使用优化的 Levenshtein 算法寻找最近的邻居

Question

我最近在posted a question 谈论优化算法以计算 Levenshtein 距离，这些回复将我带到了Levenshtein Distance 上的维基百科文章。

文章提到如果在最大距离上有一个界限k，则可能的结果可以来自给定的查询，那么运行时间可以从O(mn) 到 O(kn)，m 和 n 是字符串的长度。我查了算法，但我真的不知道如何实现它。我希望在这里得到一些线索。

优化是“可能的改进”下的#4。

让我困惑的部分是说我们只需要计算宽度为 2k+1 的对角线条纹，以主对角线为中心（主对角线定义为坐标（i ,i))。

如果有人可以提供一些帮助/见解，我将不胜感激。如果需要，我可以在这里发布书中算法的完整描述作为答案。

Answer 1

我已经做过很多次了。我这样做的方法是对可能变化的游戏树进行递归深度优先树遍历。有一个预算 k 的更改，我用它来修剪树。有了这个例程，我首先以 k=0，然后 k=1，然后 k=2 运行它，直到我获得成功或者我不想再更高了。

char* a = /* string 1 */;
char* b = /* string 2 */;
int na = strlen(a);
int nb = strlen(b);
bool walk(int ia, int ib, int k){
  /* if the budget is exhausted, prune the search */
  if (k < 0) return false;
  /* if at end of both strings we have a match */
  if (ia == na && ib == nb) return true;
  /* if the first characters match, continue walking with no reduction in budget */
  if (ia < na && ib < nb && a[ia] == b[ib] && walk(ia+1, ib+1, k)) return true;
  /* if the first characters don't match, assume there is a 1-character replacement */
  if (ia < na && ib < nb && a[ia] != b[ib] && walk(ia+1, ib+1, k-1)) return true;
  /* try assuming there is an extra character in a */
  if (ia < na && walk(ia+1, ib, k-1)) return true;
  /* try assuming there is an extra character in b */
  if (ib < nb && walk(ia, ib+1, k-1)) return true;
  /* if none of those worked, I give up */
  return false;
}

添加解释 trie-search：

// definition of trie-node:
struct TNode {
  TNode* pa[128]; // for each possible character, pointer to subnode
};

// simple trie-walk of a node
// key is the input word, answer is the output word,
// i is the character position, and hdis is the hamming distance.
void walk(TNode* p, char key[], char answer[], int i, int hdis){
  // If this is the end of a word in the trie, it is marked as
  // having something non-null under the '\0' entry of the trie.
  if (p->pa[0] != null){
    if (key[i] == '\0') printf("answer = %s, hdis = %d\n", answer, hdis);
  }
  // for every actual subnode of the trie
  for(char c = 1; c < 128; c++){
    // if it is a real subnode
    if (p->pa[c] != null){
      // keep track of the answer word represented by the trie
      answer[i] = c; answer[i+1] = '\0';
      // and walk that subnode
      // If the answer disagrees with the key, increment the hamming distance
      walk(p->pa[c], key, answer, i+1, (answer[i]==key[i] ? hdis : hdis+1));
    }
  }
}
// Note: you have to edit this to handle short keys.
// Simplest is to just append a lot of '\0' bytes to the key.

现在，为了限制预算，如果 hdis 太大就拒绝下降。