## 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 |
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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