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Section 3.1

Section 3.1. Properties of Trees. Sarah Graham. root. a. b. c. e. f. d. i. h. g. j. Tree Talk: Vocabulary.

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Section 3.1

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  1. Section 3.1 • Properties of Trees Sarah Graham

  2. root a b c e f d i h g j Tree Talk: Vocabulary • Tree: a tree is a special type of graph that contains designated vertex called a root so that there is a unique path from the root to any other vertex in the tree. Equivalently, a tree graph contains no circuits. • Rooted Tree: a directed tree graph Root = the unique vertex with in-degree of 0

  3. m f l d h b e j a i k c root g • Level Number: the length of the path from the root a to x • Parent: for any vertex except the root a parent of x is the vertex y with an edge (y,x) • Children: for any vertex say z with a connected edge with a parent (x,z) • Siblings: two vertices with the same parent children Level # 3 siblings parent Level #2

  4. Theorem 1: A tree with n vertices has n – 1 edges. Assume that the tree is rooted. Since each vertex except the root has such a unique incoming edge, there are n – 1 nonroot vertices and hence n- 1 edges. a Root = a Vertices = 6 Edges = 6 – 1 = 5 b c d e f

  5. Leaves: vertices with no children • Internal Vertices: vertices with children excluding the root • M-ary Tree: when each internal vertex of a rooted tree has m children • Binary Tree: when m = 2 • Height of a Rooted Tree: the length of the longest path to the root • Level: A set distance form the root (ex: the vertices at level 3 is the set of vertices at distance 3 from the root) • Balanced Tree (“good”): if all the leaves are at levels h and h-1 Binary tree

  6. Theorem 2: If T is an m-ary tree with n vertices, of which i vertices are internal. Then, n = mi + 1. Proof Each vertex in a tree, other than the root, is the child of a unique vertex. Each of the i internal vertices has m children, and there are a total of mi children. Adding the root gives n = mi + 1

  7. Corollary T is a m-ary tree with n vertices, consisting of i internal vertices and l leaves. • Given i, then l = (m – 1)i + 1 and n = mi + 1 • Given l, then i = (l –1)/(m – 1) and n = (ml –1)/(m – 1) • Given n, then i = (n – 1)/m and l = [(m – 1)n + 1]/m n = m + l n – m = l m(i) +1 – m = l

  8. a b c Theorem 3 • T is a m-ary tree of height h with l leaves. • l≤ mh and if all leaves are at height h, l = mh • h ≥ [logml] and if the tree is balanced, h = [logml] 5 ≤ 23 3 ≥ [log25] d e f h i g

  9. Theorem 4 There are nn-2 different undirected trees on n items. Prufer Sequence: There exists a sequence (s1, s2,…,sn-2) of length n-2. 1 Start with the smallest leaf (2). Its neighbor is 5. 5= s1Delete the edge. Take the next smallest leaf (4). Its neighbor is 3. 3= s2 Delete the edge. Continue like this. 3 5 7 2 4 6 8 (5,3,1,7,3,6)

  10. 1 6 3 8 5 7 2 4 Find the graph that goes to the Prufer Sequence (6,2,2,3,3,3) Set aside 1 4 5 2 6 7

  11. Show that any tree with more than 1 vertex has at least 2 vertices of degree 1. By the corollary of theorem 2: l= ((m-1)n+1)/m Assume l=1 and n≥2 Then: l=((m-1)n+1)/m m=(m-1)n+1 (m-1)=(m-1)n 1=n

  12. 2 1 5 1 3 8 4 6 7 4 3 6 7 5 2 (5,6,1,1,5,6) 8 Create a graph from the Prufer Sequence: Create a Prufer Sequence from the graph: (3,3,3,3,3,5)

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