By Salle R.D.

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**Sample text**

We keep an array B indexed by symbols whose entry B[c] stores the number of occurrences of c in T [1 : l]. We can keep these counters updated after a Remove by simply decreasing B[T [l]] by one. We also maintain an array R with an entry for each text position. The entry R[ j] stores the number of occurrences of symbol T [ j] in T [1 : j]. The number of elements in both B and R is no more than n, hence they take O(n) space. These two arrays are enough to correctly update the value E i after Append(wi ), which is in turn enough to estimate H0 (see Eq.

Finally we set ri = ri + 1. In this way, operation Remove requires constant time per window, hence O(log1+ε n) time overall. Append(wi ) takes constant time. The space required by the counters Ai is O(σ log1+ε n) words. Unfortunately, the space complexity of this solution can be too much when it is used as the basic-block for computing the k-th order entropy of T (see Sect. 1) as we will do in Sect. 4. In fact, we would achieve min(σ k+1 log1+ε n, n log1+ε n) space, which may be superlinear in n depending on σ and k.

S[m]#n where each S[i] is a text (called page) drawn from an alphabet σ, and #1 , #2 , . . , #n are special symbols greater than any symbol of σ. A partition of T must be page-aligned, that is it must form groups of contiguous pages S[i]#i . . S[ j]# j , denoted also S[i : j]. Our aim is to find a page-aligned partition whose cost is at most (1 + ε) the minimum possible cost, for any fixed ε > 0. We notice that this problem generalizes the table partitioning problem (Buchsbaum et al. 2003), since we can assume that S[i] is a column of the table.