RK4 Posted October 23, 2005 Share Posted October 23, 2005 (i). Let G be a simple connected cubic plane graph, and let phi_k be the number of k-sided faces. By counting the number of vertices and edges of G, prove that: 3*phi_3 + 2*phi_4 + phi_5 - phi_7 - 2*phi_8 - . . . = 12 (ii). Deduce that G has at least one face bounded by at most five edges. Link to comment Share on other sites More sharing options...
RK4 Posted October 24, 2005 Author Share Posted October 24, 2005 If G is a simple connected planar cubic graph then: | f | = 2 + | v |/2 I don't get the counting part? Why do we need to count the number of nodes and edges to come up with the number of k-sided faces' sum? Link to comment Share on other sites More sharing options...
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