Boyer Moore Algorithm

Idan Szpektor

Boyer and Moore

What It’s About

A String Matching Algorithm

Preprocess a Pattern P (|P| = n)

For a text T (| T| = m), find all of the occurrences of P in T

Time complexity: O(n + m), but usually sub-linear

Right to Left (like in Hebrew)

Matching the pattern from right to left

For a pattern abc:

↓

T: bbacdcbaabcddcdaddaaabcbcbP: abc

Worst case is still O(n m)

The Bad Character Rule (BCR)

On a mismatch between the pattern and the text, we can shift the pattern by more than one place.

Sublinearity!ddbbacdcbaabcddcdaddaaabcbcbacabc

↑

BCR Preprocessing

A table, for each position in the pattern and a character, the size of the shift. O(n |Σ|) space. O(1) access time.

a b a c b: 1 2 3 4 5

A list of positions for each character. O(n + |Σ|) space. O(n) access time, But in total O(m).

1 2 3 4 5

a 1 1 3 3 3

b 2 2 2 5

c 4 4

BCR - Summary

On a mismatch, shift the pattern to the right until the first occurrence of the mismatched char in P.

Still O(n m) worst case running time:

T: aaaaaaaaaaaaaaaaaaaaaaaaaP: abaaaa

The Good Suffix Rule (GSR)

We want to use the knowledge of the matched characters in the pattern’s suffix.

If we matched S characters in T, what is (if exists) the smallest shift in P that will align a sub-string of P of the same S characters ?

GSR (Cont…)

Example 1 – how much to move:

↓

T: bbacdcbaabcddcdaddaaabcbcbP: cabbabdbab cabbabdbab

GSR (Cont…)

Example 2 – what if there is no alignment:

↓

T: bbacdcbaabcbbabdbabcaabcbcbP: bcbbabdbabc bcbbabdbabc

GSR - Detailed

We mark the matched sub-string in T with t and the mismatched char with x

1. In case of a mismatch: shift right until the first occurrence of t in P such that the next char y in P holds y≠x

2. Otherwise, shift right to the largest prefix of P that aligns with a suffix of t.

Boyer Moore Algorithm

Preprocess(P) k := n

while (k ≤ m) do Match P and T from right to left starting at k

If a mismatch occurs: shift P right (advance k) by max(good suffix rule, bad char rule).

else, print the occurrence and shift P right (advance k) by the good suffix rule.

Algorithm Correctness

The bad character rule shift never misses a match

The good suffix rule shift never misses a match

Preprocessing the GSR – L(i)

L(i) – The biggest index j, such that j < n and prefix P[1..j] contains suffix P[i..n] as a suffix but not suffix P[i-1..n]

1 2 3 4 5 6 7 8 9 10 11 12 13 P: b b a b b a a b b c a b b L: 0 0 0 0 0 0 0 0 0 5 9 0 12

Preprocessing the GSR – l(i)

l(i) – The length of the longest suffix of P[i..n] that is also a prefix of P

P: b b a b b a a b b c a b b l: 2 2 2 2 2 2 2 2 2 2 2 1

Using L(i) and l(i) in GSR

If mismatch occurs at position n, shift P by 1

If a mismatch occurs at position i-1 in P: If L(i) > 0, shift P by n – L(i) else shift P by n – l(i)

If P was found, shift P by n – l(2)

Building L(i) and l(i) – the Z

For a string s, Z(i) is the length of the longest sub-string of s starting at i that matches a prefix of s.

s: b b a c d c b b a a b b c d dZ: 1 0 0 0 0 3 1 0 0 2 1 0 0 0

Naively, we can build Z in O(n^2)

From Z to N

N(i) is the longest suffix of P[1..i] that is also a suffix of P.

