Abstract
Let be a simple graph. A subset is a dominating set if every vertex not in is adjacent to a vertex in . The domination number of , denoted by , is the smallest cardinality of a dominating set of . The domination subdivision number of is the minimum number of edges that must be subdivided (each edge can be subdivided at most once) in order to increase the domination number. In 2000, Haynes et al. showed that for any edge with and where is a connected graph with order no less than 3. In this paper, we improve the above bound to , and furthermore, we show the decision problem for determining whether is NP-hard. Moreover, we show some bounds or exact values for domination subdivision numbers of some graphs.
1. Introduction
For terminology and notation on the graph theory not given here, the reader is referred to Xu [1]. Let be a finite, undirected, and simple graph, where is the vertex set and is the edge set of . For a vertex , let be the open set of neighbors of and be the closed set of neighbors of . The cardinality of is called the order of . The degree of vertex is the cardinality of . The maximum degree and minimum degree of are denoted and , respectively. If for graph , then is called a -regular graph. For any edge , we denote as a new graph by subdividing the edge in . For any edge , we may view as a two vertex set .
A subset is a dominating set of if every vertex in has at least one neighbor in . The domination number of , denoted by , is the minimum cardinality among all dominating sets of . A dominating set is called a minimum dominating set of if . Domination is an important and classic notion that has become one of the most widely researched topics in graph theory and is also used to study the property of networks frequently. A thorough study of domination appears in the books [2, 3] by Haynes, Hedetniemi, and Slater. Among various problems related to the domination number, some focus on graph alterations and their effects on the domination number. As for different applications, there are also many variated dominations, such as Italian domination [4], 2-rainbow domination [5], research on Zagreb indices by domination [6].
The domination subdivision number of a graph , denoted by , equals the minimum number of edges that must be subdivided in order to obtain a graph for which . Since the domination number of graph does not change when its only edge is subdivided, we must assume here that the graph is of order no less than 3. Domination subdivision number of graph has been widely studied, see [7–9] for examples.
For a graph parameter, knowing whether or not there exists a polynomial-time algorithm to compute its exact value is the essential problem. If the decision problem corresponding to the computation of this parameter is NP-hard or NP-complete, then polynomial-time algorithms for this parameter do not exist unless . The problem of determining the domination number has been proven NP-complete for chordal bipartite graphs [10]. There are many other results on complexity for variations of domination; these results can be found in the two books [3, 11] and the survey [12].
Many famous networks are bipartite graphs, such as hypercube graphs, partial cube, grid graphs, and median graphs. If we know the decision problem for the domination subdivision problem is NP-hard, then the studies on the domination subdivision number are more meaningful. So we should be concerned about the algorithmic complexity of the domination subdivision problem in bipartite graphs. In this paper, we will show that the decision problem for the domination subdivision number is NP-hard even for bipartite graphs. In other words, there are no polynomial-time algorithms to compute these parameters unless .
2. Preliminary Results
In the book [13], Garey and Johnson provide three steps to prove a decision problem to be NP-hard. We applied these three steps to prove that our decision problem is NP-hard. Our proof involved a polynomial transformation from the well-known NP-complete problem, the 3-satisfiability problem. In this section, we will recall some terms related to the 3-satisfiability problem.(i) is a set of Boolean variables.(ii)A truth assignment for is a mapping . If , then is considered “true” under ; if , then is considered “false” under .(iii) and are literals over when is a variable in . The literal (resp. ) is true under if and only if the variable is true (resp. false) under .(iv)A clause over is a set of literals over . It represents the disjunction of these literals and it is satisfied by a truth assignment if and only if at least one of its elements is true under .(v)A collection of clauses over is satisfiable if and only if there exists a truth assignment for that simultaneously satisfies all the clauses in . Such a truth assignment is called a satisfying truth assignment for .
The 3-satisfiability problem is defined as finding a satisfying truth assignment for a collection of clauses over . 3-satisfiability problem (3SAT): Instance: A collectionof clauses over a finite setof variables such thatfor. Question: Is there a truth assignment forthat satisfies all the clauses in?
Theorem 1 (Theorem 3.1 in [13]). The 3-satisfiability problem is NP-complete.
A dominating set is called an efficient dominating set of graph if for every vertex . An efficient dominating set of a graph is always a minimum dominating set [14, 15].
