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Dual Graphs of Planar Graphs

:::note Reference https://en.wikipedia.org/wiki/Dual_graph :::

A proof for Euler’s characteristic i.e. F+VE=2F + V - E = 2

Section titled “A proof for Euler’s characteristic i.e. F+V−E=2F + V - E = 2F+V−E=2”

:::info Note Note that for polyhedra, they can be reduced to planar graphs by considering one face of the polyhedra as the region outside the planar graph. :::

Planar Graph and its dualSpanning trees (called Interdigitating trees in this context)

By cut-cycle duality, if a set SS of edges in a planar graph GG is acyclic (has no cycles), then the set of edges dual to SS has no cuts, from which it follows that the complementary set of dual edges (the duals of the edges that are not in SS) forms a connected subgraph. Symmetrically, if SS is connected, then the edges dual to the complement of SS form an acyclic subgraph. Therefore, when SS has both properties – it is connected and acyclic – the same is true for the complementary set in the dual graph. That is, each spanning tree of GG is complementary to a spanning tree of the dual graph, and vice versa. Thus, the edges of any planar graph and its dual can together be partitioned (in multiple different ways) into two spanning trees, one in the primal and one in the dual, that together extend to all the vertices and faces of the graph but never cross each other.

This partition of the edges and their duals into two trees leads to a simple proof of Euler’s formula VE+F=2V - E + F = 2 for planar graphs with VV vertices, EE edges, and FF faces. Any spanning tree and its complementary dual spanning tree partition the edges into two subsets of V1V - 1 and F1F - 1 edges respectively, and adding the sizes of the two subsets gives the equation

E=(V1)+(F1)E = (V - 1) + (F - 1)

which may be rearranged to form Euler’s formula.