## Abstract

Let A equals (a//1 greater than . . . greater than a//m) and B equals (b//1 greater than . . . greater than b//n) be given ordered lists; also let there be given some order relations between a//is and b//js. Suppose that an unknown total order exists on A U B which is consistent with all these relations ( equals a linear extension of the partial order) and the author wishes to find out this total order by comparing pairs of elements a//t:b//s. If the partial order has N linear extensions, then the Information Theoretic Bound says that log//2 N steps will be required in the worst case from any such algorithm. In this paper it is shown that there exists an algorithm which will take no more than C log//2 Ncomparisons where C equals (log//2(( ROOT 5 plus 1)/2))** minus **1. The computation required to determine the pair a//t:b//s to be compared has length polynomial in (m plus n). The constant C is best possible. Many related results are reviewed.

Original language | American English |
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Pages (from-to) | 795-801 |

Number of pages | 7 |

Journal | SIAM Journal on Computing |

Volume | 13 |

Issue number | 4 |

DOIs | |

State | Published - 1984 |