Catching a Polygonal Fish with a Minimum Net
aa r X i v : . [ c s . C G ] A ug Catching a Polygonal Fish with a Minimum Net
Sepideh Aghamolaei
Sharif University of Technology, Tehran, Iran [email protected]
Abstract.
Given a polygon P in the plane that can be translated, rotated and enlargedarbitrarily inside a unit square, the goal is to find a set of lines such that at least one of themalways hits P and the number of lines is minimized.We prove the solution is always a regular grid or a set of equidistant parallel lines, whosedistance depends on P . Keywords:
Stabbing Problem · Nets.
This was an open problem of CCCG 2020 conference, posed by Joseph O’Rourke. We show thecases discussed in the conference are the only possible cases and they give the optimal solution forany convex shape.Such a net must stab any subset of copies of a shape, and the minimum number of lines forstabbing all copies is a lower bound on the minimum number of lines in the minimum net.
Algorithm.
For the case with axis-parallel lines, if the smallest enclosing square has side length a and the smallest bounding rectangle (arbitrarily oriented) has sides b and c , b ≤ c , then if a < c ,both vertical and horizontal grid lines are required with distance a from each other, otherwise a = b = c , and only one set of parallel lines is enough with distance a from each other. Example.
In the example of the slides from CCCG 2020, the 1 × √ ≈ .
8. So, this is the first case in our algorithm, and a regulargrid of side length 2 √ Lemma 1.
Rotate P to make the maximum distance of P parallel to one of the axes. Considerall translations of P in the direction of one of the axes by the width of P in the direction of thataxis, and then in the perpendicular direction until they are disjoint with respect to the perpendicularaxis as well. The number of these shapes is a lower bound on the number of lines (including theboundary lines).Proof. Any of the shapes in the statement of the lemma is a valid movement of P , so, the linesmust hit them. Since these shapes are all disjoint, each of them requires a line, if we want to use S. Aghamolaei only axis-parallel lines. If the optimal solution is in another direction, we rotate the shape until itslongest edge is parallel to one of the axes, and scale down P such that the same number of copiesfit in the direction of that axis and build a new bounding square with side ρ . This gives a scalingof the original problem by ρ , and some corners. So, the solution is at least as much as the originalproblem, if we had scaled the original instance by ρ − . So, this bound is still a lower bound on theoptimal solution. The minimum number of lines is given when the largest width (the diameter of P ) is used, since the side-length of the bounding square is fixed. ⊓⊔ Since the algorithm gives a solution with at most twice as many lines as in Lemma 1, so far weknow the algorithm is a 2-approximation. Now, we show it is an optimal solution.Now, we give a better lower bound on the number of lines.
Lemma 2.
Consider the shape R with maximum width among the smallest bounding rectangle andthe smallest bounding square of P . Build a grid with cells of shape R , with a copy of P inside eachof them. The lines of the grid (1D or 2D based on the algorithm) are the smallest subset of linesthat stab all these shapes.Proof. In the case where only one set of parallel lines are used, based on Lemma 1, that is theoptimal solution. The other case is based on the minimum enclosing square, so, if only one directionis chosen, the shapes can be translated in the perpendicular direction. Among the pairs of orthogonallines with equal distances between each set of parallel lines, the ones that are parallel to the sides ofthe bounding box have the minimum size, since other directions inside the bounding box have higherdirectional width. The minimum side-length happens when R has sides parallel to the boundingbox, which also gives the maximum number of lines required to stab all the translated shapes. ⊓⊔ The grid lines of the algorithm cover all shapes, so that is an upper bound on the number of linesrequired to solve the problem. Based on Lemma 2, the same number of lines is also the lower bound.So, it is the optimal solution:
Theorem 1.