Abstract
This paper studies the maximum-clique independence problem and some variations of the clique transversal problem such as the -clique, maximum-clique, minus clique, signed clique, and k-fold clique transversal problems from algorithmic aspects for k-trees, suns, planar graphs, doubly chordal graphs, clique perfect graphs, total graphs, split graphs, line graphs, and dually chordal graphs. We give equations to compute the -clique, minus clique, signed clique, and k-fold clique transversal numbers for suns, and show that the -clique transversal problem is polynomial-time solvable for graphs whose clique transversal numbers equal their clique independence numbers. We also show the relationship between the signed and generalization clique problems and present NP-completeness results for the considered problems on k-trees with unbounded k, planar graphs, doubly chordal graphs, total graphs, split graphs, line graphs, and dually chordal graphs.
1. Introduction
Every graph in this paper is finite, undirected, connected, and has at most one edge between any two vertices in G. We assume that the vertex set V and edge set E of G contain n vertices and m edges. They can also be denoted by and . A graph is an induced subgraph of G denoted by if and contains all the edge for . Two vertices are adjacent or neighbors if . The sets and are the neighborhood and closed neighborhood of a vertex x in G, respectively. The number is the degree of x in G. If for every , then G is k-regular. Particularly, cubic graphs are an alternative name for 3-regular graphs.
A subset S of V is a clique if for . Let Q be a clique of G. If for any other clique of G, then Q is a maximal clique. We use to represent the set is a maximal clique of . A clique is a maximum clique if for every . The number is the clique number of G. A set is a clique transversal set (abbreviated as CTS) of G if for every . The number is a CTS of is the clique transversal number of G. The clique transversal problem (abbreviated as CTP) is to find a minimum CTS for a graph. A set is a clique independent set (abbreviated as CIS) of G if or and for . The number is a CIS of is the clique independence number of G. The clique independence problem (abbreviated as CIP) is to find a maximum CIS for a graph.
The CTP and the CIP have been widely studied. Some studies on the CTP and the CIP consider imposing some additional constraints on CTS or CIS, such as the maximum-clique independence problem (abbreviated as MCIP), the k-fold clique transversal problem (abbreviated as k-FCTP), and the maximum-clique transversal problem (abbreviated as MCTP).
Definition 1
([1,2]). Suppose that is fixed and G is a graph. A set is a k-fold clique transversal set (abbreviated as k-FCTS) of G if for . The number is a k-FCTS of is the k-fold clique transversal number of G. The k-FCTP is to find a minimum k-FCTS for a graph.
Definition 2
([3,4]). Suppose that G is a graph. A set is a maximum-clique transversal set (abbreviated as MCTS) of G if for with . The number is an MCTS of is the maximum-clique transversal number of G. The MCTP is to find a minimum MCTS for a graph. A set is a maximum-clique independent set (abbreviated as MCIS) of G if for and for . The number is an MCIS of is the maximum-clique independence number of G. The MCIP is to find a maximum MCIS for a graph.
The k-FCTP on balanced graphs can be solved in polynomial time [2]. The MCTP has been studied in [3] for several well-known graph classes and the MCIP is polynomial-time solvable for any graph H with [4]. Assume that and is a function. Let for , and let be the weight of f. A CTS of G can be expressed as a function f whose domain is and range is , and for . Then, f is a clique transversal function (abbreviated as CTF) of G and is a CTF of . Several types of CTF have been studied [4,5,6,7]. The following are examples of CTFs.
Definition 3.
Suppose that is fixed and G is a graph. A function f is a -clique transversal function (abbreviated as -CTF) of G if the domain and range of f are and , respectively, and for . The number is a -CTF of is the -clique transversal number of G. The -clique transversal problem (abbreviated as -CTP) is to find a minimum-weight -CTF for a graph.
Definition 4.
