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Brooks Theorem

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					                                             Brook’s Theorem
                                           Jon Breitenbucher, Ph. D.
                                                 February 7, 2006


  1     Proof of the theorem
  Theorem 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 a
  vertex 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 and
  in both cases we can d-color G. Assume that for any graph G’ with n − 1 vertices and degree of each vertex
  at 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 a
        graph G’ with n − 1 vertices and by our assumption we can d-color G’. Since the degree of x is strictly
        less 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. Again
        remove a vertex x from G to obtain G’ with n − 1 vertices which can be d-colored. Since x had d
        neighbors we are not guaranteed to have a color left for x after coloring G’. Note that if two neighbors
        of x have the same color then we can proceed as before and use the unused color for x. So assume that
        all neighbors of x are different colors and arrange the neighbors of x as v1 , v2 , . . . , vd in a clockwise
        manner with colors c1 , c2 , . . . , cd . Define Hij (i = j, 1 ≤ i, j ≤ d) to be the subgraph of G using colors
        ci or cj and whose edges join a vertex colored ci to one colored cj . Hij does not have to be connected
        and is referred to as a Kempe chain. If vi and vj are in two disjoint connected components of Hij then
        in the component of vi we can interchange the colors of all the vertices which will make vi and vj both
        color 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 is
        adjacent to more than one vertex colored cj then there is a color other than ci not assumed by any
        vertex adjacent to vi in which case we can recolor vi with this unassumed color and color x ci . Thus
        we can assume that xi and xj have degree 1 in Cij . Say xi has a neighbor xi1 then

      Case 1 xi1 is xj
      Case 2 xi1 has degree greater than 2
      Case 3 xi1 has a unique neighbor other than xi
        We now have the same three cases for xi1 and can continue. If for each xik we end up in Case 1 or
        Case 3 then we have that Cij is a path. If this is not what happens then there will be a well-defined
        vertex y of degree three or more in Cij which is closest to xi . If y is colored ci then y is adjacent to
        three cj colored vertices and there is a color ck which is not adjacent to y. In this case we could recolor


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xi with cj , xi1 with ci , . . . , and y with ck leaving ci for vertex x. A similar argument applies if y is
colored 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 is
connected to two vertices of color ci and two vertices of color cj . As before this would leave a color ck
not 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 forces
the existence of a vertex y in Cij which is not xj and is connected to xi and colored cj . Pick a color
ck different from ci and cj and interchange the colors of Cik to color xi with ck . Now y belongs to
Cjk and y belongs to Cij but this contradicts the previous paragraph as y is not an endpoint of either
chain.
This means that all the neighbors of x are adjacent and since x was arbitrary this forces G to be a
complete graph which contradicts the assumption of the theorem. Hence it must be the case that a
d-regular graph can be d-colored and so the theorem is proven.




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