In case 2 of the master theorem, If f(n) = O(n^k) where k = logb(a), then---------------.

Question

BlackTom AI · Accepted Answer

For the recurrence T(n) = a T(n/b) + f(n) with k = log_b(a), case 2 (the balanced case) gives the extra log factor: T(n) = O(n^k log n).

1) Relevant concepts
- Master theorem compares f(n) to n^{log_b a}, the critical exponent.
- Case 2 applies when f(n) grows asymptotically the same as n^{log_b a} (i.e., f(n) = Θ(n^{log_b a})).
- Recursion-tree summation across levels yields the total cost.

2) Step-by-step reasoning
Let m = log_b a (so m = k). Assume f(n) = Θ(n^m). Evaluate costs at recursion tree levels:
- At level i the subproblem size is n/b^i and there are a^i subproblems, so level cost = a^i · f(n/b^i).
- Substitute f: level cost = a^i · Θ((n/b^i)^m) = Θ(a^i · n^m / b^{i m}).

3) Intermediate calculations
Compute the factor (a / b^m)^i:
- Since m = log_b a, b^m = a. Therefore a / b^m = 1.
- So level cost = Θ(n^m · 1^i) = Θ(n^m) for every level i.
Depth of the tree L: n/b^L ≈ 1 ⇒ L = log_b n (number of levels is Θ(log n)).
Total cost = sum_{i=0}^{L} Θ(n^m) = Θ(n^m · (L+1)) = Θ(n^m log n).

4) Why the correct answer is correct
Because each level contributes Θ(n^k) and there are Θ(log n) levels, the sum is Θ(n^k log n). Expressed in Big-O notation this is O(n^k log n), matching the provided correct answer.

5) Why other options are incorrect
- "T(n) = O(log_b(a))": dimensionally wrong; T(n) depends on n and grows at least polynomially here.
- "T(n) = O(n^k)" and "T(n) = O(n^{log_b(a)})": both omit the logarithmic factor produced by summing equal-level costs across Θ(log n) levels, so they understate the growth.

6) Conclusion
Thus, in case 2 with f(n) = Θ(n^k) where k = log_b(a), T(n) = O(n^k · log n), which matches the provided answer: T(n) = O(n^k * logn).

In case 2 of the master theorem, If f(n) = O(n^k) where k = logb(a), then---------------.单项选择题

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