# Stabbing pairwise intersecting disks by five points

We present an O(n) expected time algorithm and an O(n n) deterministic time algorithm to find a set of five points that stab a set of n pairwise intersecting disks in the plane. We also give a simple construction with 13 pairwise intersecting disks that cannot be stabbed by three points.

## Authors

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## 1 Introduction

Let be a set of pairwise intersecting disks in the plane. If every three disks in have a nonempty intersection, then, by Helly’s theorem, the whole intersection is nonempty [Hel23, Hel30, Rad21]. Thus, can be stabbed by one point. More generally, when there are three disks with empty intersection, it is known that can still always be stabbed by four points [Dan86, Sta84]. In July 1956, Danzer presented a proof at Oberwolfach (see [Dan86]). Since Danzer was not satisfied with his proof, he never published it, but he gave a new proof in 1986 [Dan86]. Previously, in 1981, Stachó published a proof for the existence of four stabbing points [Sta84], using similar arguments as in his previous construction of five stabbing points [Sta65]. Hadwiger and Debrunner showed that three points suffice for unit disks [HD55]. Danzer’s upper bound proof is fairly involved, and there seems to be no obvious way to turn it into an efficient algorithm. The two constructions of Stachó are simpler, but not enough for an easy subquadratic algorithm. We present a new argument that yields five stabbing points. Our proof is constructive and allows us to find the stabbing points in linear expected time and in deterministic time.

As for lower bounds, Grünbaum gave an example of 21 pairwise intersecting disks that cannot be stabbed by three points [Grü59]. Later, Danzer reduced the number of disks to ten [Dan86]. This example is close to optimal, because every set of eight disks can be stabbed by three points [Sta65]. It is hard to verify the lower bound by Danzer for ten disks—even with dynamic geometry software. We present a simple construction that uses 13 disks.

## 2 The geometry of pairwise intersecting disks

Let be a set of pairwise intersecting disks. A disk is given by its center and its radius . We assume without loss of generality that no disk is contained in another. The lens of two disks is the set . Let be any of the two intersection points of and . The angle is called the lens angle of and . It is at most . Three disks , , and are non-Helly if . We present some useful geometric lemmas.

###### Lemma.

Among any three non-Helly pairwise intersecting disks , , and , there are two disks with lens angle at least .

###### Proof:

By assumption, the lenses , and are pairwise disjoint. Let be the vertex of nearer to , and let , be the analogous vertices of and (see Figure 1, Left). Consider the -gon , and write , , and for the interior angles at , , and . The sum of all interior angles is . Thus, , so at least one angle is less than . It follows that the exterior angle at , , or must be larger than .

###### Lemma.

Let and be two intersecting disks with radii and lens angle . Let be the unique disk with radius and center , such that (i) the centers , , and are collinear and lies on the same side of as ; and (ii) the lens angle of and is exactly (see Figure 1, Right). Then, if lies between and , we have .

###### Proof:

Let . Then, since lies between and , the triangle inequality gives

 |xc|≤|xc2|+|c2c|=|xc2|+|c1c|−|c1c2|. (1)

Since , we get . Also, since and each have radius and form the lens angle , it follows that . Finally, by the law of cosines, . As and , we get , so

 |c1c2|2=r21+r22−2r1r2cosα≥r21+r22−2r1r2((√3−1.5)r1r2−√3+1)=(r1(√3−1)+r2)2.

Plugging this into Eq. (1), we obtain , i.e., .

###### Lemma.

Let and be two intersecting disks of equal radius with lens angle . There is a set of four points so that any disk of radius at least that intersects both and contains a point of .

###### Proof:

Consider the two tangent lines of and , and let and be the midpoints on these lines between the respective two tangency points. We set (see Figure 2, Left).

Given , we decrease its radius, keeping its center fixed, until either the radius becomes or until is tangent to or . Suppose the latter case holds and is tangent to . We move the center of continuously along the line spanned by the center of and towards , decreasing the radius of to maintain the tangency. We stop when either the radius of reaches or becomes tangent to . We obtain a disk with center so that either: (i) and intersects both and , or (ii) and is tangent to both and . Since , it suffices to show that . We introduce a coordinate system, setting the origin midway between and , so that the -axis passes through and . Then, we have , , , and .

