Multiple Circular Colouring as a Model for Scheduling

In this article we propose a new model for scheduling periodic tasks. The model is based on a variation of the circular chromatic number, called the multiple circular colouring of the conflict graph. We show that for a large class of graphs, this new model will provide better solutions than the original circular chromatic number. At the same time, it allows us to avoid the difficulty of implementation when the fractional chromatic number is used.


Introduction
We consider the scheduling problems involving tasks 1 2 .If two tasks both use a common resource, they cannot be scheduled at the same time.A valid scheduling is a mapping f from   if i and j use a common resource.Let the value of a scheduling f be  , which is the minimum length of time a task has been assigned normalized by the length of the time period.The goal is to find a scheduling f  that maximizes f .One example of this type of scheduling problem is the heavily loaded resource sharing system in computer science [1][2][3].The tasks are processes and some of them may share a common data file.Two processes that do share a common data file cannot operate at the same time.A scheduling of the processes that has the maximum value would allow the processes to operate most efficiently.The constraints of the scheduling problem can be represented by a graph, called the conflict graph.The conflict graph G has vertex set   , , , n v v v  representing the tasks where two vertices v i and v j are adjacent if and only if they use at least one common resource.Vertex colouring and chromatic numbers of the conflict graph have been used as models for scheduling problems (see for example [4]).If G can be coloured with kcolours such that no two adjacent vertices have the same colour, we can then divide [0,T] into k equal length periods.All vertices that are coloured with the same colour do not have edges between them and therefore can be assigned to one period.For this scheduling f, we have . When the tasks are periodic in nature, circular colouring of the conflict graph is a more appropriate method.Circular colouring and the circular chromatic number (also called the star chromatic number) were introduced by A. Vince in 1988 [5].A (k,d)-circular-colouring of a graph G is a mapping such The circular chromatic number of G, , is the infimum of the ratio k/d for which G has a (k,d)-circularcolouring.

 
If we assign the tasks using a (k,d)circular-colouring of the conflict graph G, we would have   , is the infimum of the ratio (k/d) for which G has a (k,d)-set-colouring.If we can consider a (k,d)-set-colouring as a mapping c from V(G) to the sets of arcs in the circle of length k such that the length of the union of arcs in f(v) is at least d for every vertex v and if . Circular chromatic number and fractional chromatic number and their variations have been extensively studied in the last two decades [6][7][8][9][10][11].More results on the circular chromatic number and fractional chromatic number can be found in the book [12] and the survey papers [13,14].It is easy to see that by their definition, we have the relations Vince proved in [5] that f can be arbitrarily large for some graphs.An example is the family of graphs called Kneser graphs.A Kneser graph ( [15,16]) , p q has all q-subsets of (p > 2q) as its vertices and two vertices are adjacent if the two subsets are disjoint.It is proved in [16] that Another example of a family of graphs where the difference between their chromatic number and fractional chromatic number is large is the graphs obtained by using Mycielski's construction.
and edge set , the Mycielskian , , , , , , For the fractional chromatic number of Mycielskians, the authors of [17] found the recurrence relation For the Grotzsch graph, since We use the notation In comparison, we have For this class of graphs, the difference between  and f  is unbounded.For the circular chromatic number, we have This shows that the difference between the circular chromatic number and the fractional chromatic number is also arbitrarily large for the Mycielskians.To achieve the optimal result for a scheduling problem in general, it appears that the fractional chromatic number of the conflict graph provides the best results.
However, there are difficulties if we use (k,d)-set colouring and the fractional chromatic number in the scheduling problem.To achieve


the optimal (k,d)-set colouring may have a very large value of d.As pointed out in [12], is an example of a graph G for which is a fraction whose denominator, when written in smallest terms, is greater than .This example shows that there is no bound on the denominator of that is a polynomial function of the number of vertices of G.If this optimal set colouring is to be used for scheduling, each task would be divided into too many fragments making it impossible in practice.To combine optimality and practicality, in the next section we propose a new colouring of the conflict graph that will provide a better solution than the circular colouring and easier to implement than the set colouring.G 

 The m-circular chromatic number, is the infimum of the ration k d for which Ghas a m-(k,d)-circular-colouring.
It is easy to see that for every positive integer m, To demonstrate this colouring indeed improves the solution of the scheduling problem in some cases, we show that even for , 2 m      c G  Recall that for a graph G, M(G) is the Mycielskian of G.There are many graphs G such that Some sufficient conditions for this equality to hold are given in, for example, [18] and [19].However, will always be strictly less than , , , , , , for each i where such that whenever and are adjacent.For each i, if

  
show that the arcs representing these vertices in the case of in Figure 3.
□ Next we show that unlike the set-chromatic number, the denominator of the m-circular chromatic number is bounded by the product of m and the number of vertices in the graph.
is the smallest.We fix a direction of the circle, say counter clockwise.
There is at least one arc l such that 1 l m  ; otherwise r could be made smaller.Let that arc be l 1 .There must be an arc l such that i) Continuing this process, some arc will have to be used more than once.Say the first time this happens is at  , the denominator is less than mn.□ Intuitively, when the value of m increases, the m-circular chromatic number will be closer to the fractional chromatic number and thus the difference between the chromatic number and m-circular chromatic number would increase.Nevertheless, our next theorem shows that the m-circular chromatic number cannot be less than one m-th of the .chromaticnumber.

Theorem 5 For every graph G
There is an m-circular colouringc such that  

Conclusion
We proved that for a large class of graphs, this multiple circular colouring and m-circular chromatic number of the conflict graph will provide better solutions for the scheduling problem than the original circular chromatic number.It is also easier to implement than the model using the fractional chromatic number.We plan to investigate more classes of graph G where  for small values of m.It would also be interesting to find out the probability for random graphs to have m-circular chromatic number strictly less than their circular chromatic number.

Figure 2
demonstrates this mapping when .5

Figure 3 . A 2 -Theorem 4
Figure 3.A 2-circular coloring of M (C 5 ).Theorem 4 Let G be a graph of n vertices, and   .


Then k and d can be integers such that d < mn.Proof: Suppose that   .

 1 :
By scaling if necessary, there is a m-circular colouring c that maps the vertices of G to sets of m arcs in a circle of length r such that each arc has length at least 1 m and if xy is an edge then    .c x c y   We assume that c is a colouring such that the set   is an arc in for some vertex and l

2   1 1 G
, ii) the left end of l 1 is the right end of and iii) circle an integer number of times.Suppose that they cover the circle s times.Since there are arcs and

6
For every graph G with n vertices, The set in the proof of Theorem 5 is an in-dependent set.We have