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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1st Edition
B0CN6VN82B, 979-8867500764
