5

For any point P in the interior of a convex polygon, the sum of the angles subtended by the edges of the polygon is obviously 2π.

  • Given a convex polygon, how does one algorithmically find the point (I don't know if it is unique) in its interior so that the least among angles subtended at this point by the edges of the polygon is maximized?

Remark 1: The question above has the property: minimizing the maximum (or maximizing the minimum) among a set of values does not automatically make all values equal unlike say, areas of n pieces into which a polygon is cut. An earlier question from the same ballpark is Cutting convex regions into equal diameter and equal least width pieces - 2

Remark 2: If the convex polygon is any triangle, the point minimizing the max subtended angle is unique - the Steiner point of the triangle.

Note 1: This question is based on this old note: https://nandacumar.blogspot.com/2007/04/triangulation-problem.html

Note 2: A couple of variants to the above question that were recorded in this very post have now been moved to a new post: Finding the point within a convex n-gon that minimizes the largest angle subtended there by an edge of the n-gon

Nandakumar R
  • 5,473
  • (i) According to these guidelines, users should "avoid trying to answer questions which [...] request answers to multiple questions". I see three questions here. (ii) You are saying "the point". I don't think such points are unique in general. – Iosif Pinelis Mar 26 '24 at 12:43
  • Thank you. Made an edit - now there is only one main question and a couple of variants to the same theme (they are but variants on the same theme) are mentioned. And corrected the indication that the desired point is unique. – Nandakumar R Mar 26 '24 at 16:39
  • When you say “not necessarily unique” you mean that you know there can be more than one or you mean that you don’t know if there can be more than one? (In fact I think this point should be unique) – Pietro Majer Mar 27 '24 at 21:39
  • Thanks. I don't know - for triangles, the points are unique, as noted in the question. So, let me edit again leaving the question of uniqueness open. – Nandakumar R Mar 27 '24 at 23:02
  • You can do a binary search. Then you have the polygon with a bunch of circles cut out from it, and you need to decide if it's not empty – Command Master Mar 28 '24 at 03:33
  • This should be solvable with a weighted voronoi diagram on the centers of the circles, where you weight points by the radius of the corresponding circle – Command Master Mar 28 '24 at 03:35

1 Answers1

1

This is about the issue of uniqueness, with a slight reformulation of the problem in view of an algorithmic approach.

Given a (closed) half-plane, a segment $e$ of its boundary, and $0<\theta<\pi$, denote $S(e,\theta)$ the set of all points $P$ of the half-plane such that the angle subtended at $P$ by the segment $e$ is not less than $\theta$. It is a disk segment cut off by the chord $e$ (and the angle subtended at $P$ is strictly larger that $\theta$ in its interior). A convex polygon with edges $e_1,\dots, e_n$ is the intersection of $n$ half-planes, each bounded by a line from an edge $e_i$, so a point $P$ sees all edges under an angle greater or equal than $\theta$ iff it is in the intersection $\bigcap_{1\le i\le n} S(e_i,\theta)$. This intersection, if not empty and not a singleton, is a strictly convex set with non-empty interior. The angles subtended at every point $P$ in its interior by the edges are all strictly larger than $\theta$, so that $\theta$ is not the maximum minimum angle. This shows that the point $P_*$ that maximises the least angle is unique.

edit. Some hints to determining the point $P_*$. Recall that the intersection of $n$ Euclidean disks is not empty iff every $3$-intersection (i.e. an intersection of three of them) is not empty. It follows that the intersection of $n$ Euclidean disks is a singleton iff all $3$-intersections are not empty, and at least one is a singleton. This means that the maximising point $P_*$ for all edges already solves the problem for some subset of $3$ of them. So a first rough procedure should be: solve the problem of the maxmin angle for all the triples of edges, and take the minimum. A smarter algorithm should avoid solving all ${n \choose 3}$ problems, of course.

Pietro Majer
  • 56,550
  • 4
  • 116
  • 260
  • 1
    Thanks! The argument for uniqueness feels very intuitive. and guess this is a nice application of Helly's theorem too. and regarding the algorithm proposed, one would guess/hope that improvement from O(n^3) would be possible via some randomized incremental approach. – Nandakumar R Mar 28 '24 at 14:28
  • 1
    Marking this as an answered question. Reg the 'variants' mentioned above, would it now be more meaningful to shift them onto a separate question ? Hope to have an opinion on this. – Nandakumar R Mar 28 '24 at 14:30
  • Yes, I think it should be OK moving the questions about variants of the problem to a nother post. – Pietro Majer Mar 28 '24 at 14:56