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# String Matching - PowerPoint PPT Presentation

String Matching. String Matching. Problem is to find if a pattern p of length m occurs within text t of length n Simple solution: Naïve String Matching Match each position in the pattern to each position in the text t = AAAAAAAAAAAAAA p = AAAAAB AAAAAB etc.

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## PowerPoint Slideshow about 'String Matching' - kineks

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### String Matching

• Problem is to find if a pattern p of length m occurs within text t of length n

• Simple solution: Naïve String Matching

• Match each position in the pattern to each position in the text

• t = AAAAAAAAAAAAAA

• p = AAAAAB

AAAAAB

etc.

• O(m(n-m+1)) worst case; average case performance is surprisingly good provided stings are neither long nor have lots of repeated letters. Unfortunately, this occurs in DNA sequences and images.

public static int search(String pattern, String text) {

int m = pattern.length();

int n = text.length();

for (int i = 0; i < n - m; i++) {

int j;

for (j = 0; j < m; j++) {

if (text.charAt(i+j) != pattern.charAt(j))

break;

}

if (j == m) return i;

}

return -1;

}

• Idea: Before spending a lot of time comparing chars for a match, do some pre-processing to eliminate locations that could not possibly match

• If we could quickly eliminate most of the positions then we can run the naïve algorithm on what’s left

• Eliminate enough to hopefully get O(n) + cost of pre-processing run-time overall. Obviously want preprocessing to also be fast.

• To get a feel for the idea say that our text and pattern is a sequence of bits.

• For example,

• p=010111

• t=0010110101001010011

• The parity of a binary value is to count the number of one’s. If odd, the parity is 1. If even, the parity is 0. Since our pattern is six bits long, let’s compute the parity for each position in t, counting six bits ahead. Call this f[i] where f[i] is the parity of the string t[i..i+5].

t=0010110101001010011

p=010111

Since the parity of our pattern is 0, we only need to check positions 2, 4, 6, 8, 10, and 11 in the text. By the way, how do we compute parity of all substrings of length m in just order n, because if we do all n-m+1 substrings separately, that will already cost us m(n-m+1) units of time.

• On average we expect the parity check to reject half the inputs.

• To get a better speed-up, by a factor of q, we need a fingerprint function that maps m-bit strings to q different fingerprint values.

• Rabin and Karp proposed using a hash function that considers the next m bits in the text as the binary expansion of an unsigned integer and then take the remainder after division by q.

• A good value of q is a prime number greater than m.

• More precisely, if the m bits are s0s1s2 .. sm-1 then we compute the fingerprint value:

• For the previous example, f[i] =

• Consider how to compute f incrementally.

For our pattern 010111, its hash value is 23 mod 7 or 2. This means that we would only use the naïve algorithm for positions where f[i] = 2.

• But we want to compare text, not bits!

• Text is represented using bits

• For a textual pattern and text, we simply use their ASCII sequences.

• We can compute f[i] in O(m) time giving us the expected runtime of O(m+n), given a good hash function. This can be a worst case of mn if we get significant hash conflicts. Of course, we could try doing just probabilistic.

public static int MonteCarloRabinKarp (String p, String t) {

int m = p.length();

int n = t.length();

int dM = 1, h1 = 0, h2 = 0;

int q = 3355439; // table size

int d = 256; // radix

for (int j = 1; j < m; j++) // precompute d^M % q

dM = (d * dM) % q;

for (int j = 0; i < m; i++) {

h1 = (h1*d + p.charAt(j)) % q; // hash of pattern

h2 = (h2*d + t.charAt(j)) % q; // hash of text

}

if (h1 == h2) return 0; // match found, we hope

for (int i = m; i < n; i++) {

h2 = (h2 - t.charAt(i-m)) % q; // remove high order digit

h2 = (h2*d + t.charAt(i)) % q; // insert low order digit

if (h1 == h2) return i – m + 1; // match found, we hope

}

}

• Las Vegas algorithms

• Expected to be fast

• Guaranteed to be correct (if halts and gives answer)

• Ex: quicksort, Rabin-Karp with match check

• Monte Carlo algorithms

• Guaranteed to be fast

• Expected to be correct

• Ex: Rabin-Karp without match check

* Monte Carlo – sometimes reports success when not true.

