Displaying similar documents to “On the Domination of Cartesian Product of Directed Cycles: Results for Certain Equivalence Classes of Lengths”

Two Graphs with a Common Edge

Lidia Badura (2014)

Discussiones Mathematicae Graph Theory

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Let G = G1 ∪ G2 be the sum of two simple graphs G1,G2 having a common edge or G = G1 ∪ e1 ∪ e2 ∪ G2 be the sum of two simple disjoint graphs G1,G2 connected by two edges e1 and e2 which form a cycle C4 inside G. We give a method of computing the determinant det A(G) of the adjacency matrix of G by reducing the calculation of the determinant to certain subgraphs of G1 and G2. To show the scope and effectiveness of our method we give some examples

On iteration digraph and zero-divisor graph of the ring ℤ n

Tengxia Ju, Meiyun Wu (2014)

Czechoslovak Mathematical Journal

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In the first part, we assign to each positive integer n a digraph Γ ( n , 5 ) , whose set of vertices consists of elements of the ring ℤ n = { 0 , 1 , ⋯ , n - 1 } with the addition and the multiplication operations modulo n , and for which there is a directed edge from a to b if and only if a 5 ≡ b ( mod n ) . Associated with Γ ( n , 5 ) are two disjoint subdigraphs: Γ 1 ( n , 5 ) and Γ 2 ( n , 5 ) whose union is Γ ( n , 5 ) . The vertices of Γ 1 ( n , 5 ) are coprime to n , and the vertices of Γ 2 ( n , 5 ) are not coprime to n . In this part, we study the structure of Γ ( n , 5 ) in detail. In the second part, we investigate...

On non-z(mod k) dominating sets

Yair Caro, Michael S. Jacobson (2003)

Discussiones Mathematicae Graph Theory

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For a graph G, a positive integer k, k ≥ 2, and a non-negative integer with z < k and z ≠ 1, a subset D of the vertex set V(G) is said to be a non-z (mod k) dominating set if D is a dominating set and for all x ∈ V(G), |N[x]∩D| ≢ z (mod k).For the case k = 2 and z = 0, it has been shown that these sets exist for all graphs. The problem for k ≥ 3 is unknown (the existence for even values of k and z = 0 follows from the k = 2 case.) It is the purpose of this paper to show that for k...