Question

Which of the given options provides the increasing order of asymptotic complexity of functions?

Answer 1

Answer:

f3, f2, f4, f1

Step by step explaination:

f1(n) = 2^n

f2(n) = n^(3/2)

f3(n) = nLogn

f4(n) = n^(Logn)

Except f3, all other are exponential. So f3 is definitely first in 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(Logn * Logn). Since Θ(n) has higher growth than Θ(Logn * Logn), f1(n) grows faster than f4(n).