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# Graphs : Coloring Maps

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Show that a planar map M = M(G) can be 2 colored iff every vertex of G has even degree.

[The map M(G) of a graph G is the collection of its faces, which are to be colored so that no adjacent faces have the same color.]

[hint: if every vertex of G has even degree then G is a union of edge-disjoint cycles. for another solution, apply induction on the number of edges, and delete the edges of a cycle forming the boundry of a face of M(G).]

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Proof : Let G we any graph of 2 face colourable. Then dual graph of G., i.e. G* having chromatic number (G)= 2. Therefore G* is a bipartite graph. Note that every graph is bipartite if and only if it has chromatic number 2. And every dual of bipartite graph is Euler graph and every Euler graph having vertex of even degree. And we also know that G is isomorphic to G**

Where G** is dual of G*
.
Thus every vertex of G has even degree.

Conversely, let G be a graph having even degree vertex. Then G** (double dual of G) is Euler graph. Therefore G* is bipartite. And since G* is bipartite therefore chromatic number of G is 2 i.e, (G*) =2 and therefore G is two face colourable.

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