KMP String Matching Prepared By: Carlens Faustin.

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KMP String Matching Prepared By: Carlens Faustin

Transcript of KMP String Matching Prepared By: Carlens Faustin.

Page 1: KMP String Matching Prepared By: Carlens Faustin.

KMP String Matching

Prepared By: Carlens Faustin

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String Matching

• String and pattern matching problems are fundamental to any computer application involving text processing.

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String Matching

• The National Institute of Standards and technology(2014) defines string matching as:

“The problem of finding occurrence(s) of a

pattern string within another string or body of text. There are many different algorithms for efficient searching.”

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String Matching

• Applications:Bioinformatics(detect pattern in DNA

sequence)Word processorsParsersDigital librariesEtc.

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String Matching

• Brute force or exhaustive searcho 2 loops.o 1 loop through the pattern(P[m]) the other

loop through the body of text(S[n])o Total of mn possibilitieso Could take up to O(mn) timeo Not a really efficient time, there is definitely

ways to get better results.

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KMP String MatchingDonald Knuth

Jim H. Morris

Vaughan Pratt

1997

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KMP String Matching

Knuth, Morris and Pratt proposed: a linear time algorithm for the string matching

problem. A matching time of O(n) is achieved by

avoiding comparisons with elements of ‘S’ that have previously been involved in comparison with some element of the pattern ‘p’ to be matched. i.e., backtracking on the string ‘S’ never occurs

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KMP String Matching

• KMP algorithm contains 2 major components:

-The longest prefix suffix function (lps), Π-The KMP matcher

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KMP String Matching• The prefix function, Π That function preprocesses the pattern by

building a prefix table(match) which encapsulates knowledge about how the pattern matches against shifts of itself.

This information can be used to avoid useless shifts of the pattern ‘p’. In other words, this enables avoiding backtracking on the string ‘S’.

Every number belongs to corresponding prefix ("a", "ab", "aba", ...) and for each prefix it represents

length of longest suffix of this string that matches prefix

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KMP String Matching

• In our context, what is prefix ? What is a suffix?

• in order to talk about the meaning, we need to know about proper prefixes and proper suffixes.

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KMP String Matching

• Proper prefix: All the characters in a string, with one or more cut off the end. “M”, “Mo”, “Mov”, and “Movi” are all the proper prefixes of “Movie”.

• Proper suffix: All the characters in a string, with one or more cut off the beginning. “ovie”, “vie”, “ie”, and “e” are all proper suffixes of “Movie”.

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KMP String Matching

• With that in mind we can try to define the meaning of the values in that match table.

• length of the longest proper prefix in the (sub)pattern that matches a proper suffix in the same (sub)pattern.

• But… what does that even mean???

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KMP String Matching

• Let’s take a string s=“bebe”• Proper prefix:”b” “be” “beb”• Proper suffix:”e” “be” “ebe”• They both have 1 element in common “be”,

that makes it a match and its length is 2• Therefore we can conclude that the length of

the longest proper prefix that matches a proper suffix is 2

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KMP String Matching

• Now that we have the match table definition, how does that help us ?

• Well let’s remember this:• If a partial match of length partial_match_length is found

and table[partial_match_length -1] > 1, we may slide ahead (partial_match_length - table[partial_match_length - 1]) characters.

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KMP String Matching

• The prefix function, ΠPrefix-Function (p)1 m length[p] //’p’ pattern to be matched2 Π[1] 0 //match table with 1st element =03 k 0 //length of match so far…4 for q 2 to m5 do while k > 0 and p[k+1] != p[q]6 do k Π[k]7 If p[k+1] = p[q]8 then k k +19 Π[q] k10 return Π

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KMP String Matching

• Let’s compute an example for the prefix function, Π

• Lets assume we have

PP xx yy xx yy xx zz xx

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KMP String Matching

qq 11 22 33 44 55 66 77

pp xx yy xx yy xx zz xx

ΠΠ 00 00

Initially: m = length[p] = 7 Π[1] = 0 k = 0

Step 1: q = 2, k=0 Π[2] = 0

Step 2: q = 3, k = 0, Π[3] = 1

Step 3: q = 4, k = 1 Π[4] = 2

qq 11 22 33 44 55 66 77

pp xx yy xx yy xx zz xx

ΠΠ 00 00 11

qq 11 22 33 44 55 66 77

pp xx yy xx yy xx zz xx

ΠΠ 00 00 11 22

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KMP String Matchingqq 11 22 33 44 55 66 77

pp xx yy xx yy xx zz xx

ΠΠ 00 00 11 22 33

qq 11 22 33 44 55 66 77

pp xx yy xx yy xx zz xx

ΠΠ 00 00 11 22 33 11

Step 4: q = 5, k =2 Π[5] = 3

Step 5: q = 6, k = 3 Π[6] = 1

Step 6: q = 7, k = 1 Π[7] = 1

After iterating 6 times, the prefix function computation is complete:

qq 11 22 33 44 55 66 77

pp aa bb aa bb aa cc aa

ΠΠ 00 00 11 22 33 11 11

qq 11 22 33 44 55 66 77

pp aa bb AA bb aa cc aa

ΠΠ 00 00 11 22 33 11 11

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KMP String Matching

• The KMP matcher

It takes as inputs a string ‘S’, pattern ‘p’ and it uses the prefix function ‘Π’.

