Independence Number and Cut-Vertices

Transcription

1 Independence Number and Cut-Vertices Ryan Pepper University of Houston Downtown, Houston, Texas 7700 Abstract We show that for any connected graph G, α(g) C(G) +1, where α(g) is the independence number of G and C(G) is the number of cut-vertices of G. The bound is sharp and the case of equality is characterized. This inequality was conjectured by the computer program Graffiti during classroom use. 1 Introduction The independence number α = α(g) of a graph G is the cardinality of a largest set of mutually non-adjacent vertices. A vertex v is a cut-vertex of G if the graph G v, formed by deleting v and all edges incident to v, has more components than G does. Similarly, an edge e is a cut-edge if G e has more components than G. In what follows, n = n(g) is the number of vertices and C = C(G) is the number of cut-vertices of the graph G. All graphs are considered to be simple and finite. The results in this paper were inspired by a conjecture of the computer program Graffiti, created by S. Fajtlowicz. This conjecture was made during classroom use in 003. (The use of Graffiti in the classroom is discussed in [1] and [].) Graffiti conjectured that, for any graph with k components, α C + k. This conjecture, which appeared in [4], is correct and follows as a corollary to the lower bound on independence number proven herein. A similar result, also conjectured by Graffiti at around the same time, concerning a lower bound on the independence number in terms of the number 1

2 of cut-edges, was also recently published [3]. In this paper we prove the inequality α(g) C(G) + 1, and give a characterization of the case of equality. Main Result First, we define the family of graphs where equality holds. Definition.1. A graph G is called a complete-tree if it is a complete graph or if it can be partitioned into m complete subgraphs, each of order at least, such that every edge with an endpoint in two different members of the partition is a cut-edge and no vertex is incident to more than two parts. The name comes from the fact that the graph formed by replacing each complete subgraph in the partition by a vertex, where two vertices of the new graph are adjacent if and only if there was a cut-edge of the original graph joining the parts, is a tree. Moreover, the structure of these graphs yields the following remarks, which we state without proof since each can easily be proven by induction and/or appeal to the definition. After the remark there is an example of a complete-tree for illustrative purposes. Remark.. 1. A complete-tree has a unique complete-subgraph partition which satisfies the definition.. If G is a complete-tree, then every vertex of G is a member of some maximum independent set and no vertex of G is a member of every maximum independent set. 3. If G is a complete-tree and {A 1, A,..., A m } is its unique complete subgraph partition, then α = m and C = (m 1).

3 Figure 1: An example of a complete tree with m = 5, C = 8, and α = 5. Theorem.3. Let G be a connected graph with independence number α and C cut-vertices. Then, α C + 1 (1) and equality holds if and only if G is a complete-tree. Proof. Proceeding by induction, notice that the theorem is true for all connected graphs with at most three vertices. Assume that the theorem is true for all connected graphs with less than n vertices and let G be a graph of order n. If G has no cut-vertices, then the theorem follows immediately, so we may assume otherwise. Not every vertex of G can be a cut-vertex, so there must be a cut-vertex with a non-cut-vertex neighbor. Let v be a cutvertex with a non-cut-vertex neighbor w. Moreover, let {G 1, G,..., G k } be the components of G the graph obtained by deletion of v where, without loss of generality, w G 1. 3

4 From inductive hypothesis we have, α(g) α(g i ) ( C(G i) + 1) = k + 1 C(G i ). () Next we will show the number of cut-vertices in each component does not decrease by more than one with the deletion of v. First observe that, for each i, any vertex of G i which is a cut-vertex of G but not of G i, must be adjacent to v. Now, if there were two such vertices in one component of G, then there must be a cycle in G containing both of them and v contradicting the fact that they were cut-vertices of G but not of the component of G containing them. Thus the number of cut-vertices decreases by at most one in each component, as claimed. Evidently, the number of cut-vertices in the component G 1 does not decrease at all. To see this, suppose p G 1 is a cut-vertex of G but not of G 1. Hence p w is adjacent to v. Now there is a cycle in G containing p, w, and v a contradiction. To find a lower bound on the number of cut-vertices of G, we may count the loss of the cut-vertex v in the component G 1 componentwise to obtain; C(G ) = and then sum C(G i ) C(G) k. (3) The lower bound is proven by combining Inequalities and 3 and then observing that k, as shown below. α(g) α(g i ) ( C(G i) + 1) C(G) + k C(G) + 1. (4) Concerning the case of equality, assume α(g) = C(G) + 1. Then Inequality 4 collapses everywhere to equality. This means that k = and, α(g 1 ) + α(g ) = ( C(G 1) + 1) + ( C(G ) + 1). 4

