# String Searching – The Knuth-Morris-Pratt Algorithm

January 3, 2011 10 Comments

We’ve seen how to do the naive approach towards pattern matching. So what about other algorithms that are much more better at doing this task? This is the Knuth-Morris-Pratt (KMP) algorithm for pattern matching.

# The Problem

The problem that we are trying to deal with, is of course, exactly the same as the naive problem. Given some pattern *p, *does it exist in some string *s*.

# Quick Example

Let’s say we have the following string and pattern,

*s*= “abra abracad abracadabra”*p*= “abracadabra”

Now if this were the naive approach, we would need quite a lot of comparisons – 154 to be exact! Before we start, it would be best to note that we’ll be using zero-based arrays for this. So since *s* is really an array of characters, then s[0] would be referencing to the character ‘a’, and p[2] would be the character ‘r’.

We also need two more variables, name *i* and *j*. Here *i* will be the indexes into *p*, and *j* will be the indexes into *s*. Now, let’s start off with the first match-up,

- j = 012345678901234567890123
- s = abra abracad abracadabra
- p = abracadabra
- i = 01234567890

Here we can see that p[0] and s[0] are equal, but then we reach s[4] and p[4]. Here we have a space character and the character ‘c’. Clearly they don’t match up. So what we need to do now is slide the pattern along. But this is the part that isn’t like the naive approach. Instead of sliding the pattern across by one, we slide the pattern across by a special value. You see, behind the scenes there is in fact a table *T*. We’ll get to this later, but for now all you need to know is that we slide the pattern across to j=3 and reset i=0.

We get the value of j=3 from j = j+i-T[i] = 0+4–1 = 3. In this equation, j=0 because if you remember, *j* will be the starting point. i=4 because this is the point of failure – the position in *p* where we found our mismatch. And T[i]=1 because, well we’ll come to this later, for now take my word for it.

With these new values, we can now slide the pattern along:

- j = 012345678901234567890123
- s = abra abracad abracadabra
- p = abracadabra
- i = 01234567890

Now we see that we find a mismatch at s[4] and p[1]. This is we slide the pattern along to j=4. If you wish to know why, it’s because j=3 and i=1 (which should be obvious) and T[i]=0, sticking them into that equation gives

j = j+i-T[i] = 3+1–0 = 4

Instead of drawing out all of the possible combinations, let’s jump ahead to:

- j = 012345678901234567890123
- s = abra abracad abracadabra
- p = abracadabra
- i = 01234567890

Now this should be better to show off why this algorithm is better than the naive approach. Here we compare the characters again, and fail at s[12] and p[7]. But because of the way the algorithm works, we slide the pattern across by a huge amount,

- j = 012345678901234567890123
- s = abra abracad abracadabra
- p = abracadabra
- i = 01234567890

I have actually skipped one step here, where we compare s[12] and p[0]. But as you can see the next step succeeds, and we find a match at j=13!

What’s more amazing is that we have only made 28 comparisons! A lot better than the naive approaches 154.

# So what’s this *T* table then!?

Well we had to get to this part sooner or later, so we may as well explain this now. Besides the obvious as to why KMP differs from the naive approach, there is another reason. You see, the KMP algorithm has a pre-processing stage, that is, that it does (or rather creates) something from the given pattern before the algorithm even starts searching for the pattern.

This pre-processing stage is the *T* table, and this is what it looked like for the above pattern:

i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

p[i] | a | b | r | a | c | a | d | a | b | r | a |

T[i] | -1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 2 | 3 |

Now this may look rather complex, but it really isn’t that hard to understand. First things first, the values of T[0] and T[1] are **always** fixed, they never change. Therefore, T[0]=-1 and T[1]=0 **all of the time**. This means that the main program begins from i=2. There is also another variable that we need to use, and we shall call this variable *sub*, with initial value of sub=0.

And here is some pseudo-code to populate the *T* table.

Remember. i=2 and sub=0.

