Fall Coloring on Product of Cycles and Powers

In this paper, we obtain necessary and sufficient conditions for the existence of fall coloring with fall achromatic number Δ(G) + 1 in the power of a cycle Ck n and in the Cartesian product of two cycles.


Introduction
A k-vertex coloring of a graph G is an assignment of k colors 1, 2, . . ., k, to the vertices.The coloring is proper if no two distinct adjacent vertices share the same color.A graph G is k-colorable if G has a proper k-vertex coloring.The chromatic number χ(G) is the minimum number r such that G is r-colorable.Each set of vertices colored with one color is an independent set of vertices of G, so a coloring is a partition of the vertex set into independent sets.Color of a vertex v is denoted by c(v).
Many graph invariants related to colorings have been defined.Most of them try to minimize the number of colors used to color the vertices under some constraints.For some other invariants, it is meaningful to try to maximize this number.The b-chromatic number is such an example.A b-coloring is a coloring of the vertices of a graph such that each color class contains a vertex that has a neighbor in all other color classes.(Mostafa Blidia, 2009, p. 1787-1793).Any such vertex is called as a colorful vertex.(Saeed Shaebani, 2009).The b-chromatic number b(G) is the largest integer k such that G admits a b-coloring with k colors.A fall coloring of a graph G is a proper coloring such that every vertex of G has neighbors in all the other color classes.(Saeed Shaebani, 2009).A fall achromatic coloring is a particular case of b-coloring in which every vertex is colorful.(Dunbar, J.E., 2000, p.257-273).We call fall achromatic number, the maximum cardinality of a fall coloring of G which we denote by ψ f (G).Not all graphs are fall colorable.Dunbar et al. have proved that the problem of deciding if a given graph admits a fall coloring is NP-complete.(Dunbar, J.E., 2000, p.257-273)For a graph G, and for any vertex v of G, the neighborhood of v is the set , where (Campos, CN., 2007, p. 585-597) Note that C k n is a 2k-regular graph and that k ≥ 1.We take (v 0 , . . ., v n−1 ) to be a cyclic order on the vertex set of G, and always perform modular operations on edge and vertex indexes.(Campos, CN., 2007, p. 585-597).(Tamizh Chelvam T., 2009, p. 56-62).In 2001, Lee has introduced a method of studying the domination parameters such as perfect and independent domination through covering projections.(Lee, J., 2001, p. 231-239).He proved the following theorem.
Theorem 1 (Lee, J., 2001, p. 231-239) Let p : G → G be a covering projection and let S be a perfect dominating set of G. Then p −1 (S ) is a perfect dominating set of G.Moreover, if S is independent, then p −1 (S ) is independent.
Tamizh Chelvam and Sivagnanam Mutharasu have studied the efficient open dominating sets through covering projections.(Tamizh Chelvam T., 2009, p. 56-62).They proved that the inverse image of an efficient open dominating set under a covering projection is an efficient open dominating set.They proved the following lemma.(Tamizh Chelvam T., 2009, p. 56-62).
Lemma 2 (Tamizh Chelvam T., 2009, p. 56-62) Let f : F → F and g : G → G be two covering projections.Then there exists a covering projection h : H → H, where H = F G and H = F G.
In this paper, we introduced a method of studying fall coloring in graphs through covering projections.We obtain a necessary and sufficient condition for the existence of fall coloring with Δ(G) + 1 colors in the power of a cycle C k n .Also, we show that the graph C k n is b-colorable.Further, we obtain a necessary and sufficient condition for the existence of fall coloring with Δ(G) + 1 colors in the Cartesian product of two cycles.

