1. Pattern Matching in String

2. Pattern Matching in String
Bài toán:
Cho:
Tập các kí tự 
xâu kí tự P (pattern), |P| = m,
văn bản T, |T| = n, n>>m.
Câu hỏi:
P  T?
Nếu P  T: vị trí xuất hiện đầu tiên của P trong T?
Example 1:
P = ABABDE
║║║║║║
T = ABABABDEAA,
i0=3
Dao Thanh Tinh 2

3. A Straightforward String Matching
Brute Force Algorithm
Input: P[1..m], T[1..n];
Output: i0 (if P  T[i0.. i0+m-1], i0 1, otherwise i0=0)
a) i=1; j= i; k=1;
b) while (jn) & (km)
if (T(j) = P(k))
{ j++; k++; }
else
{ i++; j=i; k=1;}
c) if (k>m) i0 := i
else i0 :=0;
Cpmplexity: O(mn).
Dao Thanh Tinh

4. A Straightforward String Matching
Brute Force Algorithm (*)
Input: P[1..m], T[1..n];
Output: i0 (if P  T[i0.. i0+m-1], i0 1, otherwise i0=0)
a) j= 1; k=1;
b) while (jn) & (km)
if (T(j) = P(k))
{ j++; k++; }
else
{j = j k+2; k=1;}
c) if (k>m) i0 := j  m
else i0 :=0;
Cpmplexity: O(mn). k 1 2 3 4 5 P A B E T D j 6 7 8 9 10 11 12
k=5, j=7
On the new step:
k=1, j= 7 – = 4 k 1 2 3 4 5 P A B E T D j 6 7 8 9 10 11 12
Dao Thanh Tinh 4

5. A Straightforward String Matching
Example 2:
P = ABABDE
T = ABABABDEAA
║║║║
T = A BABABDEAA
║║║║║║
T = ABABABDEAA,
successful match, i0=3
Example 3:
P = UUUUUUX
T = UUUUUUUUUUUU
║║║║║║
Dao Thanh Tinh 5

6. The Morris-Pratt Algorithm (1)
Assume that the first mismatch occurs between P(k) and T(j) with 1 < k ≤ m. Then,
P(1..k-1) = T(j-k+1... j-1) = u u P1
Pk-1 Pk Pm
Tj-1 Tj
Tj-k+1 u
P1....Pk-1 = Tj-k+1….Tj-1 =u Pk Tj P1 Pr
Pk-r
Pk-1 Pk Pm
Tj-1 Tj
Tj-k+1 v
Idea:
Shifting P on the left, expect that a prefix v of P matches some suffix of the portion u.
The longest such prefix v is called the border of u.
P1…Pr = Pk-r….Pk-1
Dao Thanh Tinh 6

7. The Morris-Pratt Algorithm (2)
The Brute Force Algorithm: T H Ủ Ư Ợ Ở N G T H Ủ Ư Ợ Ờ N G
(1)
(2)
(3)
(4)
(5)
(6)
(7) T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G
Dao Thanh Tinh 7

8. The Morris-Pratt Algorithm (3)Ủ Ư Ợ Ờ N G
(8)
(9)
(10)
(11)
(12)
(13) T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G
The Brute Force Algorithm performs on 13 steps.
Dao Thanh Tinh 8

9. The Morris-Pratt Algorithm (4)Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G
Pattern was found on the 6th step.
Dao Thanh Tinh 9

10. The Morris-Pratt Algorithm (5)
Pk-r
Pk-1 Pk Pm
Tj-1 Tj
Tj-k+1
Pr+1
P1…Pr = Pk-r….Pk-1
Set mp(k) = r+1.
Then, after a shift, the comparisons can resume between characters P(mp(k)) and T(j).
a) j= 1; k=1;
b) while (jn) & (km)
if (T(j) = P(k))
{ j++; k++; }
else
{ j=mp(k); k=1;}
c) if (k>m) i0 := j-m;
else i0 :=0;
Dao Thanh Tinh 10