Suppose that G is a graph. A function f is a signed clique transversal function (abbreviated as SCTF) of G if the domain and range of f are and , respectively, and for . If the domain and range of f are and , respectively, and for , then f is a minus clique transversal function (abbreviated as MCTF) of G. The number is an SCTF of is the signed clique transversal number of G. The minus clique transversal number of G is is an MCTF of . The signed clique transversal problem (abbreviated as SCTP) is to find a minimum-weight SCTF for a graph. The minus clique transversal problem (abbreviated as MCTP) is to find a minimum-weight MCTF for a graph.
Lee [4] introduced some variations of the k-FCTP, the -CTP, the SCTP, and the MCTP, but those variations are dedicated to maximum cliques in a graph. The MCTP on chordal graphs is NP-complete, while the MCTP on block graphs is linear-time solvable [7]. The MCTP and SCTP are linear-time solvable for any strongly chordal graph G if a strong elimination ordering of G is given [5]. The SCTP is NP-complete for doubly chordal graphs [6] and planar graphs [5].
According to what we have described above, there are very few algorithmic results regarding the k-FCTP, the -CTP, the SCTP, and the MCTP on graphs. This motivates us to study the complexities of the k-FCTP, the -CTP, the SCTP, and the MCTP. This paper also studies the MCTP and MCIP for some graphs and investigates the relationships between different dominating functions and CTFs.
Definition 5.
Suppose that is fixed and G is a graph. A set is a k-tuple dominating set (abbreviated as k-TDS) of G if for . The number is a k-TDS of is the k-tuple domination number of G. The k-tuple domination problem (abbreviated as k-TDP) is to find a minimum k-TDS for a graph.
Notice that a dominating set of a graph G is a 1-TDS. The domination number of G is .
Definition 6.
Suppose that is fixed and G is a graph. A function f is a -dominating function (abbreviated as -DF) of G if the domain and range of f are and , respectively, and for . The number is a -DF of is the -domination number of G. The -domination problem (abbreviated as -DP) is to find a minimum-weight -DF for a graph.
Definition 7.
Suppose that G is a graph. A function f is a signed dominating function (abbreviated as SDF) of G if the domain and range of f are and , respectively, and for . If the domain and range of f are and , respectively, and for , then f is a minus dominating function (abbreviated as MDF) of G. The number is an SDF of is the signed domination number of G. The minus domination number of G is is an MDF of . The signed domination problem (abbreviated as SDP) is to find a minimum-weight SDF for a graph. The minus domination problem (abbreviated as MDP) is to find a minimum-weight MDF for a graph.
Our main contributions are as follows.
- 1.
- 2.
- We prove in Section 3 that for any graph G with . Then, is polynomial-time solvable if can be computed in polynomial time. We also prove that the SCTP is a special case of the generalized clique transversal problem [8]. Therefore, the SCTP for a graph H can be solved in polynomial time if the generalized transversal problem for H is polynomial-time solvable.
- 3.
- We show in Section 4 that and for any split graph G. Furthermore, we introduce -split graphs and prove that and for any -split graph H. We prove the NP-completeness of SCTP for split graphs by showing that the SDP on -split graphs is NP-complete.
- 4.
- We show in Section 5 that for a doubly chordal graph G can be computed in linear time, but the k-FCTP is NP-complete for doubly chordal graphs as . Notice that the CTP is a special case of the k-FCTP and the -CTP when , and thus for any graph G.
- 5.
2. Suns
In this section, we give equations to compute , , , and for any sun G and show that , , , and .
Let and G be a graph. An edge is a chord if e connects two non-consecutive vertices of a cycle in G. If C has a chord for every cycle C consisting of more than three vertices, G is a chordal graph. A sun G is a chordal graph whose vertices can be partitioned into and such that (1) W is an independent set, (2) the vertices of U form a cycle, and (3) every is adjacent to precisely two vertices and , where (mod p). We use to denote a sun. Then, . If p is odd, is an odd sun; otherwise, it is an even sun. Figure 1 shows two suns.
Figure 1.
(a) The sun . (b) A sun .
Lemma 1.
For any sun , and .
Proof.
It is straightforward to see that U is a minimum 2-FCTS and is a minimum 3-FCTS of . This lemma therefore holds. □
Lemma 2.
