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MIT6_042JF10_fnl_2004_sol

MIT6_042JF10_fnl_2004_sol - 6.042/18.062J Mathematics for...

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6.042/18.062J Mathematics for Computer Science December 14, 2004 Tom Leighton and Eric Lehman Final Exam YOUR NAME: : You may use two 8 . 5 × 11 ” sheets with notes in your own handwriting on both sides, but no other reference materials. Calculators are not allowed. You may assume all results presented in class. Write your solutions in the space provided. If you need more space, write on the back of the sheet containing the problem. Please keep your entire answer to a prob- lem on that problem’s page. Be neat and write legibly. You will be graded not only on the correctness of your answers, but also on the clarity with which you express them. GOOD LUCK! Problem Points Grade Grader 1 13 2 15 3 12 4 12 5 12 6 12 7 12 8 12 Total 100
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2 Final Exam Problem 1. [13 points] Give an inductive proof that the Fibonacci numbers F n and F n +1 are relatively prime for all n 0 . The Fibonacci numbers are defined as follows: F 0 = 0 F 1 = 1 F n = F n 1 + F n 2 (for n 2 ) Solution. We use induction on n . Let P ( n ) be the proposition that F n and F n +1 are relatively prime. Base case: P (0) is true because F 0 = 0 and F 1 = 1 are relatively prime. Inductive step: Assume that P ( n ) is true where n 0 ; that is, F n and F n +1 are relatively prime. We must show that F n +1 and F n +2 are relatively prime as well. If F n +1 and F n +2 had a common divisor d > 1 , then d would also divide the linear combination F n +2 F n +1 = F n , contradicting the assumption that F n and F n +1 are relatively prime. So F n +1 and F n +2 are relatively prime. The theorem follows by induction.
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3 Final Exam Problem 2. [15 points] The double of a graph G consists of two copies of G with edges joining corresponding vertices. For example, a graph appears below on the left and its double appears on the right. Some edges in the graph on the right are dashed to clarify its structure. (a) Draw the double of the graph shown below. Solution.
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4 Final Exam (b) Suppose that G 1 is a bipartite graph, G 2 is the double of G 1 , G 3 is the double of G 2 , and so forth. Use induction on n to prove that G n is bipartite for all n 1 . Solution. We use induction. Let P ( n ) be the proposition that G n is bipartite. Base case: P (1) is true because G 1 is bipartite by assumption. Inductive step: For n 1 , we assume P ( n ) in order to prove P ( n + 1) . The graph G n +1 consists of two subgraphs isomorphic to G n with edges joining corresponding vertices. Remove these extra edges. By the assumption P ( n ) , we can color each vertex of one subgraph black or white so that adjacent vertices get different colors. If we color the corresponding vertices in the other subgraph oppositely, then adjacent vertices get different colors within that subgraph as well. And now if we add back the extra edges, each of these joins a white vertex and a black vertex. Therefore, G n +1 is bipartite.
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