Very Hard

[RK4]

(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.

[RK4]

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?