Suppose that and . Then, for any sun .
Proof.
Let such that (mod p). Since every is adjacent to precisely two vertices , is a maximal clique of . By contradiction, we can prove that there exists a minimum -CTF f of such that for . Since for , we have
Since is a nonnegative integer, .
We define a function by for every , for with odd index i and for every with even index i. Clearly, a maximal clique Q of is either the closed neighborhood of some vertex in W or a set of at least three vertices in U. If for some , then . Suppose that Q is a set of at least three vertices in U. Since , . Therefore, h is a -CTF of . We show the weight of h is by considering two cases as follows.
Case 1: p is even. We have
Case 2: p is odd. We have
Following what we have discussed above, we know that h is a minimum -CTF of and thus . □
Lemma 3.
For any sun ,
Proof.
For , is a maximal clique of . Let h be a minimum SCTF of . Then, . Note that for . We have
Since , we have . Let f be an SCTF of such that and for . The weight of f is 0. Then f is a minimum SCTF of . Hence, and . The proof for is analogous to that for . □
Theorem 1
(Lee and Chang [9]). Let be a sun. Then,
- (1)
- ;
- (2)
- ;
- (3)
- , and .
Corollary 1.
Letbe a sun. Then,
- (1)
- ;
- (2)
- for ;
- (3)
- and .
Proof.
The corollary holds by Lemmas 1–3 and Corollary 1. □
3. Clique Perfect Graphs
Let be the set of all induced subgraphs of G. If for every , then G is clique perfect. In this section, we study the -CTP for clique perfect graphs and the SCTP for balanced graphs.
Lemma 4.
Let G be such a graph that . Then, .
Proof.
Assume that D is a minimum CTS of G. Then, . Let and let f be a function whose domain is and range is , and if ; otherwise, . Clearly, f is a -CTF of G. We have .
Assume that f is a minimum-weight -CTF of G. Then, and for every . Let be a maximum CIS of G. We know that and . Therefore, . Following what we have discussed above, we know that . □
Theorem 2.
If a graph G is clique perfect, .
Proof.
Since G is clique perfect, . Hence, the theorem holds by Lemma 4. □
Corollary 2.
The -CTP is polynomial-time solvable for distance-hereditary graphs, balanced graphs, strongly chordal graphs, comparability graphs, and chordal graphs without odd suns.
Proof.
Distance-hereditary graphs, balanced graphs, strongly chordal graphs, comparability graphs, and chordal graphs without odd suns are clique perfect, and the CTP can be solved in polynomial time for them [10,11,12,13,14]. The corollary therefore holds. □
Definition 8.
Suppose that R is a function whose domain is and range is . If for every , then R is a clique-size restricted function (abbreviated as CSRF) of G. A set is an R-clique transversal set (abbreviated as R-CTS) of G if R is a CSRF of G and for every . Let is an R-CTS of . The generalized clique transversal problem (abbreviated as GCTP) is to find a minimum R-CTS for a graph G with a CSRF R.
Lemma 5.
Let G be a graph with a CSRF R. If for every , then .
Proof.
Assume that D is a minimum R-CTS of G. Then, . Let and let f be a function of G whose domain is and range is , and if ; otherwise, . Since for every , there are at least vertices in C with the function value 1. Therefore, for every , and f is an SCTF of G. Then, .
Assume that h is a minimum-weight SCTF of G. Clearly, . Since for every , C contains at least vertices with the function value 1. Let . The set D is an R-CTS of G. Therefore, . Hence, we have . □
Theorem 3.
The SCTP on balanced graphs can be solved in polynomial time.
Proof.
