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

What is the run time for this recurrence relation ?

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:
$$ T(n) = \begin{cases} \Theta(1) & \text{if } n = 0 \T\left(\frac{n}{2}\right) + \Theta(1) & \text{otherwise}\end{cases}$$

What is the run time for this recurrence relation ?

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

BlackTom AI · Accepted Answer

To understand the running time, I’ll examine the given recurrence: T(n) = T(n/2) + Theta(1) with T(0) = Theta(1).
First, identify the parameters for the Master Theorem: a = 1 (one recursive call), b = 2 (size halved), and f(n) = Theta(1).
Compute the critical exponent: n^{log_b a} = n^{log_2 1} = n^{0} = 1.
Now compare f(n) to n^{log_b a}:
- f(n) = Theta(1) is Theta(n^{0}). This matches Theta(n^{log_b a}) with k = 0 in the Master Theorem’s extended form.
- Since f(n) = Theta(n^{log_b a} log^{k} n) with k = 0, we are in the case where T(n) = Theta(n^{log_b a} log^{k+1} n) = Theta(1 * log n) = Theta(log n).
Next, evaluate the other options to see why they don’t fit:
- Theta(n^2): This would occur if the work grew with the square of n or if the combination of a, b, and f(n) produced a polynomial of degree > 0, which is not the case here because the recursion simply halves n at each step and adds a constant per level.
- Theta(n log n): This typically arises when f(n) dominates in a way that accumulates a linear factor times a logarithm, which doesn’t happen here since f(n) is constant and the number of levels is only about log n.
- Theta(n): This would imply linear growth with n, which contradicts the logarithmic depth of the recursion tree when each level costs O(1).
In summary, the structure of the recurrence—a single recursive call on n/2 plus a constant work per level—produces a recursion tree of height ~ log_2 n with constant cost per level, leading to a total cost proportional to log n.

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