Stabbing Pairwise Intersecting Disks by Five Points
I S A A C
Stabbing Pairwise Intersecting Disks by Five Points
Partially supported by a NSF AF awards CCF 1 3 4 5 7 8 9 10 11
Partially supported by DFG grant MU/ 1 3 5 7 8 9 10 11
ERC STG 1 3 5 7 8 9 10 11
Partially supported by DFG grant MU/ 1 3 5 7 8 9 10 11
Partially supported by ISF Grant 1 3 5 7 8 9 10 11
0 Institut für Informatik, Freie Universität Berlin , 14195 Berlin , Germany
1 Wolfgang Mulzer
2 Institut für Informatik, Freie Universität Berlin , 14195 Berlin, Germany https://orcid.org/0000000219485840
3 Haim Kaplan School of Computer Science, Tel Aviv University , Tel Aviv 69978 , Israel
4 Department of Computer Science, University of Illinois , Urbana, IL 61801, USA https://orcid.org/0000000326389635 , USA
5 Sariel HarPeled
6 School of Computer Science, Tel Aviv University , Tel Aviv 69978 , Israel
7 program (Center No. 4/11), by the Blavatnik Research Fund in Computer Science at Tel Aviv University, and by the Hermann MinkowskiMINERVA Center for Geometry at Tel Aviv University
8 Max Willert Institut für Informatik, Freie Universität Berlin , 14195 Berlin , Germany
9 Micha Sharir
10 Paul Seiferth
11 Liam Roditty Department of Computer Science, Bar Ilan University , Ramat Gan 5290002 , Israel
Suppose we are given a set D of n pairwise intersecting disks in the plane. A planar point set P stabs D if and only if each disk in D contains at least one point from P . We present a deterministic algorithm that takes O(n) time to find five points that stab D. Furthermore, we give a simple example of 13 pairwise intersecting disks that cannot be stabbed by three points. This provides a simple  albeit slightly weaker  algorithmic version of a classical result by Danzer that such a set D can always be stabbed by four points. 2012 ACM Subject Classification Mathematics of computing → Combinatorics
and phrases Disk graph; piercing set; LPtype problem

Related Version Also available on the arXiv as https://arxiv.org/abs/1801.03158.
Funding Work on this paper was supported in part by grant 1367/2016 from the GermanIsraeli
Science Foundation (GIF).
Acknowledgements We would like to thank Timothy Chan for pointing us to the direct LPtype
algorithm described in Lemma 4.7.
1
Introduction
Let D be a set of n disks in the plane. If every three disks in D intersect, then Helly’s theorem
shows that the whole intersection T D of D is nonempty [9, 10, 11]. In other words, there is
a single point p that lies in all disks of D, i.e., p stabs D. More generally, when we know only
that every pair of disks in D intersect, there must be a point set P of constant size such that
each disk in D contains at least one point in P . It is fairly easy to give an upper bound on
the size of P , but for some time, the exact bound remained elusive. Eventually, in July 1956
at an Oberwolfach seminar, Danzer presented the answer: four points are always sufficient
and sometimes necessary to stab any finite set of pairwise intersecting disks in the plane
(see [5]). Danzer was not satisfied with his original argument, so he never formally published
it. In 1986, he presented a new proof [5]. Previously, in 1981, Stachó had already given
an alternative proof [
15
], building on a previous construction of five stabbing points [14].
This line of work was motivated by a result of Hadwiger and Debrunner, who showed that
three points suffice to stab any finite set of pairwise intersecting unit disks [8]. In later
work, these results were significantly generalized and extended, culminating in the celebrated
(p, q)theorem that was proven by Alon and Kleitman in 1992 [1]. See also a recent paper by
Dumitrescu and Jiang that studies generalizations of the stabbing problem for translates
and homothets of a convex body [6].
Danzer’s published proof [5] is fairly involved and uses a compactness argument, and
part of it is based on an undetailed verification by computer. There seems to be no obvious
way to turn it into an efficient algorithm for finding a stabbing set of size four. The two
constructions of Stachó [
15, 14
] are simpler, but they start with three disks in D with empty
intersection and maximum inscribed circle. It is not clear to us how to find such a triple
quickly (in, say, nearlinear time). Here, we present a new argument that yields five stabbing
points. Our proof is constructive, and it lets us find the stabbing set in deterministic linear
time.
As for lower bounds, Grünbaum gave an example of 21 pairwise intersecting disks that
cannot be stabbed by three points [7]. Later, Danzer reduced the number of disks to ten [5].
