2W6B7770Ed01

2W6B7770Ed01 - ··· + c n = c n +1-1 c-1 ≤ c c-1 · c n...

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Solutions to Homework One Peng Du University of California, San Diego 1 Exercise 0.2 1.1 Part (a) g ( n ) = O (1) because g ( n ) = 1 + c + ··· + c n = 1 - c n +1 1 - c 1 1 - c = c 1 · 1 where c 1 = 1 1 - c . g ( n ) = Ω (1) because g ( n ) = 1 + c + ··· + c n c 2 · 1 where c 2 = 1. g ( n ) = O (1) and g ( n ) = Ω (1) implies that g ( n ) = Θ (1). 1.2 Part (b) g ( n ) = O ( n ) because g ( n ) = 1 + c + ··· + c n = n + 1 c 1 · n where c 1 = 2. g ( n ) = Ω ( n ) because g ( n ) = 1 + c + ··· + c n = n + 1 c 2 · n where c 2 = 1. g ( n ) = O ( n ) and g ( n ) = Ω ( n ) implies that g ( n ) = Θ ( n ). 1.3 Part (c) g ( n ) = O ( c n ) because g ( n ) = 1+ c +
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Unformatted text preview: ··· + c n = c n +1-1 c-1 ≤ c c-1 · c n = c 1 · c n where c 1 = c c-1 . g ( n ) = Ω ( c n ) because g ( n ) = 1 + c + ··· + c n ≥ c n = c 2 · c n where c 2 = 1. g ( n ) = O ( c n ) and g ( n ) = Ω ( c n ) implies that g ( n ) = Θ ( c n ). 2 Exercise 0.3 Please see the next solution for details....
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This note was uploaded on 01/31/2010 for the course CSE 101 taught by Professor Staff during the Fall '08 term at UCSD.

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