I have a database of ~150'000 words and a pattern (any single word) and I want to get all words from the database which has Damerau-Levenshtein distance between it and the pattern less than given number. I need to do it extremely fast. What algorithm could you suggest? If there's no good algorithm for Damerau-Levenshtein distance, just Levenshtin distance will be welcome as well.
Thank you for your help.
P.S. I'm not going to use SOUNDEX.
I would start with a SQL function to calculate the Levenshtein distance (in T-SQl or .Net) (yes, I'm a MS person...) with a maximum distance parameter that would cause an early exit.
This function could then be used to compare your input with each string to check the distanve and move on to the next if it breaks the threshold.
I was also thinking you could, for example, set the maximum distance to be 2, then filter all words where the length is more than 1 different whilst the first letter is different. With an index this may be slightly quicker.
You could also shortcut to bring back all strings that are perfect matches (indexing will speed this up) as these will actually take longer to calculate the Levenshtein distance of 0.
Just some thoughts....
I do not think you can calculate this kind of function without actually enumerating all rows.
So the solutions are:
Make it a very fast enumeration (but this doesn't really scale)
Filter initial variants somehow (index by a letter, at least x common letters)
Use alternative (indexable) algorithm, such as N-Grams (however I do not have details on result quality of ngrams versus D-L distance).
A solution off the top of my head might be to store the database in a sorted set (e.g., std::set in C++), as it seems to me that strings sorted lexicographically would compare well. To approximate the position of the given string in the set, use std::upper_bound on the string, then iterate over the set outward from the found position in both directions, computing the distance as you go, and stop when it falls below a certain threshold. I have a feeling that this solution would probably only match strings with the same start character, but if you're using the algorithm for spell-checking, then that restriction is common, or at least unsurprising.
Edit: If you're looking for an optimisation of the algorithm itself, however, this answer is irrelevant.
I have used KNIME for string fuzzy matching and has got very fast results. It is also very easy to make visual workflows in it. Just install KNIME free edition from https://www.knime.org/ then use "String Distance" and "Similarity Search" nodes to get your results. I have attached a small fuzzy matching smaple workflow in here (the input data come from top and the patterns to search for come from the bottom in this case):
I would recommend looking into Ankiro.
I'm not certain that it meets your requirements for precision, but it is fast.
Related
Let's say I have a dictionary (word list) of millions upon millions of words. Given a query word, I want to find the word from that huge list that is most similar.
So let's say my query is elepant, then the result would most likely be elephant.
If my word is fentist, the result will probably be dentist.
Of course assuming both elephant and dentist are present in my initial word list.
What kind of index, data structure or algorithm can I use for this so that the query is fast? Hopefully complexity of O(log N).
What I have: The most naive thing to do is to create a "distance function" (which computes the "distance" between two words, in terms of how different they are) and then in O(n) compare the query with every word in the list, and return the one with the closest distance. But I wouldn't use this because it's slow.
The problem you're describing is a Nearest Neighbor Search (NNS). There are two main methods of solving NNS problems: exact and approximate.
If you need an exact solution, I would recommend a metric tree, such as the M-tree, the MVP-tree, and the BK-tree. These trees take advantage of the triangle inequality to speed up search.
If you're willing to accept an approximate solution, there are much faster algorithms. The current state of the art for approximate methods is Hierarchical Navigable Small World (hnsw). The Non-Metric Space Library (nmslib) provides an efficient implementation of hnsw as well as several other approximate NNS methods.
(You can compute the Levenshtein distance with Hirschberg's algorithm)
I made similar algorythm some time ago
Idea is to have an array char[255] with characters
and values is a list of words hashes (word ids) that contains this character
When you are searching 'dele....'
search(d) will return empty list
search(e) will find everything with character e, including elephant (two times, as it have two 'e')
search(l) will brings you new list, and you need to combine this list with results from previous step
...
at the end of input you will have a list
then you can try to do group by wordHash and order by desc by count
Also intresting thing, if your input is missing one or more characters, you will just receive empty list in the middle of the search and it will not affect this idea
My initial algorythm was without ordering, and i was storing for every character wordId and lineNumber and char position.
