Math Fall 2011 Exam 3 Solutions - December 8, 2011

Transcription

1 Math all 2011 xam 3 Solutions - ecember 8, 2011 NM: STUNT I: This is a closed-book and closed-note examination. alculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 5 problems for a total of 100 points. POINTS: TOTL:

2 1. (10 points) Seventy carssit on a parkinglot. 30 have surround sound, 30 have heated seats and 40 have sun roofs. Given any two of these options there are exactly 15 cars which have them, and 10 have all three. How many cars on the lot have at least one of these options? How many have exactly one? Solution. enote by the set of cars which have surround sound, by the set of cars with heated seats, and by the set of cars with sun roofs. Then is the number of cars which have at least one of these options. y Inclusion-xclusion principle = Since = 30, = 30, = 40, the intersection of any two sets is 15 and = 10 we get = = 65 The number of cars which have exactly one options is +2 = = 40

3 2. (30 points) raw the simple graphs represented by the following adjacency matrices: G 1 = ,G 2 = ,G 3 = or each pair of graphs, either exhibit an isomorphism (and show that it preserves edges) or provide a rigorous argument that none exists. Solution. The graphs G 1,G 2 and G 3 are shown below. G 1 G 2 G 3 GraphsG 1 andg 3 areisomorphic. Theisomorphismf fromthesetofvertices of G 1 to the set of vertices of G 3 is given by f() =,f() =,f() =,f() =,f() =,f() = To verify that f preserves adjacency we compare the corresponding matrices: f(g 1 ) = = G

4 The graphs G 1 and G 2 (and thus G 2 and G 3 ) are not isomorphic since their subgraphs of vertices of degree 3 have different number of edges: G 1,G 3 G 2

5 3. (20 points) (a) Prove that K 3,3 has no circuits of odd length. (b) Use part (a) to show that K 3,3 is not planar. (c) Is K 5 planar? Justify your answer. (d) an 5 houses be connected to 2 utilities without connections crossing? Solution. (a) We compute the adjacency matrix for K 3,3 and n for odd n: n 1 3 n 1 3 n n 1 3 n 1 3 n 1 = , n = n 1 3 n 1 3 n 1 3 n 1 3 n 1 3 n n 1 3 n 1 3 n n 1 3 n 1 3 n Since the number of circuits of length n is represented by the elements on the diagonal of n and they are all zero, we conclude that there are no circuits of odd length in K 3,3. (b) ny planar graph without circuits of length three satisfies the inequality e 2v 4, where e is the number of edges and v is the number of vertices. Since K 3,3 has 6 vertices and 9 edges, it fails the above inequality, so it is not planar. (c) ny planar graph satisfies the inequality e 3v 6, where e is the number of edges and v is the number of vertices. Since K 5 has 5 vertices and 10 edges, it fails the above inequality, so it is not planar. (d) ive houses can be connected to 2 utilities without connections crossing since the graph K 5,2 is planar:

6 4. (20 points) What is the minimal number of weighings on a balance scale necessary to find a counterfeit coin among 12 coins if the counterfeit coin is lighter then the others? Use a tree to describe the algorithm to find the counterfeit coin using this number of weighings. Solution. We enumerate coins 1 through 12. There are three possibilities for each weighing on a balance scale. The two pans can have equal weight, the leftpancanbe lighterortherightonecanbe lighter. Thus, thedecisionthree for the sequence of weighings is a 3-ary tree with the left child representing the outcome when the left pan is lighter, the middle child representing a a outcome when the two pans are equal and the right child representing the outcome when the right pan is lighter. The total number of outcomes is 12, since any of the 12 coins could be a counterfeit one. Therefore the decision treehas12leaves. Thenumberofweighingsequalsthehightofthetreewhich has to be at least log 3 12 > 2. Therefore, we need minimum of 3 weighings to determine the counterfeit coin. The decision tree which illustrates how to find the lighter coin in three weighings is shown below

7 5. (20 points) onsider the map of several regions of a country below. I H G (a) What is the smallest number of colors necessary to color this map? (b) an a traveler start at one region, visit each region exactly once and then come back to the initial one? (c) anatravelerstartatoneregion,crosstheborderbetweenany2regions exactly once and then return to the initial region? Is it possible if the traveler does not have to come back? Solution. The dual graph to the map is shown below. I G H

8 (a) Since vertices,, H, and G are pairwise connected, they all need to be assigned different colors. Thus, we need at least four colors to color this graph. On the other hand, the four color theorem guarantees that it is possible color plane map using four colors. Thus, the smallest number of colors is four. (b) This question is equivalent to finding a Hamilton circle in the dual graph. Such circle does not exist since there is a vertex () of degree 1. (c) This question is equivalent to finding an uler circle or path in a dual graph. Since there are vertices ( and ) of odd degree, this graph does not have an uler circle. However, since there are exactly two vertices of odd degree ( and ), there is an uler path.