问题
For a Data Structures project, I must find the shortest path between two words (like \"cat\" and \"dog\"), changing only one letter at a time. We are given a Scrabble word list to use in finding our path. For example:
cat -> bat -> bet -> bot -> bog -> dog
I\'ve solved the problem using a breadth first search, but am seeking something better (I represented the dictionary with a trie).
Please give me some ideas for a more efficient method (in terms of speed and memory). Something ridiculous and/or challenging is preferred.
I asked one of my friends (he\'s a junior) and he said that there is no efficient solution to this problem. He said I would learn why when I took the algorithms course. Any comments on that?
We must move from word to word. We cannot go cat -> dat -> dag -> dog. We also have to print out the traversal.
回答1:
NEW ANSWER
Given the recent update, you could try A* with the Hamming distance as a heuristic. It's an admissible heuristic since it's not going to overestimate the distance
OLD ANSWER
You can modify the dynamic-program used to compute the Levenshtein distance to obtain the sequence of operations.
EDIT: If there are a constant number of strings, the problem is solvable in polynomial time. Else, it's NP-hard (it's all there in wikipedia) .. assuming your friend is talking about the problem being NP-hard.
EDIT: If your strings are of equal length, you can use Hamming distance.
回答2:
With a dictionary, BFS is optimal, but the running time needed is proportional to its size (V+E). With n letters, the dictionary might have ~a^n entires, where a is alphabet size. If the dictionary contains all words but the one that should be on the end of chain, then you'll traverse all possible words but won't find anything. This is graph traversal, but the size might be exponentially large.
You may wonder if it is possible to do it faster - to browse the structure "intelligently" and do it in polynomial time. The answer is, I think, no.
The problem:
You're given a fast (linear) way to check if a word is in dictionary, two words u, v and are to check if there's a sequence u -> a1 -> a2 -> ... -> an -> v.
is NP-hard.
Proof: Take some 3SAT instance, like
(p or q or not r) and (p or not q or r)
You'll start with 0 000 00 and are to check if it is possible to go to 2 222 22.
The first character will be "are we finished", three next bits will control p,q,r and two next will control clauses.
Allowed words are:
- Anything that starts with 0 and contains only 0's and 1's
- Anything that starts with 2 and is legal. This means that it consists of 0's and 1's (except that the first character is 2, all clauses bits are rightfully set according to variables bits, and they're set to 1 (so this shows that the formula is satisfable).
- Anything that starts with at least two 2's and then is composed of 0's and 1's (regular expression: 222* (0+1)*, like 22221101 but not 2212001
To produce 2 222 22 from 0 000 00, you have to do it in this way:
(1) Flip appropriate bits - e.g. 0 100 111 in four steps. This requires finding a 3SAT solution.
(2) Change the first bit to 2: 2 100 111. Here you'll be verified this is indeed a 3SAT solution.
(3) Change 2 100 111 -> 2 200 111 -> 2 220 111 -> 2 222 111 -> 2 222 211 -> 2 222 221 -> 2 222 222.
These rules enforce that you can't cheat (check). Going to 2 222 22 is possible only if the formula is satisfable, and checking that is NP-hard. I feel it might be even harder (#P or FNP probably) but NP-hardness is enough for that purpose I think.
Edit: You might be interested in disjoint set data structure. This will take your dictionary and group words that can be reached from each other. You can also store a path from every vertex to root or some other vertex. This will give you a path, not neccessarily the shortest one.
回答3:
There are methods of varying efficiency for finding links - you can construct a complete graph for each word length, or you can construct a BK-Tree, for example, but your friend is right - BFS is the most efficient algorithm.
There is, however, a way to significantly improve your runtime: Instead of doing a single BFS from the source node, do two breadth first searches, starting at either end of the graph, and terminating when you find a common node in their frontier sets. The amount of work you have to do is roughly half what is required if you search from only one end.
回答4:
You can make it a little quicker by removing the words that are not the right length, first. More of the limited dictionary will fit into the CPU's cache. Probably all of it.
Also, all of the strncmp comparisons (assuming you made everything lowercase) can be memcmp comparisons, or even unrolled comparisons, which can be a speedup.
You could use some preprocessor magic and hard-compile the task for that word-length, or roll a few optimized variations of the task for common word lengths. All of those extra comparisons can 'go away' for pure unrolled fun.
回答5:
This is a typical dynamic programming problem. Check for the Edit Distance problem.
回答6:
What you are looking for is called the Edit Distance. There are many different types.
From (http://en.wikipedia.org/wiki/Edit_distance): "In information theory and computer science, the edit distance between two strings of characters is the number of operations required to transform one of them into the other."
This article about Jazzy (the java spell check API) has a nice overview of these sorts of comparisons (it's a similar problem - providing suggested corrections) http://www.ibm.com/developerworks/java/library/j-jazzy/
回答7:
You could find the longest common subsequence, and therefore finding the letters that must be changed.
回答8:
My gut feeling is that your friend is correct, in that there isn't a more efficient solution, but that is assumming you are reloading the dictionary every time. If you were to keep a running database of common transitions, then surely there would be a more efficient method for finding a solution, but you would need to generate the transitions beforehand, and discovering which transitions would be useful (since you can't generate them all!) is probably an art of its own.
回答9:
bool isadjacent(string& a, string& b)
{
int count = 0; // to store count of differences
int n = a.length();
// Iterate through all characters and return false
// if there are more than one mismatching characters
for (int i = 0; i < n; i++)
{
if (a[i] != b[i]) count++;
if (count > 1) return false;
}
return count == 1 ? true : false;
}
// A queue item to store word and minimum chain length // to reach the word.
struct QItem
{
string word;
int len;
};
// Returns length of shortest chain to reach 'target' from 'start' // using minimum number of adjacent moves. D is dictionary
int shortestChainLen(string& start, string& target, set<string> &D)
{
// Create a queue for BFS and insert 'start' as source vertex
queue<QItem> Q;
QItem item = {start, 1}; // Chain length for start word is 1
Q.push(item);
// While queue is not empty
while (!Q.empty())
{
// Take the front word
QItem curr = Q.front();
Q.pop();
// Go through all words of dictionary
for (set<string>::iterator it = D.begin(); it != D.end(); it++)
{
// Process a dictionary word if it is adjacent to current
// word (or vertex) of BFS
string temp = *it;
if (isadjacent(curr.word, temp))
{
// Add the dictionary word to Q
item.word = temp;
item.len = curr.len + 1;
Q.push(item);
// Remove from dictionary so that this word is not
// processed again. This is like marking visited
D.erase(temp);
// If we reached target
if (temp == target)
return item.len;
}
}
}
return 0;
}
// Driver program
int main()
{
// make dictionary
set<string> D;
D.insert("poon");
D.insert("plee");
D.insert("same");
D.insert("poie");
D.insert("plie");
D.insert("poin");
D.insert("plea");
string start = "toon";
string target = "plea";
cout << "Length of shortest chain is: "
<< shortestChainLen(start, target, D);
return 0;
}
Copied from: https://www.geeksforgeeks.org/word-ladder-length-of-shortest-chain-to-reach-a-target-word/
来源：https://stackoverflow.com/questions/1521958/shortest-path-to-transform-one-word-into-another