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我正在尝试编写子字符串方法的简单验证实现。我的方法接受 2 个字符串并检查 str2 是否在 str1 中。我首先试图弄清楚为什么我的不变量不成立 - Dafny 标记该不变量可能在输入时不成立，而我的前置/后置条件失败。我对不变量的想法是有 3 个主要场景： 1. 循环未能找到索引 i 处的子字符串，并且有更多索引要探索 2. 循环未能找到索引 i 处的子字符串 - 没有更多索引可探索 3 . 循环在索引 i 处找到子字符串

代码：

method Main() {
    var str1,str2 := "Dafny","fn";
    var found,offset := FindSubstring(str1,str2);
    assert found by
    {
        calc {
            str2 <= str1[2..];
        ==>
            IsSubsequenceAtOffset(str1,str2,2);
        ==>
            IsSubsequence(str1,str2);
        ==
            found;
        }
    }
    assert offset == 2 by
    {
        assert found && str2 <= str1[2..];
        assert offset != 0 by { assert 'D' == str1[0] != str2[0] == 'f'; }
        assert offset != 1 by { assert 'a' == str1[1] != str2[0] == 'f'; }
        assert offset != 3 by { assert 'n' == str1[3] != str2[0] == 'f'; }
        assert !(offset >= 4) by { assert 4 + |str2| > |str1|; }
    }
    print "The sequence ";
    print str2;
    print " is a subsequence of ";
    print str1;
    print " starting at element ";
    print offset;
}

predicate IsSubsequence<T>(q1: seq<T>, q2: seq<T>)
{
    exists offset: nat :: offset + |q2| <= |q1| && IsSubsequenceAtOffset(q1,q2,offset)
}

predicate IsSubsequenceAtOffset<T>(q1: seq<T>, q2: seq<T>, offset: nat)
{ 
    offset + |q2| <= |q1| && q2 <= q1[offset..]
}

predicate Inv<T>(str1: seq<T>, str2: seq<T>, offset: nat, found: bool)
{
    |str1| > 0 && |str2| > 0 && |str1| >= |str2| && offset <= |str1|-|str2| &&
    (!found && offset < |str1|-|str2| ==> !(str2 <= str1[offset..])) &&
  (!found && offset == |str1| -|str2| ==> !IsSubsequence(str1, str2)) && 
    (found ==> IsSubsequenceAtOffset(str1, str2, offset))
}

method FindSubstring(str1: string, str2: string) returns (found: bool, offset: nat)

    requires |str2| <= |str1|
    ensures found <==> IsSubsequence(str1,str2)
    ensures found ==> IsSubsequenceAtOffset(str1,str2,offset)  
  {
     found, offset := str2 <= str1[0..], 0;
     assert Inv(str1,str2,offset,found);

     while !found && offset <= (|str1| - |str2|) 
        invariant Inv(str1,str2,offset,found)
     {
       if str2 <= str1[(offset + 1)..] {
         found, offset := true, offset + 1;
       } else {
         offset := offset + 1;
       }
     } 
     Lemma(str1,str2,found,offset);
  }

  lemma Lemma(str1: string, str2: string, found: bool, offset: nat)
    requires !(!found && offset <= (|str1| - |str2|))
    requires Inv(str1,str2,offset,found)
    ensures found <==> IsSubsequence(str1,str2)
    ensures found ==> IsSubsequenceAtOffset(str1,str2,offset) 
    {}

链接：http ://rise4fun.com/Dafny/QaAbU 任何帮助将不胜感激。

4

1 回答 1

2

http://rise4fun.com/Dafny/q9BG

1) |str1| > 0 && |str2| > 0ininv失败，因为此条件未添加到功能要求中FindSubstring

2）由于 while 循环体FindSubstring检查子字符串 for offset+1，因此 while 循环只需要运行 foroffset < (|str1| - |str2|)因为 if offset == (|str1| - |str2|)then 不存在任何可以满足的偏移量str2 <= str1[offset..]

3) 添加了量词，inv其中确保对于所有偏移量kwhere 0 <= k <= offset，不存在k任何str2 <= str1[offset..]

调试失败的不变量的一个小建议可能会有所帮助：用inv实际的主体替换调用，inv并且 dafny 将查明不变量中的失败条件。

希望这是有道理的。

于 2017-01-16T02:16:01.657 回答