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Knuth-Morris-Pratt

Algorithm                        Processing time        Matching time

Knuth-Morris-Pratt             O(m)                              O(n)

Algorithm

• The Knuth-Morris-Pratt (KMP) string searching algorithm differs from the brute-force algorithm by keeping track of information gained from previous comparisons.
• Afailure function (f) is computed that indicates how much of the last comparison can be reused if it fails.
• Specifically, f is defined to be the longest prefix of the pattern P[0,..,j] that is also a suffix of P[1,..,j]
- Note: not a suffix of P[0,..,j]
• Example:
- value of the KMP failure function:
j 0 1 2 3 4 5
P[j] a b a b a c
f(j) 0 0 1 2 3 0
• This shows how much of the beginning of the string matches up to the portion immediately preceding a failed comparison.
- if the comparison fails at (4), we know the a,b in positions 2,3 is identical to positions 0,1

Pros
1. Its searching complexity is O(m+n) which is faster than brute force and Rabin-Karp
2. It’s fairly easy to implement

Cons
1. It needs additional space and time – O(m) for pre-processing
2. It can be optimized a bit (Knuth-Morris-Pratt)

Final Words
Obviously this algorithm is quite useful because it improves in some very elegant manner the brute force matching. In the other hand you must know that there are faster string searching algorithms like the Boyer-Moore algorithm. However the Morris-Pratt algorithm can be quite useful in many cases, so understanding its principles can be very handy.

Worked example of the search algorithm

To illustrate the algorithm's details, we work through a (relatively artificial) run of the algorithm. At any given time, the algorithm is in a state determined by two integers, `m` and `i`, which denote respectively the position within `S` which is the beginning of a prospective match for `W`, and the index in `W` denoting the character currently under consideration. This is depicted, at the start of the run, like

``````             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W: ABCDABD
i: 0123456
```
```

We proceed by comparing successive characters of `W` to "parallel" characters of `S`, moving from one to the next if they match. However, in the fourth step, we get `S` is a space and `W = 'D'`, a mismatch. Rather than beginning to search again at `S`, we note that no `'A'` occurs between positions 0 and 3 in `S` except at 0; hence, having checked all those characters previously, we know there is no chance of finding the beginning of a match if we check them again. Therefore we move on to the next character, setting `m = 4` and `i = 0`. (m will first become 3 since `m + i - T[i] = 0 + 3 - 0 = 3` and then become 4 since `T = -1`)

``````             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:     ABCDABD
i:     0123456
```
```

We quickly obtain a nearly complete match `"ABCDAB"` when, at `W` (`S`), we again have a discrepancy. However, just prior to the end of the current partial match, we passed an `"AB"` which could be the beginning of a new match, so we must take this into consideration. As we already know that these characters match the two characters prior to the current position, we need not check them again; we simply reset`m = 8``i = 2` and continue matching the current character. Thus, not only do we omit previously matched characters of `S`, but also previously matched characters of `W`.

``````             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:         ABCDABD
i:         0123456
```
```

This search fails immediately, however, as the pattern still does not contain a space, so as in the first trial, we return to the beginning of `W` and begin searching at the next character of `S``m = 11`, reset `i = 0`. (m will first become 10 since `m + i - T[i] = 8 + 2 - 0 = 10` and then become 11 since `T = -1`)

``````             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:            ABCDABD
i:            0123456
```
```

Once again we immediately hit upon a match `"ABCDAB"` but the next character, `'C'`, does not match the final character `'D'` of the word `W`. Reasoning as before, we set `m = 15`, to start at the two-character string`"AB"` leading up to the current position, set `i = 2`, and continue matching from the current position.

``````             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:                ABCDABD
i:                0123456
```
```

This time we are able to complete the match, whose first character is `S`.

Description of and pseudocode for the search algorithm

The above example contains all the elements of the algorithm. For the moment, we assume the existence of a "partial match" table `T`, described below, which indicates where we need to look for the start of a new match in the event that the current one ends in a mismatch. The entries of `T` are constructed so that if we have a match starting at `S[m]` that fails when comparing `S[m + i]` to `W[i]`, then the next possible match will start at index `m + i - T[i]` in `S` (that is, `T[i]` is the amount of "backtracking" we need to do after a mismatch). This has two implications: first, `T = -1`, which indicates that if `W` is a mismatch, we cannot backtrack and must simply check the next character; and second, although the next possible match will begin at index `m + i - T[i]`, as in the example above, we need not actually check any of the `T[i]`characters after that, so that we continue searching from `W[T[i]]`. The following is a sample pseudocode implementation of the KMP search algorithm.

