Solving a recurrence T(n) = 2T(n/2) + n^4

Posted by user563454 on Stack Overflow See other posts from Stack Overflow or by user563454
Published on 2011-01-05T04:50:20Z Indexed on 2011/01/05 4:53 UTC
Read the original article Hit count: 176

Filed under:
|

I am studying using the MIT Courseware and the CLRS book Introduction to Algorithms.

Solving recurrence T(n) = 2T(n/2) + n4 (page 107)

If I make a recurrence tree I get: level 0 n^4 level 1 2(n/2)^4 level 2 4(n/4)^4 level 3 8(n/8)^4

The tree has lg(n) levels. Therefore the recurrence is T(n) = Theta(lg(n)n^4))

But, If I use the Master method I get.

Apply case 3: T(n) = Theta(n^4)

If I apply the substitution method both seem to hold. Which one is ri?

© Stack Overflow or respective owner

Related posts about algorithm

Related posts about mit