Circles On Arrangements of Orthogonal

On Arrangements of OrthogonalCircles

Steven Chaplick1, Henry Forster2, Myroslav Kryven1,Alexander Wolff1

1Julius-Maximilians-Universitat Wurzburg, Germany2Universitat Tubingen, Germany

GD 2019, Prague

Arrangements of Curves

[Alon et al. 2001, Pinchasi 2002],[Felsner & Scheucher 2018]

lines pseudolines circles pseudocircles

Arrangements of

[Grunbaum 1972]




Arrangements of

Classical question:How many facescan an arrangement of certain curves have?

[Grunbaum 1972]

a face




Arrangements of

Classical question:How many facescan an arrangement of certain curves have?

[Grunbaum 1972]

a face

Arrangements of Circles, Digons

Any arrangement A of n unit circles hasp2(A) = O(n4/3 log n) digonal faces;

[Alon et al. 2001]

pk(A) = # of faces of degree k in an arrangement A.



if, in addition, every pair of circles in Aintersect, then p2(A) ≤ n + 3.

[Alon et al. 2001]

[Pinchasi 2002]




if, in addition, every pair of circles in Aintersect, then p2(A) ≤ n + 3.

For any arrangement A of n circles witharbitrary radii p2(A) ≤ 20n− 2 if everypair of circles in A intersect.

[Alon et al. 2001]

[Pinchasi 2002]

[Alon et al. 2001]


Arrangements of Circles, TrianglesFor any arrangement A of (pseudo)circlesp3(A) ≤ 2

3 n2 + O(n). [Felsner & Scheucher 2018]

Arrangements of Circles, TrianglesFor any arrangement A of (pseudo)circlesp3(A) ≤ 2

3 n2 + O(n). [Felsner & Scheucher 2018]

Lower bound example A with p3(A) = 23 n2 + O(n) can be

constructed from a line arrangement A′ withp3(A′) = 1

3 n2 + O(n) . [Furedi & Palasti 1984][Felsner, S.: Geometric Graphs andArrangements, 2004]

Arrangements of Circles, Restrictions

Any arrangement A of n unit circles hasp◦2(A) = O(n4/3 log n) digonal faces;

if, in addition, every pair of circles in Aintersect, then p◦2(A) ≤ n + 3.

For any arrangement A of n circles witharbitrary radii p◦2(A) ≤ 20n− 2 if everypair of circles in A intersect.

Types of restrictions:

Orthogonal Circles

Two circles α and β are orthogonal iftheir tangents at one of theirintersection points are orthogonal.

90◦α

β

Orthogonal Circles


90◦α

β

In an arrangement of orthogonalcircles every two circles eitherare disjoint or orthogonal.

Orthogonal Circles


90◦α

β


Orthogonal Circles


90◦

No three pairwise orthogonalcircles can share the same point.

αβ


Orthogonal Circles


90◦

No three pairwise orthogonalcircles can share the same point.

αβ


InversionInversion of a point P with respect to α is apoint P′ on the ray CαP so that|CαP′| · |CαP| = r2

α.

α

Cα P P′

Properties:


α.

• each circle passing through Cα

is mapped to a line;

α

Cα P P′

Properties:


α.



α

Cα P P′

• ∃ an inversion that maps 2disjoint circles into 2 concentriccircles;

Properties:


α.



α

Cα P P′

• inversion preservesangles.

• ∃ an inversion that maps 2disjoint circles into 2 concentriccircles;

Properties:

Local PropertiesThere are no four pairwise orthogonal circles.Lem.


Proof: Assume for contradiction there exist such four circles.







can’t intersect!



can’t intersect!

Two disjoint circles can be orthogonal to at most twoother circles.

Lem.



can’t intersect!


Lem.

Proof:



can’t intersect!


Lem.

Proof:



can’t intersect!


Lem.

Proof:



can’t intersect!


Lem.

Proof:

Main Lemma

A Smallest circle in A is a circlewith the smallest radius.

Def.

Consider an arrangement Aof orthogonal circles.

Main Lemma


Def.


Main Lemma


Def.


Def. Consider a subset S ⊆ A ofmaximum cardinality such thatfor each pair of circles one isnested in the other. Theinnermost circle α in S is called adeepest circle in A.

Main Lemma


Def.



Main Lemma


Def.



Main Lemma


Def.



Main Lemma


Def.



Among the deepest circles a smallest one has atmost 8 neighbours.

Lem.

Main LemmaLet S be the set of neighbours of α s.t.S does not contain nested circles andeach circle in S has radius at leastas large as α, then |S| ≤ 6.

Lem. ?

Main Lemma

α

Let S be the set of neighbours of α s.t.S does not contain nested circles andeach circle in S has radius at leastas large as α, then |S| ≤ 6.

Lem. ?

Proof:

Main Lemma

α

at least 60◦


Lem. ?

Proof:

Main Lemma

α

at least 60◦


Lem. ?

Proof:

Main Lemma

α

at least 60◦


Lem. ?

α can have at most360◦60◦ = 6 neighbours.

Proof:

Main Lemma ProofAmong the deepest circles a smallest circle α has atmost 8 neighbours.

Lem.

Main Lemma Proof

Proof:

α

Among the deepest circles a smallest circle α has atmost 8 neighbours.

Lem.

If there are no nested circles, then, byLem.?, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

αOtherwise α is nested byat least one more circle β.


Lem.


