Question: Explain why/how the recurrences have that running time and case or answer. T(n) = 2T (n/2) + n log n T(n) = n log^2 n

Explain why/how the recurrences have that running time and case or answer.

T(n) = 2T (n/2) + n log n

T(n) = n log^2 n (Case 2)

T(n) = 3T(n/4) + n log n

T (n) = (n log n) (Case 3)

T(n) = 2T (n/2) + n/ log n

Does not apply (non-polynomial difference between f(n) and n^log_b a)

T(n) = 4T(n/2) + n/ log n

T (n) = (n^2) (Case 1)

T(n) = 3T (n/3) + n/ 2

T (n) = (n log n) (Case 2)

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