Abstract

A mapping is called negative if for every The maximum of the values of taken over all negative mappings , is called the modified negative decision number and is denoted by . In this paper, several sharp upper bounds of this number for a general graph are presented. Exact values of these numbers for cycles, paths, cliques and bicliques are found.

1. Introduction

Let be a graph with vertex set and edge set . For a vertex , the open neighbourhood of in is , and is its closed neighbourhood. For a graph , and a subset , we let denote the number of vertices in joined to . In particular, , the degree of in . For disjoint subsets and of vertices, we use for the set of edges between and , and let . Let denote the subgraph of induced by . For an integer , the bipartite graph is called a star, and the vertex of degree is called the central vertex. Let be a real-valued function. For convenience, we write for for .

In [1], we initiated the study of the negative decision number in a graph. A function is called a bad function of if for every . The maximum of the values of , taken over all bad functions , is called the negative decision number and is denoted by .

The motivation for studying this parameter may be explained from a modelling perspective. For instance, by assigning the values −1 or 1 to the vertices of a graph, one can model networks of people in which global decisions must be made (e.g., positive or negative responses). In certain circumstances, a positive decision can be made only if there are significantly more people voting for than those voting against. We assume that each individual has one vote, and each has an initial opinion. We assign 1 to vertices (individuals) which have a positive opinion and −1 to vertices which have a negative opinion. A voter votes “good” if there are at least two more vertices in its open neighbourhood with positive opinion than with negative opinion, otherwise the vote is “bad”. We seek an assignment of opinions that guarantee a unanimous decision; namely, for which every vertex votes “bad”. Such an assignment of opinions is called a uniformly negative assignment. Among all uniformly negative assignments of opinions, we are particularly interested in the minimum number of vertices (individuals) which have a negative opinion. The negative decision number is the maximum possible sum of all opinions, 1 for a positive opinion and −1 for a negative opinion, in a uniformly negative assignment of opinions. The negative decision number corresponds to the minimum number of individuals who can have negative opinions and in doing so force every individual to vote bad.

In the present paper, we study a variation of the negative decision number in a graph. A mapping is called negative if for every . The maximum of the values of , taken over all negative mappings , is called the modified negative decision number and is denoted by .

The parameter differs significantly from . For instance, for a cycle of order , ([1] see Theorem 5), while (see Corollary 2.6).

Throughout this paper, if is a negative mapping of , then we let and denote the sets of those vertices of which are assigned (under ) the values +1 and −1, respectively, and we let and . Hence, .

All graphs considered in this paper are simple, finite, and undirected. For a general reference on graph theory, the reader is referred to [2, 3]. Notice that for two disjoint graphs and . Hence, we assume that all graphs are connected in this paper.

In the next section, we present several sharp upper bounds on the modified negative decision number for a general graph. We also establish sharp upper bounds on this number for bipartite graphs and regular graphs. Exact values of these numbers for cycles, paths, cliques, and bicliques are found.

2. Main Results

In this section, we investigate the modified negative decision number of a graph. We first present an upper bound on this number for a general graph in terms of its order.

For this purpose, we define a family of graphs as follows. Let , and for , let be the graph obtained from the disjoint union of stars by adding all possible edges between the central vertices of the stars. See Figure 1 for an example of . Let .

Theorem 2.1. If is a graph of order , then and this bound is sharp.

Proof. Let be a negative mapping such that . Then . Note that every vertex in must be joined to at least one vertex in . By the pigeonhole principle, at least one vertex in is joined to at least vertices in . Hence, That is, Solving the above inequality, we obtain that Thus .
To see this bound is sharp, let . Thus, for some . Hence, has order , and so . Assigning the value −1 to each of the central vertices of the stars, and +1 to all other vertices, we define a negative mapping of satisfying . Thus, . It follows that .

Next we give an upper bound of the modified negative decision number for a general graph in terms of its order and size.

Theorem 2.2. If is a graph of order and size , then and this bound is sharp.

