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Let $f(n) = n^2 \log n$ and $g(n) = n(\log n)^{10}$ be two positive functions of $n$. Which of the following statements is correct?

  1. $f(n) = O(g(n)) \text{ and } g(n) \neq O(f(n))$
  2. $g(n) = O(f(n)) \text{ and } f(n) \neq O(g(n))$
  3. $f(n) \neq O(g(n)) \text{ and } g(n) \neq O(f(n))$
  4. $f(n) =O(g(n)) \text{ and } g(n) = O(f(n))$
in Algorithms by Veteran (52.2k points)
edited by | 3.5k views
Here for getting quick answer first do minimization on both the sides then apply log on both the sides.

For cross verification of your answer follow whatever Arun ji has mentioned in his answer.
taking log in all case  is not a good practice .

it sometimes give right answer & sometime give wrong answer

if we take log of n^2  && n^100

both are same according to log

log is not always a deciding factor

i hope it help u


First do minimization on both the sides



f(n) = n2 log n and  g(n) = n(log n)10

which means 

f(n) = n*n log n and g(n) = n(log n)10

here n is common on both sides so cancel them

then we get,

n log n for f(n) and (log n)10   for g(n)

which means f(n) is larger.







=> n*10*logn

=> 10*n*logn

=>n*logn(10 is constant)

Now, n2logn > nlogn



if it is  $\log n^{10}$ then you can do 10 logn but it is not so


6 Answers

+24 votes
Best answer

A more formal approach:

$f(n) = n^2 \log n$

$g(n) = n (\log n)^{10}$

We can use the limit definition of $O$-notation

$\displaystyle{\lim_{n \to \infty }} \frac{f(n)}{g(n)} = 0, \implies  f(n) = o(g(n))$
small $o$ implying $f$ is strictly asymptotically lower than $g$. Also by definition, $o \implies O \text{ but } O \not\implies o$.

$\displaystyle{\lim_{n\to \infty}} \frac{f(n)}{g(n)} = c,  c > 0 \implies f(n) = \Theta(g(n))$

$\displaystyle{\lim_{n \to \infty}} \frac{f(n)}{g(n)} = \infty, \implies  f(n) = \omega(g(n))$
small $\omega$ implying $f$ is strictly asymptotically higher than $g$. Also by definition, $\omega \implies \Omega, \text{ but } \Omega \not \implies \omega.$

We can use this to prove the above question

For any $k > 0$

 $\displaystyle{\lim_{n \to \infty}} \frac{(\log n)^{k}}{ n}$

Applying L'Hôpital's rule,

$\implies \displaystyle{ \lim_{n \to \infty}} \frac{ k*(\log n)^{k-1}}{ n}$

$= \displaystyle{\lim_{n \to \infty}}  \frac{ k!}{n} = 0$

So, $(\log n)^k = o(n)$

Now for large $n$, $n >> (\log n)^9$

i.e., $n^{2}\log n >> n(\log n)^{10}$ (Multiplying both sides by $n \log n)$

So, $n(\log n )^{10} = o(n^{2}\log n)$ and  $n(\log n )^{10} \neq \Theta(n^{2}\log n)$ (making LHS strictly asymptotically lower than RHS)


$n(\log n )^{10} = O(n^{2}\log n)$ and $n^{2}\log n \neq O(n(\log n )^{10})$

Option $B$.

by Active (1k points)
edited by
is it equal ? how?


k! = k * (logn)k-1
+60 votes

$$\begin{array}{|l|c|c|} \hline \text {} & \textit{f(n)} & \textit{g(n)} \\\hline \text{$n = 2^{10}$} &  \text{$10 \times 2^{10} \times 2^{10}$} & \text{$2^{10} \times 10^{10}$}\\\hline \text{$n = 2^{256}$} & \text{$256 \times 2^{256} \times 2^{256}$} &\text{$2^{256} \times 256^{10}$}  \\\hline \end{array}$$
As $n$ is going larger, $f(n)$ is overtaking $g(n)$ and the growth rate of $f$ is faster than that of $g$. So, $g(n) = O(f(n))$ and $f(n) \neq O(g(n))$.

B choice. 

by Veteran (431k points)
edited by
If we calculate log(f(n)) and log(g(n)) then we get

log(f(n))= 2*log(n)+log(log(n))= O(log(n));

log(g(n))= log(n)+10*log(log(n))=O(log(n));

Can we say that f(n)=O(g(n)) and g(n)=O(f(n)) ?
the answer should be D right?
Taking log is not a proper method -- it works for some and fails for others.

ok sir thank you, and can u please reply to my question in TOC , i really have doubt in understanding the way to approach it thank you!

@Arjun Sir .Great , taking large values of n is probably the best way to solve such questions in gate , isnt it ?

But sir then how to decide what to apply log or not when as you had used log  here

+1 vote
g(n) = O (f (n))

n (logn)^10 <= C* n^2 logn

c is stand for any constant value, it satisfy one condition of option b so by this way you can get answer ..
by Active (4.7k points)
+1 vote

# First let's do the comparison between f(n) and g(n) 


     =>      $n^{2}\log n$     &&     $n(\log n)^{10}$   

     =>     $n\log n$        &&     $(\log n)^{10}$          (Cancel out the $n$ from both side )

     =>       $n$                 &&    $(\log n)^{9}$             (Cancel out the $\log n$ from both side )

     =>      $\log n$           &&      $9\log \log n$       (Take $\log$ both side )


Now, it is clear that if you take a large value of n then f(n) is greater than g(n)

=>      $\log n$     >     $9\log \log n$   


So, $g(n) =O(f(n))$ and $f(n) \neq O(g(n))$


Option B is correct.

by (57 points)
can we cancel terms while comparing complexity between 2 functions?
Yes, you can.
0 votes

Reference:Solution of Arjun Sir


$f(n)=n^{2} logn$ 

take $n=2^{10}$  So, $f(n)=2^{20} log2^{10}$  

$f(n)=2^{20} $ *10 $

and $g(n)= n (logn)^{10}$ 

take $n=2^{10}$  So, $g(n)= 2^{10}(log2^{10})^{10}$

$g(n)= 2^{10}* 10^{10}$


And now take one more value 

take $n=2^{256}$ 

o, $f(n)=2^{512} log2^{256}$  

$f(n)=2^{512} $ *256 $

and $g(n)= n (logn)^{10}$ 

take $n=2^{256}$  So, $g(n)= 2^{256}(log2^{256})^{10}$

$g(n)= 2^{256}* 256^{10}$

Now, main problem is how to check which one is bigger than other.

Just do one simple thing divide each other 

like in case-I , if you'll divide f(n)/g(n) 

f(n)/g(n) = $2^{20} $ *10  / $2^{10}$ * $10^{10}$

f(n)/g(n) = $2^{10} $ *10  / $10^{10}$

f(n)/g(n) = 10240 / $10^{10}$

So, here f(n)< g(n)  , f(n) = O(g(n))


Now Case-II,

f(n)/g(n) = $2^{512} $ *256  / $2^{256}$ * $256^{10}$

f(n)/g(n) = $2^{256} $ *256  / $256^{10}$

Here, you can simply use Scientific calculator during GATE exam, and you will see quotient is >0.

So, here f(n)> g(n)  , g(n) = O(f(n)) and you can find n where f(n) always greater than g(n)

So, Option B

by Active (4.1k points)
–5 votes
ans is A.

for n=2 : f(n)=4, g(n)=2

for n=3 : f(n)=14, g(n)=300
by Loyal (8.1k points)
For asymptotic notations, you should always check for large values of n.

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