On (2 − d)-kernels in the cartesian product of graphs

In this paper we study the problem of the existence of (2 − d)kernels in the cartesian product of graphs. We give sufficient conditions for the existence of (2− d)-kernels in the cartesian product and also we consider the number of (2− d)-kernels.

We say that a subset D ⊆ V (G) is dominating, if every vertex of G is either in D or it is adjacent to at least one vertex of D. A subset S ⊆ V (G) is independent if no two vertices of S are adjacent in G.A subset J being dominating and independent is a kernel of G.
H. Galeana-Sánchez and C. Hernández-Cruz played an important role in researching kernels in digraphs.During the last decades they studied not only kernels in digraphs but some of its generalizations, mainly kernels by monochromatic paths and (k, l)-kernels.Most of the existing results about kernels and their generalizations in digraphs were related to operations in digraphs and how the kernels are preserved.For results concerning kernels which were obtained quite recently see [4,6,7,5,8,9,10,11].
In [17] A. Włoch introduced and studied the concept of a 2-dominating kernel (for convenience we will write shortly is independent and 2-dominating, i.e.J is independent and each vertex from V (G) \ J has at least two neighbours in J.
Not every graph possesses a (2 − d)-kernel, for example a graph P 4 is a graph without (2 − d)-kernel.In [1] it was proved that the problem of the existence of (2 − d)-kernel is N P-complete for a general graph.
Some results related to the existence of (2 − d)-kernels in graphs can be found in [1], [2] and [17].Moreover, in [1] the number of all (2 − d)-kernels (denoted by σ (2−d) (G)) in graphs was studied.In this paper we consider the problem of the existence of (2−d)-kernels in the cartesian product of graphs and also we consider the number of (2 − d)-kernels in this graph product.

Main results.
In this section we give some necessary and sufficient conditions for the existence of (2 − d)-kernels in the cartesian product of graphs.
To prove that J is independent let us assume that (x i , y j ), (x p , y q ) ∈ J and consider the following cases:

then by the definition of H and the cartesian product
G × H we have that y j , y q ∈ V 3 , so y j y q / ∈ E(H) and consequently (x i , y j )(x p , y q ) / ∈ E(G × H).If j = q, then we prove analogously with respect to the graph G. Let i = p and j = q.Then the definition of G × H immediately gives that Then we prove in the same way as in 1.1.
Since G and H are bipartite, so x i x p / ∈ E(G) and y j y q / ∈ E(H).Then by the definition of G × H we have that Consequently J is independent.Now we shall prove that J is 2-dominating.Assume that (x i , y j ) / ∈ J.It suffices to show that there are two vertices from the set J which dominate the vertex (x i , y j ).We consider the following possibilities: Because H is connected and bipartite so there is a vertex, say y q ∈ V 3 such that y j y q ∈ E(H).Consequently by the definition of G × H we have that there is Analogously with respect to graph G we can show that there exists a vertex Then we prove by using the same method as in 2.1.Finally, the set J is a (2 − d)-kernel of G × H.In the same way we can prove that J * is a (2 − d)-kernel of G × H and it is obvious that J ∩ J * = ∅, which ends the proof.
Using Theorems 1.1 and 2.1, we can prove the result for the cartesian product of n graphs.
From the above follow results for special bipartite graphs.
and a ij = 0 otherwise.For the explanation if n = 3, then Let n ≥ 5 be odd.Then the matrix A has the form Conversely suppose on the contrary that C n × C m has a (2 − d)-kernel, say J where m = n and n is odd.
Let m be even and n be odd.Then the graph C m has two disjoint = 1, a contradiction with the independence of J.
Let now m = n and m, n be odd.Since m = n, so without loss of the generality let m > n.Suppose that 2 , for all i = 1, . . ., n and J i is a subset of (2 − d)-kernel J. Using given earlier construction of (2−d)-kernel J in C m ×C n which preserve 2-domination we observe that in copy C (n) m for every maximal independent set J n the set n i=1 J i is not independent, which is a contradiction with the assumption.
Corollary 2.6 follows by the proof of Theorem 2.5.
Suppose on the contrary that a graph K n × K m has a (2 − d)-kernel J * and n > m.Clearly in each copy of a complete graph K n and K m we can choose at most one vertex to the set J * .Without loss of the generality let (x 1 , y 1 ) ∈ J * .Then in the copy K (2) m we choose an arbitrary vertex (x 2 , y i ) where i = 1 and (x 2 , y i ) ∈ J * .Analogously in the copy K (3) m we choose a vertex (x 3 , y j ) ∈ J * for j = 1 and j = i.Consequently in the copy K (i) n+1 for every vertex (x n+1 , y p ), p = 1, . . ., m there is a vertex (x r , y p ) where 1 ≤ r ≤ m.Then J * ∪ {(x n+1 , y p ), 1 ≤ p ≤ n} is not independent, which is a contradiction with the assumption that J * is a (2 − d)-kernel.
Thus the theorem is proved.
Then it is clear that in the copy K n we can choose the vertex belonging to J on n ways.Moreover, in the copy K (p) n , 2 ≤ p ≤ n we can choose the vertex belonging to J on (n−p+1) ways.This gives that σ the set J 1 × V 1 .Analogously, considering set V 2 , we can show that every vertex from J 2 × R is at least 2-dominated by the set J 2 × V 2 .All this together gives that J is a (2 − d)-kernel of G × H.In the same way we can prove that J * = J 1 × V 2 ∪ J 2 × V 1 is a (2 − d)-kernel of G × H, which ends the proof.

Conclusion and further study.
The problem of finding the characterization of the cartesian product G × H with a (2 − d)-kernel is still open.However, some sufficient conditions can be found if we add a restriction that a graph G is fixed.

Theorem 2 . 1 .
If G and H are connected bipartite graphs, then G × H has two disjoint (2 − d)-kernels.

Theorem 2 . 5 .
Let n, m ≥ 3 be integers.A graph C n × C m has a (2 − d)kernel if and only if n and m are even or n = m.Proof.Let n, m be as in the statement of the theorem.If n and m are even, then C n and C m are bipartite and C n × C m has a (2 − d)-kernel by Theorem 2.1.Assume that n = m and n is odd.We will prove that C n × C n has a (2−d)-kernel J. Let V (C n ) = {x 1 , . . ., x n }, n ≥ 3 with the numbering of vertices in the natural fashion.We can illustrate the construction of the set J in C n × C n using the matrix A = [a ij ] n×n defined as follows If a graph G is (X, Y )-(2 − d)-kernel critical and (Y, X)-(2 − d)-kernel critical, then we will write that G is an {X, Y }-(2 − d)-kernel critical.