Permutation dafny断言失败很难解释
我试图证明一个方法的正确性,这个方法决定 大小为Permutation dafny断言失败很难解释,permutation,assert,dafny,loop-invariant,post-conditions,Permutation,Assert,Dafny,Loop Invariant,Post Conditions,我试图证明一个方法的正确性,这个方法决定 大小为n的序列是0,1,…,n-1的排列。我设法证明,每当该方法返回true时,序列确实是一个有效的排列。不过,证明相反的情况(对我来说)要困难得多。我认为我有正确的循环不变量和触发器,但是Dafny引发了一个我无法理解的断言错误。以下是完整性代码: method perm_imperative(s: seq<int>) returns (status : bool) requires |s| > 0 ensures (
n
的序列是0,1,…,n-1
的排列。我设法证明,每当该方法返回true
时,序列确实是一个有效的排列。不过,证明相反的情况(对我来说)要困难得多。我认为我有正确的循环不变量和触发器,但是Dafny引发了一个我无法理解的断言错误。以下是完整性代码:
method perm_imperative(s: seq<int>) returns (status : bool)
requires |s| > 0
ensures (status == true ) ==> (forall i :: 0 <= i < |s| ==> 0 <= s[i] < |s|)
ensures (status == true ) ==> (forall i, j :: 0 <= i < j < |s| ==> (s[i] != s[j]))
ensures (status == false) ==> (exists i, j :: 0 <= i <= j < |s| && (
(s[i] < 0 ) ||
(s[i] >= |s| ) ||
(s[i] == s[j] && (i != j))))
{
var i := 0;
var used := new int[|s|];
while (i < |s|)
invariant 0 <= i <= |s|
invariant forall k :: 0 <= k < i ==> used[k] == -1;
{
used[i] := -1;
i := i + 1;
}
assert (forall k :: 0 <= k < |s| ==> used[k] == -1);
i := 0;
while (i < |s|)
invariant 0 <= i <= |s|
invariant forall k :: 0 <= k < i ==> 0 <= s[k] < |s|
invariant forall k :: 0 <= k < i ==> 0 <= used[s[k]] < i
invariant forall k, m :: 0 <= k < m < i ==> s[k] != s[m]
invariant forall k :: 0 <= k < i ==> used[s[k]] == k
invariant forall k :: 0 <= k < i ==> s[used[s[k]]] == s[k]
invariant forall k, m :: 0 <= k < m < i ==> used[s[k]] != used[s[m]]
invariant forall k :: 0 <= k < |s| ==> (0 <= used[k] < i) || (used[k] == -1)
{
if (s[i] < 0)
{
// assert (s[i] < 0);
return false;
}
else if (s[i] >= |s|)
{
// assert (s[i] >= |s|);
return false;
}
else if (used[s[i]] != -1)
{
assert (0 <= s[i] < |s|);
assert (0 <= used[s[i]] < i );
//////////////////
// //
// ASSERT ERROR //
// | //
// V //
//////////////////
assert (s[used[s[i]]] == s[i]);
return false;
}
assert (used[s[i]] == -1);
used[s[i]] := i;
i := i + 1;
}
return true;
}
方法perm_命令(s:seq)返回(状态:bool)
需要| s |>0
确保(状态==true)=>(对于所有i::0(对于所有i,j::0(s[i]!=s[j]))
确保(status==false)=>(exists i,j::0如果添加以下循环不变量,则证明将通过:
invariant forall n | 0 <= n < |s| :: used[n] != -1 ==> exists k | 0 <= k < i :: s[k] == n
如果添加以下循环不变量,则证明将通过:
invariant forall n | 0 <= n < |s| :: used[n] != -1 ==> exists k | 0 <= k < i :: s[k] == n
我现在无法深入研究这个问题,但您可以尝试将失败的assert
更改为assemption
,然后看看它是否有效,或者其他什么失败了。这可能会解除您的阻碍。我现在无法深入研究这个问题,但您可以尝试将失败的assert
更改为assemption
,然后看看它是否有效然后,或者别的什么失败了。也许这会让你解脱。