Such that V (Cn ) = x1 , x2 , . . . , xn , c c c V
Such that V (Cn ) = x1 , x2 , . . . , xn , c c c V (Cn ) = x1 , x2 , . . . , xn together with the numbering of vertices within the organic order. Let G (n) be the graph such that V ( G (n)) = V (Cn ) V (Cn ) and E( G (n)) = E(Cn ) E(Cn ) xi xic ; i 1, 2, . . . , n. Figure 21 shows an example of a (2-d)-Nitrocefin Description kernel in G (13). It can be uncomplicated to check that if n = five, then G (five) is isomorphic for the Petersen graph.G(13)Figure 21. An instance of a (2-d)-kernel in G (13).The following Theorem shows a full characterization of graphs G (n) with all the (2-d)kernel. Theorem two. Let n five be integer. The graph G (n) has a (2-d)-kernel if and only if n is odd.c c Proof. Let n 5 be odd. We’ll show that J = x2 , x3 , x1 , x4 , x6 , . . . , xn-1 will be the (2-d)kernel from the graph G (n). The independence of J is apparent. It truly is adequate to show that JSymmetry 2021, 13,9 ofis a 2-dominating set. By the definition in the graph G (n), we are able to assume that xn+1 = x1 . Suppose that y V ( G (n)) \ J. Hence, y V (Cn ) or y V (Cn ). Let y V (Cn ). Thus c / y = xk , k 2, 3, 5, . . . , n. If xk J, then there exist vertices xk-1 , xk+1 J adjacent to xk . c J, then k = 2 or k = 3. For k = two, the vertex x is adjacent to x , x c J. Furthermore, if If xk two 1 two c k = three, then the vertex x3 is adjacent to x4 , x3 J. Hence, every single vertex from the set V (Cn ) c is 2-dominated by the set J. Let now y V (Cn ). Thus y = xk , k 1, 4, 5, . . . , n. Then, c , k 5, 6, . . . , n is adjacent to x c , x c J. If k = 1, then x c x , x c x c E ( G ( n )). the vertex xk 2 three 1 1 1 three c c In addition, for k = 4 the vertex x4 is adjacent to x4 , x2 . As a result, vertices in the set V (Cn ) are 2-dominated by J and hence J can be a (2-d)-kernel of G (n). Conversely, suppose that a graph G (n) features a (2-d)-kernel J. We are going to show that n is odd. By the definition on the graph G (n), we receive that J V (Cn ) = . Otherwise, c c vertices in the set V (Cn ) are usually not 2-dominated by the set J. Let x1 J. Then either x2 J c c c or xn J. Otherwise, x2 or xn will not be 2-dominated. Therefore, | J V (Cn )| = 2. Without having loss of c , x c J. This signifies that x c , i 4, 5, . . . , n – 1 is 2-dominated generality assume that x1 two i c c c c by J and x3 , xn are dominated by J. Let J = J \ x1 , x2 . Then, J V (C ). Considering that J is ; otherwise, x c , x c usually are not 2-dominated by J. Hence, the the (2-d)-kernel, x3 , xn J 3 n graph x3 , x4 , . . . , xn G(n) Pn-2 ought to have a (2-d)-kernel to 2-dominate vertices from = V (Cn ) \ J . This indicates that n must be odd. Hence, J = x3 , x5 , . . . , xn , which ends the proof. Lastly, it turns out that if a graph G (n) has (2-d)-kernel, then the amount of (2-d)kernels depends linearly around the quantity of vertices. In addition, each and every (2-d)-kernel of G (n) has the exact same BI-0115 custom synthesis cardinality. Corollary five. If n 5 is odd, then ( G (n)) = n and (2-d) ( G (n)) = (2-d) ( G (n)) = n + two.Proof. Let n five be odd. In the building of a (2-d)-kernel described inside the proof of Theorem two, we conclude that exactly two not adjacent vertices from the set V (Cn ) V ( G (n)) belong to a (2-d)-kernel. The collection of these two vertices will establish the (2-d)-kernel in G (n). Considering the fact that two not adjacent vertices is often selected on n strategies, ( G (n)) = n. Moreover, in the building of (2-d)-kernels in G (n), it follows that all of them have the identical cardinality. Therefore, (2-d) ( G (n)) = (2-d) ( G (n)) = n + 2, which ends the 2 proof. three. Concluding Remarks In this paper, we thought of two differe.