Main Lemma Proof

β

Proof:


Consider 2 types ofneighbours of α, those that

do not intersect β


Lem.

do intersect β


Main Lemma Proof

β

Proof:



do not intersect β

• are not nested


Lem.

do intersect β


Main Lemma Proof

β

Proof:



do not intersect β

• are not nested• are larger than α


Lem.

do intersect β


Main Lemma Proof

β

Proof:



do not intersect β


• at most 6by Lem.?


Lem.

do intersect β


Main Lemma Proof

β

Proof:



do not intersect β



• are orthogonal totwo disjoint circles,that is, α and β


Lem.

do intersect β


Main Lemma Proof

β

Proof:



do not intersect β





Lem. Recall:

do intersect β

If there are no nested circles, then, byLem.?, α has at most 6 neighbours.Two disjoint circles can be orthogonal to at most twoother circles.

Lem.

Proof:

Recall:

Main Lemma Proof

β

Proof:



do not intersect β





Lem.

• at most 2

do intersect β


Main ResultEvery arrangement of n orthogonal circles has atmost 16n intersection points and 17n + 2 faces.

Thm.

Main ResultEvery arrangement of n orthogonal circles has atmost 16n intersection points and 17n + 2 faces.

Thm.

Proof: The number of intersection points follows byinductively applying the main lemma.

The bound on the number of faces follows then byEuler’s formula.

Estimating the Number of Small FacesConsider a face f with sides formed by circular arcs a, b, c.

af

b

c


Let ∠( f , a) be the angle at arc a forming a side of f .

af

∠( f , a) > 0b

c



af

∠( f , a) > 0b

∠( f , b) < 0

c



af

∠( f , a) > 0b

∠( f , b) < 0

c

∠( f , c) > 0

We call ∑ki=1 ∠( f , ai) the total angle of f .



Thm. (Gauss–Bonnet)

For every face f in an arrangement of orthogonal circles

its total angle is 2π − | f |π2 .

af

∠( f , a) > 0b

∠( f , b) < 0

c

∠( f , c) > 0


[for orthogonal circles]



Thm. (Gauss–Bonnet)

For every face f in an arrangement of orthogonal circles

its total angle is 2π − | f |π2 .

af

∠( f , a) > 0b

∠( f , b) < 0

c

∠( f , c) > 0


[for orthogonal circles]

In particular π if | f | = 2 andπ2 if | f | = 3.

Counting Small Faces

For every arrangement A of n orthogonal circles2p2(A) + p3(A) ≤ 4n.

Thm.



Thm.

Proof: • Faces do not overlap;



Thm.

Proof: • Faces do not overlap;

• each circle contributes 2π angle to faces, thus,the arrangement has 2nπ total angle availablefor faces;



Thm.

Proof:

• by Gauss–Bonnet each digonal face takes π andeach triangular face takes π/2 of the total anglesum from the arrangement.

• Faces do not overlap;




Thm.

Proof:

• by Gauss–Bonnet each digonal face takes π andeach triangular face takes π/2 of the total anglesum from the arrangement.

• Faces do not overlap;


p2(A)π + p3(A)π2 ≤ 2nπ.

Intersection GraphsOrthogonal circle intersection graph:– a vertex for each circle– an edge between two circles if they are orthogonal.

[Hlinenyand Kratochvıl, 2001]


• are K4-free and induced C4-free;Properties:



• are K4-free and induced C4-free;• always have a vertex of degree at most 8.

Properties:




∃ an orthogonal circle intersection graph thatcontains Kn as a minor for each n.

Lem.

Properties:




∃ an orthogonal circle intersection graph thatcontains Kn as a minor for each n.

Lem.

Proof (idea):

Properties:


Intersection Graphs of Unit CirclesOrthogonal unit circle intersection graph:– a vertex for each unit circle– an edge between two circles if they are orthogonal.

Intersection Graphs of Unit Circles

• are a subclass of the contact graphs of unit circlesalso known as penny graphs.

Properties:

Orthogonal unit circle intersection graph:– a vertex for each unit circle– an edge between two circles if they are orthogonal.

Intersection Graphs of Unit Circles

• are a subclass of the contact graphs of unit circlesalso known as penny graphs.

• ∃ penny graphs which are not orthogonal unitcircle intersection graphs.

an induced C4 a k-star, k ≥ 4

Properties:

Orthogonal unit circle intersection graph:– a vertex for each unit circle– an edge between two circles if they are orthogonal.

Intersection Graphs of Unit CirclesRecognizing orthogonal unit circle intersectiongraphs is NP-hard.

Thm.

Intersection Graphs of Unit CirclesRecognizing orthogonal unit circle intersectiongraphs is NP-hard.

Thm.

Proof (idea): Logic engine which emulatesNot-All-Equal-3-Sat (NAE3SAT) problem.

x ∧ ¬y ∧ ¬z¬x ∧ y ∧ ¬zx ∧ ¬y ∧ ¬z

[Di Battista et al., GD’99]

Open Problems

Bounds on the # of faces that we have so far:

upper bound

lower bound

digonal faces triangular faces all faces

2n 4n

2n− 2

17n + 2[Gauss–Bonnet] [Gauss–Bonnet] Main Thm.

Open Problems


upper bound

lower bound


2n 4n

2n− 2 3n− 3


Open Problems


upper bound

lower bound


2n 4n

2n− 2 3n− 3 ?


Open Problems


upper bound

lower bound


2n 4n

2n− 2 3n− 3 ?


Open Problems


What is the complexity of recognizing generalorthogonal circle intersection graphs?

upper bound

lower bound


2n 4n

2n− 2 3n− 3 ?


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Circles On Arrangements of Orthogonal