Using Rabin-Karp to search for multiple patterns in a string

Question

According to the wikipedia entry on Rabin-Karp string matching algorithm, it can be used to look for several different patterns in a string at the same time while still maintaining linear complexity. It is clear that this is easily done when all the patterns are of the same length, but I still don't get how we can preserve O(n) complexity when searching for patterns with differing length simultaneously. Can someone please shed some light on this?

Answer 1

I'm not sure if this is the correct answer, but anyway:

While constructing the hash value, we can check for a match in the set of string hashes. Aka, the current hash value. The hash function/code is usually implemented as a loop and inside that loop we can insert our quick look up.

Of course, we must pick m to have the maximum string length from the set of strings.

Update: From Wikipedia,

[...]
for i from 1 to n-m+1
         if hs ∈ hsubs
             if s[i..i+m-1] = a substring with hash hs
                 return i
         hs := hash(s[i+1..i+m]) // <---- calculating current hash
[...]

We calculate current hash in m steps. On each step there is a temporary hash value that we can look up ( O(1) complexity ) in the set of hashes. All hashes will have the same size, ie 32 bit.

Update 2: an amortized (average) O(n) time complexity ?

Above I said that m must have the maximum string length. It turns out that we can exploit the opposite.
With hashing for shifting substring search and a fixed m size we can achieve O(n) complexity.

If we have variable length strings we can set m to the minimum string length. Additionally, in the set of hashes we don't associate a hash with the whole string but with the first m-characters of it.
Now, while searching the text we check if the current hash is in the hash set and we examine the associated strings for a match.

This technique will increase the false alarms but on average it has O(n) time complexity.

Answer 2

It's because the hash values of the substrings are related mathematically. Computing the hash H(S,j) (the hash of the characters starting from the jth position of string S) takes O(m) time on a string of length m. But once you have that, computing H(S, j+1) can be done in constant time, because H(S, j+1) can be expressed as a function of H(S, j).

O(m) + O(1) => O(m), i.e. linear time.

Here's a link where this is described in more detail (see e.g. the section "What makes Rabin-Karp fast?")