Abstract
Given n opportunities to evaluate a function which is known to have a root in the unit interval, how should these evaluations be used to specify the smallest possible interval containing that root? If f(x) is continuous the answer is the well-known method of binary search and the smallest interval has length ( 1 2)n. The authors solve this optimal search problem in the case of a sequential and a parallel search for the class of functions whose divided differences are bounded above by a number M and below by a number m (m > 0). It is shown that in the sequential case the best interval has length { 1 2[1 - ( m M)]}n. For the optimal search in the parallel case r parallel evaluations are shown to be equivalent to approximately 1 + log r sequential evaluations.
| Original language | English |
|---|---|
| Pages (from-to) | 1-14 |
| Number of pages | 14 |
| Journal | Journal of Combinatorial Theory. Series A |
| Volume | 23 |
| Issue number | 1 |
| DOIs | |
| State | Published - Jul 1977 |
| Externally published | Yes |
ASJC Scopus subject areas
- Theoretical Computer Science
- Discrete Mathematics and Combinatorics
- Computational Theory and Mathematics
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