//whilst i doesn’t exceed the length of p

001 WHILE (i < length_of_p) DO:

//we have a continued substring in p from the start of p

002 IF (p[i-1] == p[sub]) THEN:

003 sub=sub+1

004 T[i] = sub

005 i=i+1

//if the substring stops, fall back to the start

006 ELSE IF (sub > 0) THEN:

007 sub = T[sub]

//we weren’t in a substring, so use default value

008 ELSE

009 T[i] = 0

010 i=i+1

011 ENDIF

012 ENDWHILE

I am not going to walk through the whole process. So let’s start at i=7 and sub=0.

Starting at line 002 we test to see if p[6] is equal to p[0]. Since ‘d’ is not equal to ‘a’ then we move onto the next statement. Here *sub* is not greater than 0, so we move onto the final statement. With this we make T[7]=0 and increment *i* to i=8.

Now, starting back at the first IF statement, we test to see if p[7] equals p[0], and by surprise it does since ‘a’ = ‘a’. This means that we increment *sub* to sub=1, we make T[8]=1 and increment *i* again to i=9.

Moving on again, we test p[8] to p[1] this time, since sub=1. This time we succeed again since ‘b’ = ‘b’. Therefore we increment *sub* and *i* to sub=2 and i=10. And obviously we set T[9]=2.

I’ll leave T[10] up to you, as well as the ones I missed, but it should be obvious by now how this works.

If you’re pondering why we need a –1 at the beginning, it’s so that if we fail on the very first character, we simply move along to the next character in the main string. As if this were just 0, we would not slide the pattern along.

# The Main Algorithm

Now that we understand how the T table is created, the main algorithm for the KMP is very simple.

First of all, we have to remember that we have *s* which is the string to search through, and *p* which is the pattern to find. Then of course we have *i* which is the current index into *p,* and *j* which is the current index into *s*.

And here is the pseudo-code for the KMP,

//while we don’t exceed the length of string s

001 WHILE (j+i < length_of_s) DO:

//character matching for parallel characters

002 IF (p[i] == s[j+i]) THEN:

//reached the end of the pattern, match found.

003 IF (i == length_of_p –1) THEN:

004 RETURN j

005 ENDIF

006 i=i+1

//parallel characters don’t match, slide pattern along

007 ELSE

008 j=j+i-T[i]

009 IF (T[i] > –1)

010 i = T[i]

011 ELSE

012 i = 0

013 ENDIF

014 ENDIF

015 ENDWHILE

016

//we reached the end of the string, match not found

017 RETURN length_of_s

This may look horribly daunting at first, but it really is very simple. Line 002 is simple the character by character matching process, matching parallel characters from the pattern and the string. If we reach the end of the pattern, return the position is *s* where is it (line 004), otherwise, we still have a character by character match, so increment to the next character in the pattern for testing (line 006).

The real fun beginnings inside of the ELSE statement. Now line 008 should look very familiar. Yep, we came across it earlier. This line simply sets the new start position in string *s* where we start matching the pattern against it again, and of course, you now know what the T table looks like.

The next part, from lines 009 to 013 are a little more tricky to grasp. Basically, when we slide the pattern along, we don’t always need to re-test some of the characters from the pattern. This is because of the way the T table works, and it’s called backtracking. For example, if we have the following *s* and *p*:

- j = 01234567890
- s = abcdab abcd
- p = abcdabd
- i = 0123456

We can see that the initial start point will fail at s[6] and p[6]. Now because of where is failed, T[6] is actually equal to the value 2. Therefore, we slide the pattern along to j = j + i + T[i] = 0 + 6 – 2 = 4:

- j = 01234567890
- s = abcdab abcd
- p = abcdabd
- i = 0123456

But wait a second, we don’t need to compare the first 2 characters of *p* again, because we know that they already match up due to the T table. Therefore, like is says at line 010, we start at index i = T[6] = 2. This saves us a lot of time from doing none needed comparisons!

This backtracking only works if the value at T[i] is greater than 0. Otherwise, we just start at index i=0 in the pattern.