Suppose that a graph G has n vertices and ℓ maximal cliques . Let and . Let M be an matrix such that an element of M is one if the maximal clique contains the vertex , and otherwise. We call M the clique matrix of G. If the clique matrix M of G does not contain a square submatrix of odd order with exactly two ones per row and column, then M is a balanced matrix and G is a balanced graph. We formulae the GCTP on a balanced graph G with a CSRF R as the following integer programming problem:
where is a column vector and is a column vector such that is either 0 or 1. Since the matrix M is balanced, an optimal 0–1 solution of the integer programming problem above can be found in polynomial time by the results in [15]. By Lemma 5, we know that the SCTP on balanced graphs can be solved in polynomial time. □
4. Split Graphs
Let G be such a graph that and . If I is an independent set and C is a clique, G is a split graph. Then, every maximal of G is either C itself, or the closed neighborhood of a vertex . We use to represent a split graph. The -CTP, the k-FCTP, the SCTP, and the MCTP for split graphs are considered in this section. We also consider the -DP, the k-TDP, the SDP, and the MDP for split graphs.
For split graphs, the -DP, the k-TDP, and the MDP are NP-complete [16,17,18], but the complexity of the SDP is still unknown. In the following, we examine the relationships between the -CTP and the -DP, the k-FCTP and the k-TDP, the SCTP and the SDP, and the MCTP and the MDP. Then, by the relationships, we prove the NP-completeness of the SDP, the -CTP, the k-FCTP, the SCTP, and the MCTP for split graphs. We first consider the -CTP and the k-FCTP and show in Theorems 4 and 5 that and for any split graph G. Chordal graphs form a superclass of split graphs [19]. The cardinality of is at most n for any chordal graph G [20]. The following lemma therefore holds trivially.
Lemma 6.
The k-FCTP, the -CTP, the SCTP, and the MCTP for chordal graphs are in NP.
Theorem 4.
Suppose that and is a split graph. Then, .
Proof.
Let S be a minimum k-FCTS of G. Consider a vertex . By the structure of G, is a maximal clique of G. Then, . We now consider a vertex . If , then there exists a vertex such that . Clearly, and . If , then . Hence, S is a k-TDS of G. We have .
Let D be a minimum k-TDS of G. Recall that the closed neighborhood of every vertex in I is a maximal clique. Then, D contains at least k vertices in the maximal clique for every vertex . If , D is clearly a k-FCTS of G. Suppose that . We consider three cases as follows.
Case 1: . Then, . The set D is a k-FCTS of G.
Case 2: and . Then, the set is still a minimum k-TDS and . The set is a k-FCTS of G.
Case 3: and for every . Then, . Since , there exists such that . If , then and . Otherwise, let . Then, and . The set D is a k-FCTS of G.
By the discussion of the three cases, we have . Hence, we obtain that and . The theorem holds for split graphs. □
Theorem 5.
Suppose that and is a split graph. Then, .
Proof.
We can verify by contradiction that G has a minimum-weight -CTF f and a minimum-weight -DF g of G such that and for every . By the structure of G, for every . Then, and . Since and , and .
For every , and . For every , . Therefore, the function f is also a -DF of G. We have . We now consider for the clique C. If , the function g is clearly a -CTF of G. Suppose that . Notice that g is a -DF and for every . Then, for any vertex . Therefore, g is also a -CTF of G. We have . Following what we have discussed above, we know that . □
Corollary 3.
The -CTP and the k-FCTP are NP-complete for split graphs.
Proof.
The corollary holds by Theorems 4 and 5 and the NP-completeness of the -DP and the k-TDP for split graphs [16,18]. □
A graph G is a complete if . Let G be a complete graph and let . The vertex set is the union of the sets and . Clearly, is an independent set and is a clique of G. Therefore, complete graphs are split graphs. It is easy to verify the Lemma 7.
Lemma 7.
If G is a complete graph and, then
- (1)
- for ;
- (2)
- ;
- (3)
- ;
- (4)
For split graphs, however, the signed and minus domination numbers are not necessarily equal to the signed and minus clique transversal numbers, respectively. Figure 2 shows a split graph G with . However, . We therefore introduce -split graphs and show in Theorem 6 that their signed and minus domination numbers are equal to the signed and minus clique transversal numbers, respectively. -split graphs are motivated by the graphs in [17] for proving the NP-completeness of the MDP on split graphs. Figure 3 shows an -split graph.
