Bubblesort
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 of [math]\displaystyle{ S }[/math] at the positions [math]\displaystyle{ |S| - i + 1,\dots,|S| }[/math] are placed at their correct positions 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: Step-by-step, move the maximum element seen so far to the position [math]\displaystyle{ |S| - i + 1 }[/math].
Implementation: For [math]\displaystyle{ j := 2,\dots,|S|-i+1 }[/math] (in this order): If [math]\displaystyle{ S[j-1] \gt S[j] }[/math], swap [math]\displaystyle{ S[j-1] }[/math] and [math]\displaystyle{ S[j] }[/math].
Correctness: The loop invariant of the inner loop is this: after [math]\displaystyle{ m }[/math] iterations of the inner loop, [math]\displaystyle{ S[m+1] }[/math] is the maximum out of [math]\displaystyle{ S[1],\dots,S[m+1] }[/math] To see that invariant, note that the induction hypothesis implies for an iteration on index [math]\displaystyle{ j }[/math] that [math]\displaystyle{ S[j-1] }[/math] is the maximum out of [math]\displaystyle{ S[1],\dots,S[j-1] }[/math]. So, if [math]\displaystyle{ S[j-1] \gt S[j] }[/math], [math]\displaystyle{ S[j-1] }[/math] is also the maximum out of [math]\displaystyle{ S[1],\dots,S[j] }[/math]. Therefore, swapping [math]\displaystyle{ S[j-1] }[/math] and [math]\displaystyle{ S[j] }[/math] maintains the invariant. On the other hand, if [math]\displaystyle{ S[j-1] \leq S[j] }[/math], [math]\displaystyle{ S[j] }[/math] is larger than [math]\displaystyle{ S[1],\ldots,S[j-2] }[/math] due to the induction hypothesis, so doing nothing indeed maintains the invariant in this case.
Complexity
Statement: The asymptotic complexity is in [math]\displaystyle{ \Theta(n^2) }[/math] in the best and worst case.
Proof: The asymptotic complexity of the inner loop in the [math]\displaystyle{ i }[/math]-th iteration of the outer loop is in [math]\displaystyle{ \Theta(n-i) }[/math]. Therefore, the total complexity is in [math]\displaystyle{ \Theta\left(\sum_{i=1}^{n-1} (n-i)\right) = \Theta\left(\sum_{i=1}^{n-1}i\right) = \Theta\left(\frac{n(n-1)}{2}\right) = \Theta(n^2) }[/math].