Last Updated :
21 Dec, 2018
Match the following recurrence relation in List-I to their running time complexity in List-II:
List-I |
List-II |
(a) T(4n/5) + Θ(n) |
(i) Θ(n2logn) |
(b) 16·T(n/4) +n22log3n |
(ii) Θ(√nlogn) |
(c) 4T(n/2) + Θ(n2) |
(iii) Θ(n2log4n) |
(d) 2·T(n/4) +√n |
(iv) Θ(n) |
(A) (a) – (iii), (b) – (ii), (c) – (iv), (d) – (i)
(B) (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
(C) (a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)
(D) (a) – (i), (b) – (ii), (c) – (iv), (d) – (iii)
Answer: (B)
Explanation: According to Master theorem in a recurrence relation:
- T(4n/5) + Θ(n) = Θ(n)
- 16·T(n/4) +n22log3n
Using Master Theorem, we compare n2log3n withnlog416 = n2. This is another generalized version of the theorem, so we increment the logkn for Θ(n2log4n).
- 4T(n/2) + Θ(n2) = Θ(n2logn)
- 2·T(n/4) +√n
Using the master theorem, a = 2, b = 4, and logba = log42 = 1/2. Thus, we compare nlogba = n1/2 to f(n) = √n. Since these are asymptotically equivalent. This means that we gain an additional logn factor and T(n) = Θ(√nlogn).
So, option (B) is correct.
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