N(i) is Z(i), built over P reversed.

s: d d c b b a a b b c d c a b bN: 0 0 0 1 2 0 0 1 3 0 0 0 0 1

Building L(i) in O(n)

L(i) – The biggest index j < n, such that prefix P[1..j] contains suffix P[i..n] as a suffix but not suffix P[i-1..n]

L(i) – The biggest index j < n such that: N(j) == | P[i..n] | == n – i + 1

for i := 1 to n, L(i) := 0 for j := 1 to n-1

i := n – N(j) + 1 L(i) := j

Building l(i) in O(n)

l(i) – The length of the longest suffix of P[i..n] that is also a prefix of P

l(i) – The biggest j <= | P[i..n] | == n – i + 1 such that N(j) == j

k := 0 for j := 1 to n-1

If(N(j) == j), k := j l(n – j + 1) := k

Building Z in O(n)

For calculating Z(i), we want to use the previously calculated Z(1)…Z(i-1)

For each I we remember the right most Z(j):j, such that j < i and j + Z(j) >= k + Z(k), for all k < i

Building Z in O(n) (Cont…)

↑ ↑ ↑ ↑ S i’ j i

If i < j + Z(j), s[i … j + Z(j) - 1] appeared previously, starting at i’ = i – j + 1.

Z(i’) < Z(j) – (i - j) ?

Building Z in O(n) (Cont…)

For Z(2) calculate explicitly j := 2, i := 3 While i <= |s|:

if i >= j + Z(j), calculate Z(i) explicitly else

Z(i) := Z(i’) If Z(i’) >= Z(j) – (i - j), calculate Z(i) tail

explicitly If j + Z(j) < i + Z(i), j := i

Building Z in O(n) - Analysis

The algorithm builds Z correctly

The algorithm executes in O(n) A new character is matched only once All other operations are in O(1)

Boyer Moore Worst Case Analysis

Assume P consists of n copies of a single char and T consists of m copies of the same char:

T: aaaaaaaaaaaaaaaaaaaaaaaaaP: aaaaaa

Boyer Moore Algorithm runs in Θ(m n) when finding all the matches

The Galil Rule

In a specific matching phase, We mark with k the position in T of the right end of P. We mark with s the position of last matched char in this phase.

s k k’

T: bbacdcbaabcddcdaddaaabcbcbP: abaab

abaab

The Galil Rule (Cont…)

All the chars in position s < j ≤ k are known to be matching. The algorithm doesn’t need to check them.

An extended Boyer Moore algorithm with the Galil rule runs in O(m + n) worst case (even without the bad-character rule).

Don’t Sleep Yet…

O(n + m) proof - Outline

Preprocess in O(n) – already proved

1. Properties of strings

2. Proof of search in O(m) if P is not in T, using only the good suffix rule.

3. Proof of search in O(m) even if P is in T, adding the Galil rule.

Properties of Strings

If for two strings δ, γ: δγ = γδ then there is a string β such that δ = βi and γ = βj, i, j > 0

- Proof by induction

Definition: A string s is semiperiodic with

period β if s consists of a non-empty suffix of β (possibly the entire β) followed by one or more complete copies of β.

ββ’ β β

Properties of Strings (Cont…)

A string is prefix semiperiodic if it contains one or more complete copies of β followed by a non-empty prefix of β.

A string is prefix semiperiodic iff it is semiperiodic with the same length period

Lemma 1

Suppose P occurs in T starting at position p and also at position q, q > p. If q – p ≤ n/2 then P is semiperiodic with periodα = P[n-(q-p)+1…n]

p

qα

ααα

αα

α’

α’

Proof - when P is Not Found in T

We have R rounds during the search.

After each round the good suffix rule decides on a right shift of si chars.

Σsi ≤ m

We shall use Σsi as an upper bound.

Proof (Cont…)

For each round we count the matched chars by: fi – the number of chars matched for the first

time gi –the number of chars already matched in

previous rounds.

Σfi = m

We want to prove that gi ≤ 3si ( Σgi ≤ 3m).

