This document discusses representing graphs by touching domains, specifically: - Koebe's theorem states that any 3-connected planar graph can be represented by assigning each node a circle such that two nodes are adjacent if their circles are tangent. - It defines key terms like planar graphs, 3-connected graphs, and dual graphs. - The proof of Koebe's theorem involves showing that there exists a function mapping radii to a defect vector such that there is a fixed point where the defect is zero, implying tangency between circles. This is done by showing the function maps a convex set to itself.

1. Representing Graphs by Touching
Domains
Nazli Temur

2. Koebe’s Theorem
Let G be a 3-connected planar graph.
Then, one can assign to each node i a circle Ci in the plane
So that their interiors are disjoint,
Two nodes are adjacent if and only if the corresponding circles
are tangent.

3. Definitions
Planar Graph : A graph is planar if it can be drawn in a plane without graph edges crossing.
(Wilson 1975, p. 162)
3-Connected Graph:A graph G is said to be k-connected, if there does not exist a set of k-1
vertices whose removal disconnects the graph,(Skiena 1990, p. 177).
n-Polyhedral Graph: It is a 3-connected simple planar graph on n nodes.

4. Definitions
Tangent Circles (Kissing Disks): Set of disks with non overlapping interior. Two circles with
centers at (xi, yi) with radii r_i for i=1,2 are mutually tangent if
Dual Graphs : The dual graph G* of a polyhedral graph G has graph vertices each of which
corresponds to a face of G and each of whose faces corresponds to a graph vertex of G.
Polyhedral graphs have unique dual graphs.
Planar graphs preserves the idea of nearest neighborhood in which connections are established between nearest neighborhood. This gives the opportunity to draw them on plane without
intersection. So non of the curves intersect. This means interiors are disjoint.

5. Proof Preparation
Brower's Fixed Point Theorem
x-x 0

6. Proof
http://www.cs.elte.hu/~lovasz/geomrep.pdf

7. Proof
Sum of all the faces that have a point correspondence on p equals to 2π. This is true only when
the circle is tangent.
2π
If we define the defect ratio as below.
As a part of calculation what we obtain is ,
So now we can claim there exist function Φ in which, for a set of points inputs radius
ru and outputs defect δu, then the defect equals to δu =0.
Lemma 6.1.4 If radii ru > 0 (u 2 U) are chosen so that we have ±u = 0 for all u 2 U, then
there is a double circuit representation with these radii.(tangency is present)

8. Proof
The defect vector δ(r) depends on the ratio of the ru only; hence we may restrict our attention to
positive radii satisfying
Here the assumption is that total radius is less than equal to one. Suppose not, we can always
contract the space to have points closer. Or if we reduce all the radii by half, ru (sum of r)
reduces by half, if we keep doing then the sum becomes <=1.
Lets define a new function γ takes u input elements and outputs u elements. When γ takes (ru)
elements it outputs ru+δu.
Now, if we assume that

9. Proof
Now, if we assume that and
The domain of this function is in a plane IR2 is x+y<=1 which forms a triangle. Then, because radius can not be less
than 0, rus going to be positive; If we think of higher dimensions, it will form a convex region which includes point set
inside.
So the domain is a convex shape. We have to ensure that the range is contained within the shape.
As it is mentioned at section 6.1.5 , summation of δ is referred as 0.
The range of our function is ru+ δu, by adding ∑ r+ δ <=1 which means it is contained within the region. (closed)
So this function goes from a convex set to a convex set. Hence it must have a fixed point.
It follows that there is a (unique) point r in the interior of Σ with δ(r) = 0, and the sequence r^k tends to this point.
1
1

10. The reverse case
Touching circles are useful to
obtain planar graph , as well.

11. Open Points

12. Thanks!