Fall Coloring in C k n
In this section, we obtain a necessary and sufficient condition for the existence of fall coloring with fall coloring number

be a covering projection from a graph G on to another graph H. If H has fall coloring number n, then so is G.
Proof.Assume that H admits a fall coloring with fall achromatic number n and Let us color the vertices as follows: For each j with 0 ≤ j ≤ n − 1, color of the vertex v j is denoted and defined by c(v j ) = j(mod Thus the vertex v i is a colorful vertex of G and the above coloring is a proper coloring of G. Hence G is fall colorable with 2k + 1 colors. Proof.Assume that G is fall colorable with Δ(G) + 1 = 2k + 1 colors, namely 0, 1, 2, . . ., 2k.Suppose 2k + 1 does not divide n.Then n = i(2k + 1) + j for some positive integers i, j with 1 ≤ j ≤ 2k.
Similarly, one can obtain the following: ) = 1 and so on.
Hence, we must have c(v i( 2k+1) where j 0, a contradiction to the fact that c(v 0 ) = 0.
From Lemma and Lemma , one can derive the following theorem which gives a necessary and sufficient condition for the existence of fall coloring with Δ(G) + 1 colors in the graph C k n .Theorem 6 The graph C k n is fall colorable with Δ(G) + 1 colors if and only if 2k + 1 divides n.It is conjectured that every d-regular graph with girth at least 5 has a b-coloring with d + 1 colors.(El-Sahili, A., 2006).Mostafa Blidia, Frederic Maffray and Zoham Zemira showed that the Petersen graph infirms this conjecture, and they propose a new formulation of this question and give a positive answer for small degree as given below.(Mostafa Blidia, 2009Blidia, , p. 1787Blidia, -1793)).
Further they proposed the following conjecture.
Conjecture: Every d-regular graph with girth at least 5, different from the Petersen graph, has a b-coloring with d + 1 colors.
In the next lemma, we prove that a 2k-regular graph C k n is b-colorable with 2k + 1 colors.Note that the girth of By division algorithm, one can write n = h(2k + 1) + j for some positive integers j, h with 0 ≤ j ≤ 2k and h = n 2k+1 .Case 1: Suppose 1 ≤ j ≤ k.Color the vertices as follows: For 0 ≤ g ≤ h(2k + 1), color of the vertex v g as c(v g ) = g (mod (2k + 1)) and c (v h(2k+1) Case 2: Suppose k + 1 ≤ j ≤ 2k + 1. Color the vertices as follows: One can easily verify that in both the cases, the above colorings are b-colorings of C k n with 2k + 1 colors.

