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Average-Optimal Multiple Approximate String Matching

Kimmo Fredriksson , Gonzalo NavarroACM Journal of Experimental Algorithmics,

Vol 9, Article No. 1.4,2004, Pages 1-47

Professor R.C.T LeeSpeaker K.W.Liu

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The Problem

•The approximate string matching problem:

Given text T[1...n] and pattern P[1...m] over some finite alphabet ∑ of size σ, find the approximate occurrences of P from T, allowing at most k differences ( insertion, deletion, substitution).

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For a window of size m-k, if there exists a substring s1 in this window such that its edit distance with every substring of P is greater than k, we move P

Our algorithm scans from the right as shown below:

Fig. 3

S1T:

P:

m - k

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For a window of size m-k, if there exists a suffix S

1 such that its edit distance with every substring of P is greater than k, we move P to S1

Our algorithm scans from the right as shown below:

Fig. 3

S1T:

P:

m - k

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But, how do we know that ED(S1,S2) > k?

We use a very useful lemma.

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LemmaConsider string Q and P. Let Q be divided in

to q1,q2,…,qn as shown below:

qn … q2 q1

For each qi, let pi be the substring in P such that ED(qi,pi) is the smallest, among all substrings in P.

kPQkpqn

iii

),ED( then ,),ED( If1

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Proof: Divide P into n pieces as shown below

qn … q2 q1

p'n … p'2 p'1

Q

P

smallest. theis ),ED( because ),ED(

Therefore, . ),ED( that assume We

1

1

ii

n

iii

n

iii

pqkpq

kpq

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To determine whether ED(S1,S2) > k, we mayUse the lemma.

We divide the window into small pieces: t1, t2, …,ta.

For each ti, we find the substring pi in P where ED(pi,ti) is the smallest.

T:

P:

Window W

Fig. 7

… t2 t1

p1 p2

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. ),ED( whether find togprogrammin dynamic

use tohave We.conclusionany makecannot we

, ),( window, theof end at the If

. movecan weNow

. ),ED( that know we1, Lemma

toaccording ,),( assoon As

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kPW

kpt

P

kSS

kpt

ii

ii

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In general, to find such a pi, we may use Dynamic programming [Sellers 1980].

But, we may use a special kind of small pieces.

It is customary to call a small piece with sizeL a L-gram.

Let us use the 2-gram.

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Note that for two substrings P and Q which are of length 2, the edit distance between them is equal to the Hamming distance between them.

Thus, we may use 2-grams in our algorithm.

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Our algorithm

• Make a table D to store the smallest edit

distance between each possible 2-gram from

finite alphabet set and all substrings of the

pattern P.

•The above is done in the preprocessing stage.

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Example T = ctagggaataatttacaatt P = ttaatatat k = 1

c t a g g g a a t a a t t t a c a a t t

← m-k →

Smallest edit distance between “aa” and all substrings of P = 0 Smallest edit distance between “gg” and all substrings of P = 2

∴∑ > k

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Example T = ctagggaataatttacaatt P = ttaatatat k = 1

c t a g g g a a t a a t t t a c a a t t

← m-k →

Smallest edit distance between “tt” and all substrings of P = 0

Smallest edit distance between “aa” and all substrings of P = 0

Smallest edit distance between “at” and all substrings of P = 0

Smallest edit distance between “ga” and all substrings of P = 1

∴∑ == k

← m+2k →

c t a g g g a a t a a t t t a c a a t t

← m-k →

i

i+1

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c t a g g g a a t a a t t t a c a a t t

← m-k →

← m+k →i+1

Example T = ctagggaataatttacaatt P = ttaatatat k = 1

To find the edit distance between “gaataattta” and P.

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Example T = ctagggaataatttacaatt P = ttaatatat k = 1

c t a g g g a a t a a t t t a c a a t t

← m-k →

c t a g g g a a t a a t t t a c a a t t

← m-k →

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In the preprocessing We make a D table to record the smallest edit

distance between each possible l-gram from alphabet set whose length is l and all substrings of P.

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D table : example ( step by step )

aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt

Dp 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

For example P = aacaccgaa

For P = a a c a c c g a a a

For P = a a c a c c g a a a vs “aa”

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For P = a a c a c c g a a a vs “ac”

aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt

Dp 0 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2

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For P = a a c a c c g a a a vs “ag”For P = a a c a c c g a a a with “at”

For P = a a c a c c g a a a with “ca” For P = a a c a c c g a a a with cc

aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt

Dp 0 0 1 1 0 0 2 2 2 2 2 2 2 2 2 2

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Time complexity

The average complexity of the algorithm is

mnrmk log for 121

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The end

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