`Hey,
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T(n)=8*T(n/2)+theta(n^3)
We can say theta(n^3)=c*n^3
So,
T(n)=8*T(n/2)+c*n^3
So, by master theorem
log2(8)=3
So,
n^3=theta(n^3)
So, By master theorem
T(n)=theta(n^3*log(n))
Kindly revert for any queries
Thanks.
Consider recursive divide-and-conquer algorithms with the following descriptions. For each, determine the running time in Big-Theta notation. If necessary, you may assume that the regularity conditio...
Suppose the following is a divide-and-conquer algorithm for some problem. "Make the input of size n into 3 subproblems of sizes n/2 , n/4 , n/8 , respectively with O(n) time; Recursively call on these subproblems; and then combine the results in O(n) time. The recursive call returns when the problems become of size 1 and the time in this case is constant." (a) Let T(n) denote the worst-case running time of this approach on the problem of size n....