Question-67

Question-67

What is the largest n for which one can solve within one second using an algorithm that requires f(n) bit operations, where each bit operation is carried out in 10^{-8}seconds, with these functions f(n) ?

For questions (a) to (c) , enter the exact answers. Enter brackets around exponents. Round your answers down to the nearest integer for all other parts. Note that log n = log2 n.

a)     log n

n = ?

 

b)    n

n = ?

 

c)     n. log n

n^{n} = ?

 

d)    n^{2}

n = ?

 

e)     2^{n}

n =?

 

Solution

a)log n = 10^{8}

n = 2^{10^{8}}

 

b) n = 10^{8}

 

c) n.log n = 10^{8}

Log(n^{n}) =10^{8}

n^{n} = 2^{10^{8}}

 

d) n^{2} = 10^{8}

n = \sqrt{10^{8}}

~ 10000

 

e) 2^{n} = 10^{8}

Log 2^{n} = log 10^{8}

n = 8. Log 10

~ 26

 

Leave a Reply

Your email address will not be published. Required fields are marked *