Home
About Us
Contact Us
Bookmark
Saved Bookmarks
Current Affairs
General Knowledge
Chemical Engineering
UPSEE
BSNL
ISRO
BITSAT
Amazon
ORACLE
Verbal Ability
→
GATE
→
Cs 2022 in Gate
→
The recurrence relation T(1) = 2 T(n) = 3T(n/4)+nh...
1.
The recurrence relation T(1) = 2 T(n) = 3T(n/4)+nhas the solution, T(n) equals to(A) O(n)(B) O(log n)(C) O(n^3/4)(D) None of the above
Answer»
Show Answer
Discussion
No Comment Found
Post Comment
Related InterviewSolutions
The formula used to compute an approximation for the second derivative of a function f at a point X0 is(A) A(B) B(C) C(D) D
Let Ax=b be a system of linear equations where A is an m×n matrix and b is a m×1 column vector and X is an n×1 column vector of unknowns. Which of the following is false?(A) The system has a solution if and only if, both A and the augmented matrix [Ab] have the same rank(B) If m(C) If m=n and b is a non-zero vector, then the system has a unique solution(D) The system will have only a trivial solution when m=n, b is the zero vector and rank(A) = n
Which of the following statement is false?(A) The set of rational numbers is an abelian group under addition(B) The set of integers in an abelian group under addition(C) The set of rational numbers form an abelian group under multiplication(D) The set of real numbers excluding zero is an abelian group under multiplication
Newton-Raphson iteration formula for finding 3√c, where c > 0 is.(A) a(B) b(C) c(D) d
Suppose X and Y are sets and |X| and |Y| are their respective cardinalities. It is given that there are exactly 97 functions from X to Y. From this one can conclude that(A) |X|=1,|Y|=97(B) |X|=97,|Y|=1(C) |X|=97,|Y|=97(D) None of the above
Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is(A) 1/36(B) 1/3(C) 25/36(D) 11/36
Quicksort is run on two inputs shown below to sort in ascending order taking first element as pivot,(i) 1, 2, 3,......., n(ii) n, n-1, n-2,......, 2, 1 Let C1 and C2 be the number of comparisons made for the inputs (i) and (ii) respectively. Then,(A) C1 < C2(B) C1 > C2(C) C1 = C2(D) We cannot say anything for arbitrary n
The recurrence relation T(1) = 2 T(n) = 3T(n/4)+nhas the solution, T(n) equals to(A) O(n)(B) O(log n)(C) O(n^3/4)(D) None of the above
Define for a context free language L⊆ {0, 1}* , init(L)={u ∣ uv ∈ L for some v in {0,1}∗} ( in other words, init(L) is the set of prefixes of L)Let L = {w ∣ w is nonempty and has an equal number of 0’s and 1’s}Then init(L) is(A) the set of all binary strings with unequal number of 0’s and 1’s(B) the set of all binary strings including null string(C) the set of all binary strings with exactly one more 0 than the number of 1’s or one more 1 than the number of 0’s(D) None of the above
A solution to the Dining Philosophers Problem which avoids deadlock is:(A) ensure that all philosophers pick up the left fork before the right fork(B) ensure that all philosophers pick up the right fork before the left fork(C) ensure that one particular philosopher picks up the left fork before the right fork, and that all other philosophers pick up the right fork before the left fork(D) None of the above
Reply to Comment
×
Name
*
Email
*
Comment
*
Submit Reply
Your experience on this site will be improved by allowing cookies. Read
Cookie Policy
Reject
Allow cookies