On Arrangements of Orthogonal Circles Steven Chaplick 1 , Henry F - - PowerPoint PPT Presentation

On Arrangements of Orthogonal Circles

Steven Chaplick1, Henry F¨

• rster2, Myroslav Kryven1,

Alexander Wolff1

1Julius-Maximilians-Universit¨

at W¨ urzburg, Germany

2Universit¨

at T¨ ubingen, Germany

GD 2019, Prague

Arrangements of Curves

[Alon et al. 2001, Pinchasi 2002], [Felsner & Scheucher 2018] lines pseudolines circles pseudocircles Arrangements of [Gr¨ unbaum 1972]

Arrangements of Curves

[Alon et al. 2001, Pinchasi 2002], [Felsner & Scheucher 2018] lines pseudolines circles pseudocircles Arrangements of Classical question: How many faces can an arrangement of certain curves have? [Gr¨ unbaum 1972] a face

Arrangements of Curves

[Alon et al. 2001, Pinchasi 2002], [Felsner & Scheucher 2018] lines pseudolines circles pseudocircles Arrangements of Classical question: How many faces can an arrangement of certain curves have? [Gr¨ unbaum 1972] a face

Arrangements of Circles, Digons

Any arrangement A of n unit circles has p2(A) = O(n4/3 log n) digonal faces; [Alon et al. 2001] pk(A) = # of faces of degree k in an arrangement A.

Arrangements of Circles, Digons

Any arrangement A of n unit circles has p2(A) = O(n4/3 log n) digonal faces; if, in addition, every pair of circles in A intersect, then p2(A) ≤ n + 3. [Alon et al. 2001] [Pinchasi 2002] pk(A) = # of faces of degree k in an arrangement A.

Arrangements of Circles, Digons

Any arrangement A of n unit circles has p2(A) = O(n4/3 log n) digonal faces; if, in addition, every pair of circles in A intersect, then p2(A) ≤ n + 3. For any arrangement A of n circles with arbitrary radii p2(A) ≤ 20n − 2 if every pair of circles in A intersect. [Alon et al. 2001] [Pinchasi 2002] [Alon et al. 2001] pk(A) = # of faces of degree k in an arrangement A.

Arrangements of Circles, Triangles

For any arrangement A of (pseudo)circles p3(A) ≤ 2

3n2 + O(n).

[Felsner & Scheucher 2018]

Arrangements of Circles, Triangles

For any arrangement A of (pseudo)circles p3(A) ≤ 2

3n2 + O(n).

[Felsner & Scheucher 2018] Lower bound example A with p3(A) = 2

3n2 + O(n) can be

constructed from a line arrangement A′ with p3(A′) = 1

3n2 + O(n) . [F¨

uredi & Pal´ asti 1984] [Felsner, S.: Geometric Graphs and Arrangements, 2004]

Arrangements of Circles, Restrictions

Any arrangement A of n unit circles has p◦

2(A) = O(n4/3 log n) digonal faces;

if, in addition, every pair of circles in A intersect, then p◦

2(A) ≤ n + 3.

For any arrangement A of n circles with arbitrary radii p◦

2(A) ≤ 20n − 2 if every

pair of circles in A intersect. Types of restrictions:

Orthogonal Circles

Two circles α and β are orthogonal if their tangents at one of their intersection points are orthogonal. 90◦ α β

Orthogonal Circles

Two circles α and β are orthogonal if their tangents at one of their intersection points are orthogonal. 90◦ α β In an arrangement of orthogonal circles every two circles either are disjoint or orthogonal.

Orthogonal Circles

Two circles α and β are orthogonal if their tangents at one of their intersection points are orthogonal. 90◦ α β In an arrangement of orthogonal circles every two circles either are disjoint or orthogonal.

Orthogonal Circles

Two circles α and β are orthogonal if their tangents at one of their intersection points are orthogonal. 90◦ No three pairwise orthogonal circles can share the same point. α β In an arrangement of orthogonal circles every two circles either are disjoint or orthogonal.

