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Upbrooks
 

Upbrooks

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    Upbrooks Upbrooks Document Transcript

    • Brook’s TheoremJon Breitenbucher, Ph. D.February 7, 20061 Proof of the theoremTheorem 1.1 (Brook’s Theorem).If G is not an odd length circuit or a complete graph, then χ(G) ≤ d where d is the maximum degree of avertex of G and χ(G) is the chromatic number.Proof. The restriction on G means that n must be at least four and in this case d can be two or three andin both cases we can d-color G. Assume that for any graph G’ with n − 1 vertices and degree of each vertexat most d that G can be d-colored so χ(G ) ≤ d.Let G have n vertices and use induction on n to show that G can be d-colored.Case 1: If G has a vertex x of degree strictly less than d then delete x and all of its adjacent edges. This gives agraph G’ with n − 1 vertices and by our assumption we can d-color G’. Since the degree of x is strictlyless than d we will have at least one color left for x and so G can be d-colored.Case 2: All vertices of G have degree d making G a d-regular graph and assume G cannot be d-colored. Againremove a vertex x from G to obtain G’ with n − 1 vertices which can be d-colored. Since x had dneighbors we are not guaranteed to have a color left for x after coloring G’. Note that if two neighborsof x have the same color then we can proceed as before and use the unused color for x. So assume thatall neighbors of x are different colors and arrange the neighbors of x as v1, v2, . . . , vd in a clockwisemanner with colors c1, c2, . . . , cd. Define Hij(i = j, 1 ≤ i, j ≤ d) to be the subgraph of G using colorsci or cj and whose edges join a vertex colored ci to one colored cj. Hij does not have to be connectedand is referred to as a Kempe chain. If vi and vj are in two disjoint connected components of Hij thenin the component of vi we can interchange the colors of all the vertices which will make vi and vj bothcolor cj and we can color x color ci.So we can assume that vi and vj are both in the same component Cij of Hij for all i and j. If vi isadjacent to more than one vertex colored cj then there is a color other than ci not assumed by anyvertex adjacent to vi in which case we can recolor vi with this unassumed color and color x ci. Thuswe can assume that xi and xj have degree 1 in Cij. Say xi has a neighbor xi1thenCase 1 xi1 is xjCase 2 xi1 has degree greater than 2Case 3 xi1has a unique neighbor other than xiWe now have the same three cases for xi1and can continue. If for each xikwe end up in Case 1 orCase 3 then we have that Cij is a path. If this is not what happens then there will be a well-definedvertex y of degree three or more in Cij which is closest to xi. If y is colored ci then y is adjacent tothree cj colored vertices and there is a color ck which is not adjacent to y. In this case we could recolor1
    • xi with cj, xi1with ci, . . . , and y with ck leaving ci for vertex x. A similar argument applies if y iscolored cj. Thus it must be the case that each Cij is a path from xi to xj for all i and j.Suppose w were a member of both Cij and Cjl then w must be colored cj and unless w = xj then it isconnected to two vertices of color ci and two vertices of color cj. As before this would leave a color cknot adjacent to w and we could recolor as above. Thus Cij and Cjl can only intersect at endpoints.Now assume two neighbors of x are not adjacent in G. Then they are not adjacent in G’. This forcesthe existence of a vertex y in Cij which is not xj and is connected to xi and colored cj. Pick a colorck different from ci and cj and interchange the colors of Cik to color xi with ck. Now y belongs toCjk and y belongs to Cij but this contradicts the previous paragraph as y is not an endpoint of eitherchain.This means that all the neighbors of x are adjacent and since x was arbitrary this forces G to be acomplete graph which contradicts the assumption of the theorem. Hence it must be the case that ad-regular graph can be d-colored and so the theorem is proven.2