When i try to solve this
T(n) = 2T(n/2) + 2
recurrence relation i get order N. But
http://www.geeksforgeeks.org/archives/4583
claims that it is 3/2n-2. The order is still N only but how do we get
the constants ?
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"A
answer is option A .
On Sat, Jul 2, 2011 at 8:32 PM, KK wrote:
> 3. Consider the following recurrence:
>
> T(n) = 2T(n ^ (1/2)) + 1
>
> Which one of the following is true?
> (A) T(n) = (loglogn)
> (B) T(n) = (logn)
> (C) T(n) = (sqrt(n))
> (D) T(n) = (n)
>
> Can we solve this using master theore
3. Consider the following recurrence:
T(n) = 2T(n ^ (1/2)) + 1
Which one of the following is true?
(A) T(n) = (loglogn)
(B) T(n) = (logn)
(C) T(n) = (sqrt(n))
(D) T(n) = (n)
Can we solve this using master theorem???
any other way???
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Hello all,
I have a recurrence relation of the form
T(n) = T(n-2^log n) + O(n) --- how do i solve such a recurrence
relation ?
I have tried an approach for this. Is this right ? ---
let kn = n-2^log n hence k ranges from 0 < k < [(2^log n -
2^log(n-1))/2n]
ie 0 < k < [2^(log n - 2)]