You can't prove that it's Omega(n log n) because T(n) = n satisfies the base case and the recurrence.
Recurrence of T(n) = T(n/3) + T(2n/3) [closed]
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17-06-2023 - |
Question
I've searched online for this but I only seem to find answers for a similar equation:
T(n) = T(n/3) + T(2n/3) + cn
But the one I'm trying to solve is:
T(n) = T(n/3) + T(2n/3)
Base case: We can assume T(a) = Theta(1)
for any constant a
.
I've succeeded in proving (by induction) that T(n) = O(n*log(n))
. I thought the answer should be Theta(n*log(n))
, but I cannot prove that T(n) = Omega(n*log(n))
.
So my question is - am I correct that the answer is O(n*log(n))
, and NOT Theta(n*log(n))
? IF that's true that would really be great...
If I'm wrong I will of course explain where I'm stuck in the induction process...
Thanks!
P.S. If you need to, please try to explain using induction, because I haven't learned all methods for solving these problems yet.
Solution
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