Download - Planar NonPlanar Graphs
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Planar / Non-Planar Graphs
Gabriel LadenCS146 – 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