Number Theory and Graph Theory Chapter 7 Graph Properties

1 Number Theory and Graph Theory Chapter 7 Graph properties By A. Satyanarayana Reddy Department of Mathematics Shiv Nadar University Uttar Pradesh, India E-mail: [email protected] 2 Module-1: Trees Objectives • Properties of trees • Spanning trees of a connected graph • Minimal spanning trees Recall that a graph with no cycles is called acyclic. An acyclic graph is called a forest. A forest having only one component is called a tree, i.e., a tree is a connected acyclic graph. A tree T with only one vertex is called a trivial tree and is called non trivial, otherwise. The following are 6 non isomorphic trees on 6 vertices. 3 number of number of number of number of vertices non isomorphic trees vertices non isomorphic trees 1 1 9 47 2 1 10 106 3 1 11 235 4 2 12 551 5 3 13 1301 6 6 14 3159 7 11 15 7741 8 23 16 19320 Theorem 1. Let X be a simple graph on n vertices. Then, the following statements are equivalent. 1. X is a tree (connected acyclic graph). 2. There is a unique path between any two vertices of X. 3. X is connected and has exactly n − 1 edges. 4. X contains no cycles and has n − 1 edges. 5. X is minimally connected, i.e., every edge is a bridge. Proof. 1 ) 2 Let X be a tree. Then, X is connected and there is a path between any two vertices. Suppose there exists at least two distinct paths between a pair of vertices, say u;v 2 V(X). Then, the union of these two paths produces a cycle. A contradiction. 2 ) 1 Let X be a graph with a unique path between any two vertices. Then, clearly X is connected. On the contrary assume that there is a cycle in the graph. Then, there exists at least two vertices on this cycle that have two distinct paths between them. Hence, a contradiction. 4 2 ) 3 Let X be a connected graph with a unique path between any two vertices, then clearly X is connected. We will show that X has n − 1 edges by induction on n, the number of vertices. The result is clearly true if n = 1;2;3. Now, assume that the result is true for all graphs containing less than k vertices and let X be a graph with k vertices. Since there is a unique path between any two vertices, if we remove any edge e from the graph X, then X n feg has k vertices, but has two connected components, say X1 and X2 with vertices k1 ≥ 1 and k2 ≥, respectively, with k1 + k2 = k: Thus, by induction X1 and X2 contain k1 − 1 and k2 − 1 edges, respectively. Thus, the number of edges in X is (k1 − 1) + (k2 − 1) + 1 = k − 1: 3 ) 4 Suppose X is connected and has exactly n − 1 edges. We have to show that X has no cycles. On the contrary assume that X has at least one cycle. Then, if we delete any edge from the cycle, the new graph is still connected and hence it has at least n −1 edges (see Corollary ??). Thus, we arrive at a contradiction. 4 ) 5 Suppose X has k components. We show that k = 1: Since X is acyclic, each component is acyclic. That is, each component is a tree and hence (use 1 , 2 ) 3) X contains n − k edges. Thus, k = 1 and therefore, X is a tree. Hence, there is a unique path between any two vertices in X (use 1 , 2). Hence, every edge is bridge. 5 ) 1 Let X be minimally connected. Then, X is connected and there exists a unique path between any two vertices. Thus, X it tree (use 1 , 2). Corollary 2. Every non trivial tree contains at least two vertices of degree 1. Proof. Let T be a tree on n vertices and let (d1;d2;:::;dn) be the degree sequence of T. Then, using n First Theorem on Graph Theory, we get ∑ di = 2n − 2: Now, let ` be the number of vertices of i=1 degree 1 in the T. Then, the degree of the other vertices is at least 2. Hence, 2n − 2 = ∑ di + ` ≥ 2(n − `) + `: di>1 Therefore, ` ≥ 2: 5 Note that the path graphs have exactly two pendant vertices whereas the star graph on n vertices has n − 1 pendant vertices. Theorem 3. Let X be a connected graph with degree sequence (d1;d2;:::;dn). Then, X is a tree if n and only if ∑ di = 2n − 2: i=1 n Proof. If X is a tree then using First Theorem on Graph Theory, we