Question

Suppose that \(L\) is a sorted list of 4096 elements. What is the maximum number of comparisons made by binary search to determine whether an item is in \(\mathbf{L}\) ?

Short answer

The maximum number of comparisons is 13.

Step-by-step solution

Understand Binary Search

Binary search is an efficient algorithm for finding an item from a sorted list. It divides the list into two halves with each step, eliminating one half from further consideration. This process continues until the item is found or the search space is empty.

Determine Number of Divisions

To find the maximum number of comparisons, calculate how many times we can divide the list until we are left with one element. Each division cuts the list size in half.

Calculate Using Logarithms

The maximum number of comparisons is given by \( \lfloor \log_2 n \rfloor + 1 \), where \( n \) is the number of elements. For this problem, \( n = 4096 \).

Evaluate the Expression

Calculate \( \log_2 4096 \). Since \( 4096 = 2^{12} \), \( \log_2 4096 = 12 \).

Determine the Maximum Comparisons

Substitute \( \log_2 4096 = 12 \) into the formula from Step 3: \( \lfloor \log_2 4096 \rfloor + 1 = 12 + 1 = 13 \). Thus, the maximum number of comparisons is 13.

Key concepts

Sorted List

A sorted list is a sequence of elements arranged in either ascending or descending order. The essential characteristic of a sorted list is that, if you pick any two elements, the order is predetermined based on their values. This order is typically based on numerical value, size, or any other comparable attribute.

Sorting a list before searching has significant advantages:

• It enables the use of efficient search algorithms like binary search.
• It makes the process of finding elements faster and less resource-intensive.
• It allows for the detection of duplicates or non-existing elements much faster.

Imagine going through a pile of unsorted exam papers versus those alphabetically ordered by student name; the latter is much faster when looking for a specific paper. This is why sorting is a foundational concept in computer science.

Algorithm Efficiency

Algorithm efficiency refers to how effectively an algorithm performs a particular task in terms of time and space. For any search or sorting task, understanding efficiency is critical because it impacts how quickly our programs execute.

Binary search is a prime example of an efficient algorithm when working with a sorted list. Instead of checking every item one by one, it divides the list into halves and eliminates the half that does not contain the target. This division occurs iteratively until the item is found or the list cannot be divided further.

Efficiency is crucial in large-scale applications, such as databases and search engines, where millions of items are processed quickly. The fewer resources a search operation uses, the better it can scale and the more responsive it can be to user requests.

Logarithms in Algorithms

Logarithms are mathematical functions that help us express the opposite of exponentiation. In algorithms, they are vital in analyzing time complexity, especially for operations that continually split data sets, like binary search.

Binary search divides the list into two equal parts at each step, making the logarithmic function \(\log_2(n)\) the natural way to express the number of divisions required. The base 2 indicates each division step halves the dataset size. If you have a list of 4096 elements, to understand how many halving will fit, you calculate \(\log_2(4096)\), simplifying to \(12\) since \(4096 = 2^{12}\).

Using logarithms illustrates how quickly binary search can find elements, achieving a significant level of efficiency that is obvious when compared to a linear search approach, especially as the list size increases.

Comparison Count in Search Algorithms

Calculating the number of comparisons in a search algorithm is crucial for understanding its efficiency. With binary search, the expression for the maximum number of comparisons is \(\lfloor \log_2 n \rfloor + 1\). This gives the worst-case scenario for how many checks are needed to determine the presence or absence of an element.

For a list with 4096 elements, we first find \(\log_2(4096)\), which equals \(12\). The expression then becomes \(12 + 1 = 13\), which means binary search will make a maximum of 13 comparisons. This count shows its efficiency over a linear search which would potentially require checking all 4096 elements.

By understanding the comparison count, developers can optimize their code better, foresee performance bottlenecks, and choose the best algorithm for their data size to ensure quick and reliable search operations.