5 Since each summand on the right hand side of the above equation is also a lower bound for its counterpart on the left, it follows that, α(g 1 ) = C(G 1) α(g ) = C(G ) + 1, + 1. Therefore G 1 and G are complete-trees by inductive hypothesis. Moreover, we also find from the collapsed Inequality 4, C(G) = C(G 1 ) + C(G ) = C(G ). (5) Hence, no new cut-vertices were introduced by deletion of v and the total number of cut-vertices decreases by exactly two. Since v itself was a cutvertex and the number of cut-vertices in G 1 does not decrease, let z G be the unique cut-vertex of G which is not a cut-vertex of G. It turns out that z is also the unique neighbor of v in G since, echoing previous reasoning, any other neighbor of v in G would introduce a cycle and contradict the nature of z just described. Thus, it necessarily follows that vz is a cut-edge of G. Now, since G 1 is a complete-tree, partition it into complete subgraphs satisfying the definition such that H is the part containing the vertex w. If there was a vertex p H which was not adjacent to v, then form a maximum independent set of G 1 containing p and a maximum independent set of G not containing z (that such sets exist in complete-trees is the second part of Remark.). Now take the union of these two sets together with v and we have a maximum independent set of G with α(g 1 ) + α(g ) + 1 = α(g) + 1 vertices a contradiction. Therefore, v is adjacent to every vertex of H. Moreover, suppose there was a vertex x G 1 H which was adjacent to v. Then, since G 1 is a complete-tree and x / H, any path in G 1 from x to w contains a cut-edge. Let rs be a cut-edge on a path in G 1 from x to w. Hence, r and s are cut-vertices of the component G 1. Now there is a cycle in G containing v, w, r, s, and x (or only four of these vertices if it turned out that r = x or s = x). However, by the structure of complete-trees, r is only adjacent to s and the all the vertices in the complete subgraph it is a 5

6 member of, and similarly for s. Thus it must be the case that r and s were not cut-vertices of G and only became cut-vertices after deletion of v. This contradicts Equation 5, which informs us that the the cut-vertices of G and of G are the same with the exception of the loss of v and z. Therefore no such vertex x could exist and the only neighbors v has in G 1 are the vertices of the complete subgraph H which contains w. To conclude the implication, we see that G can be partitioned into complete subgraphs satisfying the definition of a complete-tree by including v in the complete subgraph H of the partitioning of G 1 and then joining this to G by the cut-edge vz. The converse is settled quite easily by appealing to the structure of complete-trees laid out in the third part of Remark.. 3 Concluding Remarks The author would like to thank Siemion Fajtlowicz, Greg Henry, Craig Larson, and Dillon Sexton for helpful discussions about this theorem. Moreover, after the presented proof was discovered, both Henry and Larson found independent proofs of their own, though neither of these made the characterization of equality quite as simple. It should also be pointed out here that a nearly symmetric upper bound is known. Theorem 3.1. Let G be an n > 1 vertex connected graph with independence number α and C cut-vertices. Then α n C 1. (6) This theorem is much simpler in the sense that it is basically a statement about trees. It was proven independently by Fajtlowicz, Pepper, and Larson. Fajtlowicz s proof of an equivalent statement to this can be found in Ermelinda DeLaViña s list of conjectures of Graffiti and Graffiti.pc, Written on the Wall II, conjecture #1. This list can be found at: 6

7 References [1] S. Fajtlowicz, Toward Fully Automated Fragments of Graph Theory, Graph Theory Notes of New York XLII (00) [] R. Pepper, On New Didactics of Mathematics: Learning Graph Theory via Graffiti, S. Fajtlowicz, P. Fowler, P. Hansen, M. Janowitz, F. Roberts, (Eds.), Graphs and Discovery, DIMACS Series (69) in Discrete Mathematics and Theoretical Computer Science, , 005. [3] R. Pepper, G. Henry, D. Sexton, Cut-Edges and the Independence Number, MATCH Commun. Math. Comput. Chem. 56 (006) [4] R. Pepper, Binding Independence, Ph.D. Dissertation, University of Houston,