Which of the given options provides the increasing order of asymptotic complexity of functions f1, f2, f3, and f4?
f1(n) = 2n
f2(n) = n(3/2)
f3(n) = n*log(n)
f4(n) = nlog(n)
(A)
f3, f2, f4, f1
(B)
f3, f2, f1, f4
(C)
f2, f3, f1, f4
(D)
f2, f3, f4, f1
Answer: (A)
Explanation:
f1(n) = 2^n
f2(n) = n^(3/2)
f3(n) = n*log(n)
f4(n) = n^log(n)
Except for f3, all other are exponential. So f3 is definitely first in the output. Among remaining, n^(3/2) is next. One way to compare f1 and f4 is to take log of both functions. Order of growth of log(f1(n)) is Θ(n) and order of growth of log(f4(n)) is Θ(log(n) * log(n)). Since Θ(n) has higher growth than Θ(log(n) * log(n)), f1(n) grows faster than f4(n).
Following is another way to compare f1 and f4. Let us compare f4 and f1.
Let us take few values to compare
n = 32, f1 = 2^32, f4 = 32^5 = 2^25
n = 64, f1 = 2^64, f4 = 64^6 = 2^36
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