MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 11, Wednesday, September 24

1. MATH 310, FALL 2003(Combinatorial Problem Solving)Lecture 11, Wednesday, September 24

2. A Spanning Tree • Each connected graph has a spanning tree. • For finite graphs the proof is easy. [Keep removing edges that belong to some circuit]. • For infinite graphs this is not a theorem but an axiom that is equivalent to the renowned axiom of choice from set theory. • Note: A spanning subgraph H of G contains all vertices of G.

3. How many spanning trees are there in Kn? • For example, on the right we see that K3 has 3 spanning trees! • Let t(Kn) denote the number of spanning trees in the completre graph Kn. We know that t(K3) = 3. • Theorem 4 (Cayley, 1889):t(K n) = nn-2 • Proof: Bijective proof (principle of equality): Prüfer sequences (codes)!

4. Two Remarks • Remark 1: We are counting all spanning trees. [Some may be isomorphic]. • Remark 2: Counting all spanning trees in a complete graph Kn is the same as counting all labeled trees on n vertices.

5. Prüfer Sequence From a Labeled Tree • We are given a tree whose vertices are labeled (from 1 to n). • Following the rule on the left we form a sequence s1, s2, ..., sn-2 of length n-2. Each element of the sequence is a number from 1 to n. • The sequence [s1, s2, ..., sn-2] is called the Prüfer code or sequence. 7 3 2 4 8 1 6 5 Delete the leaf with the largest label and print the label of its neighboor!

6. The Steps of Prüfer Recipe • Delete 7, print 1. • Delete 6, print 8. • Delete 5, print 1. • Delete 4, print 1. • Delete 2, print 3. • Delete 3, print 8. • The sequence: • [1,8,1,1,3,8] 7 3 2 4 8 1 6 5 Delete the leaf with the largest label and print the label of its neighboor!

7. 7 6 4 2 3 8 From the Sequence to the Tree. • Examle: [2,3,2,8,2,1]. • Numbers (between 1 and 8), missing in the sequence represent the leaves of a tree. ki v kodi manjkajo, so listi drevesa. • Maximal leaf: 7 • First sequence element: 2 • Edge: 7~2. Leaves: [4,5,6] • Edge: 6~3. Leaves: [3,4,5] • Edge: 5~2. Leaves: [3,4] • Edge: 4~8. Leaves: [3,8] • Edge: 8~2. Leaves: [2,3] • Edge: 3~1. Leaves: [1,2] • Final edge: 1~2. 1 5 Add an edge between the vertex labeled by the current sequence element and the leaf with maximal label.

8. Rooted Trees, Binary Trees • Some new terms: • root • level number • parent • child • sibling • leaf (leaves) • internal vertex • m-ary tree • binary tree

9. Theorem 2 • Let T be an m-ary tree with n vertices, of which i vertices are internal. Then, • n = m i + 1. • On the left we have • n = 25, • #leaves = l = 13, • i = 12. • n = 2i + 1. • Proof: Each internal vertex gives rise to m edges. The tree has therefore m i edges. It hase m i + 1 vertices.

10. Corollary • Let T be an m-ary tree with n vertices, i internal vertices, and l leaves. Then: • (a) Given i, l = (m-1)i + 1, n = mi + 1 • (b) Given l, i = (l – 1)/(m – 1), n = (ml – 1)/(m-1) • (c) Given n, i = (n – 1)/m, l = [(m – 1)n + 1]/m.

11. Example 1 • If 56 people sign up for a tennis tournament, how many matches will be played? [Hint: binary tree]. • Answer: 55 matches. • [Every player but winner played exactly one match in which he or she lost and these are all the matiches]

12. Rooted Trees – More New Terms • More new terms: • The height of a rooted tree • balanced rooted tree • Note: the binary tree on the left is not balanced but the right subtree is.

13. Theorem 3 • Let T be an m-ary tree of height h with l leaves. Then • (a) l · mh, and if all leaves are at height h, l = mh. • (b) h ¸d logml e, and if the tree is balanced, h = d logml e.

14. 3.2 Search Trees and Spanning Trees • Homework (MATH 310#4W): • Read 4.1. • Do Exercises 3.2: 2,4,8,10,12,16,22,28 • Volunteers: • ____________ • ____________ • Problem: 28. • On Monday you will also turn in the list of all new terms (marked).