The.Algorithm.Design.Manual.Springer-Verlag.1998

Longest Common Substring
than edit distance.
Issues arising include:
● Are you looking for a common substring or scattered subsequence? - In detecting plagiarism or
attempting to identify the authors of anonymous works, we might need to find the longest phrases
shared between several documents. Since phrases are strings of consecutive characters, we need
the longest common substring between the texts.
The longest common substring of a set of strings can be identified in linear time using suffix trees,
discussed in Section . The trick is to build a suffix tree containing all the strings, label each
leaf with the set of strings that contain it, and then do a depth-first traversal to identify the deepest
node that has descendents from each input string.
For the rest of this discussion, we will restrict attention to finding common scattered
subsequences. Dynamic programming can be used to find the longest common subsequence of
two strings, S and T, of n and m characters each. This algorithm is a special case of the edit-
distance computation of Section . Let M[i,j] denote the number of characters in the longest
common substring of and . In general, if , there is no way the
last pair of characters could match, so . If S[i] = T[j], we have
the option to select this character for our substring, so
. This gives a recurrence that computes M,
and thus finds the length of the longest common subsequence in O(nm) time. We can reconstruct
the actual common substring by walking backward from M[n,m] and establishing which
characters were actually matched along the way.
● What if there are relatively few sets of matching characters? - For strings that do not contain too
many copies of the same character, there is a faster algorithm. Let r be the number of pairs of
positions (i,j) such that . Thus r can be as large as mn if both strings consist entirely of the
same character, but r = n if the two strings are permutations of . This technique treats the
pairs of r as defining points in the plane.
The complete set of r such points can be found in O(n + m + r) time by bucketing techniques. For
each string, we create a bucket for each letter of the alphabet and then partition all of its characters
into the appropriate buckets. For each letter c of the alphabet, create a point (,t) from every pair
and , where and are buckets for c.
A common substring represents a path through these points that moves only up and to the right,
never down or to the left. Given these points, the longest such path can be found in
time. We will sort the points in order of increasing x-coordinate (breaking ties in favor of
increasing y-coordinate. We will insert these points one by one in this order, and for each k,
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