Abstract
A set of vertices of a graph is a dominating set of if every vertex in is adjacent to some vertex in . A minimum dominating set in a graph is a dominating set of minimum cardinality. The cardinality of a minimum dominating set is called the domination number of and is denoted by . Let and be disjoint graphs, be a subgraph of , be a subgraph of , and be an isomorphism from to . The amalgamation (the glued graph) of and at and with respect to is the graph obtained by forming the disjoint union of and and then identifying and with respect to . In this paper, we determine the domination numbers of the amalgamations of two cycles at connected subgraphs.
1. Introduction
Studying on several graph parameters is an interesting topic in graph theory. The domination number is one of the most importance parameter which was introduced from 1958 by Berge [1], called as “coefficient of external stability.” In 1962, Ore [2] studied the same concept and used the name “dominating set” and “domination number” for a graph. In 1977, Cockayane and Hedetneimi [3] gave a survey of the results about dominating sets and used the notation for the domination number of a graph. In 1998, a text devoted to this subject was introduced by Haynes et al. [4]. Over 2000 articles on graph domination numbers have been studied extensively (see, for example, [2, 3, 5–16]), in particular, the study of the domination number of product graphs such as the Cartesian product of two cycles [12], the cross product of two paths [8], and the lexicographic product of two graphs [16]. It is natural to investigate the domination number of the amalgamation of two graphs, especially, the domination number of the amalgamation of two paths or two cycles.
Let and be disjoint graphs and and such that . Let be an isomorphism from to . The amalgamation (the glued graph) of and at and with respect to is the graph obtained by forming the disjoint union of and and then identifying and with respect to . Equivalently, is the graph such that
Note that if and , then and . As an example, Figure 1 illustrates the amalgamation with respect to the isomorphism defined by , and . The amalgamation or the glued graph of two graphs was defined from 2003 by Uiyyasathian [17] for solving the maximal-clique partition problem. In 2006, Promsakon and Uiyyasathian [18] gave an upper bound of the chromatic number of glued graphs in terms of the chromatic numbers of their original graphs. Here, we are interested in finding the domination number of an amalgamation of two cycles at connected subgraphs.
2. Basic Definitions and Results
A graph is a subgraph of a graph if and . In this case, we write , and we say that contains . When but , we write and call a proper subgraph of . For a vertex of a graph , a neighbor of is a vertex adjacent to in . The neighborhood (or open neighborhood) of is the set of neighbors of . The closed neighborhood is defined as . A vertex in a graph is said to dominate itself and each of its neighbors, that is, dominates the vertices in its closed neighborhood. Therefore, dominates vertices. For a set of vertices of a graph , the closed neighborhood is defined as . A set of vertices of a graph is said to dominate the vertices in . A set of vertices of a graph is a dominating set of if every vertex of is dominated by some vertex in , i.e., every vertex in is adjacent to some vertex in . A minimum dominating set in a graph is a dominating set of minimum cardinality. The cardinality of a minimum dominating set is called the domination number of , and is denoted by . A dominating set of a graph with minimum cardinality is called a set of .
Since the cardinality of the vertex set of a graph is finite, the number of dominating sets of with minimum cardinality is finite too. This gives, for a given graph of order , the domination number can have a value from the following range: . In particular, if and only if , where is the maximum degree of . Let denote a path of order such that and . Let denote a cycle of order () such that and , where is the addition modulo . It is easy to obtain that , where is the least integer greater than or equal to .
A graph is isomorphic to a graph if there is a bijection such that if and only if . If such a function exists, it is called an isomorphism from to and written by . A graph automorphism is simply an isomorphism from a graph to itself. Let denote the set of all isomorphisms from a graph to a graph and denote the set of all automorphisms on a graph . It is easy to see that if is a connected subgraph of , then either or . Moreover, such that and for all . Let be a connected proper subgraph of and a connected proper subgraph of such that , , , and . It follows that if , then and are paths of order for some such that and , for some and . We thus get such that and , for all . Figures 2 and 3 illustrate and , respectively.
It is easily seen that . Moreover, if and , then for all and . This implies the following lemma.
Lemma 1. Let , , , and be connected such that . Then for all and .
The next lemma gives the domination number of for the case .
Lemma 2. If , then and .
Proof. Let and . Since , it follows easily that and . We thus get .
