The running time of the following algorithm Procedure A(n)
If n<=2 return (1) else return (A([$$\sqrt n $$])); is best described by

The solution to the recurrence equation
T(2^k) = 3 T(2^k-1) + 1, T (1) = 1, is:

In $$2’s$$ complement addition, the overflow

Sign extension is the step in