Chapter 10.7 Planar Graphs These class notes are based on material from our textbook, Discrete Mathematics and Its Applications, 7th ed., by Kenneth H. Rosen, published by McGraw Hill, Boston, MA, 2011. They are intended for classroom use only and are not a substitute for reading the textbook. The House-and-Utilities Problem Planar 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 K 3,3, can K3,3 be drawn in the plane so that no two of its edges cross? K3,3
Planar 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. Example A 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. Example A 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
R1 Regions 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 v6 Regions 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 v5
R2 v R1 v Regions 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 v5 R21 R2 R1
v3 R22 v v v v
Regions 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 v5 R21 v3
R1 R22 Regions 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 v4 R12 v2 v3 Regions 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 v4 R12
v2 v3 Planar Graphs Consequently, the graph K3,3 must be nonplanar. K3,3 Regions 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. Eulers 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. R4
R3 R2 R1 # of edges, e = 6 # of vertices, v = 4 # of regions, r = e - v + 2 = 4 Eulers 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. (no proof)
Is K5 planar? K5 Eulers 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. K5 Eulers 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 from Corollary 1: e 3v 6. 2e deg(v ) vV Eulers 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. (no proof) Is K3,3 planar? Eulers 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 nonplanar. K3,3