Consequences of Euler s Formula


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1 Consequences of Euler s Formula Theorem 1 If G is a connected planar graph with V, then 2F E V 6 Proof Let G be planar and connected, with or more vertices Case 1 Suppose that G has a face bounded by fewer than edges, in this case since V, G must be the path graph P and here V =, E = 2, F = 1, so that 2 F = 2, and V 6 =, so that 2F E V 6 Case 2 Suppose that every face of G is bounded by or more edges, numbering the faces of G from 1 to F, we have # of edges bounding face 1 # of edges bounding face 2 # of edges bounding face F and adding these inequalities, we have F D = # of edges bounding face f Since each face is bounded by at least 2 edges, we must have D 2E, so that F 2E, that is, 2 F E
2 To obtain the other inequality, we have F 2 E, and adding V E to both sides of this inequality, we have V E + F 2 E + V E = V 1 E, and from Euler s formula we have so that 2 = V E + F V 1 E, E V 6 Note In this theorem, we need V, since there are only two planar and connected graphs with V <, namely, K 1 and K 2, and the theorem is false for both of them Also, the theorem is false for multigraphs, for example the multigraph below has V =, E = 4, F =, and 4 = E V 6 = 9 6 = Example 2 The graph K 5 is not planar Proof Suppose that K 5 is a planar graph, then V = 5, E = ( 5 2) = 10, so that which is a contradiction Therefore K 5 is nonplanar E = 10 > 9 = 15 6 = V 6, Theorem If G is a connected planar graph with V, and G is triangle free, then 2F E 2V 4 Proof Let G be a connected planar graph with V, such that G is triangle free Case 1 Suppose G has a face bounded by fewer than four edges, then G must be one of the graphs below P P4 T 4 and the theorem holds for all three of these graphs
3 Case 2 Suppose that every face of G is bounded by four or more edges, then we have 4 # of edges bounding face 1 4 # of edges bounding face 2 4 # of edges bounding face F and adding these inequalities, we have 4F D = # of edges bounding face f Since each face is bounded by at least 2 edges, we must have D 2E, so that 4F 2E, that is, F 1 2 E From Euler s formula, that is, Therefore, 2 = V E + F V E E = V 1 2 E, E 2V 4 2F E 2V 4 Example 4 K, is nonplanar Proof Note that the graph K, is connected and has no triangles, K, and assuming that K, is planar, the previous theorem implies that E = 9 8 = = 2V 4, which is a contradiction Therefore K, is nonplanar
4 Corollary 5 If G is a planar graph, then G has vertex of degree less than or equal to 5 Proof We may assume that G is connected, otherwise just work with the connected components Suppose that every vertex v of G has degree deg(v) 6, numbering the vertices of G from 1 to V, we have 6 deg(1) 6 deg(2) 6 deg(v ), and adding these inequalities, from the parity theorem we have 6V so that V E From Theorem 1 we have V deg(v) = 2E, V E V 6, which is a contrdiction Therefore, G has at least one vertex of degree less than or equal to 5 v=1 Exercise 6 Show that if G is a connected planar graph with less than 12 vertices then G has a vertex of degree at most 4 Hint Assume that every vertex has degree at least 5 to obtain a lower bound on E, then use the upper bound in Theorem 1 to show that V 12 Theorem 7 Let G be a connected planar graph with V vertices and E edges such that in a plane drawing of G every face has at least five edges on its boundary, then E 5V 10 Proof Euler s polyhedral formula for a plane drawing of a connected planar graph having V vertices, E edges, and F faces, is given by V E + F = 2 Let G be a connected planar graph with V vertices and E edges such that in a plane drawing of G every face has at least five edges on its boundary If we number the faces from 1 to F, then we can say 5 # of edges bounding face 1 5 # of edges bounding face 2 5 # of edges bounding face F so that 5F # of edges bounding face f
5 On the other hand, each edge bounds exactly two different faces, so the sum is equal to 2E, since each edge would be counted exactly twice, once for each of the two faces it borders, so we have 5F # of edges bounding face f = 2E From Euler s polyhedral formula F = E V + 2, so that 5F = 5E 5V E, that is, E 5V 10 Example 8 The Petersen graph P shown below is nonplanar Proof Note that the smallest cycle in P has length 5, so that if P were a planar graph then each face would be bounded by a least 5 edges, so we may apply the result in the previous theorem Now, P has V = 10 vertices and E = 15 edges, so if P were a planar graph, then we would have which is a contradiction Therefore, P is nonplanar 15 = = ,
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