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MT-F97.sol - COSC3101.03 Solutions to Mid-Term Test 1 a b c...

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COSC3101.03 Fall 1997-98 Solutions to Mid-Term Test 1 a. True. f ( n ) cg ( n ) implies lgf ( n ) lgc + lgg ( n ) ( lgc + 1) lgg ( n ), since lgg ( n ) 1. b. False. For example, 2 n + 2 = O ( n + 2), but 2 2 n + 2 = Θ (4 n ) O (2 n + 2 ). c. True. n 5 lg n = O ( n 6 / lg n ). d. True. Compare the logarithm of each term. The largest term is 2 n lg n = n n . e. True. Both are Θ (1). See chapter 3 of [CLR]. f. True. For a similar recurrence, see pages 59-61 of [CLR]. g. False. It is Θ ( n 2 ). h. True. See your lecture notes or section 10.2 of [CLR]. i. False. It is n - 1. j. False. It is Θ ( n lg n ). See your lecture notes or section 35.4 of [CLR]. 2 (a) Here the recurrence is of the form T ( n ) = aT ( n / b ) + Θ ( n d ). Apply the Master Theorem: since log b a = log 4 8 = 3/2 > d = 1/2, the solution is T ( n ) = Θ ( n log b a ) = Θ ( n f8e5 f8e5 n ). (b) Similar to part (a), but now log b a = d = 3/2. Thus, T ( n ) = Θ ( n d lg n ) = Θ ( n f8e5 f8e5 n lg n ). (c) f ( n ) is not a polynomial, so we cannot apply the Master Theorem. Instead, we start with the general solution formula: T ( n ) = Θ ( n log b a [1 + k i = 1 Σ f ( b i )/ a i ] ) , where k = log b n = Θ ( lg n ) = Θ ( n f8e5
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