This example is close to optimal, because every set of eight disks can be stabbed by three
points [14]. It is hard to verify Danzer’s lower bound example – even with dynamic geometry
software, the positions of the disks cannot be visualized easily. Here, we present a simple
construction that needs 13 disks and can be verified by inspection.
2
The Geometry of Pairwise Intersecting Disks
Let D be a set of n pairwise intersecting disks in the plane. A disk Di ∈ D is given by its
center ci and its radius ri. To simplify the analysis, we make the following assumptions:
(i) the radii of the disks are pairwise distinct; (ii) the intersection of any two disks has a
nonempty interior; and (iii) the intersection of any three disks is either empty or has a
nonempty interior. A simple perturbation argument can then handle the degenerate cases.
D2
c2
L1;2
D1
u
w
v
c1
L2;3
c3
D3
L1;3
`
D1
c1
D2
The lens of two disks Di, Dj ∈ D is the set Li,j = Di ∩ Dj . Let u be any of the two
intersection points of ∂Di and ∂Dj . The angle \ciucj is called the lens angle of Di and Dj .
It is at most π. A finite set C of disks is Helly if their common intersection T C is nonempty.
Otherwise, C is nonHelly. We present some useful geometric lemmas.
I Lemma 2.1. Let {D1, D2, D3} be a set of three pairwise intersecting disks that is nonHelly.
Then, the set contains two disks with lens angle larger than 2π/3.
Proof. Since {D1, D2, D3} is nonHelly, the lenses L1,2, L1,3 and L2,3 are pairwise disjoint.
Let u be the vertex of L1,2 nearer to D3, and let v, w be the analogous vertices of L1,3
and L2,3 (see Figure 1, left). Consider the simple hexagon c1uc2wc3v, and write \u, \v,
and \w for its interior angles at u, v, and w. The sum of all interior angles is 4π. Thus,
\u + \v + \w < 4π, so at least one angle is less than 4π/3. It follows that the corresponding
exterior angle at u, v, or w must be larger than 2π/3. J
I Lemma 2.2. Let D1 and D2 be two intersecting disks with r1 ≥ r2 and lens angle at least
2π/3. Let E be the unique disk with radius r1 and center c, such that (i) the centers c1, c2,
and c are collinear and c lies on the same side of c1 as c2; and (ii) the lens angle of D1 and
E is exactly 2π/3 (see Figure 1, right). Then, if c2 lies between c1 and c, we have D2 ⊆ E.
Proof. Let x ∈ D2. Since c2 lies between c1 and c, the triangle inequality gives
xc ≤ xc2 + c2c = xc2 + c1c − c1c2.
(1)
Since x ∈ D2, we get xc2 ≤ r2. Also, since D1 and E have radius r1 each and lens angle
2π/3, it follows that c1c = √3 r1. Finally, c1c2 = pr12 + r22 − 2r1r2 cos α, by the law
of cosines, where α is the lens angle of D1 and D2. As α√ ≥ 2π/3 and r1 ≥ r2, we get
cos α ≤ −1/2 = (√3 − 3/2) − √3 + 1 ≤ (√3 − 3/2)r1/r2 − 3 + 1, As such, we have
c1c22 = r12 + r22 − 2r1r2 cos α ≥ r12 + r22 − 2r1r2
√3 − 3/2
r1
r2 −
√3 + 1
= r12 − 2 √3 − 3/2 r12 + 2(−√3 + 1)r1r2 + r22
= (1 − 2√3 + 3)r12 + 2(−√3 + 1)r1r2 + r22 = r1(√3 − 1) + r2 2.
Plugging this into Eq. (1) gives xc ≤ r2 + √3r1 − (r1 √3 − 1) + r2 = r1, i.e., x ∈ E.
J
I Lemma 2.3. Let D1 and D2 be two intersecting disks with equal radius r and lens angle
2π/3. There is a set P of four points so that any disk F of radius at least r that intersects
both D1 and D2 contains a point of P .
D12
D1
c1
r1
q
Q
Proof. Consider the two tangent lines of D1 and D2, and let p and q be the midpoints
on these lines between the respective two tangency points. We set P = {c1, c2, p, q} (see
Figure 2, left).
