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By Shaidurov V. V., Timmerman G.

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Indeed, if not, then we arrived at v one more time than we departed from it, each time using a new edge, and finding no edges remaining at the end. Thus there were an odd number of edges of G incident with v , a contradiction. Hence we are indeed back at our starting point when the walk terminates. Let W denote the sequence of edges along which we have so far walked. If W includes all edges of G then we have found an Euler tour and we are finished. Else there are edges of G that are not in W . Erase all edges of W from G, thereby obtaining a (possibly disconnected multi-) graph G .

Now let’s put together the results of the two lemmas above. Let P (K; G) denote the number of ways of properly coloring the vertices of a given graph G. 2 assert that P (K; G − {e}) = P (K; G/{e}) + P (K; G) 45 or if we solve for P (K; G), then we have P (K; G) = P (K; G − {e}) − P (K; G/{e}). 4) The quantity P (K; G), the number of ways of properly coloring the vertices of a graph G in K colors, is called the chromatic polynomial of G. We claim that it is, in fact, a polynomial in K of degree |V (G)|.

Suppose G is a graph, and that we have a certain supply of colors available. To be exact, suppose we have K colors. 6). If we don’t have enough colors, and G has lots of edges, this will not be possible. For example, suppose G is the graph of Fig. 4, and suppose we have just 3 colors available. Then there is no way to color the vertices without ever finding that both endpoints of some edge have the same color. On the other hand, if we have four colors available then we can do the job. Fig. 4 There are many interesting computational and theoretical problems in the area of coloring of graphs.

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