Question

我最近posted a question关于优化算法以计算Levenshtein距离，回复引导我阅读Levenshtein Distance上的维基百科文章。

文章提到如果最大距离上存在绑定的 k ，可能的结果可能来自给定的查询，那么运行时间可以从 O（mn）< / em>到 O（kn）， m 和 n 是字符串的长度。我查了算法，但我无法弄清楚如何实现它。我希望在这里获得一些线索。

“可能的改进”下的优化是＃4。

困惑我的部分是那个说我们只需要计算宽度 2k + 1 的对角条纹，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，我会非常感激。如果需要，我可以在书中发布算法的完整描述作为答案。

Answer 1

我做过很多次了。我这样做的方法是使用可能更改的游戏树的递归深度优先树。有一个预算 k 的更改，我用来修剪树。有了这个例程，首先我用k = 0运行它，然后k = 1，然后k = 2，直到我得到一个命中或者我不想再高一点。

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，要将其限制为预算，如果hdis太大，则拒绝下降。

使用优化的Levenshtein算法寻找最近邻居

1 个答案: