Ms_1
Consider the recurrence relation for an algorithm $T(n)$ given by:
$$ T(n) = \begin{cases} \Theta(1) & \text{if } n = 0 \T\left(\frac{n}{2}\right) + \Theta(1) & \text{otherwise}\end{cases}$$

The Master Theorem is provided below. Use it as you see fit:

Question

Ms_1
Consider the recurrence relation for an algorithm $T(n)$ given by:
$$ T(n) = \begin{cases} \Theta(1) & \text{if } n = 0 \T\left(\frac{n}{2}\right) + \Theta(1) & \text{otherwise}\end{cases}$$

The Master Theorem is provided below. Use it as you see fit:

BlackTom AI · Accepted Answer

Let’s parse the recurrence and identify the parameters for the Master Theorem. The recurrence is T(n) = T(n/2) + Theta(1) with the base case T(0) = Theta(1). Here, a = 1, b = 2, so n^{log_b a} = n^{log_2 1} = n^0 = 1. The function f(n) in the recurrence is Theta(1), which matches Theta(n^{log_b a}) since both are Theta(1).

Option 1: Theta(n^2). This would correspond to a scenario where the work per level or the accumulation across levels grows quadratically with n, which is not supported by the parameters a = 1, b = 2 and f(n) = Theta(1). The recurrence does not introduce any n factor that would lead to n^2 growth.

Option 2: Theta(n). A linear Theta(n) solution would imply a larger growth rate than the n^{log_b a} baseline combined with f(n) = Theta(1). Since the number of levels in a/ b split is logarithmic in n, and each level does only constant work, you don’t accumulate a linear n factor across levels.

Option 3: Theta(n log n). This would typically arise in a divide-and-conquer pattern where there are both a logarithmic number of levels and linear work per level, or a similar combination. In this recurrence, the per-level work is constant (Theta(1)), not Theta(n), so there’s no n factor to multiply by the number of levels.

Option 4: Theta(log n). The Master Theorem compares f(n) with n^{log_b a} = 1. Since f(n) = Theta(1) = Theta(n^{log_b a}) with k = 0 in the Master Theorem’s extended form, we land in the case where the overall solution is Theta(n^{log_b a} log^{k+1} n) = Theta(1 · log n) = Theta(log n). Intuitively, each level costs a constant amount, and there are about log_2 n levels until the base case is reached, yielding a logarithmic total.

To recap the intuition behind the incorrect options: Theta(n^2) would require a much larger accumulation than provided by the single subproblem plus constant work per level; Theta(n) would demand either linear work per level or a linear number of levels with nontrivial per-level cost; Theta(n log n) would require a combination of linear work per level with logarithmic levels or similar, which is not the case here. The recurrence’s structure yields a logarithmic total work because the subproblem size halves each step and only constant work is performed at each level.

Ms_1 Consider the recurrence relation for an algorithm \(T(n)\) given by: \[ T(n) = \begin{cases} \Theta(1) & \text{if } n = 0 \\T\left(\frac{n}{2}\right) + \Theta(1) & \text{otherwise}\end{cases}\] The Master Theorem is provided below. Use it as you see fit:

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