Orthogonal Circles

Two circles α and β are orthogonal if their tangents at one of their intersection points are orthogonal. 90◦ No three pairwise orthogonal circles can share the same point. α β In an arrangement of orthogonal circles every two circles either are disjoint or orthogonal.

Inversion

Inversion of a point P with respect to α is a point P′ on the ray CαP so that

|CαP′| · |CαP| = r2

α.

α Cα P P′ Properties:

Inversion

Inversion of a point P with respect to α is a point P′ on the ray CαP so that

|CαP′| · |CαP| = r2

α.

• each circle passing through Cα

is mapped to a line; α Cα P P′ Properties:

Inversion

Inversion of a point P with respect to α is a point P′ on the ray CαP so that

|CαP′| · |CαP| = r2

α.

• each circle passing through Cα

is mapped to a line; α Cα P P′

• ∃ an inversion that maps 2

disjoint circles into 2 concentric circles; Properties:

Inversion

Inversion of a point P with respect to α is a point P′ on the ray CαP so that

|CαP′| · |CαP| = r2

α.

• each circle passing through Cα

is mapped to a line; α Cα P P′

• inversion preserves

angles.

• ∃ an inversion that maps 2

disjoint circles into 2 concentric circles; Properties:

Local Properties

There are no four pairwise orthogonal circles. Lem.

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles.

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles.

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles.

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles. can’t intersect!

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles. can’t intersect! Two disjoint circles can be orthogonal to at most two

• ther circles.

Lem.

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles. can’t intersect! Two disjoint circles can be orthogonal to at most two

• ther circles.

Lem. Proof:

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles. can’t intersect! Two disjoint circles can be orthogonal to at most two

• ther circles.

Lem. Proof:

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles. can’t intersect! Two disjoint circles can be orthogonal to at most two

• ther circles.

Lem. Proof:

Local Properties

There are no four pairwise orthogonal circles. Lem. Proof: Assume for contradiction there exist such four circles. can’t intersect! Two disjoint circles can be orthogonal to at most two

• ther circles.

Lem. Proof:

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.
• Def. Consider a subset S ⊆ A of

maximum cardinality such that for each pair of circles one is nested in the other. The innermost circle α in S is called a deepest circle in A.

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.
• Def. Consider a subset S ⊆ A of

maximum cardinality such that for each pair of circles one is nested in the other. The innermost circle α in S is called a deepest circle in A.

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.
• Def. Consider a subset S ⊆ A of

maximum cardinality such that for each pair of circles one is nested in the other. The innermost circle α in S is called a deepest circle in A.

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.
• Def. Consider a subset S ⊆ A of

maximum cardinality such that for each pair of circles one is nested in the other. The innermost circle α in S is called a deepest circle in A.

Main Lemma

A Smallest circle in A is a circle with the smallest radius. Def. Consider an arrangement A

• f orthogonal circles.
• Def. Consider a subset S ⊆ A of

maximum cardinality such that for each pair of circles one is nested in the other. The innermost circle α in S is called a deepest circle in A. Among the deepest circles a smallest one has at most 8 neighbours. Lem.

Main Lemma

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

• Lem. ⋆

Proof:

Main Lemma

α at least 60◦ Let S be the set of neighbours of α s.t. S does not contain nested circles and each circle in S has radius at least as large as α, then |S| ≤ 6.

• Lem. ⋆

Proof:

Main Lemma

α at least 60◦ Let S be the set of neighbours of α s.t. S does not contain nested circles and each circle in S has radius at least as large as α, then |S| ≤ 6.

• Lem. ⋆

Proof:

Main Lemma

α at least 60◦ Let S be the set of neighbours of α s.t. S does not contain nested circles and each circle in S has radius at least as large as α, then |S| ≤ 6.

