15 String Matching

7/28/2019 15 String Matching

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Jim Anderson Comp 750, Fall 2009 String Matching - 1

Chapter 32: String Matching

Given: Two strings T[1..n] and P[1..m] over alphabet .

Want to find all occurrences of P[1..m] the pattern in T[1..n] the text.

Example: = {a, b, c}a b c a b a a b c a a b a c

a b a a

text T

pattern P s=3

Terminology:

- P occurs with shift s.

- P occurs beginning at position s+1.

- s is a valid shift.

Goal: Find all valid shifts.

Applications: Text editors, search for patterns in DNA sequences

(actually, this is stretching the truth a little),


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Notation and Terminology

w pre x -- w is a prefix of x.

Example: aba pre abaabc.

w suf x -- w is a suffix of x.

Example: abc suf abaabc.

Note: In the book the symbol is used instead of pre,

and the symbol is used instead of suf.

I couldnt figure out an easy way to reproduce these

symbols in Powerpoint.


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Lemma 32.1

Lemma 32.1: Suppose x suf z and y suf z. If |x| |y| thenx suf y. If |x| |y| then y suf x. If |x| = |y| then x = y.

x

z

y

x

y

x

z

y

x

y

x

z

y

x

y


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More Notation

Pk= P[1..k] where k m.

Thus, P0 = , Pm = P[1..m] = P.

Similarly Tk= T[1..k], where k n.

Our Problem: Find all s, where 0 s nm such that P suf Ts+m.

Assumption:We assume the test x = y takes (t + 1) time,

where t is the length of the longest string z such that z pre xand z pre y.


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Nave Brute-Force Algorithm

Nave(T, P)n := length[T];

m := length[P];

for s := 0 to nm do

ifP[1..m] = T[s+1..s+m] thenprintpattern occurs with shift s

fi

od

Running time is ((nm + 1)m).

Bound is tight. Consider: T = an, P = am.


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Example

a c a a b c

a a bs = 0


7/45Jim Anderson Comp 750, Fall 2009 String Matching - 7

Example

a c a a b c

a a bs = 0



Example

a c a a b c

a a bs = 0



Example

a c a a b c

a a bs = 1


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Example

a c a a b c

a a bs = 2


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Example

a c a a b c

a a bs = 2


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Example

a c a a b c

a a bs = 2


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Example

a c a a b c

a a bs = 2

match!


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Example

a c a a b c

a a bs = 3


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Example

a c a a b c

a a bs = 3


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Example

a c a a b c

a a bs = 3


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Rabin-Karp Algorithm

Suppose = {0, 1, 2, , 9}.

Let us view P as a decimal number.

Example: View P = 31415 as 31,415.

Can also view substrings of T as decimal numbers.

Let ts = the decimal number corresponding to T[s+1..s+m].

Letp = the decimal number corresponding to P.

We want to know all s such that ts = p.

We can compute p in O(m)timeusing Horners Rule:

p = P[m] + 10(P[m1] + 10(P[m2] + + 10(P[2] + 10 P[1]) ))

Can similarly compute t0 in O(m) time.


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RK Algorithm (Continued)

Can compute t1, t2, , tn-m in O(nm) time as follows:ts+1 = 10(ts10

m-1T[s+1]) + T[s+m+1].

Example:T = 314152

ts+1 = 10(31415100003) + 2= 14152

Time Complexity:

O(n+m) + O(nm) = O(n+m)to compute

p and t0,,tn-m

to perform

nm+1 comparisons


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Two Problems

Might have || = d 10. Solution: Use radix-d arithmetic.

Numbers may be very large.

Solution: Perform computations modulo-q for some q.


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What is q?

Select q to be a large prime such that dq fits in one memory word.

all computations can be performed using single-precisionarithmetic.

To summarize, p is computed using

p = (P[m] + d(P[m1] + d(P[m2] + + d(P[2] + d P[1]) ))) mod q

t0 is computed similarly.

Other tis are computed usingts+1 = (d(tsT[s+1]h) + T[s+m+1]) mod q, where h dm-1 (mod q)

Unfortunately, we have a new problem: spurious hits.


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Example

pattern P

2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2 1

3 1 4 1 5

7

mod 13

text T

7mod 13

7

mod 13

valid

match

spurious

hit

Al i h


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Algorithm

We deal with spurious

hits by performing an

explicit checkwheneverthere is a potential match.