My main problem was that i want to search
with ee to find 'elephant'
with eleant to find 'elephant'
with antph to find 'elephant'
Every words was actually a line from file, so it's often was very long
And number of files and lines was big
I wanted quick search for directories with more than 1gb text files
So it was a problem even store them in memory, for this idea you need 3 parts
function to fill your cache
function to find by char from input
function to filter and maybe order results (i didn't use ordering, as i was trying to fill my cache in same order as i read the file, and i wanted to put lines that contains input in the same order upper )
I hope it make sense
I'm trying to find near duplicates in a large list of names by computing the metaphone key for each string, and then, within each set of possible duplicates, use something like Levenshtein distance to get a more refined estimate of duplicate likelihood.1
However, I'm finding that metaphone is heavily determined by the first characters in the strings, and so if I feed it a long list of people's names, I get huge buckets where everyone's name is "Jennifer X" or "Richard Y", but otherwise haven't got much in common.
If I reverse the string before generating the key, the results are more sensible, in that they group by last name, but still I find that the first names aren't particularly similar.
So is there a similar algorithm that samples more of the input string to produce a sound key, perhaps by using a longer key string?
[1] Ideally, I'd compute the string distances directly, but if my list has 10,000 names, that would mean 100,000,000 computations, which is why I'm trying to divide and conquer by sound keying each name first and only checking for similarities within the buckets. But if there's a better way, I'd love to hear about it!
Try eudex.
It's described as "A blazingly fast phonetic reduction/hashing algorithm."
There are many easy ways to use it, as it encodes a word into a 64-bit integer with most discriminating features towards the MSB. The hamming difference between hashes is also useful as a difference metric between words and spellings.
A group of amusing students write essays exclusively by plagiarising portions of the complete works of WIlliam Shakespere. At one end of the scale, an essay might exclusively consist a verbatim copy of a soliloquy... at the other, one might see work so novel that - while using a common alphabet - no two adjacent characters in the essay were used adjacently by Will.
Essays need to be graded. A score of 1 is assigned to any essay which can be found (character-by-character identical) in the plain-text of the complete works. A score of 2 is assigned to any work that can be successfully constructed from no fewer than two distinct (character-by-character identical) passages in the complete works, and so on... up to the limit - for an essay with N characters - which scores N if, and only if, no two adjacent characters in the essay were also placed adjacently in the complete works.
The challenge is to implement a program which can efficiently (and accurately) score essays. While any (practicable) data-structure to represent the complete works is acceptable - the essays are presented as ASCII strings.
Having considered this teasing question for a while, I came to the conclusion that it is much harder than it sounds. The naive solution, for an essay of length N, involves 2**(N-1) traversals of the complete works - which is far too inefficient to be practical.
While, obviously, I'm interested in suggested solutions - I'd also appreciate pointers to any literature that deals with this, or any similar, problem.
CLARIFICATIONS
Perhaps some examples (ranging over much shorter strings) will help clarify the 'score' for 'essays'?
Assume Shakespere's complete works are abridged to:
"The quick brown fox jumps over the lazy dog."
Essays scoring 1 include "own fox jump" and "The quick brow". The essay "jogging" scores 6 (despite being short) because it can't be represented in fewer than 6 segments of the complete works... It can be segmented into six strings that are all substrings of the complete works as follows: "[j][og][g][i][n][g]". N.B. Establishing scores for this short example is trivial compared to the original problem - because, in this example "complete works" - there is very little repetition.
Hopefully, this example segmentation helps clarify the 2*(N-1) substring searches in the complete works. If we consider the segmentation, the (N-1) gaps between the N characters in the essay may either be a gap between segments, or not... resulting in ~ 2*(N-1) substring searches of the complete works to test each segmentation hypothesis.
An (N)DFA would be a wonderful solution - if it were practical. I can see how to construct something that solved 'substring matching' in this way - but not scoring. The state space for scoring, on the surface, at least, seems wildly too large (for any substantial complete works of Shakespere.) I'd welcome any explanation that undermines my assumptions that the (N)DFA would be too large to be practical to compute/store.
A general approach for plagiarism detection is to append the student's text to the source text separated by a character not occurring in either and then to build either a suffix tree or suffix array. This will allow you to find in linear time large substrings of the student's text which also appear in the source text.
I find it difficult to be more specific because I do not understand your explanation of the score - the method above would be good for finding the longest stretch in the students work which is an exact quote, but I don't understand your N - is it the number of distinct sections of source text needed to construct the student's text?
If so, there may be a dynamic programming approach. At step k, we work out the least number of distinct sections of source text needed to construct first k characters of the student's text. Using a suffix array built just from the source text or otherwise, we find the longest match between the source text and characters x..k of the student's text, where x is of course as small as possible. Then the least number of sections of source text needed to construct the first k characters of student text is the least needed to construct 1..x-1 (which we have already worked out) plus 1. By running this process for k=1..the length of the student text we find the least number of sections of source text needed to reconstruct the whole of it.
(Or you could just search StackOverflow for the student's text, on the grounds that students never do anything these days except post their question on StackOverflow :-)).