Given the disk F that intersects both D1 and D2, we shrink its radius, keeping its center
fixed, until either the radius becomes r or until F is tangent to D1 or D2. Suppose the
latter case holds and F is tangent to D1. We move the center of F continuously along the
line spanned by the center of F and c1 towards c1, decreasing the radius of F to maintain
the tangency. We stop when either the radius of F reaches r or F becomes tangent to D2.
We obtain a disk G ⊆ F with center c = (cx, cy) so that either: (i) radius(G) = r and G
intersects both D1 and D2; or (ii) radius(G) ≥ r and G is tangent to both D1 and D2. Since
G ⊆ F , it suffices to show that G ∩ P 6= ∅. We introduce a coordinate system, setting the
origin o midway between c1 and c2, so that the yaxis passes through p and q. Then, as in
Figure 2 (left), we have c1 = (−√3 r/2, 0), c2 = (√3 r/2, 0), q = (0, r), and p = (0, −r).
For case (i), let D12 be the disk of radius 2r centered at c1, and D22 the disk of radius
2r centered at c2. Since G has radius r and intersects both D1 and D2, its center c has
distance at most 2r from both c1 and c2, i.e., c ∈ D12 ∩ D22. Let Dp and Dq be the two disks
of radius r centered at p and q. We will show that D12 ∩ D22 ⊆ D1 ∪ D2 ∪ Dp ∪ Dq. Then it is
immediate that G ∩ P 6= ∅. By symmetry, it is enough to focus on the upperright quadrant
Q = {(x, y)  x ≥ 0, y ≥ 0}. We show that all points in D12 ∩ Q are covered by D2 ∪ Dq.
Without loss of generality, we assume that r = 1. Then, the two intersection points of
D12 and Dq are r1 = ( 5√3−282√87 , 38+238√29 ) ≈ (−0.36, 1.93) and r2 = ( 5√√3+282√87 , 38−238√29 ) ≈
(0.98, 0.78), and the two intersection points of D12 and D2 are s1 = ( 23 , 1) ≈ (0.87, 1) and
√
s2 = ( 23 , −1) ≈ (0.87, −1). Let γ be the boundary curve of D12 in Q. Since r1, s2 6∈ Q and
since r2 ∈ D2 and s1 ∈ Dq, it follows that γ does not intersect the boundary of D2 ∪ Dq and
hence γ ⊂ D2 ∪ Dq. Furthermore, the subsegment of the yaxis from o to the start point of γ
is contained in Dq, and the subsegment of the xaxis from o to the endpoint of γ is contained
in D2. Hence, the boundary of D12 ∩ Q lies completely in D2 ∪ Dq, and since D2 ∪ Dq is
simply connected, it follows that D12 ∩ Q ⊆ D2 ∪ Dq, as desired.
For case (ii), since G is tangent to D1 and D2, the center c of G is on the perpendicular
bisector of c1 and c2, so the points p, o, q and c are collinear. Suppose without loss of
generality that cy ≥ 0. Then, it is easily checked that c lies above q, and radius(G) + r =
c1c ≥ oc = r + qc, so q ∈ G. J
I Lemma 2.4. Consider two intersecting disks D1 and D2 with r1 ≥ r2 and lens angle at
least 2π/3. Then, there is a set P of four points such that any disk F of radius at least r1
that intersects both D1 and D2 contains a point of P .
Proof. Let ` be the line through c1 and c2. Let E be the disk of radius r1 and center c ∈ `
that satisfies the conditions (i) and (ii) of Lemma 2.2. Let P = {c1, c, p, q} as in the proof of
Lemma 2.3, with respect to D1 and E (see Figure 1, right). We claim that
D1 ∩ F 6= ∅ ∧ D2 ∩ F 6= ∅ ⇒
E ∩ F 6= ∅.
(*)
Once (*) is established, we are done by Lemma 2.3. If D2 ⊆ E, then (*) is immediate, so
assume that D2 6⊆ E. By Lemma 2.2, c lies between c1 and c2. Let k be the line through c
perpendicular to `, and let k+ be the open halfplane bounded by k with c1 ∈ k+ and k−
the open halfplane bounded by k with c1 6∈ k−. Since c1c = √3 r1 > r1, we have D1 ⊂ k+
(see Figure 2, right). Recall that F has radius at least r1 and intersects D1 and D2. We
distinguish two cases: (i) there is no intersection of F and D2 in k+, and (ii) there is an
intersection of F and D2 in k+.
For case (i), let x be any point in D1 ∩ F . Since we know that D1 ⊂ k+, we have x ∈ k+.