Lemma 2 (Berge [16]). For any graph ,
Lemma 3 (Huang and Xu [17]). Let be a -regular graph. Then,with equality if and only if has an efficient dominating set. In addition, if has an efficient dominating set, then every efficient dominating set must be a minimum dominating set, and vice versa.
3. Bounds
Let be a simple graph. Let and . The private neighborhood of with respect to is defined as the set
For any edge, and , if , then we denote .
Theorem 4. Let be an edge in . Subdivide by a new vertex . Then, iff , , and if for any minimum dominating set of .
Proof. Assume . Suppose to the contrary that . Let be a minimum dominating set of . Then, and (if , then is also a dominating set of contradicts ). Let . Then, is a dominating set of with cardinality , which is a contradiction with . Hence,Let be a minimum dominating set of . If or and , then is also a minimum dominating set of . So and if for any minimum dominating set of .
Assume , , and if for any minimum dominating set of . Suppose to the contrary that . Let be a minimum dominating set of . Then, . If , then we can assume without loss of generality that , and hence, is also a minimum dominating set of , a contradiction with for any minimum dominating set of . Thus, . If , then is a minimum dominating set of , which is a contradiction with . If or , then is a minimum dominating set of and assume without loss of generality . Note that and is also a minimum dominating set of since . So , a contradiction with .
Theorem 5. For any connected graph of order , and for any two adjacent vertices and , where and ,
Proof. Letand letwhere and . Letand let be the graph that results from subdividing all edges in . We will show . Let the subdivided vertex of be , and the subdivided vertex of is and the subdivided vertex of is for each and . Let be a minimum dominating set of . Clearly, and we can assume if . Let and , and .
Assume . Then, is a dominating set of with cardinality . Thus, . Next, assume . We consider the following three cases. Case 1: . Then, is a dominating set of with cardinality . Thus, . Case 2: . If , then is a dominating set of with cardinality no more than . Thus, . Suppose . Then, should be dominated by some vertex in . Therefore, is a dominating set of with cardinality . Thus, . Case 3: . Then, should be dominated by some vertex in which implies . Then, is a dominating set of with cardinality no more than . Thus, .Note that . So
Corollary 6 (Haynes et al. [18]). For any connected graph and edge , where and ,
Proposition 7. Let be an edge in and be the inserted vertex in . If belongs to every -set of and , then .
Proof. Since belongs to every -set of , . Then, is a dominating set of . Since , .
Proposition 8. Let be a -regular graph and it has an efficient dominating set. Then, .
Proof. By Lemma 3, . Let be a graph by subdividing any edge of . Since is -regular and where , . By Lemma 2,Hence, which implies that .
An efficient dominating set is also known as perfect codes in coding theory. There are many classical graphs that have efficient dominating sets, such as cycle where , star graph, and pancake graph [19], some Circulant graphs, and Harary graphs [17], some Möbius ladders [20]. The domination subdivision numbers of these graphs are 1.
Proposition 9. Let be a graph and let be a support vertex that has at least two leaves. Then, .
Proof. Let where is a leaf and let be another leaf corresponding to . Let be the graph from subdividing the edge and let be a minimum dominating set of . Note that and . Since , we can without loss of generality assume . Then, is a dominating set of with cardinality at most . Therefore,which implies that .
Proposition 10. Let be a graph and be two adjacent support vertices. Then, .
Proof. Let where and are both support vertices. Let and be two leaves corresponding to and , respectively. Let be the graph from subdividing three edges of , where the inserted vertices are , and , respectively. Let be a minimum dominating set of . To dominate in , we need , , . Note thatis a dominating set of . We havewhich implies that .
Theorem 11. Let be a nonempty graph. Then, .
Proof. Denote and the corresponding vertex of is for . Clearly, . Let be a graph by subdividing any two edges of . If the two edges both belong to , then is also a dominating set of . If the two edges are both pendent edges and for some , then is also a dominating set of where and are the inserted vertices. If one of the two edges is in and the other is a pendent edge for some , then is also a dominating set of where is the inserted vertices of . Thus .
Let be an edge in . We subdivide the three edges , , and . Let the inserted vertices are , and and be the resulting graph. To dominate all the pendent vertices, we need at least vertices except in . To dominate , we need at least one vertex in . Therefore, and hence, .
