selection sort

The Problem

See the Sorting Problem.

The Algorithm

Suppose $L=\\left\\{x_{1},x_{2},\\dots,x_{n}\\right\\}$ is the initial list of unsorted elements.The selection sort algorithm sorts this list in $n$ steps. At each step $i$ , find the largestelement $L[j]$ such that $j<n-i+1$ , and swap it with the element at $L[n-i+1]$ .So, for the first step, find the largest value in the list and swap it with the last element in the list.For the second step, find the largest value in the list up to (but not including) the last element, and swap itwith the next to last element. This is continued for $n-1$ steps. Thus the selection sort algorithm is avery simple, in-place sorting algorithm.

Pseudocode

Algorithm Selection_Sort(L, n)
Input: A list $L$ of $n$ elements
Output: The list $L$ in sorted order

begin
for $i\\leftarrow n\\mbox{ downto }2$ do

begin

$temp\\leftarrow L[i]$

$max\\leftarrow 1$

for $j\\leftarrow 2\\mbox{ to }i$ do

if $L[j]>L[max]$ then

$max\\leftarrow j$

$L[i]\\leftarrow L[max]$

$L[max]\\leftarrow temp$

end
end

Analysis

The selection sort algorithm has the same runtime for any set of $n$ elements, no matterwhat the values or order of those elements are. Finding the maximum element of a list of $i$ elements requires $i-1$ comparisons. Thus $T(n)$ , the number of comparisons required tosort a list of $n$ elements with the selection sort, can be found:

$\\displaystyle T(n)$	$\\displaystyle=$	$\\displaystyle\\sum_{i=2}^{n}(i-1)$
	$\\displaystyle=$	$\\displaystyle\\sum_{i=1}^{n}i-n-2$
	$\\displaystyle=$	$\\displaystyle\\frac{(n^{2}-n-4)}{2}$
	$\\displaystyle=$	$\\displaystyle\\mathcal{O}(n^{2})$

However, the number of data movements is the number of swaps required, which is $n-1$ . Thisalgorithm is very similar to the insertion sort algorithm. It requires fewer data movements,but requires more comparisons.