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Number Theory and Graph Theory Chapter 6

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Number Theory and Graph Theory Chapter 6 1 Number Theory and Graph Theory Chapter 6 Basic concepts and definitions of graph theory By A. Satyanarayana Reddy Department of Mathematics Shiv Nadar University Uttar Pradesh, India E-mail: [email protected] 2 Module-4: Connectedness of a graph Objectives • Connected, disconnected graphs and connected components • Connectedness in directed graphs • Few properties of connected graphs Let X = (V;E) be a graph. Before proceeding further, we recall the following definitions. Connectedness: A vertex u 2 V is said to be connected to a vertex v 2 V if there is a path in X from u to v. Connected Graph: X is said to connected if any two vertices of X are connected. Note that Complete graphs, complete bipartite graphs, wheel graphs, cyclic graphs, path graphs and Johnson graphs etc. are a few examples of connected graphs. A graph that is not connected is the union of two or more connected subgraphs, each pair of which has no vertex in common. These disjoint connected subgraphs are called the connected components of the graph. The number of components of a graph X is denoted by C(X). If X is connected then C(X) = 1. Let e be an edge of a graph X then it can be easily observed that C(X) ≤ C(X n feg) ≤ C(X) + 1. A direct application of the definition of a connected/disconnected graph gives the following result and hence the proof is omitted. Theorem 1. A graph is disconnected if and only if its vertex set V can be partitioned into two nonempty disjoint subsets V1 and V2 such that there exists no edge in X whose one end vertex is in V1 and other is in V2. Theorem 2. If a graph (connected or disconnected) has exactly two vertices of odd degree then there must be a path joining these vertices. 3 Proof. Let X be a graph that has exactly two vertices, say u and v of odd degree. Let C be a connected component of X containing the vertex u. Since, the number of vertices of odd degree in any graph is even, C has to contain another vertex of odd degree. But, then v is the only other vertex in X of odd degree and hence v lies in the component C. Thus, there is a path from u to v: Before proving the next theorem, we need the following inequality. Lemma 3. Let n1;n2;:::;nk 2 N. Then k k k 2 2 ∑ ni ≤ (∑ ni) − (k − 1)(2 ∑ ni − k): i=1 i=1 i=1 Proof. We have (n1 − 1) + (n2 − 1) + ··· + (nk − 1) = (n1 + n2 + ··· + nk) − k: Squaring on both sides, we get 2 2 [(n1 − 1) + (n2 − 1) + ··· + (nk − 1)] = [(n1 + n2 + ··· + nk) − k] k k k k k 2 2 2 ∑(ni − 1) + ∑ ∑ (ni − 1)(n j − 1) = (∑ ni) − 2k(∑ ni) + k i=1 i=1 j=1; j6=i i=1 i=1 k k k k k k 2 2 2 ∑ ni − 2(∑ ni) + k + ∑ ∑ (ni − 1)(n j − 1) = (∑ ni) − 2k(∑ ni) + k i=1 i=1 i=1 j=1; j6=i i=1 i=1 | {z } positive Hence, we have k k k k 2 2 2 ∑ ni ≤ 2(∑ ni) − k + (∑ ni) − 2k(∑ ni) + k : i=1 i=1 i=1 i=1 Or equivalently, k k k 2 2 ∑ ni ≤ (∑ ni) − (k − 1)(2 ∑ ni − k): i=1 i=1 i=1 4 Theorem 4. A simple graph with n vertices and k components can have at most have (n − k)(n − k + 1)=2 edges. th Proof. Let X be a graph with k components. Let ni be the number of vertices in the i component, where 1 ≤ i ≤ k. Then, the number of edges in the graph is equal to sum of the edges in each of its components. Thus, X has maximum number of edges if each component is a complete graph. k ni(ni−1) Hence, the maximum possible number of edges in the graph X is ∑ 2 : But from Lemma 3 i=1 we have k k k k k 1 2 1 1 2 1 ∑ ni − ∑ ni ≤ [(∑ ni) − (k − 1)(2 ∑ ni − k)] − ∑ ni 2 i=1 2 i=1 2 i=1 i=1 2 i=1 1 = (n − k)(n − k + 1) 2 As a direct application of Theorem 4, we have the following result. Corollary 5. Any simple graph with n vertices and more than (n − 1)(n − 2)=2 edges is connected. Theorem 6. Let X be a simple graph with diameter d(X). If d(X) ≥ 3 then show that d(Xc) is ≤ 3: Proof. Since d(X) ≥ 3, there exist two non-adjacent vertices, say u;v in X, such that u and v have no common neighbor. Then, for each x 2 V(X) n fu;vg, either u or v or both is a non-neighbor. So, x is adjacent at least one of u or v in Xc. Also, note that u;v are adjacent in Xc. Therefore, for any x;y 2 V(Xc) = V(X), there is path of length at most 3 in Xc through u;v, namely either x−u−v−y or x − v − u − y. Thus, the required result follows. Connectedness in Directed graphs: • Weakly connected: A digraph is weakly connected if it is connected as an undirected graph (a graph in which the direction of the edges is neglected). • Unilaterally connected: if for any pair of nodes of the graph at least one of the nodes of the pair is reachable from the other. 