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Chapter 5.2

Chapter 5.2. String Searching - Part 2 Boyer-Moore Algorithm Rabin-Karp Algorithm. The Boyer-Moore String Algorithm. This method can give substantially faster searches where the language contains a large number of symbols

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Chapter 5.2

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  1. Chapter 5.2 String Searching - Part 2 Boyer-Moore Algorithm Rabin-Karp Algorithm

  2. The Boyer-Moore String Algorithm • This method can give substantially faster searches where the language contains a large number of symbols • E.g. Normal text (128 or 256 character alphabet) rather than binary strings • BM method incorporates two main ideas • start matching at the right of the pattern so as to find the rightmost mismatch • use information about the possible alphabet of the text, as well as the characters in the pattern

  3. Example Search for LEAN in CARPETS NEED CLEANING REGULARLY CARPETS NEED CLEANING REGULARLY LEAN N and P mismatch. Furthermore, P does not occur anywhere in the string LEAN. Hence move string all the way past P and compare with N again. CARPETS NEED CLEANING REGULARLY LEAN LEAN LEAN LEAN N and E mismatch, but E occurs in LEAN, so we move the E of LEAN to this position

  4. Boyer-Moore preprocessing • In order to implement the above idea, consider the characters in the alphabet which makes up the text. • C0,C1,…,Ck (k+1 characters in the alphabet) • Initialise an array skip such that • for each Cj in the pattern string set skip[j] to the distance of Cj from the right hand end of the pattern • skip[j] = M otherwise, where M is the length of the pattern.

  5. The skip array - example • Suppose pattern is LEAN and alphabet is <blank>, A,B,…,Z (C0,C1,…,C26). • skip[12] = 3 (L) • skip[5] = 2 (E) • skip[1] = 1 (A) • skip[14] = 0 (N) • skip[X] = 4 (otherwise) • skip[C]is the number of characters to move the pattern to the right after a mismatch in the text with character with index C

  6. Using the skip array • Try to match the pattern from right to left • Mismatch occurs between Cn with index n and the (M-j)th position of the pattern. • Get value of skip[n] • If (M-j)> skip[n] then shift pattern by 1 (since we have already passed the rightmost occurrence of Cn in the pattern). • Else shift pattern skip[n]-j positions, to try to align Cn in the text with the rightmost occurrence of Cn in the pattern.

  7. Example - shifting using skip Pattern X X A X X X Z Z Z Z M = 10 (length of pattern) skip[1] = 7 (distance of rightmost A from right) mismatch at position 10-4 Y Y Y Y Y A Z Z Z Z Z Z Z Z Z text X X A X X X Z Z Z Z mismatched pattern X X A X X X Z Z Z Z shift 3 positions Shift pattern by 7-4 = 3 positions

  8. Boyer-Moore Algorithm (1) int boyermoore1(String P, String T){ int 1,j,t,M=P.length(),N=T.length(); initskip(P); // initialise skip array i = M-1; j = M-1; while (j > 0){ while (T[i] != P[j]){ t = skip[index(T[i])]; if ((M-j)>t) {i=i+M-j;} else {i=i+t;} if (I >= N) return N; // no match j = M-1;} i--; j--;} return i; } // successful match

  9. Refinement to B-M Algorithm • We can apply the KMP algorithm “right-to-left” • Sometimes this gives a larger skip value than the skip index used above • E.g. Pattern BBAAA • skip[1] = 0 (skip value for A) • skip[2] = 3 (skip value for B) AAAAAAA BBAAA mismatch on A in text boyermoore1 algorithm shifts only one position However it’s clear that AAA does not occur anywhere to the left of positions 3,4,5

  10. Boyer-Moore Refinement (2) • Build KMP next array from right to left j = position of mismatch (from right) next[j] = no. of positions to shift pattern to right j next[j] BBAAA 2 1 BBAAA 3 1 BBAAA 4 5 BBAAA 5 5 BBAAA Using the next array, a mismatch on B results in a shift of 5 positions

  11. Refined Boyer-Moore Algorithm • Initialise both the skip and the next arrays (right-to-left). • Whenever a mismatch occurs, get the skip value for the mismatched character and the next value for the position of the mismatch. • Shift the pattern right by whichever gives the greater value.

  12. Rabin-Karp String Matching • Consider a text and pattern consisting of characters represented by b bits each • e.g. 7-bit ASCII characters • We can regard a sequence of characters as a (large) binary number (as with keys when using hash tables) • Idea - compute a hash value for an M -character pattern and compare it successively with the hash values of each successive sequence of M characters in the text.

  13. Rabin-Karp matching - basic idea • Example. Consider the string CARPETS NEED CLEANING and the search string LEAN. • Then we compare h(LEAN) first against h(CARP), then against h(ARPE), h(RPET), h(PETS), and so on. • Clearly h(LEAN) need be computed only once. • The key to efficient comparison is to compute the successive hash values efficiently. • We can exploit the fact that successive keys overlap, e.g. ARPE and RPET share 3 characters.

  14. Rabin-Karp - computing hash values • Let us use h(K) = K mod P as our hash function as before, where P is a large prime number • Let d = max number of characters (e.g. d=2b) • Suppose K = C1,…,Cn where C1,…,Cn is a sequence of characters in the text, and h(K) = X • It can be shown that h(C2,…,Cn+1)= h((XC1*dn-1)*d + Cn+1), since • C2,…,Cn+1 can be rewritten as (C1,…,Cn - C1* dn-1)*d + Cn+1) • E.g. (d=10): 45678 = (34567 - (3*104))*10 + 8 • Then use some properties such as h(X+Y) = h(h(X) +Y) and h(X*Y) = h(h(X) * Y) • Hence h(45678) = h((h(34567) - (3*104))*10 + 8 • Thus, successive values for h are efficiently computed, since we can reuse the previous has value to compute the next one.

  15. Rabin-Karp Algorithm int rabinkarp(String P, String T){ int q=33554393 // a large prime int d=32 // size of alphabet int i,dM=1, h1=0, h2=0; int M=P.length(), N=T.length(); for (i=0;i<M;i++){dM=(d*dM) mod q;} for (i=0;i<M;i++){ h1=(h1*d+val(P[i])) mod q; // hash P h2=(h2*d+val(T[i])) mod q; } for (i=0; h1 != h2; i++){ h2=(h2+d*q-val(T[i]))*dM) mod q; h2=(h2*d + val(a[i+M])) mod q; if (i > N-M) return N;} \\ not found return i; }

  16. Rabin-Karp - analysis • In the above algorithm, val(P[i]) is the number corresponding to the character P[i]. • h1 is the hash value of the pattern • h2 takes the hash value of successive sequences of M characters in the text. • Strictly, if h1=h2, we might not have a match, since a hash collision could occur. We still need to make a final comparison on the strings themselves. • We can use a very large prime since we do not actually have to store the hash table; this makes collisions extremely unlikely. • Average number of comparisons = N+M

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