Fast and easy Levenshtein distance using a trie (2011)

If we picked the threshold to be one, couldn't you come up with two words that are levenstein distance one, and then come up with a third word that's distance one to one of the words and not the other?

In that case, the first two would be in one bucket, where does the third go?

You are kidding right ? I cant imagine anyone seriously suggesting PCA on a half a million by half a million matrix, or evaluating this matrix in the first place.

This actually a fairly actively researched topic: whether edit distance can be embedded, even if approximately in a vector space. It has been surprisingly resistant to such efforts. Yes there are some results but nothing very satisfactory.

Edit distance is a metric, so it was believed it would easy to map individual strings to vectors, such that the distance between two vectors was same as, or close to the edit distance between the corresponding strings. What we want, then, is an (approximate) isomorphism between these two metric spaces. In addition, in the interest of size we would like these vectors to be of small dimension. If the required dimension is high it would defeat the purpose if the string to vector map needs to be stored. Unfortunately it seems edit distance is resistant to isomorphism to a low-dimensional vector space. There is also this question which metric do we use in the vector space, Euclidean, Manhattan...? there are some results on the L1, i.e. embedding on a vector space with Manhattan distances.

Fun story: was an intern in Google's spelling correction team once upon a time long ago. So I was explaining what I was doing, the person I was explaining it to was thoroughly unimpressed: "So whats the big deal, you know/hear the word and write down the spelling". It made me realize how difficult spelling in English is compared to some of the other languages. The variability in the map from phonemes to spelling is quite high in English. This person was Italian, he told me Italian has a lot less variability in this regard.

Wait, what? Doesn't this transitively result in the whole dictionary hashing to the same value?

People are pointing out there is no known hash function that will map every word within a same distance D to the same hash. That is correct, and I was just meaning to explain the process at a high level. Levenshtein-locality sensitive hashing is not a solved problem in CS.

And you know what else in the algorithm I described is not a solved problem? Efficient nearest neighbor search. In a space with more than a few dimensions, we actually don't know any better way to perform exact NNS than brute forcing the search space.

And you know what? None of that matter, since we have inexact yet statistically excellent processes to perform both of these tasks. That's the really cool thing about engineering as opposed to math or computer science: it doesn't have to be exact to be useful.

A trivial way to build an inexact, yet useful levenshtein hash function is to pick a random sample of words from your dictionary, define them as hash values, and assign to any input the closest word from your sample. Now work with a large family (100+) of such grossly approximate hashes and you can build a statistically excellent algorithm. (and of course this is just an example, there are countless better ways to do the same thing).

And I wouldn't go as far as to say it's useless. Basic Levenshtein only misses transpositions in comparison to Damerau-L, but it still catches them as insert/delete combos.