• Lem. ⋆

α can have at most

360◦ 60◦ = 6 neighbours.

Proof:

Main Lemma Proof

Among the deepest circles a smallest circle α has at most 8 neighbours. Lem.

Main Lemma Proof

Proof:

α Among the deepest circles a smallest circle α has at most 8 neighbours. Lem.

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β.

Among the deepest circles a smallest circle α has at most 8 neighbours. Lem.

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β Among the deepest circles a smallest circle α has at most 8 neighbours. Lem. do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β

• are not nested

Among the deepest circles a smallest circle α has at most 8 neighbours. Lem. do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β

• are not nested
• are larger than α

Among the deepest circles a smallest circle α has at most 8 neighbours. Lem. do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β

• are not nested
• are larger than α
• at most 6

by Lem.⋆ Among the deepest circles a smallest circle α has at most 8 neighbours. Lem. do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β

• are not nested
• are larger than α
• at most 6

by Lem.⋆

• are orthogonal to

two disjoint circles, that is, α and β Among the deepest circles a smallest circle α has at most 8 neighbours. Lem. do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β

• are not nested
• are larger than α
• at most 6

by Lem.⋆

• are orthogonal to

two disjoint circles, that is, α and β Among the deepest circles a smallest circle α has at most 8 neighbours. Lem. Recall: do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Two disjoint circles can be orthogonal to at most two

• ther circles.

Lem. Proof: Recall:

Main Lemma Proof

β

Proof:

α

Otherwise α is nested by at least one more circle β. Consider 2 types of neighbours of α, those that

do not intersect β

• are not nested
• are larger than α
• at most 6

by Lem.⋆

• are orthogonal to

two disjoint circles, that is, α and β Among the deepest circles a smallest circle α has at most 8 neighbours. Lem.

• at most 2

do intersect β

If there are no nested circles, then, by Lem.⋆, α has at most 6 neighbours.

Main Result

Every arrangement of n orthogonal circles has at most 16n intersection points and 17n + 2 faces. Thm.

Main Result

Every arrangement of n orthogonal circles has at most 16n intersection points and 17n + 2 faces. Thm. Proof: The number of intersection points follows by inductively applying the main lemma. The bound on the number of faces follows then by Euler’s formula.

Estimating the Number of Small Faces

Consider a face f with sides formed by circular arcs a, b, c. a f b c

Estimating the Number of Small Faces

Consider a face f with sides formed by circular arcs a, b, c. Let ∠( f, a) be the angle at arc a forming a side of f. a f

∠( f, a) > 0

b c

Estimating the Number of Small Faces

Consider a face f with sides formed by circular arcs a, b, c. Let ∠( f, a) be the angle at arc a forming a side of f. a f

∠( f, a) > 0

b

∠( f, b) < 0

c

Estimating the Number of Small Faces

Consider a face f with sides formed by circular arcs a, b, c. Let ∠( f, a) be the angle at arc a forming a side of f. a f

∠( f, a) > 0

b

∠( f, b) < 0

c

∠( f, c) > 0

We call ∑k

i=1 ∠( f, ai) the total angle of f.

Estimating the Number of Small Faces

Consider a face f with sides formed by circular arcs a, b, c. Let ∠( f, a) be the angle at arc a forming a side of f.

• Thm. (Gauss–Bonnet)

For every face f in an arrangement of orthogonal circles its total angle is 2π − | f |π 2 . a f

∠( f, a) > 0

b

∠( f, b) < 0

c

∠( f, c) > 0

We call ∑k

i=1 ∠( f, ai) the total angle of f.

[for orthogonal circles]

Estimating the Number of Small Faces

Consider a face f with sides formed by circular arcs a, b, c. Let ∠( f, a) be the angle at arc a forming a side of f.

• Thm. (Gauss–Bonnet)

For every face f in an arrangement of orthogonal circles its total angle is 2π − | f |π 2 . a f

∠( f, a) > 0

b

∠( f, b) < 0

c

∠( f, c) > 0

We call ∑k

i=1 ∠( f, ai) the total angle of f.

