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Graph Theory (Connected Graphs)

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Graph Theory (Connected Graphs) Graph Theory (Connected Graphs) Dr. Shubh N. Singh Department of Mathematics Central University of South Bihar May 14, 2020 2 Lecture-01: 0.1 Walks, Paths, and Cycles Let G = (V; E). If we think of the vertices of G as locations and the edges of G as roads between certain pairs of locations, then the graph G can be considered as modeling some community. There is a variety of kinds of trips that can be taken in the community. Let's start at some vertex u of a graph G. If we proceed from u to a neighbor v of u and then to a neighbor of v and so on, until we finally come to a stop at a vertex w, then we have just described a walk from u to v in G. More formally, Definition 0.1.1. Let G = (V; E), and let u; v 2 V .A walk from u to v (in short, a u − v walk) in G is a nonempty finite sequence/list u = v0; v1; : : : ; vk = v of vertices in G, where k ≥ 0 and fvi; vi+1g 2 E for all i = 0; 1; 2; : : : ; k − 1. If the vertices in a walk are all distinct, we call it a path in G. Note that all vertices or edges lie on (or, belong to ) a u − v walk need not be distinct; in fact u and v are not required to be distinct. Provided we continue to proceed from a vertex to one of its neighbors (and eventually stop), there is essentially no condi- tions on a walk in a graph. However, there will be occasions when we want to place restrictions on certain types of walks. Definition 0.1.2. Let G be a graph, and let W be a u − v walk in G. 0.1. WALKS, PATHS, AND CYCLES 3 (1) The number of edges encountered in W (including multiple occurrences of an edge) is called the length of W , and is denoted by l(W ). (2) If l(W ) = 0, then W is called a trivial walk. So, W = (v) is a trivial walk. (3) If u = v, then W is called a closed walk. (4) The u − v walk W is called a u − v trail in G if no edge in W is traversed more than once. (5) A closed trail of length at least 3 is called a circuit. (6) A closed path of length at least 3 is called a cycle. (7) A cycle of length k ≥ 3 is called a k-cycle. In particular, a 3-cycle is also referred to as a triangle. (8) A cycle of odd length is called an odd cycle; while, not surprisingly, a cycle of even length is called an even cycle. Remark that every path is a trail, but the converse is not true, in general, as shown in the following example. Example 0.1.1. Consider the following 5-vertex graph G with vertex set fu; v; w; x; yg. u v w x v Next consider the following walk W in G. W := u; w; y; x; w; v: Clearly W is a u − v trail in G. Notice that this trail W repeats the vertex w. Therefore this trail W is not a u − v path in G 4 Exercise 0.1.1. Justification should either be a proof or a coun- terexample. 1. Let G and H be graphs such that G ∼= H. Show that (i) if G contains a path of length k so does H. (ii) if G contains a cycle of length k so does H. 0.2. CONNECTED GRAPHS 5 Lecture-02: 0.2 Connected Graphs We will now have a special interest in a graph G in which it is possible to travel from each vertex of the graph G to any other vertex of G. Definition 0.2.1. Let G = (V; E). (1) Two vertices u and v of G are said to be connected if there exists a u − v path in G. (2) G is said to be connected if every pair of its points are connected. Otherwise, it is said to be disconnected. (3) A connected subgraph of G that is not a proper subgraph of any other connected subgraph of G is called a component of G. The number of components of G is denoted by k(G). Thus, Remark 0.2.1. A graph G is connected if and only if k(G) = 1. Example 0.2.1. Consider the following graph G. There are three components G1;G2;G3 in G with the vertex set V (G1) = fA; B; Cg, V (G2) = fDg and V (G3) = fE; F g. A D E B C F Problem 0.2.1. If G is a disconnected graph, then show that the complement G is connected. 6 Solution. Let u; v 2 V (G). If u and v belong to different com- ponents of G, then uv2 = E(G) =) uv 2 E(G): In this case, there is a u − v path of length one between u and v. Let us assume that u and v belong to the same component. Since G is disconnected, G has at least two components. Choose a vertex w that belongs to a component that does not contain u; v. Then uw; vw2 = E(G) =) uw; vw 2 E(G): Thus, in this case, there a u − v path "u − w − v" of length two between u and v. Hence G is connected. Problem 0.2.2. If G has exactly two vertices u and v of odd degree, then show that G has a u − v path. Solution. By contradiction, suppose that there is no u−v path in G. By definition, G is a disconnected graph and the vertices u and v must lie in two separate connected components H and K of G. Say u 2 H and v 2 K. Since G has exactly two vertices of odd degree, all other vertices in H (which, of course, are also vertices in G) have even degree. Therefore, the sum of the degrees of the vertices in H is an odd number. So we have a contradiction and our assumption, that there is no u; v path in G, must be false. 0.2. CONNECTED GRAPHS 7 Lecture-03: Problem 0.2.3. Let G be an n-vertex graph. If deg(u) + deg(v) ≥ n − 1 for every two non-adjacent vertices u and v of G, then show that G is connected. Solution. Let u; v 2 V (G). If u and v are adjacent, then we are done. Otherwise, if u and v are not adjacent, we claim that u and v have a common neighbor. By contradiction, suppose that N(u) \ N(v) = ;. Since u and v are not adjacent, we see that both u and v are not in the union N(u)[N(v) of neighbour of u and v, and so jN(u)[N(v)j ≤ n−2. Note that deg(x) = jN(x)j for every vertex x 2 V (G). Then deg(u) + deg(v) = jN(u)j + jN(v)j = jN(u) [ N(v)j (since N(u) \ N(v) = ;) ≤ n − 2: This is a contradiction since deg(u) + deg(v) ≥ n − 1. This completes the proof. (n−1) Problem 0.2.4. If G is an n-vertex graph with δ(G) ≥ 2 , then show that G is connected. Solution. Let u; v 2 V (G). If u and v are adjacent, then we are done. Otherwise, if u and v are not adjacent, we claim that u and v have a common neighbor. Observe that (n − 1) jN(u)j = deg(u) ≥ δ(G) ≥ ; 2 (n−1) and similarly jN(v)j = deg(v) ≥ δ(G) ≥ 2 . If u and v are not adjacent, we see that both u and v are not in the union N(u) [ N(v), and so jN(u) [ N(v)j ≤ n − 2: 8 Now jN(u) \ N(v)j = jN(u)j + jN(v)j − jN(u) [ N(v)j (n − 1) (n − 1) ≥ + − (n − 2) 2 2 = 1: Therefore vertices u and v have a common neighbour, say w. Hence u; w; v is a u−v path in G. This completes the proof. Theorem 0.2.1. Let G be a nontrivial connected graph. Then G is bipartite if and only if G has no odd cycles. Proof. See class note for the proof. 0.2. CONNECTED GRAPHS 9 Lecture-04: Exercise 0.2.1. Be sure to explain your answer. 1. Determine whether the statements below are true or false. (i) If every vertex of a graph G has degree 2, then G is a cycle. (ii) Every disconnected graph has an isolated vertex. (iii) The complement of a connected graph always connected. (iv) A graph is connected if and only if some vertex is ad- jacent to all other vertices. 2. Prove or disprove: (i) If every vertex of a graph has degree at least 2, then G contains a cycle. (ii) If every vertex of a graph has degree exactly 2, then G is a cycle. (iii) If every vertex of a connected graph has degree exactly 2, then G is a cycle. (n−1)(n−2) 3. If an n-vertex graph G has at least 2 +1 edges, show that G is connected. 4. Let G and H be graphs such that G ∼= H. Show that (i) if G is connected, then H is also connected.
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