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THE COMPLEXITY OF WORD CIRCUITS

机译:单词电路的复杂性

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A word circuit [1] is a directed acyclic graph in which each edge holds a w-bit word (i.e., some x ∈ {0,1}~ω) and each node is a gate computing some binary function g : {0, 1}~ω x{0, 1}~ω→ {0, 1}~ω. The following problem was studied in [1]: How many binary gates are needed to compute a ternary function f : ({0, 1}~ω)~3 → {0,1}~ω. They proved that (2 + 0(1))2" binary gates are enough for any ternary function, and there exists a ternary function which requires word circuits of size (1-o(1))2~ω. One of the open problems in [1] is to gef 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))2~ω. 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, i}~ω)~n → {0, 1}~ω (here n ≥ k). We show that for any fixed n, (1-0(1))2~((n-k))~ω basic gates are necessary and (1 + o(1))2~((n-k))~ω 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 (1 ± o(1))2~(ωn)/(k-1)n basic gates are necessary and sufficient in this case.
机译:字电路[1]是有向无环图,其中每个边沿都包含一个w位字(即x∈{0,1}〜ω),每个节点是计算某些二进制函数g的门:{0, 1}〜ωx {0,1}〜ω→{0,1}〜ω。在[1]中研究了以下问题:计算三元函数f需要多少个二进制门:({0,1}〜ω)〜3→{0,1}〜ω。他们证明了(2 + 0(1))2“二元门足以满足任何三元函数的需求,并且存在一个三元函数,它需要大小为(1-o(1))2〜ω的字电路。 [1]中的问题是在一个低阶项中严格限制这些边界,本文通过构造大小为(1-o(1))2〜ω的三元函数的新字电路来解决此问题。在一般情况下:计算一个n输入字函数f需要多少个k输入字门:({0,i}〜ω)〜n→{0,1}〜ω(这里n≥k)。我们证明对于任何固定的n,(1-0(1))2〜((nk))〜ω基本门是必要的,而(1 + o(1))2〜((nk))〜ω门是足够的(假设w足够大。)由于字电路是布尔电路的自然概括,因此我们也考虑w为常数且输入数n足够大的情况,我们证明(1±o(1))在这种情况下,需要2〜(ωn)/(k-1)n个基本门。

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