For case (i), let be the disk of radius centered at , and the disk of radius centered at . Since has radius and intersects both and , its center has distance at most from both and , i.e., . Let and be the two disks of radius centered at and . We will show that . Then it is immediate that . By symmetry, it is enough to focus on the upper-right quadrant . We show that all points in are covered by . Without loss of generality, we assume that . Then, the two intersection points of and are and , and the two intersection points of and are and . Let be the boundary curve of in . Since and since and , it follows that does not intersect the boundary of and hence . Furthermore, the subsegment of the -axis from to the startpoint of is contained in , and the subsegment of the -axis from to the endpoint of is contained in . Hence, the boundary of lies completely in , and since is simply connected, is follows that , as desired.

For case (ii), since is tangent to and , the center of is on the bisector of and , so the points , , and are collinear. Suppose without loss of generality that . Then, it is easily checked that lies above , and , so .

###### Lemma.

Consider two intersecting disks and with radii , having lens angle at least . Then, there is a set of four points such that any disk of radius at least that intersects both and contains a point of .

###### Proof:

Let be the line through and . Let be the disk of radius and center that satisfies the conditions (i) and (ii) of Lemma 2. Let be the point set specified in the proof of Lemma 2, with respect to and (see Figure 1, Right). We claim that

 D1∩F≠∅∧D2∩F≠∅⇒E∩F≠∅. (*)

Once (*) is established, we are done by Lemma 2. If , then (*) is immediate, so assume that . By Lemma 2, lies between and . Let be the line through perpendicular to , and let be the open halfplane bounded by with and the open halfplane bounded by with . Since , we have (see Figure 2, Right). Recall that has radius at least and intersects and . We distinguish two cases: (i) there is no intersection of and in , and (ii) there is an intersection of and in .

For case (i), let be any point in . Since we know that , we have . Moreover, let be any point in . By assumption (i), is not in , but it must be in the infinite strip defined by the two tangents of and . Thus, the line segment intersects the diameter segment . Since is convex, the intersection of and is in , so .

For case (ii), let be any point in . Consider the triangle . Since , the angle at is at least (Figure 2, Right). Thus, . Moreover, since , we know that . Hence, we have so and (*) follows, as .

## 3 Stabbing disks in (near) linear time

###### Lemma.

For a set of pairwise intersecting disks, we can decide in deterministic time if the intersection . In the same time, we can compute a point in , if it exists.

###### Proof:

For , let be the unique point with smallest -coordinate in . If , then is a point at infinity. The point is determined by at most three disks. It can be shown that is an LP-type problem with combinatorial dimension whose violation test needs constant time (see [SW92]). Furthermore, let be the set of all disks of that violate a basis , that is for all we have . Then, is the underlying range space for the LP-type problem, and it has clearly constant VC-dimension. Thus, we can use the deterministic solution by Chazelle and Matoušek [CM96] to compute in time and thereby decide whether is empty.

###### Lemma.

Given a set of pairwise intersecting disks with . In expected time or in deterministic time, we can find a non-Helly triple with , such that .

###### Proof:

For the deterministic algorithm, we sort the disks by increasing radius in time, and we find the first disk with using binary search and Lemma 3, in time. Among all disks with , we again use binary search and Lemma 3 to find the first disk such that , again in time. Finally, we find a disk with and by a simple scan, in time.

For the randomized algorithm, we use Chan’s framework [Cha99]. We want to find a non-Helly triple where is minimum. Using Lemma 3, given and a radius , we can decide if contains a non-Helly triple with smaller maximum radius in time. Furthermore, by partitioning into four sets of size roughly and setting , we can obtain in time four subproblems of size at most , one of which contains a non-Helly triple whose maximum radius is minimized. Thus, Chan’s framework applies, and yields a randomized algorithm with expected running time [Cha99, Lemma 2.1].