Moreover, let y be any point in D2 ∩ F . By assumption (i), y is not in k+, but it must be
in the infinite strip defined by the two tangents of D1 and E. Thus, the line segment xy
intersects the diameter segment k ∩ E. Since F is convex, the intersection of xy and k ∩ E is
in F , so E ∩ F 6= ∅.
For case (ii), fix x ∈ D2 ∩ F ∩ k+ arbitrarily. Consider the triangle Δxcc2. Since x ∈ k+,
the angle at c is at least π/2 (Figure 2, right). Thus, xc ≤ xc2. Also, since x ∈ D2, we
know that xc2 ≤ r2 ≤ r1. Hence, xc ≤ r1, so x ∈ E and (*) follows, as x ∈ E ∩ F . J
3
Existence of Five Stabbing Points
With the tools from Section 2, we can now show that there is a stabbing set with five points.
I Theorem 3.1. Let D be a set of n pairwise intersecting disks in the plane. There is a set
P of five points such that each disk in D contains at least one point from P .
Proof. If D is Helly, there is a single point that lies in all disks of D. Thus, assume that D
is nonHelly, and let D1, D2, . . . , Dn be the disks in D ordered by increasing radius. Let i∗
be the smallest index with Ti≤i∗ Di = ∅. By Helly’s theorem [9, 10, 11], there are indices
j, k < i∗ such that {Di∗ , Dj , Dk} is nonHelly. By Lemma 2.1, two disks in {Di∗ , Dj , Dk}
have lens angle at least 2π/3. Applying Lemma 2.4 to these two disks, we obtain a set
P 0 of four points so that every disk Di with i ≥ i∗ contains at least one point from P 0.
Furthermore, by definition of i∗, we have Ti<i∗ Di 6= ∅, so there is a point q that stabs every
disk Di with i < i∗. Thus, P = P 0 ∪ {q} is a set of five points that stabs every disk in D, as
desired. J
4
Algorithmic Considerations
Theorem 3.1 leads to a simple O(n log n) time deterministic algorithm for finding a stabbing
set of size 5: we sort the disks in D by radius, and we insert the disks one by one, while
maintaining their intersection. Once the intersection becomes empty, we can use the method
from Theorem 3.1 to find the stabbing set (otherwise, D is Helly, and we have a single
stabbing point). As we will see next, there is also a deterministic linear time algorithm, using
the LPtype framework by Sharir and Welzl [13, 3].
The LPtype framework. An LPtype problem (H, w, ≤) is an abstract generalization of a
lowdimensional linear program. It consists of a finite set of constraints H, a weight function
w : 2H → W, and a total order (W, ≤) on the weights. The weight function w assigns a
weight to each subset of constraints. It must fulfill the following three axioms:
D1
D4
Figure 3 left: The disks D3 and D4 are destroyers of the set {D1, D2}. Moreover, D3 is the
smallest destroyer of the whole set {D1, D2, D3, D4}. right: The disks without D∞ form a Helly
set C. Adding D∞ leads to the nonHelly set C = C ∪ {D∞} with smallest destroyer D∞. The point
v is the extreme point for C and D∞, i.e., dist(C) = d(v, D∞).
Monotonicity: for any H0 ⊆ H and H ∈ H, we have w H0 ∪ {H} ≤ w(H0);
Finite Basis: there is a constant d ∈ N such that for any H0 ⊆ H, there is a subset
B ⊆ H0 with B ≤ d and w(B) = w(H0); and
Locality: for any B ⊆ H0 ⊆ H with w(B) = w(H0) and for any H ∈ H, we have that if
w B ∪ {H} = w(B), then also w H0 ∪ {H} = w(H0).
Given a subset H0 ⊆ H, a basis for H0 is an inclusionminimal set B ⊆ H0 with w(B) = w(H0).
The FiniteBasisaxiom states that any basis has at most d constraints. The goal in an
LPtype problem is to determine w(H) and a corresponding basis B for H.
A generalization of Seidel’s algorithm for lowdimensional linear programming [12] shows
that we can solve an LPtype problem in expected time O(H), provided that an O(1)time
violation test is available: given a set B ⊆ H and a constraint H ∈ H, we say that H violates
B if and only if w B ∪ {H} < w(B). In a violation test, we are given B and H, and we must
determine (i) whether B is a valid basis for some subset of constraints; and (ii) whether H
violates B.5 Here and below, the constant factor in the Onotation may depend on d.