It then finds the occurrence of ‘p’ in ‘S’ and returns the number of shifts of ‘p’ after which occurrence is

found.

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KMP String Matching• The KMP matcher

KMP-Matcher(S,p)1 n length[S] 2 m length[p]3 Π Compute-Prefix-Function(p)4 q 0 //number of characters matched 5 for i 1 to n //scan S from left to right6 do while q > 0 and p[q+1] != S[i]7 do q Π[q] //next character does not match8 if p[q+1] = S[i]9 then q q + 1 //next character matches10 if q = m //is all of p matched?11 then print “Pattern occurs with shift” i – m12 q Π[ q] // look for the next matchNote: KMP finds every occurrence of a ‘p’ in ‘S’. That is why KMP does not terminate in step 12, rather it

searches remainder of ‘S’ for any more occurrences of ‘p’.

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KMP String MatchingExample with string “S” and pattern “P”

S

P xx yy xx yy xx zz xx

y x z y x y x y x y x z x z x

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Initially: n = size of S = 15; m = size of p = 7

Step 1: i = 1, q = 0 comparing p[1] with S[1]

S

pP[1] does not match with S[1]. ‘p’ will be shifted one position to the right.

S

p

Step 2: i = 2, q = 0 comparing p[1] with S[2]

P[1] matches S[2]. Since there is a match, p is not shifted.

y x z y x y x y x y x z x z x

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

y x z y x y x y x y x z x z x

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Step 3: i = 3, q = 1Comparing p[2] with S[3]

S

p

S

p

S

p

p[2] does not match with S[3]

Backtracking on p, comparing p[1] and S[3]Step 4: i = 4, q = 0 comparing p[1] with S[4] p[1] does not match with S[4]

Step 5: i = 5, q = 0 comparing p[1] with S[5] p[1] matches with S[5]

y x z y x y x y x y x z x z x

y x z y x y x y x y x z x z x

y x z y x y x y x y x z x z x

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

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Step 6: i = 6, q = 1Step 6: i = 6, q = 1

S

p

Comparing p[2] with S[6] p[2] matches with S[6]

S

p

Step 7: i = 7, q = 2Step 7: i = 7, q = 2Comparing p[3] with S[7] p[3] matches with S[7]

Step 8: i = 8, q = 3Step 8: i = 8, q = 3Comparing p[4] with S[8] p[4] matches with S[8]

S

p

y x z y x y x y x y x z x z x

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

y x z y x y x y x y x z x z x

y x z y x y x y x y x z x z x

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Step 9: i = 9, q = 4Step 9: i = 9, q = 4Comparing p[5] with S[9]

Comparing p[6] with S[10]

Comparing p[5] with S[11]

Step 10: i = 10, q = 5Step 10: i = 10, q = 5

Step 11: i = 11, q = 4Step 11: i = 11, q = 4

S

S

S

p

p

p

p[6] doesn’t match with S[10]

Backtracking on p, comparing p[4] with S[10] because after mismatch q = Π[5] = 3

p[5] matches with S[9]

p[5] matches with S[11]

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

y x z y x y x y x y x z x z x

y x z y x y x y x y x z x z x

y x z y x y x y x y x z x z x

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Step 12: i = 12, q = 5Step 12: i = 12, q = 5Comparing p[6] with S[12]

Comparing p[7] with S[13]

S

S

p

p

Step 13: i = 13, q = 6Step 13: i = 13, q = 6

p[6] matches with S[12]

p[7] matches with S[13]

Pattern ‘p’ has been found to completely occur in string ‘S’. The total number of shifts that took place for the match to be found are: i – m = 13 – 7 = 6 shifts.

xx yy xx yy xx zz xx

xx yy xx yy xx zz xx

y x z y x y x y x y x z x z x

y x z y x y x y x y x z x z x

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Running - time analysis• Prefix-Function (Π)1 m length[p] //’p’ pattern to be matched2 Π[1] 0 3 k 04 for q 2 to m5 do while k > 0 and p[k+1] != p[q]6 do k Π[k]7 If p[k+1] = p[q]8 then k k +19 Π[q] k10 return Π

The for loop from step 4 to step 10 runs ‘m’ times. Step 1 to step 3 take constant time. Hence the running time of compute prefix function is Θ(m).

• KMP Matcher1 n length[S] 2 m length[p]3 Π Prefix-Function(p)4 q 0 5 for i 1 to n 6 do while q > 0 and p[q+1] != S[i]7 do q Π[q] 8 if p[q+1] = S[i]9 then q q + 1 10 if q = m 11 then print “Pattern occurs with shift” i – m12 q Π[ q]

The for loop beginning in step 5 runs ‘n’ times, i.e., as long as the length of the string ‘S’. Since step 1 to step 4 take constant time, the running time is dominated by this for loop. Thus running time of matching function is Θ(n).

Total running time of: O(m + n)

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References

• http://xlinux.nist.gov/dads/HTML/stringMatching.html• https://www.my46.org/sites/www.my46.org/files/pictures/

07My46_Gen101_DNAseq_final.png• https://www.cs.ubc.ca/~hoos/cpsc445/Handouts/kmp.pdf• http://www.cs.princeton.edu/~rs/

AlgsDS07/21PatternMatching.pdf• http://cs.indstate.edu/~kmandumula/presentation.pdf• http://www.mif.vu.lt/cs2/courses/ds99fa6.pdf

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Questions