We now turn to the case . So, , for some . Assume, without loss of generality, that , i.e., . For simplicity of notation, we write instead of . Based on the result of Lemma 1, from now on, we can assume that and , and the isomorphism is defined by , for all . So, . Now, we consider with .
Lemma 3. If , then .
Proof. Suppose that . Then, (see Figure 4(a)).
Let . We check at once that is a dominating set of and . This gives . Let be a set of . Thus, since otherwise is not a set of . In order to dominate the vertices in , then must contain at least vertices. This gives . It follows that .
Now, we consider with and define sets and as follows: Note that and .
(a)
(b)
(c)
Lemma 4. Let . Then,
Proof. Suppose that . Then, (see Figure 5(a)).
If , then is the graph obtained from by joining and with a new edge. It follows easily that . For , we consider two cases.
(a)
(b)
Lemma 5. If , then
Proof. Suppose that . It is easy to check that where and (see Figures 4(b) and 5(a)). By Lemma 4, the result holds.
Next, we will give the domination number of such that (). We define three sets , , and as follows: . . .Note that if , for some , then .
Lemma 6. If and , then .
Proof. Suppose that . Then, (see Figure 4(c)). Note that if , for some , then the set defined above is a dominating set of , so . We now prove that . Let be a set of . If , then, in order to dominate vertices , there are at least vertices of in . Thus, . For , we consider two cases.
Case 1. . In this case, we consider two subcases. Case 1.1. . Consider the vertices in . We see that there are at least vertices in . This gives . Case 1.2. . Consider the vertices in . We see at once that there are at least vertices in . It follows that . Case 2. . In order to dominate , one vertex of must be in . Case 2.1. . Consider the vertices in . In order to dominate and , there are at least vertices in . It follows that . Case 2.2. . If , then this case is the same as Case 2.1. If , then, in order to dominate in , contains at least vertices. This gives . Case 2.3. . By the same argument as in Case 2.2, we also get .Lemma 7. If , then .
Proof. Suppose that . It is easy to check that , where and (see Figures 4(c) and 5(b)). By Lemma 6, the result holds.
By Lemma 3–7, we know the domination number of , for all such that . We also know the domination number of , for all such that . We now consider the case and .
For , let us denote by a graph with and . Figure 6 illustrates .
Lemma 8. If , for some and , then where , , and .
Proof. Suppose that , for some and . Let , , and . Define byIt is easy to check that is an isomorphism. Then, .
3. Domination Numbers of Amalgamations of Cycles at Connected Subgraphs
In this section, we calculate the domination number . Then, by Lemma 8, we thus get the domination number for the case and . The following two lemmas provide upper bounds of .
Lemma 9. Let and . Then, is a dominating set of and .
Proof. Let . If , for some , then there is such that . If , for some , then there is such that . If , for some , such that and , then and thus . It follows that there is such that . If , for some , such that and , then and thus . Hence, there is such that . Therefore, is a dominating set of . It is easily seen that .
Lemma 10. Let and . If and , for some , then is a dominating set of and .
Proof. Suppose that and , for some . Without loss of generality, we can assume that . We will prove that is a dominating set of . There are four cases for and . We give the proof only for the case and ; the proofs of the other cases are similar. Let . If , for some , then there is such that . If , for some such that , then and so . Thus, there is such that . If , for some such that and , then , and thus, . Consequently, there is such that . If , then there is such that . Therefore, is a dominating set of . It is easy to check that .
The following lemmas about set will be used in Theorem 1 to determine lower bounds of .
Lemma 11. Let such that , for all and , for some , and let be a set of . Then, the following hold:(1)(2)If , then for some such that (3)If , then for some such that
Proof. Without loss of generality, we can assume that .(1)There are four cases for and . We give the proof only for the case and ; the proofs of the other cases are similar. On the contrary, suppose that . This gives , and thus, every vertex in is dominated by . Obviously, (see Figure 7). In order to dominate the vertices in , then must contain at least vertices. This gives . From Lemma 10, we know that contains a dominating set of order . This contradicts the fact that is a set of .(2)Let . By (1), . Suppose, by contrary, , for all such that . In order to dominate , then there exists , for some such that and . There are two cases for . We give the proof only for the case ; the proof of the case is similar. Since , every vertex in is dominated by . It is clear that (see Figure 8). In order to dominate the vertices in , then must contain at least vertices. We thus get . From Lemma 10, we know that contains a dominating set of order . This contradicts the fact that is a set of .(3)The proof is similar to that for 2.
