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### Boyer Moore Algorithm

Idan Szpektor

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:bbacdcbaabcddcdaddaaabcbcb

P: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!

ddbbacdcbaabcddcdaddaaabcbcb

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 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’

T:bbacdcbaabcddcdaddaaabcbcb

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

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: 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.
- “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

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