Slide 1The House-and-Utilities ProblemPlanar GraphsSlide 4Slide 5ExampleSlide 7Slide 8RegionsSlide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Euler’s FormulaEuler’s Formula (Cont.)Slide 20Slide 21Slide 22Slide 23Chapter 9.7Planar GraphsThese class notes are based on material from our textbook, Discrete Mathematics and Its Applications, 6th ed., by Kenneth H. Rosen, published by McGraw Hill, Boston, MA, 2006. They are intended for classroom use only and are not a substitute for reading the textbook.The House-and-Utilities ProblemPlanar Graphs•Consider the previous slide. Is it possible to join the three houses to the three utilities in such a way that none of the connections cross?Planar Graphs•Phrased another way, this question is equivalent to: Given the complete bipartite graph K3,3, can K3,3 be drawn in the plane so that no two of its edges cross?K3,3Planar Graphs•A graph is called planar if it can be drawn in the plane without any edges crossing.•A crossing of edges is the intersection of the lines or arcs representing them at a point other than their common endpoint.•Such a drawing is called a planar representation of the graph.ExampleA graph may be planar even if it is usually drawn with crossings, since it may be possible to draw it in another way without crossings.ExampleA graph may be planar even if it represents a 3-dimensional object.Planar Graphs•We can prove that a particular graph is planar by showing how it can be drawn without any crossings.•However, not all graphs are planar.•It may be difficult to show that a graph is nonplanar. We would have to show that there is no way to draw the graph without any edges crossing.Regions•Euler showed that all planar representations of a graph split the plane into the same number of regions, including an unbounded region.R4 R3 R2 R1Regions•In any planar representation of K3,3, vertex v1 must be connected to both v4 and v5, and v2 also must be connected to both v4 and v5. v1 v2 v3 v4 v5 v6Regions•The four edges {v1, v4}, {v4, v2}, {v2, v5}, {v5, v1} form a closed curve that splits the plane into two regions, R1 and R2. v1 v5R2R1 v4 v2Regions•Next, we note that v3 must be in either R1 or R2. •Assume v3 is in R2. Then the edges {v3, v4} and {v4, v5} separate R2 into two subregions, R21 and R22. v1 v5 v1 v5R21R2R1 → v3 R22 v4 v2 v4 v2Regions•Now there is no way to place vertex v6 without forcing a crossing:–If v6 is in R1 then {v6, v3} must cross an edge–If v6 is in R21 then {v6, v2} must cross an edge–If v6 is in R22 then {v6, v1} must cross an edge v1 v5R21 v3 R1R22 v4 v2Regions•Alternatively, assume v3 is in R1. Then the edges {v3, v4} and {v4, v5} separate R1 into two subregions, R11 and R12. v1 v5 R11 R2 R12 v3 v4 v2Regions•Now there is no way to place vertex v6 without forcing a crossing:–If v6 is in R2 then {v6, v3} must cross an edge–If v6 is in R11 then {v6, v2} must cross an edge–If v6 is in R12 then {v6, v1} must cross an edge v1 v5 R11 R2 R12 v3 v4 v2Planar Graphs•Consequently, the graph K3,3 must be nonplanar.K3,3Regions•Euler devised a formula for expressing the relationship between the number of vertices, edges, and regions of a planar graph.•These may help us determine if a graph can be planar or not.Euler’s Formula•Let G be a connected planar simple graph with e edges and v vertices. Let r be the number of regions in a planar representation of G. Then r = e - v + 2.# of edges, e = 6# of vertices, v = 4# of regions, r = e - v + 2 = 4R4 R3 R2 R1Euler’s Formula (Cont.)•Corollary 1: If G is a connected planar simple graph with e edges and v vertices where v  3, then e  3v - 6.•Is K5 planar?K5Euler’s Formula (Cont.)•K5 has 5 vertices and 10 edges. •We see that v  3. •So, if K5 is planar, it must be true that e  3v – 6.•3v – 6 = 3*5 – 6 = 15 – 6 = 9.•So e must be  9. •But e = 10.•So, K5 is nonplanar. K5Euler’s Formula (Cont.)•Corollary 2: If G is a connected planar simple graph, then G must have a vertex of degree not exceeding 5.If G has one or two vertices, it is true; thus, we assume that G has at least three vertices.If the degree of each vertex were at least 6, then by Handshaking Theorem,2e ≥ 6v, i.e., e ≥ 3v,but this contradicts the inequality fromCorollary 1: 1 e ≤ 3v – 6.Vvve )deg(2Euler’s Formula (Cont.)•Corollary 3: If a connected planar simple graph has e edges and v vertices with v  3 and no circuits of length 3, then e  2v - 4.•Is K3,3 planar?Euler’s Formula (Cont.)•K3,3 has 6 vertices and 9 edges. •Obviously, v  3 and there are no circuits of length 3. •If K3,3 were planar, then e  2v – 4 would have to be true. •2v – 4 = 2*6 – 4 = 8 •So e must be  8.•But e = 9.•So K3,3 is