Abstract
The conditional matching preclusion number of a graph , denoted by , is the minimum number of edges whose deletion results in the graph with no isolated vertices that has neither perfect matching nor almost-perfect matching. In this paper, we first give some sharp upper and lower bounds of conditional matching preclusion number. Next, the graphs with large and small conditional matching preclusion numbers are characterized, respectively. In the end, we investigate some extremal problems on conditional matching preclusion number.
1. Introduction
All graphs are undirected, finite, and simple in this paper, refer to the book [1] for notation and terminology not described here. If a cycle contains every vertex of exactly once, then we called it is a Hamiltonian cycle of . A connected graph is Hamiltonian if it exists a Hamiltonian cycle in . Furthermore, if there exists a Hamiltonian path between any two vertices of , then is said Hamiltonian connected. Denoted to be a edges set has one endpoint in and another in . If an edge subset satisfies has neither perfect matching nor almost-perfect matching, then is a matching preclusion set (MP for short) of . Denoted is the minimum number of edges of all MP set in . The concept of matching preclusion was introduced in [2] and further studied in [3–18]. Some distributed algorithms require each vertex of the system to be matched by a neighbour vertex, and the matching preclusion number measures the robustness of a graph as a communications network topology. Meanwhile, matching preclusion number has a theoretical connection with conditional connectivity and “changing and unchanging of invariants.” In a network, a vertex with a special matching vertex after edge failure any time implies that tasks running on a fault vertex can be change into its matching vertex. Therefore, under this fault assumption, larger mp (G) signifies higher fault tolerance. However, in the network, the probability that the adjacent vertices of the same vertex fail at the same time is very small. So the question is if we delete edges, what are the basic obstructions to a perfect matching or an almost-perfect matching in the resulting graph if no isolated vertices are created. This motivates our next definition. If an edge subset satisfies has no isolated vertices and has neither perfect matching nor almost-perfect matching, then is a conditional matching preclusion set (CMP for short). Denoted is the minimum number of all CMP set. At present, there have been some discussions about the conditional matching preclusion number of special graphs. We mainly want to discuss the conditional matching preclusion number of general graphs. We consider the following three problems in this paper. Problem 1. Compute the minimum integer , where be the set of all graphs of order with conditional matching preclusion number Problem 2. Compute the minimum integer such that for every connected graph of order , if , then Problem 3. Compute the maximum integer such that for every connected graph of order , if , then
A basic obstruction to a perfect matching of even graph with no isolated vertex will be the existence of a path where the degrees of and are 1. Define : and are distance 2 apart}, if and are adjacent, then , and 0 otherwise.
Proposition 1 (see [7]). An even graph with , then .
The basic obstruction to an almost-perfect matching of odd graph with no isolated vertex will be the existence of a three vertices of degree 1 with a common neighbour. Define has no isolated vertices and has 3 leaves adjacent to the same vertex}.
Proposition 2 (see [7]). An odd graph has .
For odd vertices bipartite graph , we have has no isolated vertices and has 2 leaves adjacent to the same vertex in }.
Proposition 3 (see [7]). Let be a bipartite graph with an odd number of vertices. Then, .
If is a spanning subgraph of , we want to know whether we have holds.
Remark 1. For an even number , let be a graph obtained from a and a clique by adding three edges , where . Let be a graph obtained from by deleting three edges of . It is clear that is a spanning subgraph of . By deleting three edges in , we can see that there is no perfect matching in the resulting graph; hence, . From the definition of , for any , to avoid isolated vertices in , we have and , so .
Remark 2. For an odd number , let be a graph obtained from a clique with vertex set and a clique by adding four edges between them, where . Let be a graph obtained from by deleting six edges of the . By deleting and three edges in , we can find there no perfect matching in the resulting graph; hence, . For any , to avoid isolated vertices in , we have and , so .
2. Sharp Bounds for CMP Number
Lemma 1 (see [2]). Let . Then, , is super matched.
Lemma 2. (see [19]). Let . Then, , is conditionally super matched.
Theorem 1. (see [19])(1)Let be even, then(2)Let be odd, then
Proposition 4. Let be a graph with an even number of vertices and . Then, .
Proof. Let be the minimum-degree vertex in . Since , it exists a path in . Let , then exists a path and . By Proposition 1, .
By Lemma 2, , , we get for .
Proposition 5. Let be a graph with an odd number of vertices and . Then, .
Proof. Let be the minimum-degree vertex of and . Since , there exist two vertices , and two edges . Let , then has no isolated vertices and has 3 leaves adjacent to the same vertex . By Proposition 2, .
