Boyer Moore Algorithm

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# Boyer Moore Algorithm - PowerPoint PPT Presentation

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).

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### Boyer Moore Algorithm

Idan Szpektor

• 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:

P:abc

• Worst case is still O(n m)
• On a mismatch between the pattern and the text, we can shift the pattern by more than one place.

Sublinearity!

acabc

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).
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: aaaaaaaaaaaaaaaaaaaaaaaaa

P: 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:

P: cabbabdbab

cabbabdbab

GSR (Cont…)
• Example 2 – what if there is no alignment:

T: bbacdcbaabcbbabdbabcaabcbcb

P: bcbbabdbabc

bcbbabdbabc

GSR - Detailed
• We mark the matched sub-string in T with t and the mismatched char with x
• 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
• 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: bb a c d c b b a a b b c d d

Z: 1 0 0 0 0 31 0 0 21 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 bb

N: 0 0 0 12 0 0 13 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: aaaaaaaaaaaaaaaaaaaaaaaaa

P: 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’

P: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).
O(n + m) proof - Outline
• Preprocess in O(n) – already proved
• Properties of strings
• Proof of search in O(m) if P is not in T, using only the good suffix rule.
• 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.

β’

β

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 isprefix 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

α

α’

α

α

q

α’

α

α

α

• 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: bbacdcbaabcbbabdbabcaabcbcb

P: 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 δγ = γδ.
• 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.
• 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