Question

The average number of comparisons performed by linear search to find an item in an array of \(\mathrm{N}\) elements is ___________.

Short answer

Answer: The average number of comparisons performed by the linear search algorithm to find an item in an array of N elements is \(\frac{(N + 1)}{2}\).

Step-by-step solution

Understand the linear search algorithm

Linear search is an algorithm that finds a target value in a given array by checking each element until the target is found or the end of the array is reached. It performs comparisons sequentially to identify if an item is present in the array.

Find the possible number of comparisons for each element

The first element will be found with 1 comparison, the second with 2 comparisons, the third with 3 comparisons, and so on. The Nth element will be found with N comparisons. Thus, for each element in the array, we have a different number of comparisons that need to be made.

Calculate the average number of comparisons for the entire array

To find the average number of comparisons, we can add up the number of comparisons for each individual element and then divide by the total number of elements, which is N. This can be represented mathematically as: \(\text{Average number of comparisons} = \frac{1 + 2 + 3 + ... + (N-1) + N}{N}\)

Apply the formula for sum of first N natural numbers

We can simplify the above expression by using the formula to find the sum of first N natural numbers: \(\sum_{i=1}^N i = \frac{N (N + 1)}{2}\) Substituting this formula into the expression for the average number of comparisons, we get: \(\text{Average number of comparisons} = \frac{\frac{N (N + 1)}{2}}{N}\)

Simplify the expression

Now we can simplify the expression by canceling out the N from the numerator and denominator: \(\text{Average number of comparisons} = \frac{(N + 1)}{2}\) The average number of comparisons performed by linear search to find an item in an array of N elements is \(\frac{(N + 1)}{2}\).

Key concepts

Average Number of Comparisons

When performing a linear search, an important concept is the average number of comparisons needed to find a target element. This can vary depending on the target's position in the array.
To calculate this average, consider each potential position for the target from 1 to N, where N is the total number of elements. For example, if the target is the first element, only 1 comparison is needed. If it is the second, 2 comparisons are required, and so on, up to N comparisons for the last element.
To find the average, sum up the number of comparisons for all elements and divide by N. In mathematical terms, this means calculating the sum of integers from 1 to N and dividing by N. This gives us the formula \[\text{Average number of comparisons} = \frac{1 + 2 + 3 + \ldots + N}{N}\]Simplifying using the formula for the sum of the first N natural numbers, we get:\[\frac{N(N + 1)}{2N} = \frac{N + 1}{2}\]Thus, on average, a linear search requires \(\frac{N + 1}{2}\) comparisons.

Sum of First N Natural Numbers

The sum of the first N natural numbers is crucial in simplifying our average calculations in linear search. It is given by the formula:\[\sum_{i=1}^N i = \frac{N(N + 1)}{2}\]This formula is derived from adding consecutive numbers up to N, making it possible to quickly calculate the total sum.
For instance, if N is 5, the sum of the numbers 1 through 5 would be \(1 + 2 + 3 + 4 + 5 = 15\). The formula confirms this by substituting N with 5, resulting in:\[\frac{5(5 + 1)}{2} = 15\]
This formula not only simplifies our calculation of average comparisons in linear search, but also has various applications in mathematics, aiding quick computation of sums in polynomial equations or whenever summing consecutive numbers is needed.

Array Traversal

In a linear search, array traversal refers to visiting each element of the array one by one. This process is essential since the algorithm checks each position sequentially to find a match for the target value.
Traversal begins from the start of the array and proceeds to the end, comparing each element with the target. If a match is found, the search stops. If not, the search continues until all elements are checked. This simplicity makes linear search easy to implement but also inefficient for larger datasets.
Because of this step-by-step comparison, linear search's time complexity is \(O(N)\), reflecting its straightforward method of checking each element in an array with N elements.
This traversal technique, while simple, forms the backbone of algorithms that require visiting each element in a data structure, such as certain sorting or filtering operations.

Sequential Search Process

The linear search follows a straightforward sequential search process. It begins by comparing the first element of the array with the target value. If they are not equal, the search progresses to the next element, and comparisons continue until the target is found or the array ends.
This process resembles checking each lock in a row of lockers to find one that opens with a specific key. Each 'check' or comparison involves matching the current array element with the target. If they match, the search is successful and stops. If not, it moves on to the next 'lock' or element.
The sequential nature of linear search means every element could potentially be compared, making it less efficient for large arrays compared to more advanced algorithms like binary search. Despite this, its simplicity allows it to excel in cases where data is unsorted or the dataset is small, making it an essential tool in basic search operations.