3 votes 3 votes Algorithms algorithms asymptotic-notation test-series + – Sayan Das 1 asked Sep 21, 2016 • retagged Jul 13, 2022 by makhdoom ghaya Sayan Das 1 804 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes It is clear that here O(nlogn) is the biggest fn. Confusion is between f(n) and g(n) who is bigger? Take an asymptoticaly larger value for i.e. n=>$2^{10^{9}}$ now f(n)= $2^{100}$ and g(n) = $10^{9}$ =$2^{30}$ Hence g(n)<f(n)<h(n) So finally Option D is Correct ans. Rajesh Pradhan answered Sep 21, 2016 • selected Nov 3, 2016 by Sayan Das 1 Rajesh Pradhan comment Share Follow See all 2 Comments See all 2 2 Comments reply Sayan Das 1 commented Sep 21, 2016 reply Follow Share I checked it with n=10^99 but for that f(n) is coming 1 and g(n)=328 so how a larger value i can choose? 0 votes 0 votes Rajesh Pradhan commented Sep 21, 2016 reply Follow Share Asymptotic means very very large Value..So We may take 2 to the power values due to log in g(n) so that we can cancel 2. And here I took 2^10^9 bcz n^10^-7 was there .. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Take logarithm of all functions and then compare. f = 0.0000001 * log(n) g = log( log(n) ) h = log(n) + log( log(n) ) So, order is g< f <= h (equal to is used bacuse we are considering asymptotic notation) So, option (D) is true. Sushant Gokhale answered Oct 31, 2016 Sushant Gokhale comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes In your question , $h\left ( n \right )> g\left ( n \right )> f\left ( n \right )$ $\Rightarrow 0\leq f\left ( n \right )\leq c*g\left ( n \right )\Rightarrow f\left ( n \right )=O\left ( g\left ( n \right ) \right )$ $\Rightarrow0\leq f\left ( n \right )\leq c*h\left ( n \right )\Rightarrow f\left ( n \right )=O\left ( h\left ( n \right ) \right )$ Answer C sourav. answered Sep 21, 2016 sourav. comment Share Follow See all 0 reply Please log in or register to add a comment.