Proposition 1. The edges of an even graph can be split (partitioned) into cycles, no two of which have an edge in common.

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1 Math 3116 Dr. Franz Rothe June 5, SUM\3116_2012t1.tex Name: Use the back pages for extra space 1 Solution of Test 1.1 Eulerian graphs Proposition 1. The edges of an even graph can be split (partitioned) into cycles, no two of which have an edge in common. Figure 1: Split this graph into edge disjoint cycles. 10 Problem 1.1. Illustrate the above proposition for the graph from the figure on page 1. Use different colors for your cycles. Draw all different partitions into cycles. How many are there? Answer. 1

2 Figure 2: This graph has five different partitions into edge-disjoint cycles. 10 Problem 1.2. Under which conditions for the numbers a and b is the complete bipartite graph K a,b Eulerian? What are the degrees of its vertices? Answer. There are a vertices of degree b and b vertices of degree a. Hence the complete bipartite graph K a,b is Eulerian if and only if a and b are both even. 10 Problem 1.3. For which values of the number n is the complete graph K n Eulerian? What are the degrees of the vertices? Answer. All vertices have degree n 1. The complete graph K n is Eulerian if and only if n is odd. 2

3 1.2 Rooted trees Figure 3: What is Venn diagram for the nested set at the root? 10 Problem 1.4. Draw a Venn diagram for the nested set you get at the root. Use different colors for the primary elements and all other nested subsets occurring in between. Answer. Figure 4: Venn diagram for the nested set at the root. 3

4 1.3 Instant insanity 10 Problem 1.5. Solve the instant insanity game for the four cubes below. Draw your stack of cube with all four colors on top of each other, seen from left and right; and from front and back. Y G G R B R B G R R Y G Y R Y Y G B Y B B G Y B 4

5 Figure 5: How to get the graphs for the four cubes. Figure 6: How to stack the four colored cubes. Answer. 5

6 1.4 Labeled trees Proposition 2 (Cayley s Theorem counting labeled trees). There are n n 2 different labeled trees with n vertices. Corollary 1 (Spanning trees of the complete graph). The complete graph K n has n n 2 different spanning trees. For the proof we construct a bijection between the labeled trees and their Prüfer codes. Each Prüfer code is a list of length n 2 with integers 1... n as entries. Construction of the Prüfer code for a given labeled tree. Find the end-vertex (leaf) of the tree with the lowest label. Put the label of its neighbor as next item into the Prüfer code. Pull out the end-vertex which is best indicated by deleting its label, and repeat the process until only two vertices are left. Construction of the labeled tree from a given Prüfer code. We write down three lines underneath each other. The second line is the list [1, 2,..., n]. The third line is the given Prüfer code. Into the first line, we successively put the adjacencies we are going to find for the labeled vertices. Find the lowest label in the remaining list which is not in the remaining Prüfer code. Delete this label from the list. Delete the first item from the remaining Prüfer code and take it up into the first line directly above the item of the list that has just been deleted. Indicate the new adjacency. The above steps are repeated until a list of two labels and an empty Prüfer code is left over. These two remaining labels are adjacent to each other. Finally one can produce a figure of the labeled tree using the adjacencies between labeled vertices in the first and the second line underneath each other. 10 Problem 1.6. Reconstruct and draw the trees with the Prüfer codes (3, 2, 1, 1) and (3, 1, 1, 2) This is an example for two non-isomorphic trees, the Prüfer codes of which are permutations of each other. Answer. We begin the reconstruction with the three lines: Adjacencies : and get by successive deletion of the lowest term from List \ Code: Adjacencies : 3 Adjacencies : 2 3 6

7 Adjacencies : Adjacencies : This is the reconstruction for the second example: Adjacencies : and get by successive deletion of the lowest term from List \ Code: Adjacencies : 3 Adjacencies : 1 3 Adjacencies : Adjacencies : Now we draw the labeled trees, which is done in the figure on page 7. Figure 7: Labeled trees with Prüfer codes (3, 2, 1, 1) and (3, 1, 1, 2). 10 Problem 1.7. Give a reason why the two graphs G and H in the figure on page 8 are not isomorphic. Answer (Reason). In graph G, there exist only two 4-cycles, but in graph H there exist three 4-cycles. 7

8 Figure 8: Are these two graphs isomorphic? Definition 1 (The chromatic polynomial). The chromatic polynomial P G (x) counts the number of different proper vertex-coloring of a graph G, using at most x colors. 10 Problem 1.8. How many different proper colorings are there for a 4-cycle which use exactly three colors. Draw one coloring and explain how the other eleven ones are obtained from there. Answer. One color appears on two opposite vertices of the square. There are three ways to choose the color appearing twice. There are two ways to place the diagonal where they appear. There are two ways to color the remaining two vertices, using both remaining colors. Hence there are = 12 different ways to color a 4-cycle with exactly three colors. 10 Problem 1.9. We have shown in the lecture that the chromatic polynomial for the n-cycle is P (C n ) = (x 1) n +( 1) n (x 1). How many different proper colorings are there for a 4-cycle with at most three colors. for a 10-cycle with at most four colors. Answer. A 4-cycle can be properly colored with at most three colors in ( 1) 4 2 = 18 ways. A 10-cycle can be properly colored with at most four colors in ( 1) 10 3 = ways. 8