Boyer Moore Algorithm Understanding and Example?

Question

I am facing issues in understanding Boyer Moore String Search algorithm.

I am following the following document. Link

I am not able to work out my way as to exactly what is the real meaning of delta1 and delta2 here, and how are they applying this to find string search algorithm. Language looked little vague..

Kindly if anybody out there can help me out in understanding this, it would be really helpful.

Or, if you know of any other link or document available that is easy to understand, then please share.

Thanks in advance.

Answer 1

The insight behind Boyer-Moore is that if you start searching for a pattern in a string starting with the last character in the pattern, you can jump your search forward multiple characters when you hit a mismatch.

Let's say our pattern p is the sequence of characters p1, p2, ..., pn and we are searching a string s, currently with p aligned so that pn is at index i in s.

E.g.:

s = WHICH FINALLY HALTS.  AT THAT POINT...
p = AT THAT
i =       ^

The B-M paper makes the following observations:

(1) if we try matching a character that is not in p then we can jump forward n characters:

'F' is not in p, hence we advance n characters:

s = WHICH FINALLY HALTS.  AT THAT POINT...
p =        AT THAT
i =              ^

(2) if we try matching a character whose last position is k from the end of p then we can jump forward k characters:

' 's last position in p is 4 from the end, hence we advance 4 characters:

s = WHICH FINALLY HALTS.  AT THAT POINT...
p =            AT THAT
i =                  ^

Now we scan backwards from i until we either succeed or we hit a mismatch. (3a) if the mismatch occurs k characters from the start of p and the mismatched character is not in p, then we can advance (at least) k characters.

'L' is not in p and the mismatch occurred against p6, hence we can advance (at least) 6 characters:

s = WHICH FINALLY HALTS.  AT THAT POINT...
p =                  AT THAT
i =                        ^

However, we can actually do better than this. (3b) since we know that at the old i we'd already matched some characters (1 in this case). If the matched characters don't match the start of p, then we can actually jump forward a little more (this extra distance is called 'delta2' in the paper):

s = WHICH FINALLY HALTS.  AT THAT POINT...
p =                   AT THAT
i =                         ^

At this point, observation (2) applies again, giving

s = WHICH FINALLY HALTS.  AT THAT POINT...
p =                       AT THAT
i =                             ^

and bingo! We're done.