see that ∑ di = 2n − 2: i=1 n Conversely, assume that ∑ di = 2n − 2: Then, we prove the result by induction on n. If n = 2, i=1 then d1 +d2 = 2 and since X is connected, one has d1;d2 ≥ 1. Thus, d1 = d2 = 1 and hence X = K2, a tree. So, let us assume that the result is true for all connected graphs on n = k ≥ 3 vertices. Now, consider a connected graph X on n = k + 1 vertices with degree of vertices as d1 ≥ d2 ≥ ··· ≥ k+1 dk ≥ dk+1. Then, by assumption, ∑ di = 2(k + 1) − 2. As di ≥ di+1, for each i = 1;2;:::;k, one i=1 concludes that d1 ≥ 2 and dk = dk+1 = 1. Let u be the vertex with deg(u) = dk+1. As u is a pendant vertex, it is adjacent to a unique vertex, say v, with deg(v) = d`. Note that d` ≥ 2 as the graph is connected and k ≥ 3. So, now consider the graph Y = X n fug. This graph is a connected graph on k vertices with degree of its vertices being d1;:::;d`−1;d` − 1;d`+1;:::;dk. Also, note that k+1 d1 + ··· + d`−1 + (d` − 1) + d`+1 + ··· + dk = ∑ di − 1 − dk+1 = 2(k + 1) − 2 − 2 = 2k − 2: i=1 Hence, by the induction hypothesis, the graph Y represents a tree. Thus, the graph X is indeed a tree on k + 1 vertices. Theorem 4. Every tree is a bipartite graph. Proof. As we know, a graph is bipartite if and only if every cycle in it is of even length. And a tree has no cycles, hence every cycle in a tree of even(zero) length. true. Alternate proof Let T = fV;Eg be a tree on n vertices. Let v 2 V. Now we construct two sets V1;V2 such that V = V1 [V2 and every edge in T contains one end vertex in V1 and the other in V2. We built these 6 sets recursively. Initially set V1 = V2 = /0: In the first step V1 = fvg and V2 = /0: In the second step, V1 = fvg and V2 = N(v), the neighborhood of v. In the third step, V1 = fvg [u2V2 N(u) and V2 = N(v). We continue this process till all vertices are traversed. Note that the process will end in at most n steps. Theorem 5. Every tree has either one or two centers. Proof. The result is clearly true if T has at most 3 vertices. So, consider a tree T with four or more vertices. Then, by Corollary 2, T has at least two pendant vertices, and they are not centers. Let T 0 be the graph obtained from T by deleting all pendant vertices of T. Then T 0 is still a tree and both T and T 0 have the same centers. Thus, we can use induction argument to prove the required result. Definition 6. A spanning tree of a graph is a spanning subgraph which is also a tree. Theorem 7. Every finite graph is a connected graph if and only if it has a spanning tree. Proof. If a graph contains a spanning tree, then clearly it is connected. So, it is sufficient to prove that every finite connected graph has a spanning tree. We will prove this by induction on the number of cycles, say c, in the graph. If the graph has no cycles and is connected, then, by definition, the graph is a tree and hence is its own spanning tree. This makes a good base step for a proof by induction on the number of cycles of the graph that every connected graph has a spanning tree. So, let the result be true for every graph X that has fewer than c > 0 cycles. Now, consider a graph X that has c cycles. Choose a cycle of X and choose an edge, say e, of that cycle. Then, X n feg has fewer than c cycles and hence by the inductive hypothesis, the resulting graph has a spanning tree. But then this spanning tree is also a spanning tree of X. Therefore, by the principle of mathematical induction, every finite connected graph has a spanning tree. Definition 8. Circuit rank of X: Let X be a connected graph on n vertices and m edges. As a spanning tree has n − 1 edges, the number m − (n − 1) = m − n + 1 is called the circuit rank of 7 X. That is, the circuit rank of X is the number of edges that must be removed from X to obtain a spanning tree of X. Let T be a spanning tree of a graph X. Then, an edge of X that is also in T is called a branch of X. An edge of X that is not in T is called either a chord or a tie or a link of X with respect to T.

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