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The prefix S n of S is defined as the first n characters of S. [5] For example, the prefixes of S = (AGCA) are S 0 = S 1 = (A) S 2 = (AG) S 3 = (AGC) S 4 = (AGCA). Let LCS(X, Y) be a function that computes a longest subsequence common to X and Y. Such a function has two interesting properties.
Compute a longest common subsequence of these two strings, and let , be the random variable whose value is the length of this subsequence. Then the expected value of λ n , k {\displaystyle \lambda _{n,k}} is (up to lower-order terms) proportional to n , and the k th Chvátal–Sankoff constant γ k {\displaystyle \gamma _{k}} is the constant ...
The set ret is used to hold the set of strings which are of length z. The set ret can be saved efficiently by just storing the index i, which is the last character of the longest common substring (of size z) instead of S[(i-z+1)..i]. Thus all the longest common substrings would be, for each i in ret, S[(ret[i]-z)..(ret[i])].
Longest common subsequence (LCS) distance is edit distance with insertion and deletion as the only two edit operations, both at unit cost. [1]: 37 Similarly, by only allowing substitutions (again at unit cost), Hamming distance is obtained; this must be restricted to equal-length strings. [1]
The total length of all the strings on all of the edges in the tree is (), but each edge can be stored as the position and length of a substring of S, giving a total space usage of () computer words. The worst-case space usage of a suffix tree is seen with a fibonacci word , giving the full 2 n {\displaystyle 2n} nodes.
A more efficient method would never repeat the same distance calculation. For example, the Levenshtein distance of all possible suffixes might be stored in an array , where [] [] is the distance between the last characters of string s and the last characters of string t. The table is easy to construct one row at a time starting with row 0.
In the Robinson–Schensted correspondence between permutations and Young tableaux, the length of the first row of the tableau corresponding to a permutation equals the length of the longest increasing subsequence of the permutation, and the length of the first column equals the length of the longest decreasing subsequence. [3]
The length of the longest decreasing subsequence is equal to the length of the first column of P. Now, it is not possible to fit (r − 1)(s − 1) + 1 entries in a square box of size (r − 1)(s − 1), so that either the first row is of length at least r or the last row is of length at least s.