Fall coloring on Cartesian product of two cycles with achromatic number Δ + 1
The Cartesian product G H of two graphs G and H, is the graph with vertex set (Haynes, T.W., 2000).In this section, a necessary and sufficient condition for the existence of fall coloring with Δ(G) + 1 colors in the Cartesian product of two cycles has been obtained.The vertex set of the cycle C i is taken as V(C i ) = {0, 1, . . ., (i − 1)}.
Lemma 10 Let m, n be integers which are multiples of 5. Then the graph G = C m C n is fall colorable with fall achromatic number Δ(G) + 1.
Claim 1: The graph C 5 C 5 is fall colorable with 5 colors.
Consider the vertex subsets A 1 , A 2 , A 3 , A 4 and A 5 as given in Remark .
Color the vertices of C 5 C 5 as follows: For each i with 1 ≤ i ≤ 5, c(v) = i if and only if v ∈ A i .Since A i 's are independent, to prove the claim, it is enough to prove that each vertex is colorful.
Claim 2: The graph C m C n is fall colorable with fall achromatic number 5.
Define Proof.Suppose the graph C m C n is fall colorable with 5 colors.Since each vertex of C m C n is adjacent with exactly four vertices, all these adjacent vertices must be colored with different colors.
In this lemma, by five consecutive vertices of C m C n , we mean that {(a, b ⊕ n i) : 0 ≤ i ≤ 4}.
Claim 1: Any five consecutive vertices receive different colors.
On the contrary, assume that there are two vertices x and y in a set of five consecutive vertices have the same color.
Case 2: Suppose x = (a, b ⊕ n i) and y = (a, b ⊕ n (i + 2)) for 0 ≤ i ≤ 2. In this case, the two neighbors of the vertex (a, b ⊕ n (i + 1)), namely x y will have the same color and so the vertex (a, b ⊕ n (i + 1)) is not a colorful vertex.Now consider the vertex i. Clearly we can not color the vertex i with colors 1 and 3. Also, when c(i) = 2, the vertex h will not be a colorful vertex and when c(i) = 4, the vertex f will not be a colorful vertex.Hence c(i) = 5.
Consider the vertex g.Clearly we can not color the vertex g with colors 1 and 4. Also, when c(g) = 2 or 5, the vertex h will not be a colorful vertex.Hence c(g) = 3.Now, consider the vertex a.We cannot use the colors 1, 2, 3, 4 and 5 to color the vertex a, a contradiction to the fact that C m × C n is fall colorable with 5 colors.
Thus in all the cases, we get a contradiction and hence Claim 1 is true.Hence n must be a multiple of 5.
[Figure 2] Similarly, by considering {(a ⊕ m i, b) : 0 ≤ i ≤ 4} as five consecutive vertices, one can prove that any five consecutive vertices receive different colors and hence m is also a multiple of 5.
From Lemma and Lemma , one can conclude the following theorem.
Theorem 12 Let m, n ≥ 5 be integers.Then the graph C m C n is fall colorable with Δ(G) + 1 colors if and only if 5 divides m and n.
by g(x) = x (mod 5) for all x ∈ V(C n ).Then f and g are covering projections respectively from C m and C n onto the graph C 5 .Then by Lemma , there exists a covering projection from C m C n onto the graph C 5 C 5 .By Claim 1 and by Lemma , one can conclude that the graph G = C m C n is fall colorable with Δ(G) + 1 colors.Lemma 11 Let m, n ≥ 5 be integers.Suppose the graph C m C n is fall colorable with fall achromatic number Δ(G) + 1, then m and n are multiples of 5.

Case 3 :
Suppose x = (a, b ⊕ n i) and y = (a, b ⊕ n (i + 3)) for 0 ≤ i ≤ 1.Without loss of generality, assume x = (a, b) and y = (a, b ⊕ n 3) as shown in Figure 1.Without loss of generality, assume that c(x) = c(y) = 1.Consider the vertex d and their uncolored neighboring vertices b, e and g.Without loss of generality, assume that c(d) = 2, c(b) = 4, c(e) = 3 and c(g) = 5.Consider the vertex c.Obviously, we can not use colors 3 and 4 to color the vertex c.If c(c) = 2, then the vertex b is not a colorful vertex, a contradiction.If c(c) = 1, then i is not a colorful vertex, a contradiction.Hence c(c) = 5.Now, consider the vertex h.Clearly, we can not use the colors 3 and 5 to color the vertex h.If c(h) = 2, then the vertex g is not a colorful vertex.If c(h) = 1, then the vertex e is not a colorful vertex.Hence c(h) = 4.Consider the vertex f .Clearly, we can not use the colors 1 and 5 to color the vertex f .If c( f ) = 2 or 4, then the vertex g is not a colorful vertex.Hence c( f ) = 3.Now, consider the vertex a.We cannot use the colors 1,2,3,4 and 5 to color the vertex a, a contradiction to the fact that C m C n is fall colorable with 5 colors.Case 4: Suppose x = (a, b) and y = (a, b ⊕ n 4) as shown in Figure 2. Assume that c(x) = c(y) = 1.By Case 1 and Case 2, we cannot use the color 1 to color the vertices d, e and f .Consider the vertex e and their uncolored neighboring vertices b, d, f and h.Without loss of generality, assume that c(b) = 5, c(d) = 4, c( f ) = 3 and c(h) = 1.Consider the vertex c.Obviously, we can not use the colors 3 and 5 to color the vertex c.If c(c) = 1, then the vertex j is not a colorful vertex, a contradiction.If c(c) = 2, then b is not a colorful vertex, a contradiction.Hence c(c) = 4.