String Matching

1
String Matching

• The problem
• Input a text T (very long string) and a pattern
  P (short string).
• Output the index in T where a copy of P begins.

2
Some Notations and Terminologies

• P and T the lengths of P and T.
• Pi the i-th letter of P.
• Prefix of P a substring of P starting with P1.
• P1..i the prefix containing the first i
  letters of P.
• Example abcabbccaa.
• prefix a, ab, abc, abca, abcab, abcabb, .

3
Some Notations and Terminologies

• suffix of P1..i a substring of P1..i ending
  at Pi, e.g. P3..i, P5..i (igt4).
• Example P1..5abcaa.
• Suffix of P1, 3 c, bc, abc.
• Suffix of P1..4 a, ca, bca, abca.

4
Straightforward method

• Basic idea
• 1. i1
• 2. Start with Ti and match P with
• Ti,Ti1, ... TiP-1
• P1 P2 PP
• 3. whenever a mismatch is found,
• ii1 and goto 2 until
  iP-1ltT.
• Example 1 TABABABCCA and PABABC
• P ABABC A ABABC
• T ABABABCCA ABABABCCA ABABABCCA

5
Analysis

• Step 2 takes O(P) comparisons in the worst
  case.
• Step 2 could be repeated O(T) times.
• Total running time is O(TP).

6
Knuth-Morris-Pratt Method (linear time algorithm)

• A better idea
• In step 3, when there is a mismatch we move
  forward one position (ii1).
• We may move more than one position at a time when
  a mismatch occurs. (carefully study the pattern
  P).
• For example
• P ABABC ABA
• T ABABABCCA ABABABCCA

7

• Questions
• How to decide how many positions we should jump
  when a mismatch occurs?
• How much we can benefit? O(TP).
• Example 2
• P abcabcabcaa
• T abcabcabcabcaa
• abcabcab
• back here

8

• We can move forward more than one position.
  Reason?
• Study of Pattern P
• P1..7 abcabca
• P1..10 abcabcabca (when trying to P11, we
  have a mismatch)
• P1..7 abcabca
• P1..4 abca
• P1..7 is the longest prefix that is also a
  suffix of P1..10.
• P1..4 is a prefix that is a suffix of P1..10,
  but not the longest.
• Key When mismatch occurs at Pi1, we want to
  find the longest prefix of P1..i which is also
  a suffix of P1..i.

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Failure function

• f(i) is the largest r with (rlti) such that
• P1 P2 ...Pr Pi-r1Pi-r2, ...,
  Pi.
• Prefix of length r Suffix of
  P1P2Pi of length r
• That is, P1..f(i) is the longest prefix that is
  a suffix of P1..i.
• Example 3 Pababaccc and i5.
• P1 P2 P3
• a b a
• a b a b a
• P3 P4 P5 (r3) f(5)3.

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• Example 4
• Pabcabbabcabbaa
• It is easy to verify that
• f(1)0, f(2)0, f(3)0, f(4)1, f(5)2,
• f(6)0, f(7)1, f(8)2, f(9)3, f(10)4,
• f(11)5, f(12)6, f(13)7, f(14)1.

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The Scan Algorithm(draw a figure to show)

• i indicates that Ti is the next character in T
  to be compared with the right end of the
  pattern.
• q indicates that Pq1 is the next character in
  P to be compared with Ti.
• i1 and q0
• Compare Ti with Pq1case 1
  TiPq1 ii1qq1 if q1P then
  print "P occurs at i1-P"case 2 Ti?Pq1
  and q?0 qf(q) case 3 Ti?Pq1 and
  q0 ii1
• Repeat step2 until iT.

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• Example 5 Pabcabbabcabbaa
• Tabcabcabbabbabcabbabcabbaa
• abcabb
• abcabbabc
• abc
• a(ii1)
• abcabbabcabbaa(q1p)

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Running time complexity(hard)

• The running time of the scan algorithm is O(T).
• Proof
• There are two pointers i and p.
• i the next character in T to be compared.
• p the position of P1. (See figure below)
• p i
• Pabcabcabcaa
• Tabcabcabcabcaa
• P abcabcaa
• p

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• Facts
• 1 When a match is found, move i forward.
• 2 When a mismatch is found, move p forward until
  p and i are the same. (When pi and a mismatch
  occur, move both i and p forward)
• From facts 1 and 2, it is easy to see that the
  total number of comparisons is at most 2T.
• Thus, the time complexity is O(T).

