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    Mixing Circular Colourings
 always rational? Given a proper 
-colouring 
 of a graph 
, consider the following 'mixing process:'
-  \item choose a vertex 
 
; \item change the colour of 
 (if possible) to yield a different 
-colouring 
 of 
.  A natural problem arises: Can every 
-colouring of 
 be generated from 
 by repeatedly applying this process? If so, we say that 
 is 
-mixing. 
The problem of determining if a graph is 
-mixing and several related problems have been studied in a series of recent papers [1,2,4-6]. The authors of [4] provide examples which show that a graph can be 
-mixing but not 
-mixing for integers 
. For example, given 
 consider the bipartite graph 
 which is obtained by deleting a perfect matching from 
. It is an easy exercise to show that for 
 is 
-mixing if and only if 
 and 
. This example motivates the following definition.
 to be 
 An analogous definition can be made for circular colouring. Recall, a 
-colouring of a graph is a mapping 
 such that if 
, then 
. As with ordinary colourings, we say that a graph 
 is 
-mixing if all 
-colourings of 
 can be generated from a single 
-colouring 
 by recolouring one vertex at a time.
 to be 
 Several bounds on the circular mixing threshold are obtained in [3], including the following which relates the circular mixing threshold to the mixing threshold.
] For every graph 
, 
 As a corollary, we have the following: if 
, then 
. However, examples in [3] show that the ratio 
 can be arbitrarily large in general.
Regarding the problem of determining if 
 is rational, it is worth mentioning that there are no known examples of graphs 
 for which 
 is not an integer. 
Other problems are also given in [3]. One can check that 
 and 
. However, the only graphs which are known to satisfy 
 are in some sense related to 
, eg. trees and complete bipartite graphs. 
 such that 
? Using an example from [4], it is shown in [3] that if 
 is an integer, then there is a graph 
 such that 
 if and only if 
. A natural question to ask is whether a similar result holds for the circular chromatic number. Again, certain bipartite graphs are an exception.
 such that 
? Also, the example of 
 shows that the circular mixing threshold is, in general, not attained. However, the following problem is open.
For more precise versions of the last three questions, see [3].
Bibliography
[1] P. Bonsma and L. Cereceda. Finding paths between graph colourings: PSPACE-completeness and superpolynomial distances. Theoret. Comput. Sci. 410 (2009), (50): 5215--5226.
[2] P. Bonsma, L. Cereceda, J. van den Heuvel, and M. Johnson. Finding paths between graph colourings: Computational complexity and possible distances. Electronic Notes in Discrete Mathematics 29 (2007): 463--469.
*[3] R. C. Brewster and J. A. Noel. Mixing Homomorphisms and Extending Circular Colourings. Submitted. pdf.
[4] L. Cereceda, J. van den Heuvel, and M. Johnson. Connectedness of the graph of vertex-colourings. Discrete Math. 308 (2008), (5-6): 913--919.
[5] L. Cereceda, J. van den Heuvel, and M. Johnson. Mixing 3-colourings in bipartite graphs. European J. Combin. 30 (2009), (7): 1593--1606.
[6] L. Cereceda, J. van den Heuvel, and M. Johnson. Finding paths between 3-colorings. Journal of Graph Theory, 67 (2011), (1): 69--82.
[7] J. A. Noel. "Jonathan Noel - Mixing Circular Colourings." Webpage.
* indicates original appearance(s) of problem.
          
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