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The fall 2005 exam for cmsc 203, a computer science course focusing on mathematical induction and algorithms. It includes multiple choice questions and problems that require finding common sequences, writing pseudocode, ranking algorithm complexities, and proving theorems using mathematical induction and contradiction.
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CMSC 203 - Fall 2005 - Exam 2
1. Circle T if the corresponding statement is True or F if it is False. T F The sequence {1, 1, 2, 3, 5, 8, 13, 21,...} satisfies the Fibonacci relation. T F The First (Weak) Principle of Mathematical Induction implies the Second (Strong) Principles of Mathematical Induction. T F A recursive algorithm is one which never terminates. T F 2 + 4 + 6 + 8 + ... + 200 = 10100. T F Functions that are Logarithmic order grow faster than functions that are Linear order. T F In costing algorithms, the sum of the orders equals the maximum order. T F In costing algorithms, the product of the orders equals the order of the product. T F In general, Linear Search algorithms are just as fast Binary Search algorithms. 2. Let { an } and { bn } be the sequences defined, for n > 0, by:
an = n^2 − 2, and bn = n^3 + 2. Find c 0 , c 1 , c 2 , and c 3 when c (^) n = ( an )( bn ).
3. Given positive integer inputs, A and B, write out in pseudocode the algorithm to calculate (A MOD B) and (A DIV B). You may assume B < A. 4. Rank from 1 (least complex) to 10 (most complex) the complexity of algorithms with the following orders:
(b) Find the Big-Oh of the algorithm with complexity: ( n^4 + 1)(2 n + n^2 ) + (3 n^2 + 4 n^4 + 5)( n^2 ).
5. Use the Euclidean Algorithm to find GCD(1100, 35). 6. Prove one of the two Theorems below using Mathematical Induction.
Theorem 2: If a 0 = 0 and a 1 = 10, then an = an − 1 + a (^) n − 2 is divisible by 10.
7. Prove one of the two Theorems below: Theorem 1: If an integer divides two other integers, then it divides their sum.
Theorem 2: If a and b are odd, then ( a^2 + b^2 ) is even.
8. Prove one of the two Theorems below by Contradiction. Theorem 1: If a prime divides an integer, then it cannot divide the integer’s successor. Theorem 2: There does not exist a largest integer.
a i
i = 0
n
a n + 1