RK(T, P, d, q)

n := length[T];

m := length[P];h := dm-1 mod q;

p := 0;

t0 := 0;

for i := 1 to m do

p := (dp + P[i]) mod q;

t0 := (dt0 + P[i]) mod qod;

for s := 0 to nm do

ifp = tsthen

ifP[1..m] = T[s+1..s+m] then

printpattern occurs with shift s

fifi;

ifs < n-m then

ts+1 := (d(tsT[s+1]h) + T[s+m+1]) mod q

fi

od


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Running TimeWorst-Case:((nm + 1)m). (Again, consider P = am, T = an.)

Average-Case:

Some assumptions

Assume O(1) valid shifts.

Think of 0, 1, , q 1 like hash buckets.

Assume each bucket is equally likely.

We expect O(n/q) spurious hits.

Expected running time is:

O(n) + O(m(number of valid shifts + n/q))= O(n+m) choosing q m


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Finite Automata Algorithm

0 1 2 3 4 5 6 7a b a b a c a

a

a

a

a

b

b

a b c P0 1 0 0 a

1 1 2 0 b

2 3 0 0 a

3 1 4 0 b

4 5 0 0 a

5 1 4 6 c

6 7 0 0 a

7 1 2 0

state

input

i -- 1 2 3 4 5 6 7 8 9 10 11

T[i] -- a b a b a b a c a b a

state (i) 0 1 2 3 4 5 4 5 6 7 2 3

Processing time takes (n).But have to first construct FA.

Main Issue: How to construct FA?


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Need some Notation

(w) = state FA ends up in after processing w.

Example:(abab) = 4.

(x) = max{k: Pksuf x}. Called the suffix function.

Examples: Let P = ab.

() = 0(ccaca) = 1(ccab) = 2

Note: If |P| = m, then (x) = m indicates a match.T: a b a b b a b b a c

States:0 1..m.m.

Note Also: x suf y (x) (y).match match


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FA ConstructionGiven: P[1..m]

Let Q = states = {0, 1, , m}.

Define transition function as follows:

(q, a) = (Pqa) for each q and a.

Example:(5, b) = (P5b)=(ababab)= 4

Intuition:Encountering a b in state 5 means the current substring

doesnt match. But, you know this substring ends with abab -- and

this is the longest suffix that matches the beginning of P. Thus, we

go to state 4 and continue processing abab .

initial final


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Time Complexity

FA takes O(m||) time to construct.

(Book only gives a O(m3||) algorithm.)

Total time is O(n + m||).


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Correctness

Lemma 32.2:(xa) (x) + 1.

Proof:

Let r = (xa).

Case: r = 0. Clearly (xa) (x) + 1.

Case: r > 0.

a

Pr

x

Pr-1

We have:

Prsuf xa. Pr-1 suf x r1 (x).


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Another Lemma

Lemma 32.3: q = (x) (xa) = (Pqa) .

Proof:

Let q = (x). Pq suf x P

q

a suf xa

Let r = (xa). By Lemma 32.2, r q + 1. We have:

a

Pq

x

a

Pr

(Pqa) = r.


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Main Theorem

Theorem 32.4:(Ti) = (Ti) for all i = 0, 1, , n.

Implies: in accepting state

if and only if

string processed so far has a match at position (length of string) m

Proof:

Induction on i.

Basis: i = 0. T0 = .(T0) = (T0) = 0.

Step: Assume (Ti) = (Ti).

Let q = (Ti), a = T[i+1].

Then, (Ti) = (Ti) = q, which byLemma 32.3, implies

(Tia) = (Pqa). (**)


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Proof Continued

(Ti+1) = (Tia) , Ti+1 = Tia= ((Ti), a) , (wa) = ((w), a)= (q, a) , q = (Ti)

= (Pqa) , (q, a) = (Pqa)= (Tia) , by (**)= (Ti+1) , Ti+1 = Tia


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Knuth-Morris-Pratt Algorithm

Achieves (n + m) by avoiding precomputationof.

Instead, we precompute [1..m] in O(m) time.

As T is scanned, [1..m] is used to deduceinformation given by in FA algorithm.


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Motivating Example

b a c b a b

a b as

a b a a b c b a b

b a c a

T

P

q

Shift s is discovered to be invalid because of mismatch

of 6th character of P.

By definition of P, we also know s + 1 is an invalid shift

However, s + 2 may be a valid shift.

i i l


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Motivating Example

b a c b a b

a b as + 2

a b a a b c b a b

b a c a

T

P

k

The shift s + 2. Note that the first 3 characters of T starting

at s + 2 dont have to be checked again -- we already know

what they are.

i i l


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Motivating Example

a b

a b a

a b a

Pk

The longest prefix of P that is also a proper suffix of P5 is P3.

We will define [5] = 3.

