45 votes 45 votes Consider the following two functions: $g_1(n) = \begin{cases} n^3 \text{ for } 0 \leq n \leq 10,000 \\ n^2 \text{ for } n > 10,000 \end{cases}$ $g_2(n) = \begin{cases} n \text{ for } 0 \leq n \leq 100 \\ n^3 \text{ for } n > 100 \end{cases}$ Which of the following is true? $g_1(n) \text{ is } O(g_2(n))$ $g_1(n) \text{ is } O(n^3)$ $g_2(n) \text{ is } O(g_1(n))$ $g_2(n) \text{ is } O(n)$ Algorithms gate1994 algorithms asymptotic-notation normal multiple-selects + – Kathleen asked Oct 4, 2014 • recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 15.6k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 71 votes 71 votes For asymptotic complexity, we assume sufficiently large $n$. So, $g_1(n) = n^2$ and $g_2(n) = n^3$. Growth rate of $g_1$ is less than that of $g_2$, i.e., $g_1(n) = O(g_2(n)).$ Options $A$ and $B$ are TRUE here. Arjun answered Oct 22, 2014 • edited Jun 24, 2018 by Shikha Mallick Arjun comment Share Follow See all 14 Comments See all 14 14 Comments reply Show 11 previous comments Supriya Bhide commented Jan 28, 2021 reply Follow Share Won't C also be the answer along with A and B? 0 votes 0 votes rhl commented Sep 3, 2021 i edited by rhl May 30, 2023 reply Follow Share For sufficiently large n , $g_2 =n^3$ and $g_1 = n^2$, $n^2$ is not big oh of $n^3$. Hence, $g_2$ is not O($g_1$) 1 votes 1 votes KpKarenga_01 commented Dec 20, 2023 reply Follow Share Please explain in a manner way.! Why g2 is not O(g1)..?? 0 votes 0 votes Please log in or register to add a comment.
17 votes 17 votes The answer is given... prithatiti answered Jun 26, 2018 prithatiti comment Share Follow See all 2 Comments See all 2 2 Comments reply Vink7389 commented Apr 8, 2021 reply Follow Share What should be the value of n here so onwards n, g1(n) = O(g2(n)) always? 2 votes 2 votes prithatiti commented Nov 18, 2022 reply Follow Share I think 101 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes Yes. Both (a) and (b) are correct. $n^{2}$ is $O(n^{3})$. gatecse answered Sep 20, 2014 • edited Oct 16, 2017 by kenzou gatecse comment Share Follow See all 2 Comments See all 2 2 Comments reply Marv Patel commented Sep 20, 2014 reply Follow Share I though so but when paper says select the correct(only one) choice then it creates doubt! 0 votes 0 votes air1ankit commented Nov 24, 2018 reply Follow Share question has only one answer whether you look from left or right. Before 2000 GATE had questions with multiple correct answers and you were given mark only if all are marked. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Index Condition $g_{1}(n)$ $g_{2}(n)$ Time Complexity($B$) Time Complexity($A$) 1 $0 \leq n \leq 100$ $n^{3}$ $n$ $O(n^{3})$ $O(g^{2}(n))$ -- Fails 2 $101 \leq n \leq 10000$ $n^{3}$ $n^{3}$ $O(n^{3})$ $O(g^{2}(n))$ 3 $n \geq 10001$ $n^{2}$ $n^{3}$ $O(n^{3})$ $O(g^{2}(n))$ Thus the right option should be B Avik10 answered Jan 26, 2017 • edited Oct 16, 2017 by kenzou Avik10 comment Share Follow See all 4 Comments See all 4 4 Comments reply Syedarshadali commented Jan 7, 2018 reply Follow Share Make this best answer 0 votes 0 votes Arjun commented Feb 22, 2018 reply Follow Share This is wrong; big-O cares for only large $n$. 2 votes 2 votes lokeshsolanki17 commented May 10, 2020 reply Follow Share They are not asking about time complexity in question. They are asking about growth of functions g1(n) and g2(n). g1( n ) = O( g2(n) ) g1(n) <= c.g2(n) ; where n >= N and c > 0 here N value is 10,000. so option A is correct. because after N, g1(n) = n2, and g2(n) = n3 according to option B, g1( n ) = O(n3) is not tight but true because g1(n) belongs to set O(n3).so it is also true 1 votes 1 votes MohanK commented Dec 31, 2020 reply Follow Share @lokeshsolanki17, @Arjun Sir, what does it actually meant by “They are not asking about time complexity in question. They are asking about growth of functions g1(n) and g2(n)” ? Isn’t Time complexity inherently uses order of functions ? Is both Time complexity & growth of functions different ? 0 votes 0 votes Please log in or register to add a comment.