Chazelle and Matoušek [4] showed that an LPtype problem can be solved in O(H)
deterministic time if (i) we have a constanttime violation test and (ii) the range space
(H, {vio(B)  B is a basis for some H0 ⊆ H}) has bounded VCdimension [3]. Here, for a
basis B, the set vio(B) ⊂ H consists of all constraints that violate B. We will now show
that the problem of finding a nonHelly triple as in Theorem 3.1 is LPtype and fulfills the
requirements for the algorithm of Chazelle and Matoušek.
Geometric observations. The distance between two closed sets A, B ⊆ R2 is defined as
d(A, B) = min{d(a, b)  a ∈ A, b ∈ B}. From now on, we assume that all points in S D
have positive ycoordinates. This can be ensured with linear overhead by an appropriate
translation of the input. We denote by D∞ the closed halfplane below the xaxis. It is
5 Here, we follow the presentation of Chazelle [3]. Sharir and Welzl [13] do not require property (i) of
a violation test. Instead, they need an additional basis computation primitive: given a basis B and a
constraint H ∈ H, find a basis for B ∪ {H}. Given a violation test with property (i), a basis computation
primitive can easily be implemented by brute force enumeration.
D1
D3
v
D2
D3
D4
D1
w v
D2
D4
interpreted as a disk with radius ∞ and center at (0, −∞). For C ⊆ D we set C = C ∪ {D∞}.
Observe that for C1 ⊆ C2 ⊆ D, if C1 is nonHelly, then C2 is nonHelly. Furthermore, for
r ∈ R>0 ∪ {∞} and C ⊆ D, we define C≤r (resp., C<r) as the set of all disks in C with radius
at most (resp., smaller than) r. Let C ⊆ D be Helly. A disk D ∈ D is a destroyer of C if
C ∪ {D} is nonHelly. Observe that D∞ is a destroyer for every Helly subset of D. Now, let
C ⊆ D be an arbitrary subset of D (either Helly or nonHelly). We say D ∈ C is the smallest
destroyer of C if C<r is Helly and C≤r is nonHelly, where r is the radius of D. Note that
D is the unique largest disk in C≤r. Furthermore, D is the smallest disk in C that causes a
nonHelly triple. If C is Helly, then D = D∞. See Figure 3 for an example. We can make
the following two observations.
I Lemma 4.1. Let C ⊆ D be Helly and D ∈ D a destroyer of C. Then, the point v ∈ T C
with minimum distance to D is unique.
Proof. Suppose there are two distinct points v 6= w ∈ T C with d(v, D) = d T C, D =
d(w, D). Since T C is convex, the segment vw lies in T C. Now, if D 6= D∞, then every
point in the relative interior of vw is strictly closer to D than v and w. If D = D∞, then
all points in vw have the same distance to D, but since T C is strictly convex, the relative
interior of vw lies in the interior of T C, so there must be a point in T C that is closer to
D than v and w. In either case, we obtain a contradiction to the assumption v 6= w and
d(v, D) = d T C, D = d(w, D). The claim follows. J
The unique point v ∈ T C with minimum distance to a destroyer D is called the extreme
point for C and D (see Figure 3).
I Lemma 4.2. Let C1 ⊆ C2 ⊆ D be two Helly sets and D ∈ D a destroyer of C1 (and thus of
C2). Let v ∈ T C1 be the extreme point for C1 and D. We have d T C1, D ≤ d T C2, D .
In particular, if v ∈ T C2, then d T C1, D = d T C2, D and v is also the extreme point for
C2. If v 6∈ T C2, then d T C1, D < d T C2, D .
Proof. The first claim holds trivially: let w ∈ T C2 be the extreme point for C2 and D. Since
C1 ⊆ C2, it follows that w ∈ T C1, so d T C1, D ≤ d(w, D) = d T C2, D . If v ∈ T C2, then
d T C1, D ≤ d T C2, D ≤ d(v, D) = d T C1, D , so v = w, by Lemma 4.1. If v ∈/ T C2,
then d T C1, D < d T C2, D , by Lemma 4.1 and the fact that C1 ⊆ C2. See Figure 4. J
D1
D3
v
F
D
Let C be a subset of D. The radius of the smallest destroyer D of C is denoted by rad(C).