I claim that repeatedly moving along the target string from left to right, using a suffix array or tree to find the longest match at any time, will find the smallest number of different strings from the source text that produces the target string. I originally found this by looking for a dynamic programming recursion but, as pointed out by Evgeny Kluev, this is actually a greedy algorithm, so let's try and prove this with a typical greedy algorithm proof.
Suppose not. Then there is a solution better than the one you get by going for the longest match every time you run off the end of the current match. Compare the two proposed solutions from left to right and look for the first time when the non-greedy solution differs from the greedy solution. If there are multiple non-greedy solutions that do better than the greedy solution I am going to demand that we consider the one that differs from the greedy solution at the last possible instant.
If the non-greedy solution is going to do better than the greedy solution, and there isn't a non-greedy solution that does better and differs later, then the non-greedy solution must find that, in return for breaking off its first match earlier than the greedy solution, it can carry on its next match for longer than the greedy solution. If it can't, it might somehow do better than the greedy solution, but not in this section, which means there is a better non-greedy solution which sticks with the greedy solution until the end of our non-greedy solution's second matching section, which is against our requirement that we want the non-greedy better solution that sticks with the greedy one as long as possible. So we have to assume that, in return for breaking off the first match early, the non-greedy solution gets to carry on its second match longer. But this doesn't work, because, when the greedy solution finally has to finish using its first match, it can jump on to the same section of matching text that the non-greedy solution is using, just entering that section later than the non-greedy solution did, but carrying on for at least as long as the non-greedy solution. So there is no non-greedy solution that does better than the greedy solution and the greedy solution is optimal.
Have you considered using N-Grams to solve this problem?
http://en.wikipedia.org/wiki/N-gram
First read the complete works of Shakespeare and build a trie. Then process the string left to right. We can greedily take the longest substring that matches one in the data because we want the minimum number of strings, so there is no factor of 2^N. The second part is dirt cheap O(N).
The depth of the trie is limited by the available space. With a gigabyte of ram you could reasonably expect to exhaustively cover Shakespearean English string of length at least 5 or 6. I would require that the leaf nodes are unique (which also gives a rule for constructing the trie) and keep a pointer to their place in the actual works, so you have access to the continuation.
This feels like a problem of partial matching a very large regular expression.
If so it can be solved by a very large non deterministic finite state automata or maybe more broadly put as a graph representing for every character in the works of Shakespeare, all the possible next characters.
If necessary for efficiency reasons the NDFA is guaranteed to be convertible to a DFA. But then this construction can give rise to 2^n states, maybe this is what you were alluding to?
This aspect of the complexity does not really worry me. The NDFA will have M + C states; one state for each character and C states where C = 26*2 + #punctuation to connect to each of the M states to allow the algorithm to (re)start when there are 0 matched characters. The question is would the corresponding DFA have O(2^M) states and if so is it necessary to make that DFA, theoretically it's not necessary. However, consider that in the construction, each state will have one and only one transition to exactly one other state (the next state corresponding to the next character in that work). We would expect that each one of the start states will be connected to on average M/C states, but in the worst case M meaning the NDFA will have to track at most M simultaneous states. That's a large number but not an impossibly large number for computers these days.
The score would be derived by initializing to 1 and then it would incremented every time a non-accepting state is reached.
It's true that one of the approaches to string searching is building a DFA. In fact, for the majority of the string search algorithms, it looks like a small modification on failure to match (increment counter) and success (keep going) can serve as a general strategy.
For example, starting with the set of english words, is there a structure/algorithm that allows one fast retrieval of strings such as "light" and "tight", using the word "right" as the query? I.e., I want to retrieve strings with small Levenshtein distance to the query string.
The BK-tree data structure might be appropriate here. It's designed to efficiently support queries of the form "what are all words within edit distance k or less from a query word?" Its performance guarantees are reasonably good, and it's not too difficult to implement.
Hope this helps!
Since calculating Levenshtein distance is O(nm) for strings of length n and m, the naive approach of calculating all Levenshtein distances L(querystring, otherstring) is very expensive.
However, if you visualize the Levenshtein algorithm, it basically fills an n*m table with edit distances. But for words that start with the same few letters (prefix), the first few rows of the Levenshtein tables will be the same. (Fixing the query string, of course.)
This suggests using a trie (also called prefix tree): Read the query string, then build a trie of Levenshtein rows. Afterwards, you can easily traverse it to find strings close to the query string.
(This does mean that you have to build an new trie for a new query string. I don't think there is a similarly intriguing structure for all-pairs distances.)