3. Graphs with Given CMP Numbers
Theorem 2 (see [20]). For a graph with order , if , then is Hamiltonian.
Lemma 3 (see [19]). A graph has a perfect matching if and only if satisfies Tutte’s condition, that is, .
Lemma 4 (see [19]). A graph has an almost-perfect matching if and only if every subset vertices satisfies Berge’s condition, that is, .
The graphs with can be characterized completely.
Proposition 6. The even graph has order of , and . Then, if and only if .
Proof. From Theorem 1, we know when is a complete graph. Conversely, if and , then there exist . If is a common neighbour vertex of , let . So there exists a path in satisfies . Since , has no isolated vertices and no perfect matching exist. Since , it means , which contradicts to . If no exist a common neighbour vertex for and , we choose a shortest path between and , say . Let , then there exists a path in and . Since , then . So .
Theorem 3. An even graph has order and . Then, if and only if and .
Proof. From Proposition 6, we only need to consider . When , we first suppose and choose a minimum-degree vertex . Since , there exists a path in . Let , then there exists a path in and . If , then , and we have . If , then , and we have . Thus, . Let , and let and . By induction on , we prove that if and only if . From Proposition 6, the result holds for . Suppose that the argument is true for every integer , that is, if and only if . Since and , then , it contradicts to . So .
Conversely, suppose . We prove that . Let . Since , it follows that there exists a path in . Let , then exists a path and . If , then , and we have . If , then , and we have . Thus, .
Now, we prove that . Suppose for every , . By Theorem 2, has a Hamiltonian cycle, so there is a perfect matching in . It exists a vertex and . Since has no isolated vertices, there has a vertex and one edge . Let , then .
If has an isolated vertex, say , then . Since has no isolated vertices, it means must adjacent to at least one of and . If , let , then . Clearly, for any vertex pairs , and then, contains a Hamiltonian cycle and a perfect matching . Furthermore, is a perfect matching of . If , then . If there is a vertex and such that , we let . Clearly, () for any vertex pairs , and then, has a Hamiltonian cycle and a perfect matching in . Furthermore, is a perfect matching of . If there exists no vertex (except ) such that , let , we have and . So , it means , which contradicts to .
If has no isolated vertices, suppose for every vertex , then . Then, contains a Hamiltonian cycle, so a perfect matching in , and then, is a perfect matching in . Suppose that there exist two vertices with such that . Let , then . Since for each vertex , has no isolated vertices. Suppose for every vertex , then . By Theorem 2, contains a Hamiltonian cycle and a matching in , so is a perfect matching in . Suppose that there exists a vertex and and . Let . . It follows that for any vertex pair , . . So contains a Hamiltonian cycle, and hence, there is a perfect matching in . Clearly, is a perfect matching of . From the above argument, we get .
Next, we characterize the odd graphs with , respectively.
Proposition 7. Let be an order and . Then, if and only if is .
Proof. Suppose , but . It exists two vertices and such that . If they have a common neighbourhood , then there is a vertex such that by . Let , then and are three vertices of degree 1 with common neighbour in . From , that has no isolated vertices, and is a basic obstruction set to an almost-perfect matching of . Thus, , and it contradicts to . If and no common neighbourhood, choose a shortest path () between and . By , there exists another vertex such that . Let , then and are three vertices of degree 1 with the common neighbour in . Thus, , which contradicts to .
Proposition 8. Let be an odd graph of order and . Then, if and only if , where .
Proof. If , . Choosing another two vertices from , we know . Let , then and are three vertices of degree 1 with a common neighbour in . It means that . For every with , is the graph which is obtained from by deleting at most edges. By Proposition 7, has an almost-perfect matching, so . By above argument, .
Conversely, it shows that when . If , then contains a path or two independent edges as its subgraph. For the former case, if have a common neighbour , let , we have and are three vertices of degree 1 in , and has no isolated vertices since . It follows that . If have no common neighbour, it means that for any vertex . Since , there exists a vertex such that ; furthermore, there are two vertices such that . Let , then and are three vertices of degree 1 with common neighbour in . From , we know has no isolated vertices. Thus, , a contradiction.
For the case of two independent edges , there are three situations to be considered. If there are three vertices in with a common neighbour , without loss of generality, they are have a common neighbour. Let , then and are three vertices of degree 1 in . If there are two vertices in with a common neighbour , then there is a vertex such that . Clearly, is connected to at most two of vertices in ; thus, . Let , so are three vertices of degree 1 with common neighbour in and . If any two vertices of have no common neighbour, it means for every vertex is connected to at most one of vertices in , so . Choose a vertex () and , then must has two other adjacent vertices . Let , then and are three vertices of degree 1 in . It is a contradiction. Together with the above argument, we can find .
