Abstract

A word circuit [1] is a directed acyclic graph in which each edge holds a w-bit word (i.e., some x ∈ {0, 1}w) and each node is a gate computing some binary function g : {0, 1}w × {0, 1}w → {0, 1}w. The following problem was studied in [1]: How many binary gates are needed to compute a ternary function f : ({0, 1}w)3 → {0, 1}w. They proved that (2 + o(1))2w binary gates are enough for any ternary function, and there exists a ternary function which requires word circuits of size (1 - o(1))2w. One of the open problems in [1] is to get these bounds tight within a low order term. In this paper we solved this problem by constructing new word circuits for ternary functions of size (1 + o(1))2w. We investigate the problem in a general setting: How many k-input word gates are needed for computing an n-input word function f : ({0, 1}w)n → {0, 1}w (here n ≥ k). We show that for any fixed n, (1 - o(1))2(n - k)w basic gates are necessary and (1 + o(1))2(n - k)w gates are sufficient (assume w is sufficiently large). Since word circuit is a natural generalization of boolean circuit, we also consider the case when w is a constant and the number of inputs n is sufficiently large. We show that [Formula: see text] basic gates are necessary and sufficient in this case.

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