Definition A tree is a connected, acyclic graph. Alternately, a tree is a graph in which any two vertices are connected by exactly one simple path.

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1 11.5 Trees

2 Examples

3 Definitions Definition A tree is a connected, acyclic graph. Alternately, a tree is a graph in which any two vertices are connected by exactly one simple path. Definition In an undirected tree, a pendent vertex is a vertex of degree one. Definition A forest is the disjoint union of trees.

4 Applications of Trees - Mail Sorting Starting with root, sort by most significant digit (10 branches), then sort by next digit (10 branches), etc

5 Applications of Trees - Binary Search Tree Properties 1 all of the values to the left are less than the root 2 all of the values on the right are greater than the root 3 each value only appears once in the tree 4 the left and right subtrees are binary search trees themselves

6 Properties of Trees Every tree with at least one edge has at least two pendent vertices.

7 Properties of Trees Every tree with at least one edge has at least two pendent vertices. Proof: Let P = {v 1, v 2,..., v n } be a maximal path in a tree T. Suppose one of its endpoints, say v 1, has degree greater than 1. Then v 1 is adjacent to vertex v 2 on path P and also some other vertex w. If w is different from all of the vertices v i then P could be extended, contradicting its maximality. w v 1 v 2... vn

8 Proof (cont.) On the other hand, if w is one of the vertices v i on the path, then the acyclic property of T would be contradicted. v 1 v 2 v 3 w... vn Thus, both endpoints of P must be pendent vertices of T.

9 Proof (cont.) On the other hand, if w is one of the vertices v i on the path, then the acyclic property of T would be contradicted. v 1 v 2 v 3 w... vn Thus, both endpoints of P must be pendent vertices of T. Corollary If the degree of every vertex of a graph is at least two, then that graph contains a cycle.

10 Number of Edges Every tree on n vertices contains exactly n 1 edges. Proof: A tree with one vertex is a trivial tree and has no edges.

11 Number of Edges Every tree on n vertices contains exactly n 1 edges. Proof: A tree with one vertex is a trivial tree and has no edges. Assume for some integer k 1, as an inductive hypothesis, that every tree on k vertices has exactly k 1 edges.

12 Number of Edges Every tree on n vertices contains exactly n 1 edges. Proof: A tree with one vertex is a trivial tree and has no edges. Assume for some integer k 1, as an inductive hypothesis, that every tree on k vertices has exactly k 1 edges. Next consider any tree T on k + 1 vertices. By our last theorem, T contains a pendent vertex, say v. Then, the graph T v is acyclic, since deleting a vertex from an acyclic graph cannot create a cycle. Moreover, T v is connected, since the vertex v had degree 1 in T. Thus, T v is a tree on k vertices, and hence T v has k 1 edges, by the inductive hypothesis. But since deg(v) = 1, it follows that T v has one fewer edge than T. Therefore, T has a total of k edges, which completes the proof.

13 Bridges Every edge of a tree T is a bridge.

14 Bridges Every edge of a tree T is a bridge. Proof: A tree T n has n 1 edges, which is the minimum required for any connected graph. Let e be any edge of T. Then, T e still has n vertices, but n 2 edges, making it impossible to be connected.

15 No Cycles A graph G is a tree iff it contains no cycles.

16 No Cycles A graph G is a tree iff it contains no cycles. Proof: Suppose G is a tree. Let u and v be vertices of G. Since G is connected, there must be a u v path in G. Suppose there are two distinct u v paths P 1 and P 2 in G. Then, we could traverse P 1 from u to v and then return to u via P 2, implying that G has a cycle. But, since G is a tree, we know each edge is a bridge, contradicting that the paths are unique. This, there is only one u v path in G and thus G is acyclic.

17 No Cycles A graph G is a tree iff it contains no cycles. Proof: Suppose G is a tree. Let u and v be vertices of G. Since G is connected, there must be a u v path in G. Suppose there are two distinct u v paths P 1 and P 2 in G. Then, we could traverse P 1 from u to v and then return to u via P 2, implying that G has a cycle. But, since G is a tree, we know each edge is a bridge, contradicting that the paths are unique. This, there is only one u v path in G and thus G is acyclic. Now suppose G has no cycles. Then there is exactly one path between any two vertices, say u and v. So, each edge of this path would therefore be a bridge, making G a tree.

18 Spanning Trees Definition A spanning tree of a graph G with order n is a connected subgraph of a graph G with the same order as G and n 1 edges.

19 Spanning Trees Definition A spanning tree of a graph G with order n is a connected subgraph of a graph G with the same order as G and n 1 edges. One application of spanning trees would be the power grid. For example, the electric company would want to minimize the amount of lines they would need in order to get to every house, so they would want to only have each house have one way to get the electricity. If we imagine a grid, the spanning tree would require the least number of edges (lines) to reach all houses (vertices).

20 Spanning Trees A spanning tree for this neighborhood could be

21 Spanning Trees A spanning tree for this neighborhood could be Another use of spanning trees would be the decisions about where to put highways and their on/off ramps.

22 One Spanning Graph Every connected graph has a spanning tree.

23 One Spanning Graph Every connected graph has a spanning tree. Proof: Suppose that G is connected. Every graph G has a spanning subgraph H of G with a minimum number of edges. Now, H can have no circuits. For if C is a circuit of H, removal of any edge of C (without removing the incident vertices) leaves a spanning subgraph of G that is still connected but has one less edge than H. This is impossible by the choice of H. Thus, H has no circuits. Also, by choice, H is connected. Thus, H is a spanning tree.

2. Determine each of the 11 nonisomorphic graphs of order 4, and give a planar representation of each.

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