GATE | Gate IT 2008 | Question 42

When n = 2^2k for some k ≥ 0, the recurrence relation
T(n) = √(2) T(n/2) + √n, T(1) = 1
evaluates to :

(A)

√(n) (log n + 1)

(B)

√(n) (log n )

(C)

√(n) log √(n)

(D)

n log √(n)

Answer: (A)
Explanation:

Please note that the question is asking about exact solution. Master theorem provides results in the form of asymptotic notations. So we can\’t apply Master theorem here. We can solve this recurrence using simple expansion or recurrence tree method.

T(n) = √(2) T(n/2) + √n
     = √(2) [√(2) T(n/4) + √(n/2) ] + √n
     = 2 T(n/4) + √2 √(n/2) +√n
     = 2[ √2 T(n/8) + √(n/4) ]+√2 √(n/2)+√n
     = √(2^3) T(n/8)+ 2 √(n/4) + √2 √(n/2) +√n
     = √(2^3) T(n/8)+√n +√n +√n
     = √(2^3) T(n/(2^3))+3√n
     .............................................
     = √(2^k) T(n/(2^k))+k√n
     = √(2^logn) + logn √n
     = √n + logn √n
     = √n(logn +1)

Alternate Solution : This question can be easily done by substitution method look: T(1)= 1; GIVEN. Now use n=2 in the given recurrence relation which gives 2*(1.414) (since value of root over 2 is 1.414) now by looking at the options use n=2 which satisfies option A.

Quiz of this Question
Please comment below if you find anything wrong in the above post