Proposition 9. Let has and . Then, if and only if one of the following conditions holds.(1) and (2)
Proof. Suppose . Then, we claim . Assume, on the contrary, that . Then, there exists a vertex in such that . Choose and , since . Let , so are three vertices of degree 1 with common neighbour in and . Thus, , which contradicts . By Propositions 7 and 8 and , we have and or .
When , we need to show . Assume . It means there exist two vertices such that and for every vertex . Assume, on the contrary, that . Except , there exists two vertices , in and . If , then . Choose and ; clearly, . Let , so are three vertices of degree 1 with common neighbour in and . , So . Since , choose and . Let , so are three vertices of degree 1 with common neighbour in and . , which is a contradiction. In summary, we have , or and , as required.
Conversely, by Propositions 7 and 8, we have . We show . If , for every with , it follows that is a graph obtained from by deleting at most edges. By Proposition 7, has an almost-perfect matching, as desired. If and , then , where is a matching of of size at least 2. It suffices to prove that has an almost-perfect matching for every with such that has no isolated vertices. Suppose any vertex has , then . By Theorem 2, contains a Hamiltonian cycle. So an almost-perfect matching in , a contradiction. Now, we suppose that there exists a vertex and . Since has no isolated vertices, it follows that there exists a vertex and . Let , then .
If has no isolated vertices, suppose every vertex has . Then, . By Theorem 2, contains a Hamiltonian cycle. So an almost-perfect matching in and is an almost-perfect matching in , a contradiction.
We suppose that there exists a vertex and and . Let , then . Recall that , where is a matching of of size at least 2; thus, . It means removing up to edges from . Since , . By Proposition 7, has an almost-perfect matching , and is an almost-perfect matching of .
If has two isolated vertices and , then . Since has no isolated vertices, the vertices and must be at least adjacent to one of , respectively. Since , we can assume and . Let . Since , and . So has an almost-perfect matching , and is an almost-perfect matching of .
If has one isolated vertex , then . Let , clearly, . Suppose that every vertices have , then . By Theorem 2, contains a Hamiltonian cycle. So there is a perfect matching in , and is an almost-perfect matching in missing . We suppose that there exists a vertex and . Since has only one isolated vertex , it follows that there exists one vertex and . Let , then . Clearly, and is even, and we have . Since means from remove at most edges and (), has a perfect matching , and is an almost-perfect matching of missing , a contradiction. Thus, .
4. Extremal Problems on CMP Number
Consider the three extremal problems in the introduction.
Lemma 5. Let be two positive integers and be odd. Then,(1)(2)(3)
Proof. (1)Let be a no edges with order graph. Clearly, . So .(2)Let , where is a graph obtained from a star by adding one edge between two leaves. Clearly, and has edges. Then, . Conversely, assume that , it means exists an odd graph of order with thus . Assume that are the connected components in . If two of are odd components in , then . So there is exactly one odd component in , suppose it is . We may assume . If , then , it contradicts with . Then, and . If , then , a contradiction. So or . If , we claim . Otherwise, and . We have in , and . So or . For two cases, we know . If , we claim . Otherwise, or . If , then and . So or . For two cases, we know . If , then , which is a contradiction. So , and . Thus, or . For two cases, we know . So .(3)Let . Clearly, and has edges. Then, . Since , it follows that . Now, we prove . If we assume that , then it exists an odd graph of order with such that . So is not connected. Let be the connected components in . If two of are odd components, then . So there is exactly one odd component in , and we can assume that is odd and is even for . Assume . If , then , but . So . Now, we consider three cases of the number of . First assume , if , then , a contradiction. All even components are , and . By and , the structure of must be or add one pendant edge or add two pendant edges. For above three structure of , we have or . Next assume , if , then . Since , So and . Since has an almost-perfect matching, so must be or add one pendant edge. For any structure of , . At last, If , then . Since . So and . So only have , other even components are . If , then . Since . It follows that , so must be or add one pendant edge or add two pendant edges. For above structure of , or . If , then . So and , or and . It follows that or or . For any case, we know or . So .
