Homework Answers
The order will be
12 < log(n) < (n)^(1/3) < sqrt(n) < (n)^3 <
(2)^(log(n)) < (2)^(sqrt(n)) < (2)^(n) < (3)^(n)
i.e,
f3(n) < f7(n) < f2(n) < f5(n) < f9(n) < f4(n) <
f8(n) < f6(n) < f1(n)
Here
(1)initially 12 is constant so it will give O(1) constant time so
it is the least of all
(2)log(n) is the O(log n) which is logarithmic time
(3)(n)^(1/3) since it is third root it will be less that
sqrt(n)
(4) sqrt(n) it is O(sqrt(n))
(5)(n^3) it is O(n^3) which is cubic time
(6)(2)^(log(n)) it is exponential since it is logarithmic it is
smaller than remaining
(7)(2)^(sqrt(n)) it is exponential since the exponent has sqrt(n)
it is larger that logarithmic
(8)(2)^(n) it is exponential time
(9)(3)^(n) it is also exponential since the base is high it will be
high
If you have any doubts please comment and please don't dislike.
2^log(n) = n.
So that should be in front of n^3. Except for that the answer is correct.
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