4. NP-Hardness of Domination Subdivision Number
In this section, we show the algorithmic complexity of the problem for determining the domination subdivision number of a graph. We first state the problem as the following decision problem. Domination subdivision problem: Instance: A nonempty graph. Question: Is?
For proving the algorithmic complexity of domination subdivision problem is NP-hard, we follow the method introduced in [21] which is to prove the algorithmic complexity of bondage problem is NP-hard. The steps are similar but the constructed graph and details are different.
Theorem 12. The domination subdivision problem is NP-hard for bipartite graphs.
Proof. We start the proof by using 3SAT problem which is a well-known NP-complete problem by Theorem 1. Let Boolean variables set and clauses set be an arbitrary 3SAT instance where for each . To reduce the above instance of 3SAT to an instance of domination subdivision problem, we will construct a graph from the above instance, and then prove is satisfiable if and only if .
For each variable , we create a graph with vertex set and edge set . We then create a single vertex for each and add three edges . Finally, we add a path with length 2, and join and to vertex for every . Figure 1 shows an example of constructed where and , where .
Note that contains vertices and edges; this construction can be done in polynomial time. To demonstrate that this is truly a transformation, it is necessary to establish that if and only if when there exists a truth assignment for which satisfies all the clauses in .
Assume be a minimum dominating set of . Note that , and for every . Hence,Suppose that . Then, , for every , and for all . Because should be dominated by , for every . Since all vertices in can only be dominated by and , .
For any edge , we claim that . If for some and the inserted vertex is , thenis a dominating set of with cardinality . If , thenis a dominating set of with cardinality . If for some and or (assume without loss of generality ), thenis a dominating set of with cardinality .
We then claim that if and only if is satisfiable. Assume . Let be a minimum dominating set of . Define a function byBy the above discussions, for every . Hence, the definition of is well-defined. Recall that . For any clause where , there exists some integer with such that should be dominated by or , without loss of generality we say . This implies by (19), which deduces that the clause is satisfied. Therefore, is satisfiable. Conversely, assume that is satisfiable by , where be a satisfying truth assignment for . We create a subset as follows. Put (resp. ) to when is true (resp. flase) under . Because is a satisfying truth assignment for , at least one literal in is true under for each which implies . Hence is a dominating set of with cardinality . By (15), , and then .
Finally, we claim if and only if . Suppose . We subdivide and let the inserted vertex be . By contradiction, assume and be a minimum dominating set of . By the similar discussions as above, for every . Since is a path in , at least 2 vertices of should be in . So , and then . Suppose . Let be an edge with . By the above discussions, . Hence, . Therefore, .
By the above discussions, we see that if and only if there exists a truth assignment for which satisfies all the clauses of . We complete the proof.
5. Domination Subdivision Number for Some Cartesian Product Graphs
Let and be two undirected graphs. The Cartesian product of and is an undirected graph, denoted by , where , two distinct vertices and , where and , are linked by an edge in if and only if either and , or and . Throughout this paper, the notation and denote a path with vertex set . For integers and , the Cartesian product of with order and with order is that has vertex set
Let and where and are called the layers of and , respectively. Figure 2 is the Cartesian product of and .
If we want to compute the domination subdivision number of a graph, we need to know the exact value of its domination number. We start with the following classical results.
Theorem 13 (Jacobson and Kinch [22]). For ,
Theorem 14 (Nandi et al. [23]). For ,
Lemma 15. For and ,
Proof. Let and . By Theorem 14, . Note that is a 3-regular graph. Since has an efficient dominating set by Lemma 3. By Proposition 8, .
Lemma 16. For and ,
Proof. Let and . Let and let be a minimum dominating set of where the inserted vertex in is . Suppose . Then can dominate the vertices in . Note that . By Theorem 13,Hence,In the following, assume .
Suppose . Then, can dominate the vertices in . Note that . By Theorem 14,Hence,Suppose finally , and suppose without loss of generality . Then, or (say ) must be in to dominate . Note that any vertex of from dominates three vertices of including itself and any vertex of from dominates one vertex of in . Since and dominate a common vertex in ,Since cannot dominate in ,By summing (30) and (31), we haveIf the equality holds in (32), then the equalities hold in (30) and (31) which implies contradicts is an integer. HenceBy Theorem 14, . So which implies that .
