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# T 0 cg n f n n n0 fn is bounded below by gn fn f

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Unformatted text preview: n) f(n) f ( n ) = O( g( n )) ≈ f ( n) ≤ g ( n) O(n 2 ) : n 2 , n 2 + n,10000n 2 + 999n n, n /1000, n1.99999 , n 2 / lg lg lg n n n0 NCKU IIM NCKU 資料結構 Chapter 3 資料結構 6 Asymptotic Lower Bound Ω( g (n)) = { f (n) | ∃c > 0, n0 >0 s.t. 0 ≤ cg (n) ≤ f (n) ∀n ≥ n0 } f(n) is bounded below by g(n), f(n) f ( n ) = Ω( g( n )) ≈ f ( n) ≥ g ( n) cg(n) Ω(n 2 ) : n 2 , n 2 + n,10000n 2 + 999n n3 , n 2.00001 , n 2 lg lg lg n, 2n n n0 NCKU IIM NCKU 資料結構 Chapter 3 資料結構 7 Properties Thm f (n) = O( g (n)) ⇔ g (n) = Ω( f (n)) ⎧ f (n) = O( g (n)) f (n) = Θ( g (n)) ⇔ ⎨ ⎩ f (n) = Ω( g (n)) ‧In general p ( n ) = ∑ i = 0 ai n i where ai are constant with ad > 0. d Then p ( n) = Θ(n d ). NCKU IIM NCKU 資料結構 Chapter 3 資料結構 8 Example of Θ Show that ¼ n3 + 4n2 + 24 is Θ(n3). By definition, we must show that c1 n3 ≤ ¼ n3 + 4n2 + 24 ≤ c2 n3 (1) Divide both sides by n3 to get c1 ≤ ¼ + 4/n + 24/n3 ≤ c2 (2) 1. Notice that for any n ≥ 0, equation (2) will never go below ¼. So pick c1 = ¼. Here, we have shown (1) is (1) Ω(n3). 2. Now see that, for n = 1, rhs of (2) reduces to ¼ + 4 + 24 < 29, which is an upper bound for n ≥ 1. So pick c2 =29 and n0 =1. This shows that (1) is Ο(n3). (1) 3. By thm 3.1, ¼ n3 + 4n2 + 24 is Θ(n3). NCKU IIM NCKU 資料結構 Chapter 3 資料結構 9 Asymptotic Notation in Equations • When on rhs: O(n 2 ) stands for some anoymous function in the set O(n 2 ) 2n 2 + 3n + 1 = 2n 2 + Θ(n) means ∃f (n) ∈ Θ(n) such that 2n 2 + 3n + 1 = 2n 2 + f (n) • When on lhs: no matter how the anonymous functions are chosen on the lhs, there is a way to choos...
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