I have a huge list with about 100 000 lines like this:
ipadnews
abcipad
cddeeffipad
hellworld
iworldthis
.. and so on
And would like to find popular substrings, in this case "ipad" would be the most popular and "world" would be on second place. Minimum length should be three or four chars.
I can't predict the substrings so using a dictionary is a no no.
This is a relatively complicated problem ... but it's tractable using prefix/suffix trees. It's essentially a variation of the longest common subsequence and longest common substring problems. - which is where I would start.
There's actually quite a bit of research on problems on this form - you should be able to use the terms above to narrow your search.
You can solve this using a generalized suffix tree which can be built in O(n) time. This is effectively a play on the LCS problem.
I would go about this problem using the following flow of logic:
Extract the set of suffixes for each word. So from 'ipadnews' we get: 'ipadnews', 'padnews', 'adnews', and so on. This way, 'news' will be one of the suffixes, but not 'ipad'.
To make up for the missing substrings in the above step, extract the prefixes as well. We get 'ipadnew', 'ipadne', and so on, including 'ipad'.
For each of the substrings above, hash them towards a count, e.g. $hash{$substr}++.
At the end we will have a long hashtable with frequency of words as values. Instead of an expensive sorting, suppose you only want 10 most popular words. Keep a set from the beginning whose criteria is that any word in it must have a score more than the current min score. You can keep track of the word with min score and when you add the 11th item with score more than the min score, bump out the word with the min score and update the min score pointer.
The max number of keys in the hashtable will be 2*k*n where k is the average length of the words and n is total number of words.
Related
I am looking for an algorithm, but I don't know the name of the problem so I can't find anything. Hopefully my explanation of the problem makes sense!
Let's say you have a long list of phrases, where each phrase is a set of words. The user inputs a list of words, and their list "matches" a phrase every word in the phrase is found in their list. A list's "score" is the number of phrases it matches. The goal is to provide the user with a list of words that would most improve their list's score.
Here's a simple example. We have ten phrases:
wood cabin
camping in woods
camping cabin
fun camping
bon fire
camping fire
swimming hole
fun cabin
wood fire
fire place
And the user provides this list:
wood
fun
camping
We match phrases 1 and 4, so the score is 2. But if the user adds "cabin" to their list, they will match 3 more phrases and get a score of 5. "fire" would add 2 to the score.
With the trivially short list, there isn't any complicated problem, as you can just iterate through the options in almost no time. But as the list grows to the hundreds of thousands, it starts taking hundreds of milliseconds. It feels like there should be a way to build an index to make the process faster, but I can't think of what the index's structure would be.
Anyone who took the time to read all this, thank you! Hopefully someone knows what I'm talking about.
You need to map words to number of occurrences. If you use a hash table you can do it very quickly (O(N) - with N being the number of words in the phrases) - loop over all phrases, break them into words, if the word is already in the map increment its count, if not - add it to the map with count 1.
To compute the score of the input, just loop over the input words and accumulate the number of occurrences. O(M) - this time M being the number of input words.
I doubt you can get better complexity (you need to scan the phrases at least once), and with a proper implementation of a map (available in almost all modern languages) - it will be fast as well.
Suffix tree.
They're rather fiddly and complicated things, but basically we store a node for each character (26 * 2), then we store the suffixes for each character, so entries for th and an and so on, but presumably not for qj or other combinations which won't occur. Then you get the suffixes for those, (so the, thr, and and so on, but plenty of combinations of three letters disallowed).
It allows for very fast searching, which doesn't have to be exact. If we want to match a*d we simply follow all the suffixes of a, then only d suffixes, then we insist on nul.
Given a dictionary as a hashtable. Find the minimum # of
deletions needed for a given word in order to make it match any word in the
dictionary.
Is there some clever trick to solve this problem in less than exponential complexity (trying all possible combinations)?
