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GATE | GATE CS 1996 | Question 37

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The recurrence relation

 T(1) = 2
 T(n) = 3T(n/4)+n

has the solution, T(n) equals to

(A)

O(n)

(B)

O(log n)

(C)

O(n^3/4)

(D)

None of the above


Answer: (A)

Explanation:

As the recurrence equation is given:
    = 3 (3t(n/4^2)+ n/4) +n

   = 3^2t(n/4^2 ) + n(1+3/4)

  = 3^2(3t(n/4^3) +n/4^2) + n(1+3/4)

 = 3^3t(n/4^3 ) + n(1+3/4 + 3^2/4^2)

= 3^it(n/4^i) +  n(1+3/4 + 3^2/4^2 + ……..+3^(i-1)/4^(i-1))

we will stop at  i = log4n, n/4^i = 1

Therefore,

t(n) = 3^ log4nt(1) +n . Σ(k-= to log4 n-1) (3/4)^k

   <= 3log4n. Θ(1) + n  Î£( k= 0 to infinity) (3/4)^k = Θn^log4(3)+4n = o(n)+4n

t(n) = O(n)


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Last Updated : 01 Nov, 2017
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