York University EECS 3101 October 8 2015 Solution to...

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York University EECS 3101 October 8, 2015 Solution to Homework Assignment #4 1. Yes. Consider any n e . Then, log n log e = 1 (since log and · are both increasing functions. So, p log n p log n · p log n = log n = 2 · 1 2 log n = 2 · log n. (In the definition of big-O notation, we are using n 0 = e and c = 2.) 2. No. Consider any c, n 0 . We must show there is an n n 0 such that n ! > c · 2 n . For n 3, n ! = n Y i =1 i = 2 · n Y i =3 i 2 · n Y i =3 3 = 2 · 3 n - 2 . Now, 2 · 3 n - 2 > c · 2 n 3 n - 1 > 3 c · 2 n - 1 (1 . 5) n - 1 > 3 c n > log(3 c ) log(1 . 5) + 1 . So, let n = max { n 0 , 3 , log(3 c ) log(1 . 5) +2 } . Then, n n 0 and, by the reasoning above, n ! > c · 2 n . 3. (a) It’s easy to write a programme to compute this value. It is 298/32 = 9.3125. 1 over. . .
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EECS 3101 Assignment 4 Solutions October 8, 2015 (b) We will use n 0 = 1. So, we want to prove that, for some constant c , T ( n ) cn for all n 1. Since T is defined using a recurrence, induction is a natural way to prove such a claim. Before we do the proof, let’s do some rough work to figure out which
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