Papers
Topics
Authors
Recent
Search
2000 character limit reached

Searching, Sorting, and Cake Cutting in Rounds

Published 1 Dec 2020 in cs.DS and cs.GT | (2012.00738v5)

Abstract: We study searching and sorting in rounds motivated by a fair division question: given a cake cutting problem with nn players, compute a fair allocation in at most kk rounds of interaction with the players. Rounds interpolate between the simultaneous and the fully adaptive settings, also capturing parallel complexity. We find that proportional cake cutting in rounds is equivalent to sorting with rank queries in rounds. We design a protocol for proportional cake cutting in rounds, while lower bounds for sorting in rounds with rank queries were given by Alon and Azar. Inspired by the rank query model, we then consider two basic search problems: ordered and unordered search. In unordered search, we get an array x⃗=(x1,…,xn)\vec{x}=(x_1, \ldots, x_n) and an element zz promised to be in x⃗\vec{x}. We have access to an oracle that receives queries of the form "Is zz at location ii?" and answers "Yes" or "No". The goal is to find the location of zz with success probability at least pp in at most kk rounds of interaction with the oracle. We show the expected query complexity of randomized algorithms on a worst case input is np(k+12k)±O(1)np\bigl(\frac{k+1}{2k}\bigr) \pm O(1), while that of deterministic algorithms on a worst case input distribution is np(1−k−12kp)±O(1)np \bigl(1 - \frac{k-1}{2k}p \bigr) \pm O(1). These bounds apply even to fully adaptive unordered search, where the ratio between the two complexities converges to $2-p$ as the size of the array grows. In ordered search, we get sorted array x⃗=(x1,…,xn)\vec{x}=(x_1, \ldots, x_n) and element zz promised to be in x⃗\vec{x}. We have access to an oracle that gets comparison queries. Here we find that the expected query complexity of randomized algorithms on a worst case input and deterministic algorithms on a worst case input distribution is essentially the same: pk⋅n<sup>1k</sup>±O(1+pk)p k \cdot n<sup>{\frac{1}{k}}</sup> \pm O(1+pk).

Citations (1)

Summary

No one has generated a summary of this paper yet.

Paper to Video (Beta)

No one has generated a video about this paper yet.

Whiteboard

No one has generated a whiteboard explanation for this paper yet.

Open Problems

We haven't generated a list of open problems mentioned in this paper yet.

Continue Learning

We haven't generated follow-up questions for this paper yet.