11. The Morris-Pratt Algorithm (6)
The value of mp(1) is set to 0. P1 Pr
Pk-r
Pk-1 Pk Pm
Tj-1 Tj
Tj-k+1
Pr+1 k 5 P A B E k 1 2 3 4 5 P A B E T D j 6 7 8 9 10 11 12
k>1:
r = k-2;
while (r>0) && (P1..Pr ≠ Pk-r..Pk-1)
do r--;
mp(k) = r+1;
On the next step:
k=mp(5) =3, j= 7 (giữ nguyên)
r= k-2 = 3
P[1..3] ? P[2..4]: ABA ≠ BAB
r = 2
P[1..2] ? P[4..5]: AB = AB
mp(5)= r+1 = 3 k 1 2 3 4 5 P A B E T j 6 7 8 9 10 11 12
Dao Thanh Tinh 11

12. The Morris-Pratt Algorithm (7)
k>1:
r = k-2;
while (r>0) && (P1..Pr ≠ Pk-r..Pk-1)
do r--;
mp(k) = r+1; k 1 2 3 4 5 6 7 P U X
k=7
r= k-2 = 5
P[1..5] ? P[2..6]: UUUUU =UUUUU
mp(5)= r+1 = 6 k 1 2 3 4 5 6 7 P U X T j 8 9 10 11 12 13 14 15
On the next step:
k=mp(7) =6, j= 7 (giữ nguyên) k 1 2 3 4 5 6 7 P U X T j 8 9 10 11 12 13 14 15
Dao Thanh Tinh 12

13. The Morris-Pratt Algorithm (8)k 1 2 3 4 5 6 7 P T H Ủ Ư
k>1:
r = k-2;
while (r>0) && (P1..Pr ≠ Pk-r..Pk-1)
do r--;
mp(k) = r+1;
k=7
r= k-2 = 5
P[1..5] ? P[2..6]: THU_T ≠ HỦ_TH
r= 4
P[1..4] ? P[3..6]: THU_ ≠ Ủ_TH
r= 3
P[1..3] ? P[4..6]: THU ≠ _TH
r= 2
P[1..2] ? P[5..6]: TH = TH
mp(7)= r+1 = 3 k 1 2 3 4 5 6 7 P T H Ủ Ư Ợ Ờ N G j 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
On the next step:
k=mp(7) =3, j= 7 (giữ nguyên) k 1 2 3 4 5 6 7 P T H Ủ Ư Ợ Ờ N G j 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Dao Thanh Tinh 13

14. The Morris-Pratt Algorithm (9)k 1 2 3 4 5 6 7 P T H Ủ Ư
k>1:
r = k-2;
while (r>0) && (P1..Pr ≠ Pk-r..Pk-1)
do r--;
mp(k) = r+1;
k=5
r= k-2 = 3
P[1..3] ? P[2..4]: THU ≠ HỦ_
r= 2
P[1..2] ? P[3..4]: TH ≠ Ủ_
r= 1
P[1..1] ? P[4..4]: T ≠ _
r= 0
mp(5)= r+1 = 1 k 1 2 3 4 5 6 7 P T H Ủ Ư Ỉ N G Y Ê j 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
On the next step:
k=mp(5) =1, j= 5 (giữ nguyên) k 1 2 3 4 5 6 7 P T H Ủ Ư Ỉ N G Y Ê j 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Dao Thanh Tinh 14

15. The Morris-Pratt Algorithm (10)
k=1:
r = -1
mp(k) = 0 ? P1 Pm Tj
Tj+1
k=1:
mp(1) = 0
comparisons can resume between characters P(mp(k)) = P(0) and T(j),
but P(0) is not existent.
In this case, comparisons can resume between P(1) and T(j+1).
Then, set mp(1) = 1, j= j+1.
Dao Thanh Tinh 15