Figure 2.
A split graph G with .
Figure 3.
A split graph G with one of its partitions indicated.
Definition 9.
Suppose thatis a split graph withvertices. Let U, S, X, and Y be pairwise disjoint subsets ofsuch that,,, and. The graph G is an-split graph ifand G entirely satisfies the following three conditions.
- (1)
- and .
- (2)
- for .
- (3)
- and for .
Theorem 6.
For any -split graph , and .
Proof.
We first prove . Let be an -split graph. As stated in Definition 9, I can be partitioned into and , and C can be partitioned into and . Assume that f is a minimum-weight SDF of G. For each , and is adjacent to only the vertex . Then, for . Since and , we know that . Notice that and for every . Therefore, f is also an SCTF of G. We have .
Assume that h is a minimum-weight SCTF of G. For each , and is adjacent to only the vertex . Then, for . Consider the vertices in I. Since for every , . We now consider the vertices in C. Recall that . Let . Since and , we know that . Let . Then, . Therefore, h is also an SDF of G. We have .
Following what we have discussed above, we have . The proof for is analogous to that for . Hence, the theorem holds for any -split graphs. □
Theorem 7.
The SDP on -split graphs is NP-complete.
Proof.
We reduce the (3,2)-hitting set problem to the SDP on -split graphs. Let and let such that and for . A (3,2)-hitting set for the instance is a subset of U such that for . The (3,2)-hitting set problem is to find a minimum (3,2)-hitting set for any instance . The (3,2)-hitting set problem is NP-complete [17].
Consider an instance of the (3,2)-hitting set problem. Let , , and . We construct an -split graph by the following steps.
- (1)
- Let be an independent set and let be a clique.
- (2)
- For each vertex , a vertex is connected to if .
- (3)
- For , the vertex is connected to the vertex .
- (4)
- For , the vertex is connected to the vertex .
Let be the minimum cardinality of a (3,2)-hitting set for the instance . Assume that is a minimum (3,2)-hitting set for the instance . Then, . Let f be a function whose domain is and range is , and if and if . Clearly, f is an SDF of G. We have .
Assume that f is minimum-weight SDF of G. For each , and is adjacent to only the vertex . Then, for . For any vertex , no matter what values the function f assigns to the vertices in U or in S. Consider the vertices in S. By the construction of G, and for . There are at least three vertices in with the function value 1. If , then there exists an SDF g of G such that and for every . Then, . It contradicts the assumption that the weight of f is minimum. Therefore, there exists a minimum-weight SDF h of G such that for . Notice that for . There are at least two vertices in with the function value 1. Then, the set is a (3,2)-hitting set for the instance . We have .
Following what we have discussed above, we know that . Hence, the SDP on -split graphs is NP-complete. □
Corollary 4.
The SCTP and the MCTP on split graphs are NP-complete.
Proof.
The corollary holds by Theorems 6 and 7 and the NP-completeness of the MDP on split graphs [17]. □
5. Doubly Chordal and Dually Chordal Graphs
Assume that G is a graph with n vertices . Let and let be the subgraph . For every , let . In , if there exists a vertex such that for every , then the ordering is a maximum neighborhood ordering (abbreviated as MNO) of G. A graph G is dually chordal [21] if and only if G has an MNO. It takes linear time to compute an MNO for any dually chordal graph [22]. A graph G is a doubly chordal graph if G is both chordal and dually chordal [23]. Lemma 8 shows that a dually chordal graph is not necessarily a chordal graph or a clique perfect graph. Notice that the number of maximal cliques in a chordal graph is at most n [20], but the number of maximal cliques in a dually chordal graph can be exponential [24].
Lemma 8.
For any dually graph G, , but G is not necessarily clique perfect or chordal.
Proof.
Brandstädt et al. [25] showed that the CTP is a particular case of the clique r-domination problem and the CIP is a particular case of the clique r-packing problem. They also showed that the minimum cardinality of a clique r-dominating set of a dually chordal graph G is equal to the maximum cardinality of a clique r-packing set of G. Therefore, .
