What is the maximum number of passes a binary search would require to determine if a specific number N is in a sorted list of 500 numbers?

Steps to Solve

1. Understand Binary Search

Binary search is an efficient algorithm for finding a target value within a sorted list. It works by repeatedly dividing the search interval in half. If the middle element is the target, the search is successful. If the target is less than the middle element, the search continues in the lower half. If the target is greater, the search continues in the upper half.

2. Determine the Worst-Case Scenario

The worst-case scenario occurs when the target value is not in the list, or when it's found in the very last step. This means we keep dividing the list in half until we're left with a sublist of size 1.

3. Calculate the Number of Iterations

We need to find the smallest integer $n$ such that $2^n \geq 500$. This is because each iteration of binary search roughly halves the search space. After $n$ iterations, the size of the remaining list is approximately $500 / 2^n$. We want to find $n$ such that $500 / 2^n \approx 1$, or equivalently, $2^n \approx 500$.

4. Find the Value of $n$

Let's test some values of $n$:

• $2^8 = 256$
• $2^9 = 512$

Since $2^8 = 256 < 500$ and $2^9 = 512 > 500$, $n = 9$ is the smallest integer that satisfies $2^n \geq 500$.

Therefore, the maximum number of iterations required is 9.