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

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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

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

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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

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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.

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Examples of Counting Regions

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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

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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

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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…

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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

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(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

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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

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(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

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

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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

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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’

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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)

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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

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Peterson Graph

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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)

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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)

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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)

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My References

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