Note that rad(C) ∈ R>0 ∪ {∞}. Moreover, let dist(C) be the distance between D and the set
T C<rad(C), i.e., dist(C) = d T C<rad(C), D . Then, C is Helly if and only if rad(C) = ∞. In
this case, dist(C) is the distance between T C and the xaxis. We define the weight w(C) of
C as w(C) = (rad(C), − dist(C)), and we denote by ≤ the lexicographic order on R2. Chan
observed, in a slightly different context, that (D, w, ≤) is LPtype [2]. However, Chan’s paper
does not contain a detailed proof for this fact. Thus, in the following lemmas, we show that
the three LPtype axioms hold.
I Lemma 4.3. For any C ⊆ D and E ∈ D, we have w C ∪ {E} ≤ w(C).
Proof. Set C∗ = C ∪ {E}. Let D be the smallest destroyer of C, and let r = rad(C) be the
radius of D. Then, D is the largest disk in C≤r. The set C≤r is nonHelly. Adding E does
not change this, i.e., C≤r is also nonHelly. Thus, the smallest destroyer of C∗≤r is either D or
∗
some smaller disk in C<∗r. In the latter case, we have rad(C∗) < rad(C). In the former case, we
have rad(C∗) = rad(C), and Lemma 4.2 gives − dist(C∗) = −d T C<∗r, D ≤ −d(T C<r, D) =
− dist(C). In either case, w C∗) ≤ w(C). See Figure 5 for an illustration. J
I Lemma 4.4. For any C ⊆ D, there is a set B ⊆ C with B ≤ 3 and w(B) = w(C).
Proof. Let D be the smallest destroyer of C. Let r = rad(C) be the radius of D, and let
v ∈ T C<r be the extreme point for C<r and D. By general position, there are at most two
disks E, F ∈ C<r with v ∈ ∂(E ∩ F ). Note that E and F may be the same disk.
Set B = {D, E, F } \ {D∞}. There are three possibilities. If C is nonHelly, then D 6= D∞
and B is a nonHelly triple (indeed, as the disks in D are pairwise intersecting, the extreme
point v must lie at the intersection of two disk boundaries). If C is Helly, then D = D∞ and
B ≤ 2. If B = 2, then v is the vertex of ∂(E ∩ F ) with minimum ycoordinate. If B = 1,
then v is the point on ∂E with minimum ycoordinate. In either case, dist(B) is the value of
the smallest ycoordinate of a point in T B. See Figure 6 for an illustration.
We claim that w(B) = w(C). Firstly, rad(B) = rad(C), because B and C have the same
smallest destroyer. Secondly, we show dist(B) = dist(C): since B<r ⊆ C<r, by Lemma 4.2, we
get dist(B) = d T B<r, D) ≤ d T C<r, D) = dist(C). Suppose that dist(B) < dist(C). Then,
there is a point w ∈ E ∩ F with d(w, D) < d(v, D). Furthermore, by general position and
since v is the intersection of two disk boundaries, there is a relatively open neighborhood
N around v in T C<r such that N is also relatively open in E ∩ F . Since E ∩ F is convex,
there is a point x ∈ N that also lies in the relative interior of the line segment wv. Then,
d(x, D) < d(v, D) and x ∈ T C<r, which is impossible, as v is the extreme point.
The set B is actually a basis for C: if B is a nonHelly triple, then removing any disk from
B creates a Helly set and increases the radius of the smallest destroyer to ∞. If B ≤ 2, then
D∞ is the smallest destroyer of B and the minimality follows directly from the definition. J
I Lemma 4.5. Let B ⊆ C ⊆ D with w(B) = w(C) and let E ∈ D. Then, if w B∪{E} = w(B)
we also have w C ∪ {E} = w(C).
Proof. Set C∗ = C ∪ {E}, B∗ = B ∪ {E}. Let r = rad(C) and D the smallest destroyer of C.
Since w(C) = w(B) = w(B∗), we have that D is also the smallest destroyer of B and of B∗. If
E has radius > r, then E cannot be the smallest destroyer of C∗, so w C∗ = w(C). Assume
that E has radius < r. Let v be the extreme point of C<r and D. Since w(B∗) = w(B),
we know that d T B<r, D = d T B<∗r, D = d(v, D). Now, Lemma 4.2 implies v ∈ E,
since E ∈ B<∗r. Thus, the set C<∗r = C<r ∪ {E} is Helly. Furthermore, C∗≤r is nonHelly,
because the subset C≤r is nonHelly. Therefore, D is also the smallest destroyer of C∗, so
rad(C∗) = r = rad(C). Finally, since B<∗r ⊆ C<r we can use Lemma 4.2 to derive
∗
d \ C<r, D
= d \ B<∗r, D
≤ d \ C<∗r, D
≤ d(v, D) = d \ C<r, D .
