Acta Informatica

, Volume 11, Issue 4, pp 363–372 | Cite as

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

Article

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 2k+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.

Keywords

Information System Operating System Data Structure Communication Network Information Theory 
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

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References

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Copyright information

© Springer-Verlag 1979

Authors and Affiliations

  • R. Kemp
    • 1
  1. 1.Fachbereich Angewandte Mathematik und InformatikUniversität des SaarlandesSaarbrückenGermany (Fed. Rep.)

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