Pq

In general, if q characters have matched successfully at shifts, the next potentially valid shift is s = s + (q[q]).

Th P fi F i


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The Prefix Function

is called the prefix function for P.

: {1, 2, , m} {0, 1, , m1}

[q] = length of the longest prefixof P that is a proper suffixof Pq, i.e.,

[q] = max{k: k < q and Pksuf Pq}.

Compute-(P)1 m := length[P];

2 [1] := 0;3 k := 0;

4 for q := 2 to m do

5 while k > 0 and P[k+1] P[q] do6 k := [k]

od;

7 ifP[k+1] = P[q] then

8 k := k + 1

fi;

9 [q] := kod;

10 return

E l


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Example

i 1 2 3 4 5 6 7

P[i] a b a b a c a

[i] 0 0 1 2 3 0 1

Same as our

FA example

P7 = a b a b a c aa = P1

P6 = a b a b a c

= P0

P5 = a b a b a

a b a = P3

P4 = a b a ba b = P2

P3 = a b a

a = P1

P2 = a b

= P0

P1 = a = P0

A h E l i


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Another Explanation

0 1 2 3 4 5 6 7a b a b a c a

Essentially KMP is computing a FA with epsilon moves. The spine

of the FA is implicit and doesnt have to be computed -- its just thepattern P. gives the transitions. There are O(m) such transitions.

Recall from Comp 455 that a FA with epsilon moves is

conceptually able to be in several states at the same time (in

parallel). Thats whats happening here -- were exploring

pieces of the pattern in parallel.

A h E l


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Another Example

i 1 2 3 4 5 6 7 8 9 10

P[i] a b b a b a a b b a[i] 0 0 0 1 2 1 1 2 3 4

P7 = a b b a b a aa = P1

P6 = a b b a b a

a = P1

P5 = a b b a b

a b = P2

P4 = a b b aa = P1

P3 = a b b

= P0

P2 = a b

= P0

P1 = a = P0

P10 = a b b a b a a b b a a b b a = P4

P9 = a b b a b a a b b

a b b = P3

P8 = a b b a b a a b

a b = P2

Ti C l it


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Time Complexity

Amortized Analysis --

0 loop q = 2 (1st iteration)1 loop q = 3 (2nd iteration)2 loop q = 4 (3rd iteration)

loop q = m ((m1)st iteration)m-1

= potential function = value of k

Amortized cost:i = ci + ii-1

iteration actual loop cost

Ti C l it (C ti d)


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Time Complexity (Continued)

Total amortized cost:

1m

1i01-mi

1m

1i

1iii

1m

1i

i

c

)(c

c

Ifm-1 0, then amortized cost upper bounds real cost.

We have 0

= 0 (initial value of k)

m-1 0 (final value of k).

We show i = O(1).

Ti C l it (C ti d)


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Time Complexity (Continued)

The value ofi obviously depends on how many times statement

6 is executed.

Note that k > [k]. Thus, each execution of statement 6 decreasesk by at least 1.

So, suppose that statements 5..6 iterate several times, decreasingthe value of k.

We have: number of iterations koldknew. Thus,

i O(1) + 2(k

oldk

new) +

i

i-1

Hence, i = O(1). Total cost is therefore O(m).

for statements

other than 5 & 6= knew = kold

R t f th Al ith


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Rest of the Algorithm

KMP(T, P)

n := length[T];

m := length[P];

:= Compute-(P);q := 0;

for i := 1 to n do

while q > 0 and P[q+1] T[i] doq := [q]

od;

ifP[q+1] = T[i] then

q := q + 1

fi;

ifq = m thenprintpattern occurs with shift i m;

q := [q]fi

od

Time complexity

of loop is O(n)

(similar to the

analysis of

Compute-).

Total time is

O(m + n).

E l


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Examplei 1 2 3 4 5

P[i] a b a b c[i] 0 0 1 2 0

P = a b a b c

1 2 3 4 5 6 7 8 9 10

T = a b b a b a b a b c

Start of 1st loop: q = 0, i = 1 [a]

2nd loop: q = 1, i = 2 [b]

3rd loop: q = 2, i = 3 [b]

4th loop: q = 0, i = 4 [a]

5th loop: q = 1, i = 5 [b]

6th loop: q = 2, i = 6 [a]7th loop: q = 3, i = 7 [b]

mismatch

detected

8th loop: q = 4, i = 8 [a]

9th loop: q = 3, i = 9 [b]

10th loop: q = 4, i = 10 [c]

Termination: q = 5

mismatch

detected

match

detected

Please see the book for formal correctness proofs.

(Theyre very tedious.)

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15 String Matching