Assume that H is a graph obtained by connecting every vertex of a cycle of four vertices to a vertex . Clearly, the ordering is an MNO and thus H is a dually chordal graph. The cycle is an induced subgraph of H and does not have a chord. Moreover, , but and . Hence, a dually chordal graph is not necessarily clique perfect or chordal. □
Theorem 8.
Suppose that and . The k-FCTP on doubly chordal graphs is NP-complete.
Proof.
Suppose that G is a chordal graph. Let H be a graph such that and . Clearly, H is a doubly chordal graph and we can construct H from G in linear time.
Assume that S is a minimum -FCTS of G. By the construction of H, each maximal clique of H contains the vertex x. Therefore, is a k-FCTS of H. Then .
By contradiction, we can verify that there exists a minimum k-FCTS D of H such that . Let . Clearly, S is a -FCTS of G. Then . Following what we have discussed above, we have . Notice that and the CTP on chordal graphs is NP-complete [14]. Hence, the k-FCTP on doubly chordal graphs is NP-complete for doubly chordal graphs. □
Theorem 9.
For any doubly chordal graph G, can be computed in linear time.
Proof.
The clique r-dominating problem on doubly chordal graphs can be solved in linear time [25]. The CTP is a particular case of the clique r-domination problem. Therefore, the CTP on doubly chordal graphs can also be solved in linear time. By Lemmas 4 and 8, the theorem holds. □
6. k-Trees
Assume that G is a graph with n vertices . Let and let be the subgraph . For every , let . If is a clique for , then the ordering is a perfect elimination ordering (abbreviated as PEO) of G. A graph G is chordal if and only if G has a PEO [26]. A chordal graph G is a k-tree if and only if either G is a complete graph of k vertices or G has more than k vertices and there exists a PEO such that is a clique of k vertices if ; otherwise, is a clique of vertices for . Figure 4 shows a 2-tree with the PEO .
Figure 4.
A 2-tree H.
In [3], Chang et al. showed that the MCTP is NP-complete for k-trees with unbounded k by proving for any k-tree G. However, Figure 4 shows a counterexample that disproves for any k-tree G. The graph H in Figure 4 is a 2-tree with the perfect elimination ordering . The set is the minimum dominating set of H and the set is a minimum MCTS of H. A modified NP-completeness proof is therefore desired for the MCTP on k-tree with unbounded k.
Theorem 10.
The MCTP and the MCIP are NP-complete for k-trees with unbounded k.
Proof.
The CTP and the CIP are NP-complete for k-trees with unbounded k [8]. Since every maximal clique in a k-tree is also a maximum clique [27], an MCTS is a CTS and an MCIS is a CIS. Hence, the MCTP and the MCIP are NP-complete for k-trees with unbounded k. □
Theorem 11.
The SCTP is NP-complete for k-trees with unbounded k.
Proof.
Suppose that and G is a -tree with . Let . Since G is a -tree, for .
Let Q be a clique with vertices. Let H be a graph such that and . Let be a clique for . Clearly, . Let . Then, H is a -tree and for . Clearly, we can verify that there exists a minimum-weight SCTF h of H of such that for every . Then, contains at least one vertex x with for . Let and . Then, S is a CTS of G. Since , we have .
Assume that D is a minimum CTS of G. Let f be a function of H whose domain is and range is , and (1) for every , (2) for every , and (3) for every . Each maximal clique of H has at least vertices with the function value 1. Therefore, f is an SCTF. We have . Following what we have discussed above, we know that . The theorem therefore holds by the NP-completeness of the CTP for k-trees [8]. □
Theorem 12.
Suppose that the κ-FCTP is NP-complete on k-trees with unbounded k.
Proof.
Assume that and G is a -tree with . Let H be a graph such that and . Clearly, H is a -tree and we can construct H in linear time. Following the argument analogous to the proof of Theorem 8, we have . The theorem therefore holds by the NP-completeness of the CTP for k-trees [8]. □
Theorem 13.