J
Next, we describe a violation test for (D, w, ≤): given a set B ⊆ D and a disk E ∈ D
with radius r, determine (i) whether B is a basis for some subset of D, and (ii) whether E
violates B, i.e., whether w B ∪ {E} < w(B). This is done as follows:
If (i) B > 3; or (ii) B = 3 and B is Helly; or (iii) B = 2 and the yminimum of T B is
also the yminimum of a single disk of B, return “B is not a basis”.
If B = 1, then, if the yminimum in E ∩ T B differs from the yminimum in T B, return
“E violates B” ; otherwise, return “E does not violate B”.
If B = {D1, D2}, find the yminimum v of D1 ∩ D2 and return “E violates B” if v 6∈ E,
and “E does not violate B”, otherwise.
Finally, if B = {D, D1, D2} is nonHelly with smallest destroyer D.6 Let r = rad(B) be
the radius of D and r0 be the radius of E:
6 Note that since B is a subset of D and since B is nonHelly, the smallest destroyer D of B cannot be the
disk D∞.
If r0 > r, return “E does not violate B”.
If r0 < r, find the vertex v of D1 ∩ D2 that minimizes the distance to E and return
“E violates B” if v 6∈ E, and “E does not violate B”, otherwise.
The violation test obviously needs constant time. Finally, to apply the algorithm of
Chazelle and Matoušek, we still need to check that the range space (D, R) with R = {vio(B) 
B is a basis of a subset in D} has bounded VC dimension.
I Lemma 4.6. The range space (D, R) has VCdimension at most 3.
Proof. The discussion above shows that for any basis B, there is a point vB ∈ R2 such that
E ∈ D violates B if and only if vB 6∈ E. Thus, for any v ∈ R2, let R0v = {D ∈ D  v 6∈ D}
and let R0 = {R0v  v ∈ R2}. Since R ⊆ R0, it suffices to show that (D, R0) has bounded
VCdimension. For this, consider the complement range space (D, R00) with R00 = {R0v0  v ∈ R2}
and R0v0 = {D ∈ D  v ∈ D}, for v ∈ R2. It is well known that (D, R0) and (D, R00) have the
same VCdimension [3], and that (D, R00) has VCdimension 3 (e.g., this follows from the
classic homework exercise that there is no planar Venndiagram for four sets). J
Finally, the following lemma summarizes discussion so far.
I Lemma 4.7. Given a set D of n pairwise intersecting disks in the plane, we can decide
in O(n) deterministic time whether D is Helly. If so, we can compute a point in T D in
O(n) deterministic time. If not, we can compute the smallest destroyer D of D and two
disks E, F ∈ D<r that form a nonHelly triple with D. Here, r is the radius of D.
Proof. Since (D, w, ≤) is LPtype, the violation test needs O(1) time, and the VCdimension
of (D, R) is bounded, we can apply the deterministic algorithm of Chazelle and Matoušek [4]
to compute w(D) = (rad(D), − dist(D)) and a corresponding basis B in O(n) time. Then, D
is Helly if and only if rad(D) = ∞. If D is Helly, then B ≤ 2. We compute the unique point
v ∈ T B with d(v, D∞) = d T B, D∞ . Since B ⊆ D and d T B, D∞ = d T D, D∞ , we
have v ∈ T D by Lemma 4.2. We output v. If D is nonHelly, we simply output B, because
B is a nonHelly triple with the smallest destroyer D of D and two disks E, F ∈ D<r, where
r is the radius of D. J
I Theorem 4.8. Given a set D of n pairwise intersecting disks in the plane, we can find in
O(n) time a set P of five points such that every disk of D contains at least one point of P .
Proof. Using the algorithm from Lemma 4.7, we decide whether D is Helly. If so, we return
the point computed by the algorithm. Otherwise, the algorithm gives us a nonHelly triple
{D, E, F }, where D is the smallest destroyer of D and E, F ∈ D<r, with r being the radius
of D. Since D<r is Helly, we can obtain in O(n) time a stabbing point q ∈ T D<r by using
the algorithm from Lemma 4.7 again. Next, by Lemma 2.1, there are two disks in {D, E, F }
whose lens angle is at least 2π/3. Let P 0 be the set of four points from the proof of Lemma 2.4.