Lemma 12. Let and let be a set of and . Then, the following hold:(1)If , then either , , or (2)If and , then (3)If and , then (4)If and , then
Proof. We give the proofs only for (1) and (2); the proofs of (3) and (4) are similar to that of (2).(1)Let . Suppose, by contrary, that neither , nor . So, in order to dominate 1 and , we get for some such that . There are nine cases for and . We give the proof only for the case and ; the proofs of the other cases are similar. Since , every vertex in is dominated by . We check at once that (see Figure 9). In order to dominate the vertices in , then must contain at least vertices. This gives . From Lemma 10, we know that contains a dominating set of order . This contradicts the fact that is a set of .(2)Let and . On the contrary, suppose that . In order to dominate 1 and , then , for some such that or . If , then the proof is similar to that for 1. Let . There are two cases for and . We give the proof only for the case ; the proof of the case is similar. Since , every vertex in is dominated by . If , then (see Figure 10). In order to dominate the vertices in , then must contain at least vertices. We thus get . If , then . It is easily seen that . In order to dominate the vertices in , then must contain at least vertices. This gives . From Lemma 10, we know that contains a dominating set of order . This contradicts the fact that is a set of .
Lemma 13. Let and let be a set of . Then, if one of the following holds:(1) and (2) and (3) and
Proof. We give the proof only for the case and ; the proofs of the other cases are similar. Suppose, by contrary, that . By Lemma 11 (1), . We thus get . So, in order to dominate 1 and , we get , for some . There are two cases: and . Following in a same manner as the proof of Lemma 12 (2), we can obtain , which contradicts the fact that is a set of .
Theorem 1. Let , , and . Then,
Case 1. , for some or . By Lemma 9, . We next prove that . Let be a set of . Case 1.1. for some . Without loss of generality, we can assume that . In order to dominate 1 and , there are sixteen cases, , for some and . We give the proofs only for the four cases , , , and ; the proofs of the other cases are similar. Case 1.1.1. . Then, every vertex in is dominated by . Thus, , so . Case 1.1.2. . It follows that every vertex in is dominated by . If , then it is easy to check that . This gives . For , . It follows that . Since , . Case 1.1.3. . Then, every vertex in is dominated by . We can see that if , and otherwise. It follows that if , and otherwise. Since , . Case 1.1.4. . We thus get every vertex in is dominated by . Since , . Case 1.2. . In order to dominate 1 and , there are sixteen cases as above. In the same manner, we can prove that . Case 2. and , for some . By Lemma 10, . We next show that . Without loss of generality, we can assume that . Let be a set of . By Lemma 11 (1), . It follows that or . Consider the following four cases. Case 2.1. . If , then or . We give the proof only for the case ; the proof of the case is similar. By Lemma 11 (2), , , or . Here, we will give the proof only for the case ; the proofs of the other two cases are similar. Hence, . As in the proof of Case 1.1.2, . If , then , , or by Lemma 12 (1). We give the proof only for the case ; the proofs of the other two cases are similar. As in the proof of Case 1.1.3, we get that if , then . If , then . It follows that . Case 2.2. and . If , then or . We give the proof only for the case ; the proof of the case is similar. By Lemma 11 (2), or . Here, we will give the proof only for the case ; the proof of the case is similar. Hence, . As in the proof of Case 2.1, . If , then by Lemma 12 (4). Thus, every vertex in is dominated by . It is clear that (see Figure 11). In order to dominate the vertices in , then must contain at least vertices. This gives . Case 2.3. and . The proof is similar to that of Case 2.2. Case 2.4. . By Lemma 13, . So, or . We give the proof only for the case ; the proof of the case is similar. By Lemma 11 (2), . Following in a same manner as the proof of Case 2.1, we can obtain .
Summarizing, we get the domination number of amalgamations of cycles at connected subgraphs . From Lemmas 2‐7 and Theorem 1, we then get the following theorem.
Theorem 2. Let , , , and where , , and . Then
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This research was supported by Chiang Mai University and Faculty of Science, Chiang Mai University, Thailand.