Proof (Cont…)

Each round don’t find P it matched a substring ti and one bad char xi in T (xiti T)

T: bbacdcbaabcbbabdbabcaabcbcbP: bdbabc

|ti|+1 ≤ 3si gi ≤ 3si (because gi + fi = |ti|+1)

For the rest of the proof we assume that for the specific round i: |ti| + 1 > 3si

Lemma 2 (|ti| + 1 > 3si)

In round i we look at the matched suffix of P, marked P*. P* = yi ti, yi≠ xi.

Both P* and ti are semiperiodic with period α of length si and hence with minimal length period β, α = βk.

Proof: by Lemma 1.

Lemma 3 (|ti| + 1 > 3si)

Suppose P overlapped ti during round i. We shall examine in what ways could P overlap ti in previous rounds.

In any round h < i, the right end of P could not have been aligned with the right end of any full copy of β in ti.

- proof:

Both round h and i fail at char xi

two cases of possible shift after round h are invalid

Lemma 4 (|ti| + 1 > 3si)

In round h < i, P can correctly match at most |β|-1 chars in ti.

By Lemma 3, P is not aligned with a right end of ti in phase h.

Thus if it matched |β| chars or more there is a suffix γ of β followed by a prefix δ of β such that δ γ = γ δ.

By the string properties there is a substring μ such that β = μk, k>1.

This contradicts the minimal period size property of β.

Lemma 5 (|ti| + 1 > 3si)

If in round h < i the right end of P is aligned with a char in ti, it can only be aligned with one of the following: One of the left-most |β|-1 chars of ti

One of the right-most |β| chars of ti

-proof: If not, By Lemma 3,4, max |β|-1 chars are matched and only

from the middle of a β copy, while there are at least |β| A shift cannot pass the right end of that β copy

Proof (Cont…)

If |ti| + 1 > 3si then gi ≤ 3si

Using Lemma 5, in previous rounds we could match only the

bad char xi, the last |β|-1 chars in ti or start from the first |β| right chars in ti.

In the last case, using Lemma 4, we can only match up to |β|-1 chars

in total we could previously match:gi = 1 + |β|-1 + (|β| + |β|-1) ≤ 3|β| ≤ 3si

Proof - Final

Number of matches = ∑(fi + gi) =∑fi + ∑gi ≤ m + ∑3si ≤ m + 3m = 4m

Proof - when P is Found in T

Split the rounds to two groups: “match” rounds –an occurrence of P in T was

found. “mismatch” rounds –P was not found in T.

we have proved O(m) for “mismatch” rounds.

Proof (Cont…)

After P was found in T, P will be shifted by a constant length s. (s = n – l(2)).

|n| + 1 ≤ 3s ∑ matches in round i ≤ ∑3s ≤ m

For the rest of the proof we assume that:|n| + 1 > 3s

Proof (|n| + 1 > 3s)

By Lemma 1, P is semiperiodic with minimal length period β, |β| = s.

If round i+1 is also a “match” round then, by the Galil rule, only the new |β| chars are compared.

A contiguous series of “match” rounds, i…i+k is called a “run”.

Proof (|n| + 1 > 3s)

∑ The length of a “run”, not including chars that where already matched in previous “runs” ≤ m

How many chars in a “run” where already matched in previous “runs”?

Lemma (|n| + 1 > 3s)

Suppose k-1 was a “match” round and k is a “mismatch” round that ends the “run”.

If k’ > k is the first “match” round then it overlaps at most |β|-1 chars with the previous “run” (ended by round k-1).

The left end of P at round k’ cannot be aligned with the left end

of a full copy of |β| at round k-1. As a result, P cannot overlap |β| chars or more with round k-1.

Proof (|n| + 1 > 3s)

By the Lemma and because the shift after every “match” round is of |β|, only the first round of a “run” can overlap, and only with the last previous “run”.

∑ The length of the chars that where already

matched in previous “runs” ≤ m

Proof (|n| + 1 > 3s) - Final

∑ The length of a “run” = ∑ The length of a “run”, not including chars that where already matched in previous “runs” + ∑ The length of the chars that where already matched in previous “runs” ≤ m + m