Dividing X into V Subsets, Each of Size M=v. We Then Apply Our Method to Each Segment

and the information needed for each is stored in O(v) space. Therefore, if v = m u for some u, 0 < u < 1, each segment occupies K 1 m 4(1?u)=3 storage locations, for some constant K 1. Occupancy is therefore K 1 m u m 4(1?u)=3 + K 2 m u in all and the actual problem is to nd the value of u that minimizes this quantity. However, this depends on the two constants K 1 and K 2 , which are obviously implementation sensitive. Consequently, though not diicult, the problem goes beyond the scope of this paper. For example, if K 1 = K 2 , we have u = 1 ? 3 4 log m 3; for m large, u 1, and this indicates that the segments have to be small. This is rather intuitive, and using a practical method would involve choosing v so that m=v 10 12. In any case, the preprocessing time remains O(m). Aknowledgments We wish to thank the anonymous referee whose comments helped us in improving the contents and readability of our paper. Exact expected case analysis of a simple heuristic for compressed array storage. Unpublished. 7] R. Sprugnoli. Perfect hash functions: a single probe retrieving method for static sets. 15 Apply the compression algorithm described in Section 2 to T: We can now solve the static dictionary problem in the following way: given an element x 2 V; we reduce it modulo p; x 0 = (x mod p) + 1; and then we search x 0 in the compressed array C (obtained from T) as described in Section 2. If x 0 2 C; then x 2 S; and, if x 0 = 2 C; then x = 2 S: The other problems are solved accordingly. Our nal result concerns the evaluation of the space occupied by the new algorithm. Strictly speaking, we have m = O(p p) and not m = o(p p) (required by the results in Section 4). However, we can apply the result in Theorem 4.2 to get a rough idea of the amount of storage required by the previous algorithm: Corollary 5.4 The space occupied by the previous algorithm is in the order of O(m 4=3): We wish to conclude our paper with a simple but meaningful example. Let: and consider the ASCII equivalent of these strings when stored in three bytes. We …