** Las Vegas – does match check

(really unlikely to take this long as most checks are cut off)

• This is a commonly used linear-time running string matching algorithm that achieves O(m+n) running time (worst and expected).

• Uses an auxiliary function pi[1..m] pre-computed from p in time O(m).

• This function contains knowledge about how the pattern shifts against itself.

• If we know how the pattern matches against itself, we can slide the pattern more characters ahead than just one character as in the naïve algorithm.

Naive

p: pappar

t: pappappapparrassanuaragh

p: pappar

t: pappappapparrassanuaragh

Smarter technique:

We can slide the pattern ahead so that the longest PREFIX of p that we have already processed matches the longest SUFFIX of t that we have already matched.

p: pappar

t: pappappapparrassanuaragh

p: pappar

t: pappappapparrassanuaragh

p: pappar

t: pappappapparrassanuaragh

p: pappar

t: pappappapparrassanuaragh

The characters mismatch so we shift over one character for both the text and the pattern:

p: pappar

t: pappappapparrassanuaragh

We continue in this fashion until we reach the end of the text.

• Runtime

• O(m) to compute the Pi values

• O(n) to compare the pattern to the text

• Total O(n+m) runtime

• KMP algorithm.

• Use knowledge of how search pattern repeats itself.

• Build DFA from pattern.

• Run DFA on text.

• DFA used in KMP has special property.

• Upon character match, go forward one state.

• Only need to keep track of where to go upon character mismatch:

• go to state next[j] if character mismatches in state j

2nd Example of KMP next

for (int i = 0, j = 0; i < n; i++) {

if (t.charAt(i) == p.charAt(j)) j++; // match

else j = next[j]; // mismatch

if (j == m) return i - m + 1; // found

}

* Monte Carlo – sometimes reports success when not true.

** Las Vegas – does match check

(really unlikely to take this long as most checks are cut off)

• It is possible in some cases to search text of length n in less than n comparisons!

• Horspool’s algorithm is a relatively simple technique that achieves this distinction for many (but not all) input patterns. The idea is to perform the comparison from right to left instead of left to right.

• Consider searching:

T=BARBUGABOOTOOMOOBARBERONI

P=BARBER

• There are four cases to consider

1. There is no occurrence of the character in T in P. In this case there is no use shifting over by one, since we’ll eventually compare with this character in T that is not in P. Consequently, we can shift the pattern all the way over by the entire length of the pattern (m):

2. There is an occurrence of the character from T in P. Horspool’s algorithm then shifts the pattern so the rightmost occurrence of the character from P lines up with the current character in T:

3. We’ve done some matching until we hit a character in T that is not in P. Then we shift as in case 1, we move the entire pattern over by m:

4. If we’ve done some matching until we hit a character that doesn’t match in P, but exists among its first m-1 characters. In this case, the shift should be like case 2, where we match the last character in T with the next corresponding character in P:

• More on case 4

• We first precompute the shifts and store them in a table. The table will be indexed by all possible characters that can appear in a text. To compute the shift T(c) for some character c we use the formula:

• T(c) = the pattern’s length m, if c is not among the first m-1 characters of P, else the distance from the rightmost occurrence of c in P to the end of P

In running only make 12 comparisons, less than the length of the text! (24 chars)

• Similar idea to Horspool’s algorithm in that comparisons are made right to left, but is more sophisticated in how to shift the pattern.

• Using the “bad symbol” heuristic, we jump to the next rightmost character in P matching the char in T:

• upon mismatch of text character c, look up index[c]

• increase offset i so that jth character of pattern lines up with text character c

• Use KMP-like suffix rule.

• effective with small alphabets

• different rules lead to different worst-case behavior

* Monte Carlo – sometimes reports success when not true.

** Las Vegas – does match check

(really unlikely to take this long as most checks are cut off)