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All tutorials on algorithms show the complexity for the linear search in the unsorted array in the average case as N/2. I understand that the average case means the items in the list are randomly distributed.

Can anyone show how I would arrive at N/2 if I have items randomly distributed? Or does it come out of randomly shuffling array bazillion times and recording the number of operations?

Raphael
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Max Koretskyi
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1 Answers1

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First assume input is uniformly distributed. More precisely it is $\frac{n+1}{2}$. When you search for a particular element $x$ in an array of size $n$, that element may be located at the position either $1$, or $2$, $\dots$ or $n$. When we search we check each element in the array and compare it with $x$, and so when we reach $k^\text{th}$ element of the array we have already done $k$ comparisons.

If it is at $1$ then you find it in 1 comparison, if it is at $2$, you find it in 2 comparisons, $\dots$, if it is at $n$ then you do $n$ comparison in order to find it. In order to average it you sum the total number of comparisons $1+2+\dots + n = \frac{(n+1)n}{2}$ and divide it by $n$ (size of the array) resulting in $\frac{n+1}{2}$.

More formal proof: Assume that the input has uniform probability and the the size of the array is $n$. Let $X$ be a random variable denoting the number of comparisons. Then $p(X=x)$ will denote the probability that we do $x$ comparisons. Since the distribution is uniform, $p(X=x) =\frac{1}{n}$. Once we formally defined our probability distribution now we can calculate the average number of comparisons in the linear search. What we do is compute the expected value $$E[X] = \sum_{x=1}^{n}{xp(x)} = \sum_{x=1}^{n}{\frac{x}{n}} = \frac{1}{n}+ \frac{2}{n} + \dots +\frac{n}{n} = \frac{n+1}{2}$$
fade2black
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