MIT6_042JF10_midterm_sol

MIT6_042JF10_midterm_sol - 6.042/18.062J Mathematics for...

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Unformatted text preview: 6.042/18.062J Mathematics for Computer Science October 27, 2010 Tom Leighton and Marten van Dijk Midterm Problem 1. [10 points] Consider these two propositions: P: ( A B ) C Q: ( C A ) ( C B ) Which of the following best describes the relationship between P and Q ? Please circle exactly one answer. 1. P and Q are equivalent 2. P Q 3. Q P 4. All of the above 5. None of the above Draw a truth table to illustrate your reasoning. You can use as many columns as you need. Solution. A B C P Q T T T T T T T F F F T F T T T T F F F T F T T T T F T F F T F F T T T F F F T T Observe from the last two columns of the table that P Q is always true, but Q P is not always true (e.g. line 4). Thus P and Q are not equivalent but P Q . 2 Midterm Problem 2. [10 points] Let G 0 = 1, G 1 = 3, G 2 = 9, and define G n = G n 1 + 3 G n 2 + 3 G n 3 (1) for n 3. Show by induction that G n 3 n for all n 0. Solution. The proof is by strong induction with hypothesis P ( n ) : = G n 3 n . Proof. Base Cases n = : G 0 = 1 = 3 . n = 1 : G 1 = 3 3 1 . n = 2 : G 2 = 9 3 2 . Inductive Step : Assume n 3 and P ( k ) for all k such that 0 k n . G n = G n 1 + 3 G n 2 + 3 G n 3 by ( 1 ) 3 n 1 + ( 3 ) 3 n 2 + ( 3 ) 3 n 3 by induction hypothesis = 3 n 2 [ 3 + 3 + 1 ] = ( 7 ) 3 n 2 < ( 9 ) 3 n 2 = 3 n 3 Midterm Problem 3. [20 points] In the game of Squares and Circles, the players (you and your computer) start with a shared finite collection of shapes: some circles and some squares. Players take turns making moves. On each move, a player chooses any two shapes from the collection. These two are replaced with a single one according to the following rule: A pair of identical shapes is replaced with a square. A pair of different shapes is replaced with a circle. At the end of the game, when only one shape remains, you are a winner if the remaining shape is a circle. Otherwise, your computer wins. (a) [5 pts] Prove that the game will end. Solution. Proof. We use induction on the number of turns to prove the statement. Let n be the number of shapes originally, and let P(k) be the proposition that if 0 k n 1 then after k turns, the number of remaining shapes is n k . Thus the game ends after n 1 steps. Base case: P ( ) is true by definition; the number of reamaining shapes after 0 turns is n 0 = n , the original number of shapes. Inductive step: Now we must show that P ( k ) implies P ( k + 1 ) for all k 0. If k > = n 1, P ( k ) implies P ( k + 1 ) would always be true, since P ( k + 1 ) would be trivially true. So we only need to prrove this for k < n 1. So assume for your inductive hypothesis that P ( k ) is true, where 0 k < n 1; that is, after k turns the number of remaining shapes in n k . Since k < n 1, the number of remaining shapes is n k > 1. Hence there are at least 2 shapes to choose from and the game has not ended yet. In the k+1st turn either least 2 shapes to choose from and the game has not ended yet....
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This note was uploaded on 01/19/2012 for the course CS 6.042J / 1 taught by Professor Tomleighton,dr.martenvandijk during the Fall '10 term at MIT.

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MIT6_042JF10_midterm_sol - 6.042/18.062J Mathematics for...

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