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// Bubble sort, where specification says the output is a permutation of its input
// Rustan Leino, 30 April 2009
const N: int;
axiom 0 <= N;
var a: [int]int;
procedure BubbleSort() returns (perm: [int]int)
modifies a;
// array is sorted
ensures (forall i, j: int :: 0 <= i && i <= j && j < N ==> a[i] <= a[j]);
// perm is a permutation
ensures (forall i: int :: 0 <= i && i < N ==> 0 <= perm[i] && perm[i] < N);
ensures (forall i, j: int :: 0 <= i && i < j && j < N ==> perm[i] != perm[j]);
// the final array is that permutation of the input array
ensures (forall i: int :: 0 <= i && i < N ==> a[i] == old(a)[perm[i]]);
{
var n, p, tmp: int;
n := 0;
while (n < N)
invariant n <= N;
invariant (forall i: int :: 0 <= i && i < n ==> perm[i] == i);
{
perm[n] := n;
n := n + 1;
}
while (true)
invariant 0 <= n && n <= N;
// array is sorted from n onwards
invariant (forall i, k: int :: n <= i && i < N && 0 <= k && k < i ==> a[k] <= a[i]);
// perm is a permutation
invariant (forall i: int :: 0 <= i && i < N ==> 0 <= perm[i] && perm[i] < N);
invariant (forall i, j: int :: 0 <= i && i < j && j < N ==> perm[i] != perm[j]);
// the current array is that permutation of the input array
invariant (forall i: int :: 0 <= i && i < N ==> a[i] == old(a)[perm[i]]);
{
n := n - 1;
if (n < 0) {
break;
}
p := 0;
while (p < n)
invariant p <= n;
// array is sorted from n+1 onwards
invariant (forall i, k: int :: n+1 <= i && i < N && 0 <= k && k < i ==> a[k] <= a[i]);
// perm is a permutation
invariant (forall i: int :: 0 <= i && i < N ==> 0 <= perm[i] && perm[i] < N);
invariant (forall i, j: int :: 0 <= i && i < j && j < N ==> perm[i] != perm[j]);
// the current array is that permutation of the input array
invariant (forall i: int :: 0 <= i && i < N ==> a[i] == old(a)[perm[i]]);
// a[p] is at least as large as any of the first p elements
invariant (forall k: int :: 0 <= k && k < p ==> a[k] <= a[p]);
{
if (a[p+1] < a[p]) {
tmp := a[p]; a[p] := a[p+1]; a[p+1] := tmp;
tmp := perm[p]; perm[p] := perm[p+1]; perm[p+1] := tmp;
}
p := p + 1;
}
}
}
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