Question

Order the given functions by increasing growth rate.

f_1(n)=n^3f
1
​
(n)=n
3


f_2(n)=n^{0.3}f
2
​
(n)=n
0.3


f_3(n)=nf
3
​
(n)=n

f_4(n)=\sqrt{n}f
4
​
(n)=
n
​


f_5(n)=\frac{n^2}{\sqrt{n}}f
5
​
(n)=
n
​

n
2

​


f_6(n)=n^2f
6
​
(n)=n
2


As an answer, provide a string of length exactly six (with no spaces or quotes) containing the numbers of six functions (the first number indicates the function with the smallest growth rate, while the last number indicates the function with the largest growth rate). For example, for a similar, but simpler problem f_1(n)=n^2f
1
​
(n)=n
2
, f_2(n)=nf
2
​
(n)=n, f_3(n)=n^3f
3
​
(n)=n
3
the answer would be 213

Answer 1

Answer:

given below

Explanation:

IDK- I don't know

Answer 2

Answer:

They have given options provides the increasing order of asymptotic complexity of functions f1, f2, f3 and f4

Explanation:

From the above question,

They have given options provides the increasing order of asymptotic complexity of functions f1, f2, f3 and f4

f1(n) = $2^n$

 f2(n) = $n^(3/2)$

 f3(n) = nLogn

 f4(n) = $n^(Logn)$

OPTION:

              (A) f3, f2, f4, f1

              (B) f3, f2, f1, f4

              (C) f2, f3, f1, f4

              (D) f2, f3, f4, f1

ANSWER:

  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).

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