Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page (16 points) Circle T if the corresponding statement is True or F if it is False.

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1 Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page ( points) Circle T if the corresponding statement is True or F if it is False T F GCD(,0) = 0 T F For every recursive algorithm, there is an equivalent sequential algorithm T F In general, sequential algorithms use memory less efficiently than their equivalent recursive representations T F If a and b are positive integers, then a = b(a div b) + (a mod b) T F = T F Algorithms whose order is O(nlogn) are less efficient than those of order O(n 2 ) T F Given two algorithms, f(n) and g(n), O(f + g) = max(o(f ), O(g)) T F GCD( p,q) = GCD(q, p mod q) 2 (8 points) Let {a n } and {b n } be the sequences defined, for n > 0, by: a n = (n + ), and b n = (n + 2) Find c, c 2, c 3, and c 4 when c n = (a n )(b n ) + n c = (a )(b ) + = ( + )( + 2) + = 2(3) + = + = 7; c 2 = (a 2 )(b 2 ) + 2 = (2 + )(2 + 2) + 2 = 3(4) + 2 = = 4; c 3 = (a 3 )(b 3 ) + 3 = (3 + )(3 + 2) + 3 = 4(5) + 3 = = 23; c 4 = (a 4 )(b 4 ) + 4 = (4 + )(4 + 2) + 4 = 5() + 4 = = 34 3 (8 points) Write out the Division Algorithm and trace its steps to calculate (54 MOD 2) INPUT(a,d) SET DIV = 0 SET MOD = a WHILE (MOD > d) DIV = DIV + MOD = MOD d ENDWHILE OUTPUT(MOD) Step DIV MOD (MOD > 2) 0 OUTPUT: 54 MOD 2 =

2 Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page 2 4 (5 points) (a) Give a Recursive Definition for the even Natural numbers Basis: 0 is an even Natural number Inductive: If n is an even Natural number, then (n + 2) is an even Natural number OR If n is an even Natural number, then 2n is an even Natural number (b) Find the Big-Oh of the algorithm with complexity: (n 3 + 3)(n + ) + [3n 3 + 2n 2 (log n)] (n 3 + 3)(n + ) + [3n 3 + 2n 2 (log n)] < (n 3 + 3n 3 )(n + n ) + [3n 3 + 2n 2 (n)] = (4n 3 )(7n ) + (3n 3 + 2n 3 ) = (4n 3 )(7n ) + (3n 3 + 2n 3 ) = 28n 9 + 5n 3 < 28n 9 + 5n 9 = 33n 9, hence (n 3 + 3)(n + ) + [3n 3 + 2n 2 (log n)] is O(n 9 ) 5 (8 points) (a) Given a = and b = , then GCD(a,b) = and LCM(a,b) = (b) List out the search intervals in applying the Binary Search algorithm to find 4 in the list: Pass : ( ) ( ) 4 = 0, with 0 = lower = upper Pass 2: (8 9 ) ( ) Pass 3: (2 3) (4 5) Pass 4: (4) (5)

3 Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page 3 (5 points) Prove ONE of the TWO Theorems below using Mathematical Induction Theorem : For all integers n >, i 2 = nn ( + ) ( 2n + ) n Theorem 2: If a 0 = 5, a = 0, and a 2 = 5, then a n = a n + a n 2 + a n 3 is a multiple of 5, for all n > 3 Theorem : (Weak Induction) Basis: Show i 2 = ( + ) ( 2 ( ) + ) Now, for n =, i 2 = 2 = nn, and ( + ) ( 2n + ) = ( )( 3) = -- =, n therefore i 2 = nn ( + ) ( 2n + ), when n = k Induction: Assuming i 2 = kk ( + ) ( 2), show i 2 ( ) ( k + 2) ( 2k + 3) = Now, i 2 = i 2 + i 2 = kk ( + ) ( 2) + ( ) 2 = ()[k(2) + ()] / k i = k+ = ()(2k 2 + k + k + ) / = ()(2k 2 + 7k + ) / = ()(k + 2)(2k + 3) / QED Theorem 2: (Strong Induction) Basis: Show a 3 is a multiple of 5 Now, a 3 = a 2 + a + a 0 = = 30 = (5), hence a 3 is a multiple of 5 Induction: Assuming a i is a multiple of 5 for i = 4, 5,, (k 2), (k ), k, show a k+ is a multiple of 5 Now, a k+ = a k + a k + a k 2 Since a k, a k and a k 2 are each a multiple of 5 by the induction hypothesis, it follows that there are integers, p, q, and r, such that a k = 5p, a k = 5q, and a k 2 = 5r Thus, a k+ = a k + a k + a k 2 = 5p + 5q + 5r = 5(p + q + r) Since p, q, and r are integers, (p + q + r) is an integer, therefore a k+ is a multiple of 5 QED

4 Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page 4 7 (5 points) Prove ONE of the TWO Theorems below: Theorem : If a, b, and m are positive integers, with (a mod m) = (b mod m), then a b mod m Theorem 2: The difference of the squares of successive positive integers is odd Theorem : Proof: Let a, b, and m be positive integers, with (a mod m) = (b mod m) Now, by the Quotient-Remainder Theorem: a = m(a div m) + (a mod m) and b = m(b div m) + (b mod m), so a m(a div m) = (a mod m) = (b mod m) = b m(b div m), allowing us to deduce that a m(a div m) = b m(b div m) Reworking this equation, we get a b = m(a div m) m(b div m) = m[(a div m) (b div m)] Observing that (a div m) and (b div m) are Integers, we conclude that [(a div m) (b div m)] is an integer, hence a b is an integer multiple of m, therefore a b mod m QED Theorem 2: Proof: Let a be a positive integer Show that [(a + ) 2 a 2 ] is odd Now, [(a + ) 2 a 2 ] = [(a + )(a + ) a 2 ] = [(a 2 + 2a + ) a 2 ] = 2a + Since a is a positive integer, we have that 2a + is an odd integer, therefore [(a + ) 2 a 2 ] is odd QED

5 Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page 5 8 (5 points) Prove ONE of the TWO Theorems below by Contradiction or Contraposition Theorem : There is no largest Natural number Theorem 2: If n is an integer and n 2 is odd, then n is odd Theorem : Proof: (Contradiction) Assume there is a largest Natural number Denote M as this number Now, since M is a Natural number, it follows that (M + ) is a Natural number However, since M is the largest Natural number, we see that: But this yields hence < 0, a contradiction (M + ) < M (M + ) M < M M Therefore, there is no largest Natural number QED Theorem 2: Proof: (Contraposition) Show that is n is an integer and n is even, then n 2 is even Let n be an integer and assume n is even; that is, assume n = 2k for some integer k Now, n 2 = (2k) 2 = (2k)(2k) = 2(2k 2 ) Since k is an integer, it follows that k 2 and 2k 2 are integers Therefore n 2 is an even integer QED