15
Another version of scan algorithm (code)

• nT
• mP
• q0
• for i1 to n
• while qgt0 and Pq1?Ti do
• qf(q)
• if Pq1Ti then
• qq1
• if qm then
• print "pattern occurs at i-m1"
• qf(q)

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Failure Function Construction

• Basic idea
• Case 1 f(1) is always 0.
• Case 2 if PqPf(q-1)1 then f(q)f(q-1)1.
• Example pabcabcc
• abc
• f(1)0 f(2)0 f(3)0 f(4)1 f(5)2 f(6)3
  f(7)0
• P4 Pf(4-1)1, f(4)f(4-1)11.
• P5 Pf(5-1)1, f(5)f(5-1)1112.
• P6 Pf(6-1)1. F(6)f(6-1)1213.

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• Case 3 if Pq?Pf(q-1)1 and f(q-1)?0 then
  consider Pq ? Pf(f(q-1))1 (Do it
  recursively)
• Case 4 if Pq ? Pf(q-1)1 and f(q-1)0 then
  fq0.
• Example abc abc abb
• abc abc f(8)5
• abc f(5)2
• a
  f(2)0
• i 1 2 3 4 5 6 7 8 9
• f(i) 0 0 0 1 2 3 4 5 0

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The algorithm (code) to compute failure function

• 1. mP
• 2. f(1)0
• 3. k0
• 4. for q2 to P do
• 5. kf(q-1)
• 6. if(kgt0 and Pk1!Pq)
• kf(k) goto 6
• 7. if(kgt0 and Pk1Pq)
• fqk1
• 8. if(k0)
• if(Pk1Pq fq1
• else fq0

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Another version

• 1. mP
• 2. f(1)0
• 3. k0
• 4. for q2 to P do
• 5. kf(q-1)
• 6. while(kgt0 and Pk1!Pq) do
• 7. kf(k)
• 8. if(Pk1Pq) then kk1
• 9. fqk

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• Example 3
• 1 2 3 4 5 6 7 8 9 10 11 12
• Pa b c a b c a b c a a c
• f(1)0 f(2)0 f(3)0 f(4)1 f(5)2
• f(6)3 f(7)4 f(8)5 f(9)6 f(10)7
• f(11)1.
• (The computation of f(11) is very interesting.)
• Question Do we need to compute f(12)?
• Yes, if you want to find ALL occurrences of P.
• No, if you just want to find the first occurrence
  of P.

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• Example
• Pabcabc
• Tabcabcabc
• abcabc
• abcabc
• When a match is found at the end of P, call
  f(p).
• Running time complexity (Fun Part, not required)
• The running time of failure function construction
  algorithm is O(P). (The proof is similar to
  that for scan algorithm.)
• Total running time complexity
• The total complexity for failure function
  construction and scan algorithm is O(PT).

22
Linear Time Algorithm for Multiple patterns (Fun
Part)

• Input a string T (very long) and a set of
  patterns P1,P2,...,Pk.
• Output all the occurrences of Pi's in T.
• Let us consider the set of patterns he, she,
  his, hers . We can construct an automata as
  follows

23
e,i,r
h
e
r
s
1
s
i
s
h
e
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• g(s,a)s' means that at state s if the next input
  letter is a then the next state is s'.
• The states of the automata is organized column by
  column.
• Each state corresponds to a prefix of some
  pattern Pi.
• F the set of final states (dark circled)
  corresponding to the ends of patterns.
• For the starting state 0, add g(0,a)0, if g(0,a)
  is originally fail.

25

• Exercise write down the g() function for the
  above automata.
• Failure function
• f(s) the state for the longest prefix of some
  pattern Pi that is a suffix of the string in the
  path from 0 (starting state) to s.
• Example
• he is the longest prefix for hers that is a
  suffix of the string she.

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The scan algorithm

• Text T1T2...Tn
• s0
• for i1 to n do
• while g(s,Ti)fail do sf(s)
• sg(s,Ti)
• if s is in F then return "yes"
• return "no"

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• Theorem The scan algorithm takes O(T) time.
• Proof Again, the two pointer argument.
• When a match is found, move the first pointer
  forward. (sg(s,Ti))
• When a mismatch is found (g(s,Ti)fail), move
  the second pointer forward. (sf(s))
• When a final state is meet, declare the finding
  of a pattern. (if s is in F then return "yes")

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• Example
• i1 2 3 4 5 6 7 8
• s h e r s h i i
• 3 4 5
• 2 8 9
• 3 4
• 1
• 0
• 0 0

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Failure function construction

• Basic idea similar to that for one pattern.
• for each state s of depth 1 do
• f(s)0
• for each depth dgt1 do
• for each state sd of depth d and character a
  such that g(sd,a)s' do
• sf(sd)
• while g(s,a)fail do
• sf(s)
• f(s')g(s,a)

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• g(0,c)?fail for any possible character c.
• The failure function for he, she, his, hers is
• Time complexity O(P1P2...Pk).
• Proof Two pointer argument.
• Leave it for assignment (optional)