Selection sort: Difference between revisions

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'''Proof:''' The asymptotic complexity of the <math>i</math>-th iteration is in <math>\Theta(T\cdot(n - i))</math>. Therefore, the total complexity is in <math>\Theta\left(T\cdot\sum_{i=1}^{n-1} (n-i) \right) = \Theta\left(T\cdot\sum_{i=1}^{n-1} i \right) = \Theta\left(T\cdot\frac{n(n-1)}{2} \right) = \Theta\left(T\cdot n^2\right)</math>.
'''Proof:''' The asymptotic complexity of the <math>i</math>-th iteration is in <math>\Theta(T\cdot(n - i))</math>. Therefore, the total complexity is in <math>\Theta\left(T\cdot\sum_{i=1}^{n-1} (n-i) \right) = \Theta\left(T\cdot\sum_{i=1}^{n-1} i \right) = \Theta\left(T\cdot\frac{n(n-1)}{2} \right) = \Theta\left(T\cdot n^2\right)</math>.
== Java Code ==
<syntaxhighlight lang="java">
/**
   
        int[] selectionSort(int [] list) {
        int n = list.length - 1;
        int left = 0;
        int min;
        int tmp;
        while(left < n) {
            min = left;
            for(int i = left + 1; i<=n; i++) {
                if (list[i] < list[min]) {
                    min = i;
                }
            }
            tmp = list[min];
            list[min] = list[left];
            list[left] = tmp;
            left = left + 1;
        }
    }
   
</syntaxhighlight>

Revision as of 15:46, 19 June 2015

General information

Algorithmic problem: Sorting based on pairwise comparison

Type of algorithm: loop

Abstract view

Invariant: After [math]\displaystyle{ i \geq 0 }[/math] iterations: The elements at positions [math]\displaystyle{ |S|-i+1,\dots,|S| }[/math] are correctly placed in sorting order.

Variant: [math]\displaystyle{ i }[/math] increases by [math]\displaystyle{ 1 }[/math].

Break condition: [math]\displaystyle{ i = |S| - 1 }[/math].

Induction Basis

Abstract view: Nothing to do.

Implementation: Nothing to do.

Proof: Nothing to show.

Induction step

Abstract view: Identify the maximum out of [math]\displaystyle{ S[1],\dots,S[|S|-i+1] }[/math] and then move it, by a swap, to position [math]\displaystyle{ |S|-i+1 }[/math].

Implementation:

  1. Set [math]\displaystyle{ m := 1 }[/math]
  2. For all [math]\displaystyle{ j=2,\dots,|S|-i+1 }[/math]: If [math]\displaystyle{ S[j] \gt S[m] }[/math] set [math]\displaystyle{ m := j }[/math].
  3. Swap [math]\displaystyle{ S[m] }[/math] and [math]\displaystyle{ S[|S|-i+1] }[/math].

Correctness: Obviously: [math]\displaystyle{ S[m] }[/math] is the maximum element out of [math]\displaystyle{ S[1],\dots,S[j] }[/math].

Complexity

Statement: The asymptotic complexity is in [math]\displaystyle{ \Theta(T\cdot n^2) }[/math] in the best and worst case, where [math]\displaystyle{ T }[/math] is the complexity of the comparison.

Proof: The asymptotic complexity of the [math]\displaystyle{ i }[/math]-th iteration is in [math]\displaystyle{ \Theta(T\cdot(n - i)) }[/math]. Therefore, the total complexity is in [math]\displaystyle{ \Theta\left(T\cdot\sum_{i=1}^{n-1} (n-i) \right) = \Theta\left(T\cdot\sum_{i=1}^{n-1} i \right) = \Theta\left(T\cdot\frac{n(n-1)}{2} \right) = \Theta\left(T\cdot n^2\right) }[/math].

Java Code

/**
	     	
        int[] selectionSort(int [] list) {

        int n = list.length - 1;
        int left = 0;
        int min;
        int tmp;

        while(left < n) {
            min = left;
            for(int i = left + 1; i<=n; i++) {
                if (list[i] < list[min]) {
                    min = i;
                }
            }

            tmp = list[min];
            list[min] = list[left];
            list[left] = tmp;
            left = left + 1;
        }
    }