The average number of registers needed to evaluate a binary tree optimally

Summary

In this paper we determine the number of binary trees with n leaves which can be evaluated optimally with less than or equal to k registers. Furthermore we obtain the result that this number is equal to the number of the binary trees with n leaves, using for traversal a maximum size of stack less than or equal to 2^k+1−1. This fact is only a connection between the numbers of the trees and not between the sets of the trees. We compute also the average number ¯R(n) of registers needed to evaluate a binary tree optimally. We get for all ɛ>0: $\bar R(n + 1) = 1d(\sqrt {n)} + C + F(n) + O(n^{ - 0.5 + \varepsilon } )$where C = 0.82574... is a constant and F(n) is a function with F(n) = F(4n) for all n>0 and −0.574<F(n)< −0.492.