How to find the recurrence relation T(N) = T(N/2) + N^2 -


what process involved in simplifying recurrence relation?

i able much:

t(n) = t(n/2) + n^2 t(n) = t(n/4) + (n/2)^2 +n^2 t(n) = t(n/8) + (n/4)^2 + (n/2)^2 + n^2 

i understand terminate when n = 1 because 1/2 = 0; c(0) = 0. past stuck on way figure out these problems.

well basic idea c(n) , c(n/2) should expression of same form. since difference simple function of n, infinite sum should pop mind, c(n)-c(n/2) becomes telescopic. each term of sum should given function n/2^k (for k=0, 1, ...) argument.

hence c(n) = n^2 + (n/2)^2 + (n/4)^2 + (n/8)^2 + ... job, , can further evaluated of identities geometric series c(n)=4/3*n^2.


Comments

Popular posts from this blog

java - WrongTypeOfReturnValue exception thrown when unit testing using mockito -

php - Magento - Deleted Base url key -

android - How to disable Button if EditText is empty ? -