What is the asymptotic behavior of T(n) = 4T(n/2) + O(1)? Select tightest bound that holds.

Question

BlackTom AI · Accepted Answer

This recurrence fits the Master Theorem with a = 4 and b = 2.

1. Relevant knowledge points and concepts
- Master Theorem for recurrences T(n) = a T(n/b) + f(n): compare f(n) to n^{log_b a}.
- If f(n) = O(n^{log_b a - ε}) for some ε > 0, then T(n) = Θ(n^{log_b a}).
- Recursion-tree method: sum work at each level; number of levels = log_b n.

2. Complete reasoning step by step
- Compute exponent from a and b: log_b a = log_2 4.
- log_2 4 = 2, so n^{log_b a} = n^2.
- Given f(n) = O(1) = Θ(1) = Θ(n^0). Compare exponents: 0 < 2, so f(n) is polynomially smaller than n^2 (take ε = 2).
- By Master Theorem case 1, T(n) = Θ(n^{log_b a}) = Θ(n^2). Thus asymptotic growth is quadratic.

3. Intermediate calculations (recursion-tree confirmation)
- Level i has 4^i subproblems, each cost O(1), so cost at level i = Θ(4^i).
- Number of levels = log_2 n (stop when subproblem size = 1).
- Total internal node cost = Σ_{i=0}^{log_2 n -1} 4^i = (4^{log_2 n} - 1)/(4 - 1) = (n^2 - 1)/3 = Θ(n^2).
- Leaves: count = 4^{log_2 n} = n^2, each constant, so leaf cost = Θ(n^2). Total cost = Θ(n^2).

4. Why the correct answer is correct
- Both Master Theorem and recursion-tree show the dominant contribution is Θ(n^2), so the tightest asymptotic bound in Big-O notation is O(n^2).

5. Why other plausible options are incorrect (brief)
- O(n) or O(n log n): these are too small because work grows as n^2 (leaves alone are n^2).
- O(n^2 log n): too loose; extra log factor does not appear in the sum.
- O(1) or sublinear bounds: impossible given growth to n^2.

6. Conclusion
Therefore T(n) = Θ(n^2), and the tightest Big-O bound is O(n2), which matches the provided answer.

What is the asymptotic behavior of T(n) = 4T(n/2) + O(1)? Select tightest bound that holds.Single choice

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