```algorithm kmp_search:
input:
an array of characters, S (the text to be searched)
an array of characters, W (the word sought)
output:
an integer (the zero-based position in S at which W is found)

define variables:
an integer, m ← 0 (the beginning of the current match in S)
an integer, i ← 0 (the position of the current character in W)
an array of integers, T (the table, computed elsewhere)

while m+i is less than the length of S, do:
if W[i] = S[m + i],
if i equals the (length of W)-1,
return m
let i ← i + 1
otherwise,
let m ← m + i - T[i],
if T[i] is greater than -1,
let i ← T[i]
else
let i ← 0

(if we reach here, we have searched all of S unsuccessfully)
return the length of S
```

Efficiency of the search algorithm

Assuming the prior existence of the table `T`, the search portion of the Knuth–Morris–Pratt algorithm has complexity O(k), where `k` is the length of `S` and the `O` is big-O notation. As except for the fixed overhead incurred in entering and exiting the function, all the computations are performed in the `while` loop, we will calculate a bound on the number of iterations of this loop; in order to do this we first make a key observation about the nature of `T`. By definition it is constructed so that if a match which had begun at `S[m]` fails while comparing `S[m + i]` to `W[i]`, then the next possible match must begin at `S[m + (i - T[i])]`. In particular the next possible match must occur at a higher index than `m`, so that `T[i] < i`.

Using this fact, we will show that the loop can execute at most `2k` times. For in each iteration, it executes one of the two branches in the loop. The first branch invariably increases `i` and does not change `m`, so that the index `m + i` of the currently scrutinized character of `S` is increased. The second branch adds `i - T[i]` to `m`, and as we have seen, this is always a positive number. Thus the location `m` of the beginning of the current potential match is increased. Now, the loop ends if `m + i = k`; therefore each branch of the loop can be reached at most `k` times, since they respectively increase either `m + i` or `m`, and `m ≤ m + i`: if `m = k`, then certainly `m + i ≥ k`, so that since it increases by unit increments at most, we must have had `m + i = k` at some point in the past, and therefore either way we would be done.

Thus the loop executes at most `2k` times, showing that the time complexity of the search algorithm is `O(k)`.

Here is another way to think about the runtime: Let us say we begin to match W and S at position i and p, if W exists as a substring of S at p, then W[0 through m] == S[p through p+m]. Upon success, that is, the word and the text matched at the positions(W[i] == S[p+i]), we increase i by 1 (i++). Upon failure, that is, the word and the text does not match at the positions(W[i] != S[p+i]), the text pointer is kept still, while the word pointer roll-back a certain amount(i = T[i], where T is the jump table) And we attempt to match W[T[i]] with S[p+i]. The maximum number of roll-back of i is bounded by i, that is to say, for any failure, we can only roll-back as much as we have progressed up to the failure. Then it is clear the runtime is 2k.

"Partial match" table (also known as "failure function")

The goal of the table is to allow the algorithm not to match any character of `S` more than once. The key observation about the nature of a linear search that allows this to happen is that in having checked some segment of the main string against an initial segment of the pattern, we know exactly at which places a new potential match which could continue to the current position could begin prior to the current position. In other words, we "pre-search" the pattern itself and compile a list of all possible fallback positions that bypass a maximum of hopeless characters while not sacrificing any potential matches in doing so.

We want to be able to look up, for each position in `W`, the length of the longest possible initial segment of `W` leading up to (but not including) that position, other than the full segment starting at `W` that just failed to match; this is how far we have to backtrack in finding the next match. Hence `T[i]` is exactly the length of the longest possible proper initial segment of `W` which is also a segment of the substring ending at `W[i - 1]`. We use the convention that the empty string has length 0. Since a mismatch at the very start of the pattern is a special case (there is no possibility of backtracking), we set `T = -1`, as discussed above.

Worked example of the table-building algorithm

We consider the example of `W = "ABCDABD"` first. We will see that it follows much the same pattern as the main search, and is efficient for similar reasons. We set `T = -1`. To find `T`, we must discover aproper suffix of `"A"` which is also a prefix of `W`. But there are no proper suffixes of `"A"`, so we set `T = 0`. Likewise, `T = 0`.

Continuing to `T`, we note that there is a shortcut to checking all suffixes: let us say that we discovered a proper suffix which is a proper prefix and ending at `W` with length 2 (the maximum possible); then its first character is a proper prefix of a proper prefix of `W`, hence a proper prefix itself, and it ends at `W`, which we already determined cannot occur in case T. Hence at each stage, the shortcut rule is that one needs to consider checking suffixes of a given size m+1 only if a valid suffix of size m was found at the previous stage (e.g. T[x]=m).

Therefore we need not even concern ourselves with substrings having length 2, and as in the previous case the sole one with length 1 fails, so `T = 0`.

We pass to the subsequent `W``'A'`. The same logic shows that the longest substring we need consider has length 1, and although in this case `'A'` does work, recall that we are looking for segments endingbefore the current character; hence `T = 0` as well.

Considering now the next character, `W`, which is `'B'`, we exercise the following logic: if we were to find a subpattern beginning before the previous character `W`, yet continuing to the current one `W`, then in particular it would itself have a proper initial segment ending at `W` yet beginning before it, which contradicts the fact that we already found that `'A'` itself is the earliest occurrence of a proper segment ending at`W`. Therefore we need not look before `W` to find a terminal string for `W`. Therefore `T = 1`.

Finally, we see that the next character in the ongoing segment starting at `W = 'A'` would be `'B'`, and indeed this is also `W`. Furthermore, the same argument as above shows that we need not look before`W` to find a segment for `W`, so that this is it, and we take `T = 2`.

Therefore we compile the following table:

 `i` `W[i]` `T[i]` 0 1 2 3 4 5 6 A B C D A B D -1 0 0 0 0 1 2

Other example more interesting and complex:

 `i` `W[i]` `T[i]` 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 P A R T I C I P A T E I N P A R A C H U T E -1 0 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 3 0 0 0 0 0

Description of pseudocode for the table-building algorithm

The example above illustrates the general technique for assembling the table with a minimum of fuss. The principle is that of the overall search: most of the work was already done in getting to the current position, so very little needs to be done in leaving it. The only minor complication is that the logic which is correct late in the string erroneously gives non-proper substrings at the beginning. This necessitates some initialization code.