###### Theorem.

Given a set of pairwise intersecting disks in the plane, we can find, in expected time or in deterministic time, a set of five points such that every disk of contains at least one point of .

###### Proof:

First, we decide if is empty, using Lemma 3. This needs linear time and returns a point in the common intersection, if it exists. Otherwise, by Helly’s theorem, there is a non-Helly triple . We find such a triple with smallest maximum radius , using Lemma 3. For the disks with , we can obtain in linear time a stabbing point by using Helly’s theorem and Lemma 3. Next, by Lemma 2, there are two disks and among , and whose lens angle is at least . Let be the set of four points, as described in the proof of Lemma 2, that stabs any disk of radius at least that intersects both and . Then is a set of five points that stabs all disks of .

## 4 13 pairwise intersecting disks requiring four stabbing points

The construction begins with an inner disk , say of radius , and three larger disks , , of equal size, so that is tangent to all three disks, and each pair of the disks are tangent to each other. Denote the contact point of and by , for .

We add six very large disks as follows. For , we draw the two common outer tangents to and , and denote by and the halfplanes that are bounded by these tangents and are openly disjoint from . For concreteness, the labels and are such that the points of tangency between and , , and , appear along in this counterclockwise order. One can show that the nine points of tangency between and the other disks and halfplanes are all distinct (see Figure 3). We regard the six halfplanes , , for , as disks; in the end, we can apply a suitable inversion to turn the disks and halfplanes into actual disks, if so desired.

Finally, we construct three additional disks , , . To construct , we slightly expand into a disk of radius , while keeping it touching at . We then roll along , by a tiny angle to obtain .

This completes the construction, giving disks. For sufficiently small and , we can ensure the following properties for each : (i) intersects all other disks, (ii) the nine intersection regions , , , for , are pairwise disjoint. (iii) .

###### Lemma.

The disks in the construction cannot be stabbed by three points.

###### Proof:

Consider any set of three stabbing points. Let be the union . If is a stabbing point in , then typically will stab all these four disks (unless lies at certain peculiar locations), but, by construction, it stabs at most one of the nine remaining disks. It is thus impossible for all three stabbing points to lie in , but at least one of them must lie there.

Assume first that contains two stabbing points. As just argued, there are at most two of the remaining disks that are stabbed by these points. The following cases can then arise.

1. The stabbed disks are both halfplanes. Then , , form a non-Helly triple, i.e. they do not have a common intersection, and none of them is stabbed. Since a non-Helly triple must be stabbed by at least two points, an unstabbed disk remains.

2. The stabbed disks are both among . Then the six unstabbed halfplanes form many non-Helly triples, e.g., , , and , and again a disk remains unstabbed.

3. One stabbed disk is , , or , and the other is a halfplane. Then, there is (at least) one disk such that it, and the two associated halfplanes , are all unstabbed. ( is a disk that is not stabbed by either of the two initial points, and neither of its two tangent halfplanes is stabbed.) Then , , and form an unstabbed non-Helly triple.

Assume then that contains only one stabbing point , so at most one of the nine remaining disks is stabbed by . Since is the only point that stabs all three disks , , , it cannot be any of , , , so the other disk that it stabs (if there is such a disk) must be a halfplane. That is, does not stab any of , , . Since , , form a non-Helly triple, they require two points to stab them all. Moreover, since we only have two points at our disposal, one of them must be the point of tangency of two of these disks, say of and . This point stabs only two of the six halfplanes (concretely, they are and ). But then , , and form an unstabbed non-Helly triple.

## 5 Conclusion

We presented simple algorithms for the computation of five stabbing points for a set of pairwise intersecting disks by using algorithms for solving LP-type problems. Nevertheless, the question remains open how to use the proofs of Danzer or Stachó (or any other technique) for an efficient construction of four stabbing points. Since eight disks can always be stabbed by three points [Sta65], it remains open whether nine disks can be stabbed by three points or not. Furthermore, it would be interesting to find a simpler construction of ten pairwise intersecting disks that cannot be stabbed by three points.

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