Final answer:
The Master Theorem is used to characterize the time complexity of different recurrence relations. Solutions follow the cases outlined by the theorem, with complexity depending on how the function f(n) compares to n^log_b(a). Solutions range from Θ(n) to Θ((n log n)⁴) based on the provided recurrence equations.
Step-by-step explanation:
The Master Theorem provides a way to determine the time complexity of recurrence relations of the form T(n) = aT(n/b) + f(n), where a ≥ 1 and b > 1 are constants, and f(n) is an asymptotically positive function. The theorem classifies the complexity based on a comparison between f(n) and nlogba.
- (a) T(n) = 2T(n/2) + log n: Here, a=2, b=2, and f(n)=log n. Since f(n) = O(nlog22°), which is O(1), it satisfies Case 1 of the Master Theorem. The solution is T(n) = Θ(nlog22) = Θ(n).
- (b) T(n) = 8T(n/2) + n²: In this case, a=8, b=2, and f(n)=n². Since f(n) = Ω(nlog28¹⁵), which suggests Case 3 applies. Hence, T(n) = Θ(n²).
- (c) T(n) = 16T(n/2) + (n log n)⁴: With a=16, b=2, and f(n)=(n log n)⁴, f(n) grows faster than nlog216 indicating Case 3 of the Master Theorem. Thus, T(n) = Θ((n log n)⁴).
- (d) T(n) = 7T(n/3) + n: Here, a=7, b=3, and f(n)=n. nlog37 grows faster than f(n), which places it in Case 1. The solution is T(n) = Θ(nlog37).
- (e) T(n) = 9T(n/3) + n³ log n: With a=9, b=3, and f(n)=n³ log n, since f(n) grows faster than nlog39, Case 3 applies. The solution is T(n) = Θ(n³ log n).