What is the asymptotic behavior of T(n) = 4T(n/2) + O(n3)? Select tightest bound that holds.

Question

BlackTom AI · Accepted Answer

Apply the Master Theorem for recurrences of the form T(n) = a T(n/b) + f(n). Here a = 4, b = 2, and f(n) = O(n^3) (we take f(n) proportional to n^3 for the asymptotic comparison). 1. Relevant concepts: - Master Theorem compares f(n) to n^{log_b a}. Compute n^{log_b a} = n^{log_2 4} = n^2. - If f(n) = Theta(n^{log_b a + epsilon}) for some epsilon > 0 and a regularity condition holds, then T(n) = Theta(f(n)) (case 3). 2. Step-by-step reasoning and intermediate calculations: - Compute exponent: log_b a = log_2 4 = 2, so n^{log_b a} = n^2. - Compare f(n) ~ n^3 with n^2: n^3 = n^{2+1} so epsilon = 1 > 0. Thus f(n) grows polynomially faster than n^{log_b a}. - Verify the regularity condition: check a f(n/b) ≤ c f(n) for some c < 1. If f(n) = C n^3, then a f(n/2) = 4 * C * (n/2)^3 = 4 * C * (n^3 / 8) = (1/2) C n^3 = (1/2) f(n), so we can take c = 1/2 < 1. Regularity condition holds. 3. Conclusion from Master Theorem: - By case 3, T(n) = Theta(f(n)) = Theta(n^3). Therefore the tight asymptotic bound is Theta(n^3), which implies O(n^3). 4. Why the correct answer is correct: - Because the work per level (f(n) = n^3) dominates the total cost compared to the tree-structure contribution n^2, the recurrence's growth is driven by f(n), yielding Theta(n^3). 5. Why other plausible bounds are incorrect: - O(n^2) or O(n^2 log n) are too small because they ignore the n^3 non-recursive work which dominates. - O(n^4) is correct as an upper bound but not tight; Theta(n^3) is the tight growth, so O(n^4) is weaker. Final confirmation: The asymptotic behavior is O(n^3), matching the provided answer.

What is the asymptotic behavior of T(n) = 4T(n/2) + O(n3)? Select tightest bound that holds.单项选择题

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