2c03-review - 00027

# 2c03-review - 00027 - 3 = 6 × n × n 2 ≥ 60n 2 Then...

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1 CS2MD3. Sample solutions to the assignment 1 (many questions have more than one solutions). Total for this assignment is 214 pts. The assignment is worth 5%. If you think your solution has been marked wrongly, write a short memo stating where marking in wrong and what you think is right, and resubmit to me during class, office hours, or just slip under the door to my office. 1.[17] Using only definition of O(f(n)) proof that the following statements are true: You should use the following definition: g(n) = O(f(n)) if and only if C>0 n 0 >0 n>n 0 g(n) Cf(n) There are many solutions to the question 1. a.[3] 6n 3 /(log n + 1) = O(n 3 ) Let c=7, n 0 =2. For all n 2, log n 1, and log n +1 2 Then, 6n 3 /(log n +1) 6n 3 /2 = 3n 3 7n 3 hold. b.[3] 23=O(1) Let c=24, n 0 =0. For all n 0, 23 24 × 1 hold. c.[3] max(6n 3 ,10n 2 )=O(n 3 ) Let c=7, n 0 =10. For all n 10, 6n
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Unformatted text preview: 3 = 6 × n × n 2 ≥ 60n 2 . Then, max(6n 3 ,10n 2 ) ≤ 6n 3 < 7n 3 hold. d.[5] log n + n 1/2 = O(n 1/2 ). [2] First, we can show (for instance by induction) that, for all m ≥ 7, m 2 ≤ 2 m-1 . [3] Let c=2, n =128 =2 7 , for all n ≥ n , 2 m-1 < n ≤ 2 m , for some m. And m ≥ 7 Therefore, (log n) 2 ≤ m 2 ≤ 2 m-1 < n ⇒ log n ≤ n 1/2 ⇒ log n + n 1/2 ≤ 2n 1/2 e.[3] log n + n 1/2 = O(n) Let c=2, n =1. For all n ≥ 1, log n ≤ n, and n 1/2 ≤ n Then, log n + n 1/2 ≤ 2n hold. 2.[17] Using only definition of O(f(n)) proof that the following statements are false: You should use the following definition: g(n) = O(f(n)) if and only if ∃ C>0 ∃ n >0 ∀ n>n g(n) ≤ Cf(n), which means: g(n) ≠ O(f(n)) if and only if ∀ C>0 ∀ n >0 ∃ n>n g(n) > Cf(n)...
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## This note was uploaded on 12/10/2009 for the course CAS 2c03 taught by Professor Janicki during the Spring '03 term at McMaster University.

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