Lemma 17. For and ,
Proof. Let and . By Theorem 14, . Let or . Thenwith cardinality is a dominating set of . Since there are only two types of edge in ,for any edge in . Hence, .
Let be the graph that results from subdividing two edges and , by adding subdivision vertices and . Let be a minimum dominating set of . Suppose . Then, can dominate the vertices in . Note that . By Theorem 13,Hence,Suppose . Then, can dominate the vertices in . Note that . By Theorem 13,Hence,In the following, assume .
Suppose . Then can dominate the vertices in . Note that . By Theorem 13,Hence,Suppose . Then, can dominate the vertices in . Note that . By Theorem 13,Hence,Suppose finally , and suppose without loss of generality . Then or must be in to dominate . Assume in to dominate the vertex and not in . Since should be dominated by , or (say since it is more efficient). Note that any vertex of from dominates three vertices of including itself and any vertex of from dominates at most one vertex of in . Since , , and dominate a common vertex in ,Since and cannot dominate vertices in corresponding to ,By summing (45) and (46), we haveAssume in dominates the vertex and not in . If , then is also a minimum dominating set of , and this case has been solved in the second paragraph. If , then is also a minimum dominating set of , and this case also has been solved in the second paragraph. The remaining case is and . Since and should be dominated by in , and must be in . Note that any vertex of from dominates three vertices of including itself and any vertex of from dominates at most one vertex of in . Since and dominate a common vertex in , and cannot dominate , we haveSince cannot dominate vertices in corresponding to and dominates which is also dominated by in ,By summing (48) and (49), we haveBy Theorem 14, . By the above discussions, which implies that .
Lemma 18. For and ,
Proof. By Theorem 14, . We first show . Assume and where . Let be the graph that results from subdividing the two edges and . If , thenwith cardinality is a dominating set of . If , thenwith cardinality is a dominating set of . By the above discussion, we can easily check that the above constructed set of is also a dominating set of when and . If and be the subdivision vertex where , thenwith cardinality is a dominating set of . If and be the subdivision vertex where , thenwith cardinality is a dominating set of .
Assume and where and . Let be the graph that results from subdividing the two edges and . If , thenwith cardinality is a dominating set of . If , thenwith cardinality is a dominating set of . If and be the subdivision vertex for , thenwith cardinality is a dominating set of . If and be the subdivision vertex for , thenwith cardinality is a dominating set of . By symmetry and the above discussions, we see that where is the graph that results from subdividing any two edges of . Hence, .
Let be the graph that results from subdividing three edges , , and , by adding subdivision vertices , , and . Let be a minimum dominating set of . In the following we prove by Theorem 14. We can easily check for . Assume below. Suppose . Then, can dominate which is isomorphic to . Hencethe last but one inequality comes from Theorem 13. In the following, assume (cannot be smaller than 3 since should be dominated by 3 distinct vertices in ).
Assume and we say for example. For dominating the vertex , either or in . Note that any vertex of from or the corresponding subdivision vertices dominates three vertices of including itself and any vertex of from dominate at most one vertex of in . Since cannot dominate vertices in ,Since and have a common neighbor , have no neighbors in .By summing (61) and (62), we haveAssume . Suppose and . Note that and should be dominated by , and also and should be dominated by . Since , and must be in . Then can dominate which is isomorphic to . Hencethe last but one inequality comes from Theorem 13. Similar discussions for and .
By the above discussions, by Theorem 14, which implies that . We complete the proof of this Lemma.
Combing Lemmas 15, 16, 17, and 18, we show the exact value for domination subdivision number of .
Theorem 19. For ,
6. Conclusions
In this paper, we prove the decision problem for determining whether is NP-hard. This result is fundamental for a mathematical parameter. This result shows it is meaningful to study the bounds and other properties of the domination subdivision number of graphs. Since it is difficult to determine whether for any edge in graph , it is obvious that the domination subdivision problem is not a NP problem. So we only prove the decision problem for determining whether is NP-hard but not NP-complete. Moreover, we show a better tight bound for domination subdivision number of connected graphs.
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.
Acknowledgments
This research was supported by Anhui Provincial Natural Science Foundation (Grant no. 2108085MA02) and University Natural Science Research Project of Anhui Province (Grant nos. KJ2020A0001 and 2023AH050060).