16. The Morris-Pratt Algorithm (11)
k=2..m:
r = k-2;
while (r>0) && (P1..Pr ≠ Pk-r..Pk-1) do r--;
mp(k) = r+1;
k=1:
mp(k)=1.
a) j= 1; k=1;
b) while (jn) & (km)
if (T(j) = P(k)) { j++; k++; }
else
{ if (k=1) j++;
k=mp(k); }
c) if (k>m) i0 := j-m; else i0 :=0; mp
k =7 T H Ủ Ư 3 6 2 5 1 4
Dao Thanh Tinh 16

17. The Morris-Pratt Algorithm (12)Ủ Ư mp 1 2 3 T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G T H Ủ Ư Ợ Ờ N G
Pattern was found on the 6th step.
Dao Thanh Tinh 17

18. The Morris-Pratt Algorithm (13)
a) j= 1; k=1;
b) while (jn) & (km)
if (T(j) = P(k)) { j++; k++; }
else
{ if (k=1) j++;
k = mp(k); }
c) if (k>m) i0 := j - m; else i0 :=0; T H Ủ Ư mp 1 2 3 T H Ủ Ư R Ờ N G T H Ủ Ư R Ờ N G T H Ủ Ư R Ờ N G T H Ủ Ư R Ờ N G T H Ủ Ư R Ờ N G T H Ủ Ư R Ờ N G
Dao Thanh Tinh 18

19. The Knuth-Morris-Pratt Algorithm (1)
Look more closely at the Morris-Pratt algorithm: P1 Pk Tj T ≠ u a
Input: P[1..m], T[1..n];
Output: i0
a) j= 1; k=1;
b) while (jn) & (km)
if (T(j) = P(k)) { j++; k++; }
else { if (k=1) j++;
k =mp(k); }
c) if (k>m) i0= j-m; else i0=0; b v P1
Pmp(k) c a v PK P1 a
k=1..m:
r = k-2;
while (r>0) && (P1..Pr ≠ Pk-r..Pk-1) do r--;
mp(k) = r+1;
Dao Thanh Tinh 19

20. The Knuth-Morris-Pratt Algorithm (2)Pk Tj T ≠ u a
Let 1< k ≤ m:
If c=a then c≠b.
The mismatch between P(mp(k)) and T(j) occurs!
To avoid another immediate mismatch, the character P(mp(k)) must be different from a=P(k). b v P1
Pmp(k) c a P1 v PK a b
k=1..m:
r = k-2;
while (r>0) && ((P1..Pr ≠ Pk-r..Pk-1) OR (Pr=Pk)) do r--;
kmp(k) = r+1;
Dao Thanh Tinh 20

21. The Knuth-Morris-Pratt Algorithm (3)
The Morris-Pratt:
The Knuth-Morris-Pratt: 1 2 3 4 5 6 7 mp
k =7 T H Ủ Ư 1 2 3 4 5 6 7
kmp
k =7 T H Ủ Ư
k=6
r= k-2 = 4: P[1..4] ? P[2..5]: THU_ ≠ HỦ_T
r= 3: P[1..3] ? P[3..5]: THU ≠ Ủ_T
r= 2: P[1..2] ? P[4..5]: TH ≠ _T
r= 1: P[1..1] ? P[5..5]: T = T,
P[r] ? P[k] H = H
r = 0
kmp(6)= r+1 = 1
Dao Thanh Tinh 21

22. The Knuth-Morris-Pratt Algorithm (4)
Example:
mp kmp
k = 8, P=“ABABABAB”
k = 7, P=“ABABABAB”
k = 6, P=“ABABABAB”
k = 5, P=“ABABABAB”
k = 4, P=“ABABABAB”
k = 3, P=“ABABABAB”
k = 2, P=“ABABABAB”
k = 1, P=“ABABABAB”
Dao Thanh Tinh 22

23. The Knuth-Morris-Pratt Algorithm (5)
Look more closely at the Morris-Pratt algorithm: P1 Pk Tj T ≠ u a
Input: P[1..m], T[1..n];
Output: i0
a) j= 1; k=1;
b) while (jn) & (km)
if (T(j) = P(k)) { j++; k++; }
else { if (k=1) j++;
k =kmp(k); }
c) if (k>m) i0= j-m; else i0=0; b v P1
Pmp(k) c a v PK P1 a
k=1..m:
r = k-2;
while (r>0) && ((P1..Pr ≠ Pk-r..Pk-1) OR (Pr=Pk)) do r--;
kmp(k) = r+1;
Dao Thanh Tinh 23

