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The image presents an exercise where you are asked to show that the recurrence relation \( T(n) = T\left(\left\lfloor \frac{n}{2} \right\rfloor\right) + 1 \) is \( O(\log n) \), meaning it grows logarithmically. The attempts in the image try to solve this using two different approaches:

### Key Points:
1. **Attempt 1:**
   - The goal is to prove that \( T(n) \leq c \log n \) for some constant \( c \).
   - The steps try to bound the recurrence relation by substituting \( T(n) \) with a logarithmic expression.
   - The step where \( \log(n+1) \) appears doesn't successfully reduce the problem, and hence, the attempt fails (as marked with the red face).
   
2. **Attempt 2:**
   - In this attempt, the approach is to subtract a constant \( \alpha \), assuming \( \alpha \in \mathbb{R} \), so that \( T(n) \leq c \log(n - \alpha) \).
   - The key idea is to use a stronger assumption: \( T(n) \leq c \log(n - \alpha) \), for some constant \( \alpha \), where \( \alpha \geq 1 \).
   - After simplification, the expression bounds the recurrence as desired, leading to a logarithmic bound.
   - A base case check is suggested to finalize the proof.

### Summary of Approaches:
- **First Attempt:** Unsuccessful due to an incorrect reduction step when trying to simplify the expression involving \( \log(n+1) \).
- **Second Attempt:** Successful, provided a constant \( \alpha \) is subtracted, and the base case holds.

### Detailed Steps to Show \( T(n) = O(\log n) \):

The recurrence \( T(n) = T\left(\left\lfloor \frac{n}{2} \right\rfloor\right) + 1 \) suggests that for each step, we reduce the size of \( n \) by half. This is characteristic of logarithmic time behavior.

1. **Base Case:** \( T(1) = 1 \), which is constant and fits within \( O(\log n) \).
2. **Recurrence Analysis:**
   \[
   T(n) \leq T\left(\frac{n}{2}\right) + 1.
   \]
   Unfolding the recurrence:
   \[
   T(n) = T\left(\frac{n}{2}\right) + 1 = T\left(\frac{n}{4}\right) + 1 + 1 = \dots = T(1) + \log_2 n = \log_2 n.
   \]
   Therefore, \( T(n) \) is bounded by \( O(\log n) \).

Would you like further details on any specific step or concept here?

### Related Questions:
1. What is the general method for solving recurrence relations using the master theorem?
2. How can we prove that a recurrence is \( O(\log n) \) using substitution methods?
3. What are some common pitfalls when trying to bound recurrences?
4. How can we verify the correctness of a base case in recursive algorithms?
5. Can the recurrence \( T(n) = T\left(\frac{n}{3}\right) + 1 \) also be \( O(\log n) \)?

### Tip:
When solving recurrence relations, it's important to carefully check the base case to ensure that the derived asymptotic behavior holds across all \( n \).

Show that T(n) = T(floor(n/2)) + 1 is O(log n).

Learn how to prove that the recurrence T(n) = T(floor(n/2)) + 1 is O(log n) using complexity analysis. The solution involves analyzing the recurrence relation and logarithmic growth, making it suitable for university-level students in algorithm design.

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