[for orthogonal circles] In particular π if | f | = 2 and

π 2 if | f | = 3.

Counting Small Faces

For every arrangement A of n orthogonal circles 2p2(A) + p3(A) ≤ 4n. Thm.

Counting Small Faces

For every arrangement A of n orthogonal circles 2p2(A) + p3(A) ≤ 4n. Thm. Proof:

• Faces do not overlap;

Counting Small Faces

For every arrangement A of n orthogonal circles 2p2(A) + p3(A) ≤ 4n. Thm. Proof:

• Faces do not overlap;
• each circle contributes 2π angle to faces, thus,

the arrangement has 2nπ total angle available for faces;

Counting Small Faces

For every arrangement A of n orthogonal circles 2p2(A) + p3(A) ≤ 4n. Thm. Proof:

• by Gauss–Bonnet each digonal face takes π and

each triangular face takes π/2 of the total angle sum from the arrangement.

• Faces do not overlap;
• each circle contributes 2π angle to faces, thus,

the arrangement has 2nπ total angle available for faces;

Counting Small Faces

For every arrangement A of n orthogonal circles 2p2(A) + p3(A) ≤ 4n. Thm. Proof:

• by Gauss–Bonnet each digonal face takes π and

each triangular face takes π/2 of the total angle sum from the arrangement.

• Faces do not overlap;
• each circle contributes 2π angle to faces, thus,

the arrangement has 2nπ total angle available for faces; p2(A)π + p3(A)π 2 ≤ 2nπ.

Intersection Graphs

Orthogonal circle intersection graph: – a vertex for each circle – an edge between two circles if they are orthogonal. [Hlinˇ en´ y and Kratochv´ ıl, 2001]

Intersection Graphs

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

• are K4-free and induced C4-free;

Properties: [Hlinˇ en´ y and Kratochv´ ıl, 2001]

Intersection Graphs

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

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

Properties: [Hlinˇ en´ y and Kratochv´ ıl, 2001]

Intersection Graphs

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

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

∃ an orthogonal circle intersection graph that

contains Kn as a minor for each n. Lem. Properties: [Hlinˇ en´ y and Kratochv´ ıl, 2001]

Intersection Graphs

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

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

∃ an orthogonal circle intersection graph that

contains Kn as a minor for each n. Lem. Proof (idea): Properties: [Hlinˇ en´ y and Kratochv´ ıl, 2001]

Intersection Graphs of Unit Circles

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 circles

also 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 circles

also known as penny graphs.

• ∃ penny graphs which are not orthogonal unit

circle 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 Circles

Recognizing orthogonal unit circle intersection graphs is NP-hard. Thm.

Intersection Graphs of Unit Circles

Recognizing orthogonal unit circle intersection graphs is NP-hard. Thm. Proof (idea): Logic engine which emulates Not-All-Equal-3-Sat (NAE3SAT) problem. x ∧ ¬y ∧ ¬z

¬x ∧ y ∧ ¬z

x ∧ ¬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

Bounds on the # of faces that we have so far: upper bound lower bound digonal faces triangular faces all faces 2n 4n 2n − 2 3n − 3 17n + 2

[Gauss–Bonnet] [Gauss–Bonnet] Main Thm.

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 3n − 3 ? 17n + 2

[Gauss–Bonnet] [Gauss–Bonnet] Main Thm.

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 3n − 3 ? 17n + 2

[Gauss–Bonnet] [Gauss–Bonnet] Main Thm.

Open Problems

Bounds on the # of faces that we have so far: What is the complexity of recognizing general

• rthogonal circle intersection graphs?

upper bound lower bound digonal faces triangular faces all faces 2n 4n 2n − 2 3n − 3 ? 17n + 2

[Gauss–Bonnet] [Gauss–Bonnet] Main Thm.