24. The Brute Force Algorithm 2
Input: P[1..m], T[1..n];
Output: i0 (if P  T[i0.. i0+m-1], i0 1, otherwise i0=0)
a) i=m; j= i; k=m;
b) while (jn) & (k>0)
if (T(j) = P(k)) { j--; k--; }
else { i++;
j=i+m; k=m; }
c) if (k=0) i0 := i
else i0 :=0;
Complexity: O(mn).
Example:
P = DEABAB
║║
T = ABDEABABAA 
T = ABDEABABAA
║║║║║║
T = ABDEABABAA,
successful match, i0=3
Dao Thanh Tinh 24

25. The Brute Force Algorithm 2*
Input: P[1..m], T[1..n];
Output: i0 (if P  T[i0.. i0+m-1], i0 1, otherwise i0=0)
a) j= m; k=m;
b) while (jn) & (k>0)
if (T(j) = P(k)) { j--;
k--; }
else {
j = j+m-k+1;
k=m; }
c) if (k=0) i0 := j-m
else i0 :=0
Complexity: O(mn).
Example:
P = DEABAB
║║
T = ABDEABABAA 
T = ABDEABABAA
║║║║║║
T = ABDEABABAA,
successful match, i0=3
Dao Thanh Tinh 25

26. The Boyer-Moore Algorithm (1)
m-k-1 u Pk
Pk+1 Pm
Tj+m-k Tj
Tj+1
T[j] P[k], T[j+1....j+m-k] = P[k+1....m] = u
Dao Thanh Tinh 26

27. The Boyer-Moore Algorithm (2)
The good-suffix shift consists in aligning the segment u with its rightmost occurrence in P
good-suffix shift
Pq-1 Pm Pq Pt
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k u
Tj-new
a) Find largest t [1..m-1] such that:
u = P[k+1..m] P[q..t], Pq-1≠Pk (q>1)
u = P[k+1..m] P[q..t], (q=1)
Then,
j-new = j + m-q+1 = j + 2m-t-k
Dao Thanh Tinh 27

28. The Boyer-Moore Algorithm (3)
b) If not exists t [1..m-1] such that:
u = P[k+1..m] P[q..t]
the shift consists in aligning the longest suffix v of P with a matching prefix of P
Find largest t [1..m-1] such that: u = P[m-t+1..m] P[1..t],
Then, j-new = j + 2m-t-k
good-suffix shift P1 Pt Pm v
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Pm-t+1
Tj-new
Dao Thanh Tinh 28

29. The Boyer-Moore Algorithm (4)
c) If not exists t [1..m-1] such that:
u = P[m-t+1..m] P[1..t]
Then,
j-new = j + 2m-k
or j-new = j + 2m-k-t, where t=0
good-suffix shift Pm P1
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Tj-new
Dao Thanh Tinh 29

30. The Boyer-Moore Algorithm (5)
d) If Tj [P1...Pm] :
then, j-new = j + m
good-suffix shift Pm P1
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Tj-new
Dao Thanh Tinh 30

31. The Boyer-Moore Algorithm (6)
a) j=m; k=m;
b) while (jn) & (k>0)
if T(j) = P(k) { j--; k--; }
else { k = m;
j = jnew; }
c) if (k=0) i0= j+1; else i0= 0;
Complexity: O(nm)
remark: Jnew and j+1 are the new components
on comparison with Brute Force Algorithm.
Dao Thanh Tinh 31