Theorem 4. Let be an odd integer, . Then,(1)If is odd, then(2)If is even, then
Proof. (1)Let be a graph obtained from , , , by arbitrarily adding edges between and , and making the join graphs: and . Let . Suppose are three vertices in , the degree of in is less than or equal to the degree of other vertices in . Clearly, we delete edges between and and three edges in of , and then, are three vertices of degree 1 with the common neighbour and the rests of have no isolated vertices. And , so .(2)Let be a graph obtained from tree cliques , , . By arbitrarily adding edges between and , then make the join graphs: and . Let . The degree of three vertices in is less than or equal to the degree of other vertices in . Clearly, when we delete edges between and and three edges in , then are three vertices of degree 1 with common neighbour , and the rest of graphs have no isolated vertices. This just deletes the edges from and no almost-perfect matching in result graph., so .
Lemma 6. Let be two positive integers and be even. Then,(1)(2)(3)(4)
Proof. (1)Let be the graph of order with no edges. Clearly, . So .(2)Let . is a star of order 4 and add one edge between two of pendent vertices. Clearly, and has edges. Then, . Conversely, assume that . Then, there exists an even graph of order with edges such that . Because exists a perfect matching, so . If , then and . If , then and . So .(3)Let . Clearly, and has edges. Then, . Since , it follows . Assume that . Then, there exists an even graph of order with edges such that . Because has a perfect matching, so , now is a graph of add two edges to a perfect match of size . So has the following situations: , , , , add one pendant edge on the middle vertex), and for any of the above, we have or . So .(4)Let . Clearly, and has edges. Then, . Since , it follows that . Assume that . Then, there exists an even graph of order with edges such that . Since has a perfect matching, so is a graph of add three edges to a perfect match of size . All possible structures of are shown as follows (Figure 1). For any structure, we have or or . So .
Theorem 5. Let be two positive integers and be even and . Then,(1)If is odd, then (2)If is even, then
Proof. (1)Let be a graph obtained from and . add edges between {} and , then make the join graph: . Let . And the degree of three vertices in is less than or equal to the degree of other vertices in . Clearly, when we delete edge and edges between {} and , then are two vertices of degree 1 with common neighbour , and the rest of the graphs have no isolated vertices. This just deletes the edges from and no perfect matching in result graph. , so .(2)Let be a graph obtained from and . Arbitrarily add edges between {} and , then make the join graph: . Let . From the construction of graph , the degree of three vertices in is less than or equal to the degree of other vertices in . Clearly, when we delete edge and edges between {} and , then are two vertices of degree 1 with common neighbour , and the rest of the graphs have no isolated vertices. This just deletes the edges from and no perfect matching in result graph. , so .
Observation 1. , be two positive integers.
Corollary 1. Let be two positive integers and be odd, . Then,(1)If is odd, then (2)If is even, then
Corollary 2. Let , be two positive integers and be even. Then,(1)If is odd, then (2)If is even, then
Theorem 6. Let be two positive integers. Then,(1)If is even and , then (2)If is odd and , then
Proof. (1)First show , we construct as follows, give three components , , . edges are connected between and , adjacents to all vertices in and , and then make a join between and . Clearly, is a connected graph on vertices, , and . So . Now to show , let be a graph with vertices such . For any , , so and . Since by Theorem 1, thus has a perfect matching, and hence, . So .(2)First prove , we construct as follows: give three components . is a star graph with central vertex and three pendant vertices . , . edges are connected between and , adjacent to all vertices in and and then make a join between and . Clearly, is a connected graph on vertices, , and . So . Now to show , let be a graph with vertices such . For any , , so and . Since , tt follows that has an almost-perfect matching, and hence, . So .
5. Conclusion
The concept of matching preclusion was introduced in [2]. The matching preclusion number measures the robustness of a graph as a communications network topology. In a network, a vertex with a special matching vertex after edge failure any time implies that tasks running on a fault vertex can be change into its matching vertex. Larger signifies higher fault tolerance. However, the probability that the adjacent vertices of the same vertex fail at the same time is very small. In the paper, we mainly want to discuss the conditional matching preclusion number of general graphs. First of all, we want to discuss the bound of the conditional matching preclusion number of a general graph. It is necessary to discuss the bounds of reaching the number of conditional matches according to the parity of the number of vertices of the graph. Next, we will draw a graph from the conditional matching preclusion number. When the number of conditional matches is a special value, what property does the obtained graph satisfy. From the perspective of graph description, this is very meaningful. Finally, we discussed three extreme value problems. It is actually an extension of the two problems discussed previously, but it is more difficult to solve than the previous problems.
Data Availability
No data are used in this article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the National Natural Science Foundation of China (12201335, 12261074) and the Science Found of Qinghai Province (Nos. 2021-ZJ-703).