GATE2008-39

Question

Consider the following functions:

$f(n) = 2^n$

$g(n) = n!$

$h(n) = n^{\log n}$

Which of the following statements about the asymptotic behavior of $f(n)$, $g(n)$ and $h(n)$ is true?

• A. $f\left(n\right)=O\left(g\left(n\right)\right); g\left(n\right)=O\left(h\left(n\right)\right)$

• B. $f\left(n\right) = \Omega\left(g\left(n\right)\right); g(n) = O\left(h\left(n\right)\right)$

• C. $g\left(n\right) = O\left(f\left(n\right)\right); h\left(n\right)=O\left(f\left(n\right)\right)$

• D. $h\left(n\right)=O\left(f\left(n\right)\right); g\left(n\right) = \Omega\left(f\left(n\right)\right)$

Comments

I understand growth rate h(n)<f(n)<g(n)

but how could g(n) is equal to best case of f(n)

g(n)=Ω(f(n))???

Answer 1

$g(n) = n!$ on expanding factorial we get $g(n) = \mathcal{O}(n^n)$ :
$$\begin{align*} n^n &> n^{\log n} \\ n^n &> 2^n \end{align*}$$
this condition is violated by option A, B and C by first statements of each Hence, they cannot be said true.

second statement of option D says that $g(n)$ is asymptotically biggest of all.

answer = option D

Comments

$h(n)$ can be rewritten as : $\Large{2^{logn^{logn}}}$ $=2^{(logn)^2}  \color{Red}< \ 2^n \ \{f(n)\}$

Answer 2

	$f(n)$	$g(n)$	$h(n)$
$n = 2^{10}$	$2^{1024}$	$1024!$	${2^{10}}^{10} = 2^{100}$
$n = 2^{11}$	$2^{2048}$	$2048!$	$2^{121}$
Increase	$2^{1024}$	$1025 \times 1026 \times \dots \times 2048$	$2^{21}$

 

So, growth rate is highest for $g(n)$ and lowest for $h(n)$. Option D. 

Answer 3

Take log of all function with base 2

log(f(n)) = Log(2^n)  = n

log(g(n)) = Log(n!) = Log(n^n) // using sterling approximation = nlogn

Log(h(n)) = Log(n ^ logn) = log(n) * log(n).

It becomes clear now that h(n) < f(n) < g(n).

Looking at options D is only option which satisfy this constraints.

Answer 4

Answer is D.

1 < loglogn < logn < n^e < n^c < n^logn < cⁿ < nⁿ < c^{c^n} < n! .

Comments

1 < loglogn < logn < n^e < n^c < n^logn < cⁿ < nⁿ < c^{c^n} < n!

isn't n! = O(n^n)?

Answer 5

Take log on both sides of the functions and compare each functions.

eg   2n   n!

taking log on both sides

nlog(base 2)  log(n!)

and put n = 2^128 or some 2^k values

 ll get

nlog(base2) < log(n!)

and simlarly for 2^n and n^logn

 

now, 2^n > n^logn

 

so n! > 2^n > n^logn

 

Answer is (D).

 

correct me if i'm wrong

2n