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Average Case Analysis of an Exact String Matching Algorithm

Advisor: Professor R. C. T. LeeSpeaker: S. C. Chen

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Problem Definition

We are given text T=t1t2…tn with length n and a pattern P=p1p2…pm with length m and we are asked to find all occurrences of P in T.

Example:

CCTAP

CCTAAGTCAGCCTAAGCTT

There are two occurrences of P in Tas shown below:

AGTCCCTAAGCTCCTAAG

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There are many rules in exact string matching algorithms. For example, the Suffix to Prefix Rule, the Substring Matching Rule, ….

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We use the idea, the substring matching rule, in this algorithm.

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The Substring Matching Rule

For any substring S in T, find a nearest S in P which is to the left of it. If such an S in P exists, move P such then the two S’s match; otherwise, we may define a new partial window.

windows

T

P

Exactly matched

S

S

windows

T

P

S

S

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windows

T

P

windows

T

P

S

S

ii-r+1i-m+1

ii-r+1i-m+1

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In this algorithm, we first check whether S=T[i-r+1…i] is a substring of P or not.

If S does not occur in P, we shift P to right m-r steps.

windows

T

P

windows

T

P

S

S

mm-r+11

mm-r+11

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• If S occurs in P, according to the Substring Matching Rule, we should slide P so that the two substrings S match as shown below.

windows

T

P

S

S

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• But, our algorithm is not that smart, instead of sliding P so that the two substrings S match, we simply examine the entire window starting from i-m+1 to 2i-r to see whether P occurs in this window, as shown below.

2m-r

T

P

S

ii-m+1 2i-ri-r+1

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• Note that our not so smart algorithm covers the case of sliding P to match the two substrings S.

2m-r

T

P

S

ii-m+1 2i-ri-r+1

S

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Algorithm•Algorithm fast-on-average;• i=m;• while i≦n do begain• if T[i-r+1…i] is a substring of P then• compute all occurrences of P whose starting positions are in T[i-m+1…i-r+1] applying KMP algorithm.• else { P does not start in T[i-m+1…i-r+1] }• i=i+m-r•end

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Analysis

First of all, let us note that in the above algorithm, we have to determine whether the suffix S occurs in P or not. This is again an exact string matching problem. Let us assume that there is a pre-processing to construct a suffix tree of P. Whether S occurs in P or not can be determined by feeding S into the suffix tree of P. Because the length of S is r, we can determine whether S occurs in P in O(r).

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For reasons which will become clear later, we assume that mr log2

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We assume that the text is a random string and the size of alphabet is α.

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There are αr possible substrings with length r consisting of α distinct characters.

There are only m-r substrings with length r in P whose length is m .

Thus, the probability that S is a substring of P is not great than

mm

rmrmrm

mr

11)(

1)(

2log2

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If S is a substring of P, we find all occurrences of P in T[i-m…2i-r] using KMP algorithm.

2m-r

T

P

S

ii-m+1 2i-ri-r+1

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2m-r

T

P

S

ii-m+1 2i-ri-r+1

Because the length of T[i-m…2i-r] is 2m-r, time complexity of Step i using KMP algorithm is O(m)

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(1)The probability that S occurs in P is .

(2)When S occurs in P, the time complexity that we use KMP algorithm to find all occurrences of P in T[i-m+1…2i-r] is O(m).

Summary of (2) and (3), the average time-complexity of applying the KMP algorithm is )1(

1O

mmO

In the above, the time complexity of checking whether S occurs in P is O(r).

Thus, the average time-complexity of applying the KMP algorithm once is O(r).

m

1

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Thus, if S does not occurs in P, the time complexity of Step i is only the checking time-complexity which is O(r) .

If does, the time complexity of Step i is O(r).

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Because there are windows with length m in T, the time complexity of this algorithm on average is .

m

nO

m

mnOr

m

nO log

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Reference

• [KMP77] Faster Pattern Matching in Strings, SIAM Journal on Computing 6 (2),1977, pp. 323–350.

• [CR2002] Section 2.2:Boyer-Moore algorithm and its variations, Jewels of Stringology, 2002, pp. 30-31.

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Thank you