Planar / Non-Planar Graphs Gabriel Laden CS146 – Spring 2004 Dr. Sin-Min Lee

Definitions Planar – graph that can be drawn without edges that intersect within a plane Non-Planar – graph that cannot be drawn without edges that intersect within a plane

Planar graphs can sometimes be drawn as non- planar graphs. It is still a planar graph, because they are isomorphic. Do Edges Intersect?

Three Houses / Three Utilities Q. Suppose we have three houses and three utilities. Is it possible to connect each utility to each of three houses without any lines crossing? Planar or Non-Planar ? This is also known as K(3,3) bipartite graph

Another definition Region – The area bounded by a subset of the vertices and edges of a graph Note: the outside area of a graph also counts as a region. Therefore a tree has one region, a simple cycle has two regions.

Examples of Counting Regions

Commonly Used Variables Variables used in following mathematical proofs G = an arbitrary graph P = number of vertices Q = number of edges R = number of regions n = number of edges that bound a region N = sum of n for all regions of G

First Theorem Let G be a connected planar graph p = vertices, q = edges, r = regions Then p – q + r = 2 Theorem is by Euler Proof can be made by induction

Second Theorem Let G be a connected planar graph p = (vertices >= 3), q = edges Then q <= 3p - 6 Proof is a little more interesting, uses first theorem to help solve…

Proof: q <= 3p – 6 For each region in graph, n = number of edges to form boundary of its region. Sum of all these n’s in graph = N N >= 3r must be true, since all regions need at least 3 edges to form them. N <= 2q must be true, since no edge can be used more than twice in forming a region

(con’t) Proof: q <= 3p – 6 3r <= N <= 2q Solve p – q + r = 2 for r, then substitute 3(-p +q + 2) <= 2q q <= 3p – 6 is simplified answer

Proof: K(3,3) is Non-Planar Proof by contradiction of theorems Since graph is bipartite, no edge connects two edges within same subset of vertices N >= 4r must be true, since graph contains no simple triangle regions of 3 edges. N <= 2q must be true, since no edge can be used more than twice in forming a region

(con’t) Proof of K(3,3) For K(3,3) p=6, q= 9, r= ?? 4r <= N <= 2q 4r <= (2q = 2 * 9 = 18) r <= 4.5 Using first theorem of planar graphs, p – q + r = 2 6 – 9 + r = 2 r = 5 Proof by contradiction: r cannot be both equal to 5 and less than 4.5 Therefore, K(3,3) is a non-planar graph

Complete Graphs Denoted by Kp All vertices are connected to all vertices q = p * (p - 1) / 2

Proof: K5 is non-planar p=5 q= p * (p – 1) / 2 = 10 Using second theorem of planar graphs: q <= 3p – 6 10 <= 3(5) – 6 10 <= 9 ??? By contradiction, K5 must be non-planar

More Definitions Isomorphic – one-to-one maping of two graphs, such that they are equivalent Subgraph – a graph which is contained as part of another equivalent or greater graph Supergraph – if G’ is a subgraph of G, then G is said to be a supergraph of G’

Subdivisions of graph G Subdivision – a graph obtained from a graph G, by inserting vertices of degree two into any edge (H is a valid subdivision of G, while F is not)

Kuratowski Reduction Theorem A graph G is planar if and only if G contains no subgraph isomorphic to K5 or any sudivision of K5 or K(3,3) Every non-planar graph is a supergraph of K(3,3) or K5

Peterson Graph

Using Kuratowski Q. Is Peterson graph non-planar? A. We can use Kuratowski theorem to pick apart the graph until we find K5 or K(3,3). (solution given on chalkboard)

Scheduling Problem Q. How many time periods are needed to offer the following courses for the set of student schedules? Course Listings: Combinatorics (C), Graph Theory (G), Linear Algebra (L), Numerical Analysis (N), Probability (P), Statistics (S), Topology(T) Student Schedules: CLT, CGS, GN, CL, LN, CG, NP, GL, CT, CST, PS, PT A. This can be drawn as a graph, then find the chromatic number (solution given on chalkboard)

Chromatic Number Rules Four Color Theorem: If G is a planar graph, then X(G) <= 4 Theorem for any graph: Where  (G) = max degree of its vertices, X(G) <= 1 +  (G)

My References