Then, P = P 0 ∪ {q} is a set of five points that stabs every disks in D. J
5
A Simple Lower Bound
We now exhibit a set of 13 pairwise intersecting disks in the plane such that no point set of
size three can pierce all of them. The construction begins with an inner disk A of radius 1
and three larger disks D1, D2, D3 of equal radius, so that A is tangent to all three disks and
so that each two disks are tangent to each other. For i = 1, 2, 3, we denote the contact point
of A and Di by ξi.
D1
A
D2
D3
We add six more disks as follows. For i = 1, 2, 3, we draw the two common outer tangents
to A and Di, and denote by Ti− and Ti+ the halfplanes that are bounded by these tangents
and are openly disjoint from A. The labels Ti− and Ti+ are chosen such that the points of
tangency between A and Ti−, Di, and Ti+, appear along ∂A in this counterclockwise order.
One can show that the nine points of tangency between A and the other disks and tangents
are pairwise distinct (see Figure 7). We regard the six halfplanes Ti−, Ti+, for i = 1, 2, 3,
as (very large) 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 A1, A2, A3. To construct Ai, we slightly
expand A into a disk A0i of radius 1 + ε1, while keeping the tangency with Di at ξi. We then
roll A0i clockwise along Di, by a tiny angle ε2 ε1, to obtain Ai.
This gives a set of 13 disks. For sufficiently small ε1 and ε2, we can ensure the following
properties for each Ai: (i) Ai intersects all other 12 disks; (ii) the nine intersection regions
Ai ∩ Dj, Ai ∩ Tj−, Ai ∩ Tj+, for j = 1, 2, 3, are pairwise disjoint; and (iii) ξi ∈/ Ai.
I Theorem 5.1. The construction yields a set of 13 disks that cannot be stabbed by 3 points.
Proof. Consider any set P of three points. Set A∗ = A ∪ A1 ∪ A2 ∪ A3. If P ∩ A∗ = ∅, we
have unstabbed disks, so suppose that P ∩ A∗ 6= ∅. For p ∈ P ∩ A∗, property (ii) implies
that p stabs at most one of the nine remaining disks Dj, Tj+ and Tj−, for j = 1, 2, 3. Thus, if
P ⊂ A∗, we would have unstabbed disks, so we may assume that P ∩ A∗ ∈ {1, 2}.
Suppose first that P ∩ A∗ = 2. As just argued, at most two of the remaining disks are
stabbed by P ∩ A∗. The following cases can then arise.
(a) None of D1, D2, D3 is stabbed by P ∩A∗. Since {D1, D2, D3} is nonHelly and a nonHelly
set must be stabbed by at least two points, at least one disk remains unstabbed.
(b) Two disks among D1, D2, D3 are stabbed by P ∩ A∗. Then the six unstabbed halfplanes
form many nonHelly triples, e.g., T1−, T2−, and T3−, and again, a disk remains unstabbed.
(c) The set P ∩ A∗ stabs one disk in {D1, D2, D3} and one halfplane. Then, there is (at
least) one disk Di such that Di and its two tangent halfplanes Ti−, Ti+ are all unstabbed
by P ∩ A∗. Then, {Di, Ti−, Ti+} is nonHelly, and at least two more points are needed to
stab it.
Suppose now that P ∩ A∗ = 1, and let P ∩ A∗ = { }
p . We may assume that p stabs all
three disks A1, A2, A3, since otherwise a disk would stay unstabbed. At most one of the
nine remaining disks is stabbed by p. Thus, p 6∈ {ξ1, ξ2, ξ3}, so the other disk that it stabs (if
any) must be a halfplane. That is, p does not stab any of D1, D2, D3. Since {D1, D2, D3} is
nonHelly, it requires two stabbing points. Moreover, since P \ {p} = 2, we may assume that
one point q of P \ A∗ is the point of tangency of two of these disks, say q = D2 ∩ D3. Then,
q stabs only two of the six halfplanes, say, T1− and T1+. But then, {D1, T2+, T3−} is nonHelly
and does not contain any point from {p, q}. At least one disk remains unstabbed. J
6
We gave a simple lineartime algorithm to find five stabbing points for a set of pairwise
intersecting disks in the plane. It remains open how to use the proofs of Danzer or Stachó [
15,
5
] (or any other technique) for an efficient construction of four stabbing points. It is also not
known whether nine disks can be stabbed by three points or not (for eight disks, this is the
case [14]). Furthermore, it would be interesting to find a simpler construction, than the one
by Danzer, of ten pairwise intersecting disks that cannot be stabbed by three points.
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