32. Computing Jnew ? The Boyer-Moore Algorithm (7) 2018-12-31
Dao Thanh Tinh 32

33. The Boyer-Moore Algorithm (8)
a) Find largest t [1..m-1] such that:
u = P[k+1..m] P[q..t], Pq-1≠Pk (q>1)
u = P[k+1..m] P[q..t], (q=1)
Then,j-new = j + m-q+1 = j + 2m-t-k
good-suffix shift
Pq-1 Pm Pq Pt
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k u
Tj-new
a) bmg(k) = 2m-k-t
t = m-1;
while (t>m-k) & (P[k+1..m] ≠P[t-m+k+1..t]) OR (Pt-m+k=Pk) t=t-1;
if (t=m-k) & P[k+1..m] ≠ P[1..t]) t=0;
remark: when t=0, bmg(k) = 2m-k;
Dao Thanh Tinh 33

34. The Boyer-Moore Algorithm (9)
b) If not exists t [1..m-1] such that: u = P[k+1..m] P[q..t]
the shift consists in aligning the longest suffix v of P with a matching prefix of P
Find largest t [1..m-1] such that: u = P[m-t+1..m] P[1..t],
Then, j-new = j + 2m-k-t
b) bmg(k) = 2m-k-t
t = m - k-1;
while (t>0) & (P[m-t+1..m] ≠P[1..t]) t = t-1;
if (t=m-k+2) & P[k+1..m] ≠ P[1..t]) t=0;
remark: when t=0, bmg(k) = 2m-k;
good-suffix shift
Tj-new Pm P1 Pt
Pk+1 Pk
Tj+1 Tj
Tj+m-k
Pm-t+1 v
Dao Thanh Tinh 34

35. The Boyer-Moore Algorithm (10)
c) If not exists t [1..m-1] such that: u = P[m-t+1..m] P[1..t]
Then, j-new = j + 2m-k
or j-new = j + 2m-k-t, where t=0
c) bmg(k) = 2m – k - t
where t = 0
good-suffix shift Pm P1
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Tj-new
Dao Thanh Tinh 35

36. The Boyer-Moore Algorithm (11)
Pq-1 Pm Pq Pt
The Boyer-Moore Algorithm (11)
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Tj-new
t = m-1;
while (t>m-k) & (P[k+1..m] ≠P[t-m+k+1..t]) OR (Pt-m+k=Pk) t=t-1;
if (t=m-k) & P[k+1..m] ≠ P[1..t]) t=0;
if (t>0) bmg(k) = 2m-k-t;
else
t = m - k-1;
while (t>0) & (P[m-t+1..m] ≠P[1..t]) t = t-1;
if (t=m-k+2) & P[k+1..m] ≠ P[1..t]) t=0;
if (t>0) bmg(k) = 2m-k-t
else bmg(k) = 2m – k
Tj-new Pm P1 Pt
Pk+1 Pk
Tj+1 Tj
Tj+m-k
Pm-t+1 Pm P1
Tj-new
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Dao Thanh Tinh 36

37. The Boyer-Moore Algorithm (12)
d) If Tj [P1...Pm] :
then, the left end of the window is aligned with the character immediately after Tj, namely Tj+1. j-new = j + m
d) bmS(Tj) = m
but Tj  {P1, ..., Pm} ?
good-suffix shift Pm P1
Pk+1 Pk Pm
Tj+1 Tj
Tj+m-k
Tj-new
Dao Thanh Tinh 37

38. bmS(c) =m The Boyer-Moore Algorithm (13) Define:
for all c  {P1,...., Pm}
good-suffix shift Pm P1
Pk+1 Pk Pm
Tj+1 c
Tj+m-k
Tj-new
Dao Thanh Tinh 38

39. The Boyer-Moore Algorithm (14)
Find
Px= b
where Px is rightmost occurrence characer’s b in {P1,...., Pm-1}
contains no b
Px+1 Pm b
Tj-new
Pk+1 Pk Pm
Tj+1
T j=b
Tj+m-k
jnew = j + m-x
Dao Thanh Tinh 39

