LCS problem?

A A B A C

A B C

…Finding a sequence of greatest possible length that can be obtained From both A and B by deleting zero or more (not necessarily adjacent) symbols.

Some boring history…Year Author Time Constants Paradigm

1975 Hirschberg O(mn) 2 Dyn. Prog.

1985 Apostolico, Guerra O(mLgm+pm) [2,logm] contours

1986 Myers O(n(n-p)) 2 Shortest path

1987 Kumar, Rangan O(n(m-p)) 3 contours

1990 Wu et al. O(n(m-p)) 2 Shortest path

1992 Apostolico, et al. O(n(m-p)) 3 contours

1992 Apostolico, et al. O(pm) 3 contours

1999 Goeman, Clausen O(min(pm, mLgm + p(n-p)])

[5,25,lgM] contours

1999 This article O(min(pm,p(n-p)]) 2 contours

Pre-Info

Divide and conquer Midpoint

Some basic terms

Ordered Pair (i,j)

A A B A C

A B C

(2,3)= (A,C)

Some basic terms

Match

A A B A C

A B C

Some basic terms

Chain

A A B A C

A B C

Rank k

A A B A C

A B C

Some basic terms

c b a b b a c a cabacbcba

Matching Matrix

Some basic terms

Dominant matches

All Upper-left matches in each rank

c b a b b a c a cabacbcba

Dominant matches

1

2

3

4

5

A A B A C

A B C

c b a b b a c a cabacbcba

c b a b b a c a c

abacbcba

Backward contours (BC)

1

2

3

4

5

Some last basic terms

FCk

BCk

c b a b b a c a cabacbcba

1

2

3

4

5

Forward contours (FC)

c b a b b a c a c

abacbcba

Backward contours (BC)

1

2

3

4

5

Let p be the length of an LCS between strings A and B. Then for every match (i,j) the following holds:

•There is an LCS containing (i,j) if and only if (i,j) is on the kth forward contour and on the (p-k+1)st backward contour.

Lemma 1

Lemma 1- proof

|BC|- (p-k+1)|FC|= (k)

P

P

K <(p-k+1)<(p-k+1)

Start calculating

FC1 BC1 FC2 BC2

Sooner or later…

Really really last terms

Define sets Mi as:

M0= M

M1= M0\FC1

M2= M1\BC1

M2i-1=M2(i-1) \FCi

M2i=M2i-1\BCi

c b a b b a c a cabacbcba

abacbcba

c b a b b a c a c

M

c b a b b a c a cabacbcba

abacbcba

c b a b b a c a c

M1M2M3M4M5

Let call the first empty Mi….

M p’

Lemma 2

The Length of an LCS is p’ and each match in M(p’-1) is a possible midpoint

Lemma 2- proof

K

M 0

K-1K-210

M 2M 1M k-1M kK=p

Little problem…

We can`t keep tracks of each set- very expensive

c b a b b a c a cabacbcba

abacbcba

c b a b b a c a c

What do we do?

Keep only dominant matches…

When we see a dominant match below- done.

c b a b b a c a cabacbcba

abacbcba

c b a b b a c a c

Lets define:

FCf’ , BCb’ the minimal indices as stated above

Lemma 3

The Length of an LCS is b’ + f’ -1.

Complexity

Finding the dominant matches each contour:

O(min(m, (n-p))

Number of contours:

P

O(Min(pm, p(n-p)

The End

Simple and fast linear space computation of longest common subsequence

Written by: Claus Rick,1999

Based on algorithm by:D.Hirschberg, 1975

Cast:

MatricesLines

ArrowsSquares

Blue Red

BrownGreyBlack

String AString B

Presentation: Uri Scheiner

No Dominant Matches were harmed during the making of this presentation

Appendix

What is the LCS

Divided And Conquer

Match

Chain

Dominant Matches

FC

BC

Lemma 1

Define M…

Lemma 2

Keep just Dominant…

Lemma 3

Complexity