BOOL(N)

Abstract

Let p be a positive integer and let N = 2^p. Here we describe a certain poset , which we call BOOL (N). Let us use p-1 steps to build BOOL(N/2) out of the first N/2 keys, K[0] through K[N/2-1] and BOOL(N/2) out of the last N/2 keys K[N/2] through K[N-1]. After that, let us use one step of N/2 comparators to compare K[i] with K[i + N/2] for i = 0, 1,…, N/2-1. We conjecture that this method of building a poset of N = 2^p in p steps minimizes the number of 0/1-cases . Here we also showed some ideas on starting a sorting network for N keys when N < 2^p. In general, for any BOOL (N) where N > 8, one can build a poset of N-2 keys by removing keys K[0] and K[N-1] and adding a comparator in each step to compare the two keys that aren’t compared with any other key in that step. The poset obtained will look like BOOL(N) except with keys K[0] and K[N-1] removed and p extra coverings added between the keys in the top rank and keys in the bottom rank. These extra coverings eliminate strangers that are far out of place.