# The Complexity

So, after all of this, what is the complexity of the KMP algorithm? Well the best part is that both the pre-processing phase and the main phase run in linear time. That is, the construction of the T table takes O(m) time, where m is the length of the pattern *p*. Then, the main program itself takes O(n) time where n is the length of the string *s*. Therefore the overall time complexity is O(m+n).

# Resources

[1] http://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm

[2] University of Manchester, Advanced Algorithms 1 lecture notes by David Rydeheard

[3] Knuth-Morris-Pratt: An Analysis by Mireille Regnier

[4] Badgerati Tutorials – Knuth-Morris-Pratt Algorithm

Excellent Badgeratri…Nice way of describing the Algorithm..Thanks…But there is one Typo.It where you first time compute the J.You have written it like this “We get the value of i=3 from j = j+i-T[i] = 0+4–1 = 3. In this equation, j=0 because if you remember, j will be the starting point. j=4 because this is the point of failure – the ….”.

But it has to be like “We get the value of j=3 from j = j+i-T[i] = 0+4–1 = 3. In this equation, j=0 because if you remember, j will be the starting point. i=4 because this is the point of failure – the ….”.I hope you got it ..you simply need to write j=3 instead of i=3 and i=4 instead of j=4…

Nicely spotted! I’ve fixed the typo (i’s and j’s, they all look the same to me…). Thanks a lot!

Thanks for fix!!!!!!

can u please find me a T table for “0010″. ?

and T[0] = -1 and T[1] = 0, is TRUE for ANY PROBLEM?

please reply. MOST URGENT.

Yes, that is correct. T[0]=-1 and T[1]=0 is true for any problem. Unless, of course, your pattern is 1 character long; in which case, you would have only T[0]=-1.

As for the T-table for the pattern “0010″, then the T-table constructed by this algorithm would be the following: T[] = {-1, 0, 1, 0}.

The first ‘-1′ and ’0′ are mandatory. After that, remembering that i=2 and sub=0 initially, the following ’1′ is because p[i-1] (=0) is equal to p[sub] (=0). So ‘sub’ is incremented by 1, and T[2]=sub (=1). The final value of ’0′ is because p[i-1] (=1) is not equal to p[sub] (=0). Therefore, due to sub being greater than 0, T[3]=T[sub] (=0).

In one of my textbooks the first two numbers i.e. p[0] and p[1] are both taken 0 unlike -1 and 0 respectively….pls help what is supposed to be followed??

This all comes down to the implementation of the KMP algorithm itself. I, personally, have always used the method described in this post; whereas I know other people may use a different implementation. I believe I did once come across a method for the KMP algorithm where the first two values of the T-table were both 0. Therefore, both the implementations described here and in your textbook are correct. It just depends on which one you personally want to follow.

The reasons as to why T[0] and T[1] have values -1 and 0 respectively here, is due to the way the pattern is slid along the string. If we take ‘j’ as the index into the main string and ‘i’ as the index into the given pattern, then take this example: what happens if we fail on the first character of the pattern when matching to the string?

Well, the way we calculate where to slide the pattern along to is calculated using ‘j = j + i – T[i]‘. Say we fail at the start of the string, so ‘j=0′. Therefore, knowing we fail at the start of the pattern with ‘i=0′, we get: ‘j = 0 + 0 – T[0]‘, where ‘T[0]=-1′. For this method of sliding, if T[0] was the value 0, then the pattern would never slide along whenever we failed on the first character of the pattern. But with it having the value of -1, we get: ‘j = 0 + 0 – -1 = 0 + 1 = 1′, so the pattern slides along one character nicely.

As for when T[0] and T[1] both have values of 0 respectively; I can’t remember this implementation properly as it has been a fair while. But just remember that both methods are perfectly acceptable. In the end, it comes down to what ever way you have been taught. If this is for University, stick to the textbook method, as that is the method they probably want you to know. Otherwise, choose which ever way feels more comfortable to yourself.

I like this post! Clean explanation.

nice explanation….:)

Thanku soon much pretty easy explanation….☺