40. The Boyer-Moore Algorithm (15)
for k=1 to m-1
t=k
for i=k+1 to m-1
if (P(t)=P(i)) t=i;
bmS(P(k)) = m-t;
bmS(P(m)) = 1;
T H Ủ T H Ư
bmS:
contains no b Px Pm b
Tj-new
Pk+1 Pk Pm
Tj+1
T j=b
Tj+m-k
Dao Thanh Tinh 40

41. The Boyer-Moore Algorithm (16)Ủ Ư
bmg 13 12 11 10 9 8 1 T H Ủ Ư Ợ Ờ N G 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 T H Ủ Ư
bmS 2 1 4 3 T H Ủ Ư Ợ Ờ N G 2 1 4 3 7 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Dao Thanh Tinh 41

42. The Boyer-Moore Algorithm (17)Ủ Ư mp 1 2 3 T H Ủ Ư Ợ Ờ N G T H Ủ Ư
bmS 2 1 4 3 T H Ủ Ư Ợ Ờ N G 3 1 4 7 2 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Dao Thanh Tinh 42

43. The Boyer-Moore Algorithm (18)S E N L D Y T M F O R 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
P =“SENSE”
bms=[ ]
bmg=[ ] S E N L D Y T M F O R 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 S E N L D Y T M F O R 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
max{bms,bmg}
Dao Thanh Tinh 43

44. The Karp-Rabin Algorithm
Giả thiết  = {1, 2, ...,9}
p = ts ?
 s {1,...,n-m+1}: ts = p ?
Dao Thanh Tinh 44

45. The Karp-Rabin Algorithm (2)
Tính p theo sơ đồ Horner’s :
p= P(m) + 10*{P(m-1)+ 10*{P(m-2) *{P(2)+10*P(1)}..}}
p=P(1)
for i=2 to m do p = P(i) + 10*p;
Thời gian tính: O(m)
Dao Thanh Tinh 45

46. The Karp-Rabin Algorithm (3)
Tính ts: ts = 10m-1T(s) + 10m-2T(s+1)+10m-3T(s+2)...+10T(s+m-2)+T(s+m-1)
ts+1 = 10m-1T(s+1) + 10m-2T(s+2) T(s+m-2)+10T(s+m-1)+T(s+m)
= 10{10m-2T(s+1) + 10m-3T(s+2) T(s+m-2)+T(s+m-1)}+T(s+m)
= 10{ ts – 10m-1T(s)} + T(s+m)
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47. The Karp-Rabin Algorithm (4)
p=P(1);
t=T(1);
a=1;
for i=2 to m
{ p = P(i) + 10*p;
t = T(i) + 10*t;
a = a*10; }
1. s=1;
2. while (s<n-m+1) &(t ≠ p)
a) t=10*( t – a*T(s))+ T(s+m)
b) s = s+1;
3. if (s=n-m) return 0
else return s;
O(m+n)
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48. The Karp-Rabin Algorithm (5)
p=P(1) mod q;
t=T(1) mod q;
a=1;
for i=2 to m
{ p = (P(i) + 10*p) mod q;
t = (T(i) + 10*t) mod q;
a = (a*10) mod q; }
defined:
a(q) = a mod q
t(q) = t mod q
p(q) = p mod q
t(q) =10*( t(q) – a(q)*T(s)mod q)+ T(s+m)mod q
t = p  t(q) = p(q)
t(q) ≠ p(q)  t ≠ p
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49. The Karp-Rabin Algorithm (6)
p=P(1) mod q;
t=T(1) mod q;
a=1;
for i=2 to m
{ p = (P(i) + 10*p) mod q;
t = (T(i) + 10*t) mod q;
a = (a*10) mod q; }
s=1;
while(s<m-n+1)
if (t(q)=p(q))
if (P=Ts) return s;
else
t(q) =10*( t(q) – a(q)*T(s)mod q)+ T(s+m)mod q
s = s+1;
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50. Conclusion Brute Force Algorithm 1: Straightforward Matching
The Morris-Pratt Algorithm
Knuth-Morris-Pratt Algorithm
Brute Force Algorithm 2: Backing
The Boyer-Moore Algorithm
The Karp-Rabin Algorithm
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