I thought I recently saw an article about this with a nice python implementation. Will add a link if I can find it. Edit: Here it is, on Steve Hanov's blog.
I'm thinking the fastest way would be to pre-build a cache of similarities which you can index and access in O(1) time. The trick would be to find common misspellings to add to your cache, which could get pretty large.
I imagine Google would do something similar using their wide range of statistical query search data.
This is a problem appeared in today's Pacific NW Region Programming Contest during which no one solved it. It is problem B and the complete problem set is here: http://www.acmicpc-pacnw.org/icpc-statements-2011.zip. There is a well-known O(n^2) algorithm for LCS of two strings using Dynamic Programming. But when these strings are extended to rings I have no idea...
P.S. note that it is subsequence rather than substring, so the elements do not need to be adjacent to each other
P.S. It might not be O(n^2) but O(n^2lgn) or something that can give the result in 5 seconds on a common computer.
Searching the web, this appears to be covered by section 4.3 of the paper "Incremental String Comparison", by Landau, Myers, and Schmidt at cost O(ne) < O(n^2), where I think e is the edit distance. This paper also references a previous paper by Maes giving cost O(mn log m) with more general edit costs - "On a cyclic string to string correcting problem". Expecting a contestant to reproduce either of these papers seems pretty demanding to me - but as far as I can see the question does ask for the longest common subsequence on cyclic strings.
You can double the first and second string and then use the ordinary method, and later wrap the positions around.
It is a good idea to "double" the strings and apply the standard dynamic programing algorithm. The problem with it is that to get the optimal cyclic LCS one then has to "start the algorithm from multiple initial conditions". Just one initial condition (e.g. setting all Lij variables to 0 at the boundaries) will not do in general. In practice it turns out that the number of initial states that are needed are O(N) in number (they span a diagonal), so one gets back to an O(N^3) algorithm.
However, the approach does has some virtue as it can be used to design efficient O(N^2) heuristics (not exact but near exact) for CLCS.
I do not know if a true O(N^2) exist, and would be very interested if someone knows one.
The CLCS problem has quite interesting properties of "periodicity": the length of a CLCS of
p-times reapeated strings is p times the CLCS of the strings. This can be proved by adopting a geometric view off the problem.
Also, there are some additional benefits of the problem: it can be shown that if Lc(N) denotes the averaged value of the CLCS length of two random strings of length N, then
|Lc(N)-CN| is O(\sqrt{N}) where C is Chvatal-Sankoff's constant. For the averaged length L(N) of the standard LCS, the only rate result of which I know says that |L(N)-CN| is O(sqrt(Nlog N)). There could be a nice way to compare Lc(N) with L(N) but I don't know it.
Another question: it is clear that the CLCS length is not superadditive contrary to the LCS length. By this I mean it is not true that CLCS(X1X2,Y1Y2) is always greater than CLCS(X1,Y1)+CLCS(X2,Y2) (it is very easy to find counter examples with a computer).
But it seems possible that the averaged length Lc(N) is superadditive (Lc(N1+N2) greater than Lc(N1)+Lc(N2)) - though if there is a proof I don't know it.
One modest interest in this question is that the values Lc(N)/N for the first few values of N would then provide good bounds to the Chvatal-Sankoff constant (much better than L(N)/N).
As a followup to mcdowella's answer, I'd like to point out that the O(n^2 lg n) solution presented in Maes' paper is the intended solution to the contest problem (check http://www.acmicpc-pacnw.org/ProblemSet/2011/solutions.zip). The O(ne) solution in Landau et al's paper does NOT apply to this problem, as that paper is targeted at edit distance, not LCS. In particular, the solution to cyclic edit distance only applies if the edit operations (add, delete, replace) all have unit (1, 1, 1) cost. LCS, on the other hand, is equivalent to edit distances with (add, delete, replace) costs (1, 1, 2). These are not equivalent to each other; for example, consider the input strings "ABC" and "CXY" (for the acyclic case; you can construct cyclic counterexamples similarly). The LCS of the two strings is "C", but the minimum unit-cost edit is to replace each character in turn.
At 110 lines but no complex data structures, Maes' solution falls towards the upper end of what is reasonable to implement in a contest setting. Even if Landau et al's solution could be adapted to handle cyclic LCS, the complexity of the data structure makes it infeasible in a contest setting.
Last but not least, I'd like to point out that an O(n^2) solution DOES exist for CLCS, described here: http://arxiv.org/abs/1208.0396 At 60 lines, no complex data structures, and only 2 arrays, this solution is quite reasonable to implement in a contest setting. Arriving at the solution might be a different matter, though.