On the largest convex subsets in Minkowski sums

doi:10.1016/j.ipl.2014.02.013

Information Processing Letters

Volume 114, Issue 8, August 2014, Pages 405-407

https://doi.org/10.1016/j.ipl.2014.02.013 Get rights and content

Highlights

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We study the size of convex subsets of the Minkowski sum of two planar point sets.
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We prove an $O (n \log n)$ upper bound when the two point sets are convex.
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Our result complements the existing upper and lower bounds.

Abstract

Given two point sets $A, B \subset R^{2}$ of n points each, the Minkowski sum $A + B$ has a quadratic number of points in general. How large can a subset $S \subseteq A + B$ be if S is required to be convex? We prove that when A and B are both convex then S can have only $O (n \log n)$ points. This complements the existing results that are known when A and B are not in convex position or when $B = A$ and A is convex.

Section snippets

Introduction and related work

Let A, B be two sets of n points each in $R^{2}$ . The Minkowski sum $A + B$ is defined as the set of sums obtained from any point in A with any other point in B. That is, $A + B = {a + b | a \in A, b \in B}$

Clearly, the size of $A + B$ will be quadratic in general. We call a point set S convexly independent if any of the points cannot be represented as a convex combination of the other points in S. Suppose we want to pick a subset of points from $A + B$ such that the points are in convex position, that is, convexly independent.

Proof of Theorem 1

A point set in $R^{2}$ in convex position is said to define a monotone chain if both the x and the y coordinates are monotone. It is easy to see that any point set in convex position can be decomposed into at most four subsets such that each subset defines a monotone chain. See Fig. 1 for example.

Therefore, we can consider each of the convex chains separately and bound the size of a convexly independent subset from each pair of these chains from A and B.

Observation 1

For any two point sets A, B, translating

Concluding remarks

The proof of Theorem 1 seems to be very generous with counting which points of the Minkowski sum may be selected to be convexly independent. We do not know whether this is really the case and the upper bound can be improved by more careful argument. The only lower bound that we are aware of, for the specific case discussed here, is the trivial linear bound. We leave this disparity between the upper and the lower bounds as an open problem.

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There are more references available in the full text version of this article.

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