Chapter 1
15) We show that n2 + 3n3 ∈ O(n3) because for n≥0,
n2 + 3n3 ≤ 4n3 ,
we can take c = 4, N = 0 to obtain our result.
We show that n2 + 3n3 ∈ Ω(n3) because for n≥0,
n2 + 3n3 ≥ 3n3 ,
we can take c = 3, N = 0 to obtain our result.
Thus, since n2 + 3n3 ∈ O(n3) and n2 + 3n3 ∈ Ω(n3), n2 + 3n3 ∈ Θ(n3).
16) 6n2 + 20n
O(n3)：c=1
6n2 + 20n ≤ n3
⇒6n + 20 ≤ n2
N=9
3
Ω (n )：c=0
6n2 + 20n ≤ 0
⇒3n + 10 ≤ 0
⇒3n ≤ -10
N=-4
18) p(n) = aknk + ak-1nk-1 + … + a1n + a0
p(n) ∈Θ(nk)
aknk + ak-1nk-1 + … + a1n + a0
let c = ak + ak+1 + … + a1 + a0
then aknk + ak-1nk-1 + … + a1n + a0
≤ aknk + ak-1nk + … + a1nk + a0 nk
= ( ak + ak-1 + … + a1 + a0)nk
=cnk
⇒aknk + ak-1nk-1 + … + a1n + a0 ∈ O(nk)
since aknk + ak-1nk-1 + … + a1n + a0 ≥ aknk
⇒aknk + ak-1nk-1 + … + a1n + a0 ∈ Ω (nk)
thus aknk + ak-1nk-1 + … + a1n + a0 ∈ Θ (nk)
19) n ㏑ n、5n2+7n、nn、nn+㏑ n、10n+n20；
5 ㏒ n、8n+12、en、√n；(㏒ n)2、㏒(n!) ；
2n!、n5/2、n! 、(㏒ n)!
Chapter 2
15)
T(n)=5T(n/3)+g(n)
16)
T(n)=10T(n/3)+ Θ(n2)
17)
If at least S(n) to finish
S(n+1) = S(n)+1+S(n)
= 2S(n)+1 ⇒ the least move times
S(1)=1
S(2)=S(1)+1+S(1)=2S(1)+1
S(3)=2S(2)+1
⇓
S(n)=2S(n-1)+1=2n-1+2n-2+…+2+1
S(n+1)=2S(n)+1=2(2n-1+2n-2+…+2+1)+1
=2n+2n-1+…+2+1
=(2n+1-1)/(2-1)
=2n+1-1
⇒S(n)=2n-1
33)
(a) T(n) = 9T(n/3)+ Θ(n)
(b) T(n) = 16T(n/4)+ Θ(n)