1. 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. 2 Some Notations and Terminologies |P| and |T|: the lengths of P and T. P[i]: the i-th letter of P. Prefix of P: a substring of P starting with P[1]. P[1..i]: the prefix containing the first i letters of P. Example: abcabbccaa. prefix: a, ab, abc, abca, abcab, abcabb, ….

3. 3 Some Notations and Terminologies suffix of P[1..i]: a substring of P[1..i] ending at P[i], e.g. P[3..i], P[5..i] (i>4). Example: P[1..5]=abcaa. Suffix of P[1.. 3]: c, bc, abc. Suffix of P[1..4]: a, ca, bca, abca.

4. 4 Straightforward method Basic idea: 1. i=1; 2. Start with T[i] and match P with T[i],T[i+1],... T[i+|P|-1] | | | P[1] P[2] P[|P|] 3. whenever a mismatch is found, i=i+1 and goto 2 until i+|P|-1<|T|. Example 1: T=ABABABCCA and P=ABABC P: ABABC A ABABC | | | T: ABABABCCA ABABABCCA ABABABCCA

5. 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(|T||P|).

6. 6 Knuth-Morris-Pratt Method (linear time algorithm) A better idea In step 3, when there is a mismatch we move forward one position (i=i+1). 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. 7 Questions: How to decide how many positions we should jump when a mismatch occurs? How much we can benefit? O(|T|+|P|). Example 2: P: abcabcabcaa | T: abcabcabcabcaa | abcabcab back here

8. 8 We can move forward more than one position. Reason? Study of Pattern P P[1..7] abcabca P[1..10] abcabcabca (when trying to P[11], we have a mismatch) P[1..7] abcabca P[1..4] abca P[1..7] is the longest prefix that is also a suffix of P[1..10]. P[1..4] is a prefix that is a suffix of P[1..10], but not the longest. Key: When mismatch occurs at P[i+1], we want to find the longest prefix of P[1..i] which is also a suffix of P[1..i].

9. 9 Failure function f(i) is the largest r with (r<i) such that P[1] P[2]...P[r] = P[i-r+1]P[i-r+2],..., P[i]. Prefix of length r Suffix of P[1]P[2]…P[i] of length r That is, P[1..f(i)] is the longest prefix that is a suffix of P[1..i]. Example 3: P=ababaccc and i=5. P[1] P[2] P[3] a b a a b a b a P[3] P[4] P[5] (r=3) f(5)=3.

10. 10 Example 4: P=abcabbabcabbaa 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.

11. 11 The Scan Algorithm (draw a figure to show) i: indicates that T[i] is the next character in T to be compared with the right end of the pattern. q: indicates that P[q+1] is the next character in P to be compared with T[i]. 1.i=1 and q=0; 2.Compare T[i] with P[q+1] case 1: T[i]==P[q+1] i=i+1;q=q+1; if q==|P| then print "P occurs at i+1-|P|“; q=f(q); case 2: T[i]≠P[q+1] and q≠0 q=f(q); case 3: T[i]≠P[q+1] and q==0 i=i+1; 3.Repeat step2 until i==|T|.

12. 12 Example 5: P=abcabbabcabbaa T=abcabcabbabbabcabbabcabbaa abcabb | | | abcabbabc | abc | a (i=i+1) abcabbabcabbaa (q+1=|p|) i1 2 3 4 5 6 7 8 9 10 11 12 13 14 f(i)0 0 0 1 2 0 1 2 3 4 5 6 7 1

13. 13 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 P[1]. (See figure below) p i P:abcabcabcaa | T:abcabcabcabcaa | P: abcabcaa p

14. 14 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 p=i 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 2|T|. Thus, the time complexity is O(|T|).

