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解三題簡單的:
a. T (n) = 3T (n/2) + n lg n
We have f (n) = nlgn and ( n^log a) = nlg 3 = n^lg3 ≒ n^1.585
b
Since n lg n = O(n^lg 3-ε) for any 0 < ε≦ 0.58
by case 1 of the master theorem, we have T(n) = θ(n^lg 3)
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b.T (n) = 4T (n/2) + n^2√n
We have f (n) = n^2(√n) = n^5/2 and n^log a = n^log 4 = n^lg 2.
b 2
Since n^5/2 = Ω(n^lg 2+3/2),
we look at the regularity condition in case 3 of the master theorem.
We have af(n/b) = 4(n/2)^2 (√n/2) = (n^5/2)/(√2) ≦ cn^5/2,for 1/√2 ≦ c < 1
Case 3 applies, and we have T (n) = θ(n^2 √n)
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g.T(n) = T(n-1) + 1/n
This recurrence corresponds to the harmonic series, so that T (n) = Hn,
where Hn = 1/1+1/2+1/3+· · ·+1/n. For the base case, we have T (1) = 1 = H1.
For the inductive step, we assume that T(n-1) = Hn-1, and we have
T (n) = T (n-1) + 1/n
= Hn-1 + 1/n
= Hn
Since Hn = θ(lgn) by equation,we have that T(n) = θ(lgn).
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◆ From: 140.112.5.53
※ 編輯: jerry771210 來自: 140.112.5.53 (03/18 00:13)
※ 編輯: jerry771210 來自: 140.112.5.53 (03/18 00:15)