Hence, the vertex ui+4+2k is just not 2-dominated, a / contradiction.+k
Therefore, the vertex ui+4+2k just isn’t 2-dominated, a / contradiction.+k+2kFigure 8. The case when ui , ui+2 , ui+4 , ui+7 J (the second subcase).2.three. vi+1 J and vi+3 J for some i (see Figure 9). / Then, vi+3+k J and vi+1+k , vi+5+k , vi+6+k , ui+3+k J. Considering that vi+4 have to be 2-dominated, / vi+4-k J or vi+4+k J. With out loss of PK 11195 manufacturer generality, assume that vi+4+k J. Therefore, ui+4+k J. Moreover, ui+2+k , ui+5+k J, which causes ui+1+k , ui+6+k , vi+2+k J. To / / 2-dominate ui+1+k , we need to have ui+k J. Then, vi+k J and vi+2k , vi+1+2k , vi+2+2k J. / From the independence from the set J, we get that ui+2k , ui+1+2k , ui+2+2k J. Therefore, the / vertex ui+1+2k just isn’t 2-dominated, a contradiction.+k+2kFigure 9. The case when ui , ui+2 , ui+4 , ui+7 J (the third subcase).2.4. vi+1 , vi+3 J for some i. Proving analogously as in subcase two.three., we obtain a contradiction using the assumption that J is usually a (2-d)-kernel. Therefore, for each n and k, it is not feasible that the vertices ui , ui+2 , ui+4 , ui+7 belong to a (2-d)-kernel of P(n, k). Therefore, for the graph together with the (2-d)-kernel, the configurations of P1 , P2 shown in the Figure 10 would be the only ones that might be probable. Now, we will show that they’re certainly feasible.ui ui+1 ui+2 ui+3 ui+ui ui+1 ui+2 ui+3 ui+4 ui+Figure 10. Probable configurations on the paths P1 , P2 for the graph P(n, k) with all the (2-d)-kernel.Suppose that ui , ui+2 , ui+4 J for some i, as in Figure 11. Then, ui+1 , ui+3 , vi , vi+2 , three. vi+4 J. /Figure 11. The case when ui , ui+2 , ui+4 J.We look at 4 subcases. three.1. vi+1 , vi+3 J for some i (see Figure 12). / Given that vi+2 has to be 2-dominated, we obtain that vi+2+k J or vi+2-k J. Without having loss of generality, assume that vi+2+k J. Furthermore, vi+1+k , vi+3+k J and ui+1+k , ui+2+k , ui+3+k / J. Therefore, the vertex ui+2+k is just not 2-dominated, a contradiction.Symmetry 2021, 13,six of+kFigure 12. The case when ui , ui+2 , ui+4 J (the very first subcase).3.2. vi+1 J and vi+3 J for some i (see Figure 13). / Then, vi+1+k J and vi+3+k J. Because vi+2 have to be 2-dominated, vi+2+k J or / vi+2-k J. With no loss of generality, assume that vi+2+k J. Hence, ui+1+k , ui+2+k J / and ui+k , ui+3+k J, which causes vi+k , ui+4+k J and vi+4+k J. Moreover, vi+1+2k , / vi+2+2k , vi+4+2k J, vi+2k J, ui+2k J, ui+1+2k J, ui+2+2k J, ui+3+2k J and / / / ui+4+2k , vi+3+2k J. Lastly, vi+2+3k , vi+3+3k , vi+4+3k J and ui+2+3k , ui+3+3k , ui+4+3k J. / / Therefore, the vertex ui+3+3k will not be 2-dominated, a contradiction.+k+2k+3kFigure 13. The case when ui , ui+2 , ui+4 J (the second subcase).3.three. vi+1 J and vi+3 J for some i. / Proving analogously as in subcase three.two., we GYY4137 Autophagy acquire a contradiction together with the assumption that J is really a (2-d)-kernel. 3.four. vi+1 , vi+3 J for some i (see Figure 14). Then, vi+1+k , vi+3+k J. Very first, we are going to show that vi+k and vi-k need to belong to a (2-d)/ kernel J. Suppose on contrary that vi+k J. Considering that vi+k must be 2-dominated, ui+k J. / Hence, ui+1+k J and ui+2+k J. Additionally, vi+2k , vi+1+2k , vi+2+2k J and ui+2k , ui+1+2k , / ui+2+2k J. Therefore, the vertex ui+1+2k is not 2-dominated, a contradiction.+k+2kFigure 14. The case when ui , ui+2 , ui+4 J (the fourth subcase).This suggests that vi+k , vi-k J and also vi+2+k , vi+4+k , ui+1+k , ui+3+k belong to a (2-d)kernel (see Figure 15).+kFigure 15. The case when ui , ui+2 , ui+4 J implies that vi+k , vi+2+k , vi+4+k , ui+1+k , ui+3+k J.Therefore, n should be even, and from the definition of P(n, k), we conclude that k must be odd, whic.