The SCTP and κ-FCTP problems can be solved in linear-time for k-trees with fixed k.
Proof.
Assume that and G is a graph. The -FCTP is the GCTP with the CSRF R whose domain is and range is . By Lemma 5, can be obtained from the solution to the GCTP on a graph G with a particular CSRF R. Since the GCTP is linear-time solvable for k-trees with fixed k [8], the SCTP and -FCTP are also linear-time solvable for k-trees with fixed k. □
7. Planar, Total, and Line Graphs
In a graph, a vertex x and an edge e are incident to each other if e connects x to another vertex. Two edges are adjacent if they share a vertex in common. Let G and H be graphs such that each vertex corresponds to an edge and two vertices are adjacent in H if and only if their corresponding edges and are adjacent in G. Then, H is the line graph of G and denoted by . Let be a graph such that and two vertices are adjacent in H if and only if x and y are adjacent or incident to each other in G. Then, is the total graph of G and denoted by .
Lemma 9
([28]). The following statements hold for any triangle-free graph G.
- (1)
- Every maximal clique of is the set of edges of G incident to some vertex of G.
- (1)
- Two maximal cliques in intersect if and only if their corresponding vertices (in G) are adjacent in G.
Theorem 14.
The MCIP is NP-complete for any 4-regular planar graph G with the clique number 3.
Proof.
Since for any planar graph G [29], the MCIP on planar graphs is in NP. Let be the class of triangle-free, 3-connected, cubic planar graphs. The independent set problem remains NP-complete even when restricted to the graph class [30]. We reduce this NP-complete problem to the MCIP for 4-regular planar graphs with the clique number 3 as follows.
Let and . Clearly, we can construct H in polynomial time. By Lemma 9, we know that H is a 4-regular planar graph with and each maximal clique is a triangle in H.
Assume that is an independent set of G of maximum cardinality. Since , for every . Let have the vertex in common for . Then, form a triangle in H. Let be the triangle formed by in H for . For each pair of vertices , is not adjacent to in G. Therefore, and in H do not intersect. The set is an MCIS of H. We have .
Assume that is a maximum MCIS of H. Then, each is a triangle in H. Let be formed by in H for . Then, are incident to the same vertex in G. For , let have the vertex in common. For each pair of , and do not intersect. Therefore, is not adjacent to in G. The set is an independent set of G. We have .
Hence, . For , we know that if and only if . □
Corollary 5.
The MCIP is NP-complete for line graphs of triangle-free, 3-connected, cubic planar graphs.
Proof.
The corollary holds by the reduction of Theorem 14. □
Theorem 15.
The MCIP problem is NP-complete for total graphs of triangle-free, 3-connected, cubic planar graphs.
Proof.
Since for a planar graph G, the MCIP on planar graphs is in NP. Let be the classes of traingle-free, 3-connected, cubic planar graphs. The independent set problem remains NP-complete even when restricted to the graph class [30]. We reduce this NP-complete problem to MCIP for for total graphs of triangle-free, 3-connected, cubic planar graphs. as follows
Let and . Clearly, we can construct H in polynomial time. By Lemma 9, we can verify that H is a 6-regular graph with .
Assume that is an independent set of G of maximum cardinality. Since , for every . Let have the vertex in common. Then, form a maximum clique in H. Let be the maximum clique formed by in H for . For each pair of vertices , is not adjacent to in G. Therefore, and in H do not intersect. The set is an MCIS of H. We have .
Assume that is a maximum MCIS of H. By the construction of H, each is formed by three edge-vertices in and their common end vertex in . Let and in H such that is formed by for . For each pair of , and do not intersect. Therefore, is not adjacent to in G. The set is an independent set of G. We have .
Hence, . For , we know that if and only if . □
Funding
This research is supported by a Taiwanese grant under Grant No. NSC-97-2218-E-130-002-MY2.
Acknowledgments
We are grateful to the anonymous referees for their valuable comments and suggestions to improve the presentation of this paper.
Conflicts of Interest
The author declares no conflict of interest
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