15. 15 Another version of scan algorithm (code) n=|T| m=|P| q=0 for i=1 to n { while q>0 and P[q+1]≠T[i] do { q=f(q) } if P[q+1]==T[i] then q=q+1 if q==m then { print "pattern occurs at i-m+1" q=f(q) }

16. 16 Basic idea: Case 1: f(1) is always 0. Case 2: if P[q]==P[f(q-1)+1] then f(q)=f(q-1)+1. Example: p=abcabcc abc f(1)=0; f(2)=0; f(3)=0; f(4)=1; f(5)=2; f(6)=3; f(7)=0; P[4]= P[f(4-1)+1], f(4)=f(4-1)+1=1. P[5]= P[f(5-1)+1], f(5)=f(5-1)+1=1+1=2. P[6]= P[f(6-1)+1]. F(6)=f(6-1)+1=2+1=3. Failure Function Construction

17. 17 Case 3: if P[q]  P[f(q-1)+1] and f(q-1)≠0 then consider P[q] ?= P[f(f(q-1))+1] (Do it recursively) Case 4: if P[q]  P[f(q-1)+1] and f(q-1)==0 then f[q]=0. 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

18. 18 The algorithm (code) to compute failure function 1. m=|P|; 2. f(1)=0; 3. k=0; 4. for q=2 to |P| do { 5. k=f(q-1); 6. if(k>0 and P[k+1]!=P[q]) { k=f(k); goto 6; } 7. if(k>0 and P[k+1]==P[q]) { f[q]=k+1; } 8. if(k==0) { if(P[k+1]==P[q] f[q]=1; else f[q]=0; }

19. 19 Another version 1. m=|P|; 2. f(1)=0; 3. k=0; 4. for q=2 to |P| do { 5. k=f(q-1); 6. while(k>0 and P[k+1]!=P[q]) do { 7. k=f(k); } 8. if(P[k+1]==P[q]) then k=k+1; 9. f[q]=k; }

20. 20 Example 3: 1 2 3 4 5 6 7 8 9 10 11 12 P=a 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.

21. 21 Example: P=abcabc T=abcabcabc 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(|P|+|T|). i1 2 3 4 5 6 f(i)0 0 0 1 2 3

22. 22 Linear Time Algorithm for Multiple patterns (Fun Part) Input: a string T (very long) and a set of patterns P 1,P 2,...,P k. Output: all the occurrences of P i 's in T. Let us consider the set of patterns { he, she, his, hers }. We can construct an automata as follows:

23. 23 0 1 54367289 hers i s s h e e,i,r s1 2 3 4 5 6 7 8 9 f(s)0 0 0 1 2 0 3 0 3

24. 24 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 P i. 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. 25 Exercise: write down the g() function for the above automata. Failure function f(s) = the state for the longest prefix of some pattern P i 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.

26. 26 The scan algorithm Text: T[1]T[2]...T[n] s=0; for i:=1 to n do { while g(s,T[i])=fail do s=f(s); s:=g(s,T[i]); if s is in F then return "yes"; } return "no"

27. 27 Theorem: The scan algorithm takes O(|T|) time. Proof: Again, the two pointer argument. When a match is found, move the first pointer forward. (s:=g(s,T[i]);) When a mismatch is found (g(s,T[i])==fail), move the second pointer forward. (s=f(s);) When a final state is meet, declare the finding of a pattern. (if s is in F then return "yes";)

28. 28 Example: i=1 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 s1 2 3 4 5 6 7 8 9 f(s)0 0 0 1 2 0 3 0 3

29. 29 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 d>=1 do for each state s d of depth d and character a such that g(s d,a)=s' do { s=f(s d ) while g(s,a)=fail do { s=f(s) } f(s')=g(s,a) }

30. 30 g(0,c)≠fail for any possible character c. The failure function for {he, she, his, hers} is Time complexity: O(|P 1 |+|P 2 |+...+|P k |). Proof: Two pointer argument. Leave it for assignment (optional) s1 2 3 4 5 6 7 8 9 f(s)0 0 0 1 2 0 3 0 3