5 • Strongly connected: If for any pair of nodes of the graph both the nodes of the pair are reachable from one to another. From the definition every strongly connected graph is unilaterally connected and every unilaterally connected graph is weakly connected graph. 3 4 3 4 3 4 1 2 1 2 1 2 X Y Z It is easy check that the graph Z is strongly connected, Y is unilaterally connected but not strongly connected, where as X is weakly connected but not unilaterally connected. Exercises 7. 1. A graph X has 20 vertices. Any two distinct vertices x and y have the property that deg x + deg y ≥ 19. Prove that X is connected. 2. Show that a connected graph with n vertices has at least n − 1 edges. 3. Let X be a graph with 15 vertices and 4 components. Then prove that at least one component will contain 4 vertices. What is the largest number of vertices that a component of X may have? 4. Let X be a graph with n vertices and d(X) ≥ (n − 1)=2. Then show that X is connected. 6 0.1 Disconnecting sets, Cut sets • A disconnecting set in a connected graph X is a set of edges whose removal disconnects X. 3 4 5 1 2 X For example, in the above example, some of the disconnecting sets are ff4;5gg;ff1;2g;f3;4gg;ff1;3g;f2;4gg;ff1;2g;f3;4g;f4;5gg are disconnecting sets. Cut set is a disconnecting set having the property that no proper subset of which is a discon- necting set. Cut set with only one edge is called a bridge. 3 4 7 5 1 2 6 X The edge f4;5g is a bridge. A graph may contain cut sets of different sizes. Edge connectivity: Let X be a connected graph. Then, its edge connectivity, denoted l(X), equals l(X) = The number of elements in the smallest cutest. = Minimum number of edges required to disconnect the graph. 7 • Separating set: It is a set of vertices whose removal disconnects the graph. Vertex connectivity of a connected graph X, denoted k(X), equals the minimum number of vertices required to remove so that the graph becomes disconnect. • A connected graph is said to be separable if its vertex connectivity is one. All other connected graphs are non-separable graphs. • In a separable graph a vertex whose removal disconnects the graph is called a cut vertex or a cut node or an articulation point. [In a tree (connected acyclic graph) every vertex with degree grater than 1 is a cut vertex. So, every tree with three or more vertices is a separable graph]. • A separable graph consists of two or more non separable sub graphs. • Each of the largest non-separable sub graph is called a block. [A non separable connected graph consists of just one block.] Lemma 8. Let X be a graph. Then deleting an edge increases the number of components by 0 or 1. Proof. Components are pairwise disjoint; no two share a vertex. Adding an edge with endpoints in distinct components combines them into one component. Thus adding an edge decreases the number of components by 0 or 1 and deleting an edge increases the number of components by 0 or 1. Lemma 9. Every graph with n vertices and k edges has at least n − k components. Proof. A n-vertex graph with no edges has n components, by Lemma 8 each edge added reduces this by at most one, so when k edges have been added, the number of components is still at least n − k. As an immediate application, we have the following result. 8 Corollary 10. Let X be a connected simple graph on n vertices. Then, the minimum number of edges in X is n − 1. Proof. By Lemma 9, every graph with n vertices and k edges has at least n − k components.
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