使用优化的Levenshtein算法查找最接近的邻居

Z时代
2024-01-10
分类：问答

我最近发布了一个有关优化算法以计算Levenshtein距离的问题，而答复使我进入Wikipedia上有关Levenshtein距离的文章。

文章提到，如果最大距离上有一个 k ，则可能来自给定查询的结果可能是运行时间从 O（mn） 减少到 O（kn） ， m 和 n

是字符串。我查了一下算法，但是我真的不知道如何实现它。我希望在这里获得一些线索。

优化是“可能的改进”下的＃4。

让我感到困惑的部分是，我们只需要计算以主对角线为中心（主对角线定义为坐标（i，i））的宽度为 2k + 1 的对角线条。

如果有人可以提供帮助/见解，我将不胜感激。如果需要，我可以在此处将关于算法的完整描述发布在书中作为答案。

回答：

我已经做过很多次了。我这样做的方式是对可能更改的游戏树进行深度优先的递归树遍历。我用来修剪树的预算为 k 。有了这个例程，首先我以k = 0，然后k =

1，然后k = 2来运行它，直到我被击中或者我不想再走高了。

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;
}

添加了解释特里搜索的说明：

// 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.