Proof. Let be a negative mapping such that . Then . As every vertex in must be joined to at least one vertex in , . For each vertex of , , and hence, Namely, Hence, Thus, the total number of edges in is So, . Hence, To see this bound is sharp, let for some and let be the negative mapping of defined in the proof of Theorem 2.1. As , . On the other hand, as has order and size , . Therefore, .

In the following theorem, we establish an upper bound of the modified negative decision number for a bipartite graph in terms of its order and we characterize the graphs attaining this bound. We define a family of bipartite graphs as follows.

For , let be the bipartite graph obtained from the disjoint union of stars with centres by adding all edges of the type , . See Figure 2 for an example of . Let .

Theorem 2.3. If is a bipartite graph of order , then The equality holds if and only if .

Proof. Let be a negative mapping of such that . Let and be the partite sets of . Further, let and be the sets of vertices in that are assigned the value +1 and −1 (under ), respectively. Let and be defined analogously. Then and . For convenience, let , , and . Hence, .
As every vertex in must be joined to at least one vertex in , by the pigeonhole principle, at least one vertex in is joined to at least vertices in . Hence, Namely, By a similar argument, one may show that Adding (2.13) and (2.14), we have that Thus, Solving the above inequality for , we obtain that Thus, .
If , then by the above analysis, we have and . Moreover, each vertex of (resp., ) has degree 1 and is joined to one vertex of (resp., ), while each vertex of is joined to all vertices of and vertices of and each vertex of is joined to all vertices of and vertices of . Thus, implies that .
If , then for some . Hence, has order , and so . Assigning to the central vertices of the stars the value −1, and to all other vertices , we define a negative mapping of satisfying . Hence, . It follows that .

Finally, we present an upper bound of the modified negative decision number for a regular graph in terms of its order in Theorem 2.5. We encourage the reader to verify our next theorem first.

Theorem 2.4. For any integer , one has

For a general -regular graph, we have the following.

Theorem 2.5. If is -regular graph of order , then This bound is sharp.

Proof. Let be any negative mapping of . As is a -regular graph, We discuss the following two cases.Case 1. is even. Since is a negative mapping, for every vertex , . By (2.20), it follows that Hence, .Case 2. is odd. In this case, is even for every . So, for each , implies that . Thus, By (2.20), it follows that . Hence, .The sharpness follows from Theorem 2.4.

Corollary 2.6. For any integer , one has

Proof. As is a 2-regular graph, by Theorem 2.5, we have . Let be the vertices of in a clockwise order. We discuss the following three cases.
Case 1. for some integer . To show that , it suffices to show that there is a negative mapping of such that . In fact, assigning starting with clockwise, and repeating, we produce a negative mapping of satisfying .Case 2. for some integer . To show that , it suffices to show that there is a negative mapping of such that . In fact, if we assign to the vertices in a clockwise order when , or assign , starting with clockwise when , then we produce a negative mapping of satisfying .Case 3. for some integer . To show that , it suffices to show that there is a negative mapping of such that . In fact, if we assign to the vertices in a clockwise order when , or assign starting with clockwise when , then we produce a negative mapping of satisfying .
To show , we let be a negative mapping of such that . So . Hence, , implying that .

Corollary 2.7. For any integer , one has

Proof. It is easy to see that Corollary 2.7 is true for . So we may assume that . Let be the vertices of with end vertices and . We only show that . The rest of the proof is similar to that of Corollary 2.6, so is omitted.
Take any negative mapping of . For or , as , we have that For ,. Hence, Thus,
Note that . So, implying that .

For the biclique , we have the following result.

Theorem 2.8. For any integer , one has

Proof. Let and be the partite sets of . As is an -regular graph, by Theorem 2.5, . Hence, . As has order , which is even,
To prove the case when is even, it suffices to show that there exists a negative mapping of such that . In fact, we can produce such by assigning , .
Now we prove the case when is odd. Let be a negative mapping of such that . Let , , , and be defined the same as in the proof of Theorem 2.3. Then and . Let , , and . Thus, . As , by (2.30), we have that
Notice that we may assume and . If , then and contradicting to the fact , where is odd. Hence, .
To show that , it suffices to show that there is a negative mapping of such that . In fact, we can produce such by assigning , for and , .