```algorithm kmp_table:
input:
an array of characters, W (the word to be analyzed)
an array of integers, T (the table to be filled)
output:
nothing (but during operation, it populates the table)

define variables:
an integer, pos ← 2 (the current position we are computing in T)
an integer, cnd ← 0 (the zero-based index in W of the next character of the current candidate substring)

(the first few values are fixed but different from what the algorithm might suggest)
let T ← -1, T ← 0

while pos is less than the length of W, do:
(first case: the substring continues)
if W[pos - 1] = W[cnd],           let cnd ← cnd + 1, T[pos] ← cnd, pos ← pos + 1

(second case: it doesn't, but we can fall back)
otherwise, if cnd > 0, let cnd ← T[cnd]

(third case: we have run out of candidates.  Note cnd = 0)
otherwise, let T[pos] ← 0, pos ← pos + 1
```

Efficiency of the table-building algorithm

The complexity of the table algorithm is `O(n)`, where `n` is the length of `W`. As except for some initialization all the work is done in the `while` loop, it is sufficient to show that this loop executes in `O(n)` time, which will be done by simultaneously examining the quantities `pos` and `pos - cnd`. In the first branch, `pos - cnd` is preserved, as both `pos` and `cnd` are incremented simultaneously, but naturally, `pos` is increased. In the second branch, `cnd` is replaced by `T[cnd]`, which we saw above is always strictly less than `cnd`, thus increasing `pos - cnd`. In the third branch, `pos` is incremented and `cnd` is not, so both `pos` and `pos - cnd` increase. Since `pos ≥ pos - cnd`, this means that at each stage either `pos` or a lower bound for `pos` increases; therefore since the algorithm terminates once `pos = n`, it must terminate after at most `2n`iterations of the loop, since `pos - cnd` begins at `1`. Therefore the complexity of the table algorithm is `O(n)`.

Efficiency of the KMP algorithm

Since the two portions of the algorithm have, respectively, complexities of `O(k)` and `O(n)`, the complexity of the overall algorithm is `O(n + k)`.

These complexities are the same, no matter how many repetitive patterns are in `W` or `S`.

Variants

real-time version of KMP can be implemented using a separate failure function table for each character in the alphabet. If a mismatch occurs on character in the text, the failure function table for character is consulted for the index in the pattern at which the mismatch took place. This will return the length of the longest substring ending at matching a prefix of the pattern, with the added condition that the character after the prefix is . With this restriction, character in the text need not be checked again in the next phase, and so only a constant number of operations are executed between the processing of each index of the text. This satisfies the real-time computing restriction.

The Booth algorithm uses a modified version of the KMP preprocessing function to find the lexicographically minimal string rotation. The failure function is progressively calculated as the string is rotated.