chapter0 - Chapter 0 Solutions January 30, 2007 2 0.1. a)...

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Chapter 0– Solutions January 30, 2007
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2 0.1. a) n 100 = Θ( n 200) b) n 1 / 2 = O ( n 2 / 3 ) c) 100 n + log n = Θ( n + (log n ) 2 ) d) n log n = Θ(10 n log 10 n ) e) log 2 n = Θ(log 3 n ) f) 10 log n = Θ(log( n 2 )) g) n 1 . 01 = Ω(log 2 n ) h) n 2 / log n = Ω( n (log n ) 2 ) i) n 0 . 1 = Ω((log n ) 10 ) j) (log n ) log n = Ω( n/ log n ) k) n = Ω((log n ) 3 ) l) n 1 / 2 = O (5 log 2 n ) m) n 2 n = O (3 n ) n) 2 n = Θ(2 n +1 ) o) n ! = Ω(2 n ) p) (log n ) log n = O (2 (log 2 n ) 2 ) q) n i =1 i k = Θ( n k +1 ) 0.2. By the formula for the sum of a partial geometric series, for c n = 1: g ( n ) = 1 c n +1 1 c = c n +1 1 c 1 . a) 1 > 1 c n +1 > 1 c . So: 1 1 c > g ( n ) > 1. b) For c = 1, g ( n ) = 1 + 1 + ··· + 1 = n + 1. c) c n +1 > c n +1 1 > c n . So: c 1 c c n > g ( n ) > 1 1 c c n . 0.3. a) Base case: F 6 = 8 2 6 / 2 = 8. Inductive Step: for n 6, F n +1 = F n + F n 1 2 n/ 2 +2 ( n 1) / 2 = 2 ( n 1) / 2 (2 1 / 2
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This note was uploaded on 04/30/2010 for the course L&S 101 taught by Professor Chow during the Spring '10 term at University of California, Berkeley.

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chapter0 - Chapter 0 Solutions January 30, 2007 2 0.1. a)...

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