We try to simulate this code. Starting with i = 0, we will match the pattern and text one by one starting from the first character of the text. How to create an Apple eCommerce website – Apple website designs – HTML and CSS Tutorials e-commerce store. If all the characters match then we get the pattern. If I can’t get to one place, I will break the loop. Check out the picture below:
Going to character number 5 we get a mismatch. The loop inside the brute force algorithm will break at line 6, then continue searching at index i=1.
In this way, we have to loop through each index of text to find the pattern, so the time complexity of this approach is O(n∗m).
. Is there any way we can avoid looping through each index? Before that we need to know what are suffixes and prefixes.
Prefix: When zero or more characters are removed from the end of a string, whatever is left is the prefix of the string. For example, the string “ABC” is prefixed with “A”, “AB” and “ABC”. Among these, “A” and “AB” are proper prefixes because they are not equal to the original string.
Suffix: The suffix of a string is what is left when zero or more characters are removed from the beginning of a string. For example, the suffixes of the string “ABC” are “C”, “BC” and “ABC”. Among these “C” and “BC” are proper suffixes because they are not equal to the original string.
Assume the pattern we are looking for is abxyabcd. Now look at the picture below (Figure 1.3):
In the image (Figure 1.3) we found a mismatch at one place in pattern matching with text. There is no need to worry about what the mismatched character is or what the next characters are. Now if we try to bruteforce match the pattern by shifting it one position to the left, is there any point at all?
If I shift 1 house then whatever is in the question mark places is of no use. How much house shifting can be profitable depends on what? It depends on how many prefixes in the pattern match the text, in this case that prefix is “ABXYAB”. If we shift as follows we can get a match:
That means we have to shift the pattern in such a way that we get a ‘partial’ matching of the prefix of the pattern with the suffix of the pattern itself. So what happens is, we get a partial match of the input text with the pattern prefix, and then we go back and do a character-by-character match to see if the entire text matches.
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Another example will make it clear. Suppose now the pattern is “ABABAC” and we get a partial matching like this:
Now how much we shift will depend on the matching prefix “ABABA”. See the picture below:
In Figure 1.7, I have shifted the pattern to the right by 2 spaces. This will give us a partial match of pattern prefix with pattern suffix. In this case the last 3 characters match the first 3 characters of the pattern. That means the first 3 characters of the pattern will match the text as well. Then we will go forward again and look at the rest of the characters.
Now suppose unfortunately we get a mismatch again:
How much shift now? It depends on “ABA”. We need to shift ABA to the right such that the suffix of ABA partially matches the prefix of ABA after the shift. In this case the shift should be as shown in Figure 1.9:
The purpose of giving so many examples is to make one thing clear, how many prefixes in the pattern match the text will depend on how far the pattern should be shifted.
Assume that the string name P is as many prefixes as match
. P
By how many shifts will we get a partial match “with the suffix of the already matched text”?
To know that we need to find P
The largest prefix of the string that is also a suffix of P. This line makes me cringe myself, so let’s look at another picture:
Hope the meaning of the line is clear after seeing the picture. The maximum number of suffixes that match the prefixes should be found out and shifted. The string “ABABAC” has 7 prefixes:
Now find the length of the largest prefix for all strings which is also a suffix. Since we need a partial match in this case, we will only care about the proper suffixes and prefixes (ie the length of the suffix/prefix will be less than the stringer).
This table has a name, it is called Failure table. By looking at this table, we can tell how many prefixes of the pattern are matched and if a mismatch is found, where to start matching again.
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For example I got a failed match after matching up to ABABA above. From row number 6 of the table in Figure 1.12, we find that the prefix with the first 3 characters of the pattern is the same as the suffix with the last 3 characters of the pattern. That is, the first 3 characters of the pattern are equal to the suffix of “the text that already matched”. Then we skip the first 3 characters of the pattern and start matching again from the next character.
If you have taken any theory of computation course then you understand that the failure table is actually a kind of “finite state automata” (don’t worry if you haven’t read it). Automation is a “set of states” and some rules on how to move from one state to another. In our case the state is how many matches the pattern has.
And how to go from one state to another depends on whether the next character in the text and pattern matches. If it matches, then it is easy, I will try to match again by going to the next character. And if it does not match then how many suffixes are matched with the prefix (Figure 1.10) I will find out by looking at the failure table. For easy understanding we can draw the failure table as below:
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