Correct bound

T ( n ) = 2 T ( n 2 ) + n , T ( 0 ) = T ( 1 ) = 1 T(n)=2T\left(\frac{n}{2}\right) +n,\quad T(0)=T(1)=1

Based on the above relation, which of the following answer choices is false ?

T ( n ) = Ω ( n 2 ) T(n)=\Omega (n^{2}) T ( n ) = Θ ( n l o g n ) T(n)=\Theta (nlogn) T ( n ) = O ( n log n ) T(n)=O(n\log n) T ( n ) = O ( n 2 ) T(n)=O(n^{2})

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Beakal Tiliksew
Dec 3, 2015

We can use Master theorem to show that the above recurrence relation is:

T ( n ) = ( O ( n l o g n ) ) T(n)=(O(nlogn))

Also by definition of big-Oh we have:

Θ ( n L o g n ) = O ( n L o g n ) = O ( n 2 ) \Theta(nLogn) = O(nLogn) = O(n^{2})

but

Θ ( n l o g n ) \Theta(nlogn) is not equal to Ω ( n 2 ) \Omega(n^{2})

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...