For starters, suppose that you have a single word w in the the hash table and that your word is x. You can delete letters from x to form w if and only if w is a subsequence of x, and in that case the number of letters you need to delete from x to form w is given by |x - w|. So certainly one option would be to just iterate over the hash table and, for each word, to see if x is a subsequence of that word, taking the best match you find across the table.
To analyze the runtime of this operation, let's suppose that there are n total words in your hash table and that their total length is L. Then the runtime of this operation is O(L), since you'll process each character across all the words at most once. The complexity of your initial approach is O(|x| · 2|x|) because there are 2|x| possible words you can make by deleting letters from x and you'll spend O(|x|) time processing each one. Depending on the size of your dictionary and the size of your word, one algorithm might be better than the other, but we can say that the runtime is O(min{L, |x|·2|x|) if you take the better of the two approaches.
You can build a trie and then see where your given word fits into it. The difference in the depth of your word and the closest existing parent is the number of deletions required.
I am thinking about writing a program to collect for me the most common phrases in a large volume of the text. Had the problem been reduced to just finding words than that would be as simple as storing each new word in a hashmap and then increasing the count on each occurrence. But with phrases, storing each permutation of a sentence as a key seems infeasible.
Basically the problem is narrowed down to figuring out how to extract every possible phrase from a large enough text. Counting the phrases and then sorting by the number of occurrences becomes trivial.
I assume that you are searching for common patterns of consecutive words appearing in the same order (e.g. "top of the world" would not be counted as the same phrase as "top of a world" or "the world of top").
If so then I would recommend the following linear-time approach:
Split your text into words and remove things you don't consider significant (i.e. remove capitalisation, punctuation, word breaks, etc.)
Convert your text into an array of integers (one integer per unique word) (e.g. every instance of "cat" becomes 1, every "dog" becomes 2) This can be done in linear time by using a hash-based dictionary to store the conversions from words to numbers. If the word is not in the dictionary then assign a new id.
Construct a suffix-array for the array of integers (this is a sorted list of all the suffixes of your array and can be constructed by linear time - e.g. using the algorithm and C code here)
Construct the longest common prefix array for your suffix array. (This can also be done in linear-time, for example using this C code) This LCP array gives the number of common words at the start of each suffix between consecutive pairs in the suffix array.
You are now in a position to collect your common phrases.
It is not quite clear how you wish to determine the end of a phrase. One possibility is to simply collect all sequences of 4 words that repeat.
This can be done in linear time by working through your suffix array looking at places where the longest common prefix array is >= 4. Each run of indices x in the range [start+1...start+len] where the LCP[x] >= 4 (for all except the last value of x) corresponds to a phrase that is repeated len times. The phrase itself is given by the first 4 words of, for example, suffix start+1.
Note that this approach will potentially spot phrases that cross sentence ends. You may prefer to convert some punctuation such as full stops into unique integers to prevent this.
I need to match a series of user inputed words against a large dictionary of words (to ensure the entered value exists).
So if the user entered:
"orange" it should match an entry "orange' in the dictionary.
Now the catch is that the user can also enter a wildcard or series of wildcard characters like say
"or__ge" which would also match "orange"
The key requirements are:
* this should be as fast as possible.
* use the smallest amount of memory to achieve it.
If the size of the word list was small I could use a string containing all the words and use regular expressions.
however given that the word list could contain potentially hundreds of thousands of enteries I'm assuming this wouldn't work.
So is some sort of 'tree' be the way to go for this...?
Any thoughts or suggestions on this would be totally appreciated!
Thanks in advance,
Matt
Put your word list in a DAWG (directed acyclic word graph) as described in Appel and Jacobsen's paper on the World's Fastest Scrabble Program (free copy at Columbia). For your search you will traverse this graph maintaining a set of pointers: on a letter, you make a deterministic transition to children with that letter; on a wildcard, you add all children to the set.
The efficiency will be roughly the same as Thompson's NFA interpretation for grep (they are the same algorithm). The DAWG structure is extremely space-efficient—far more so than just storing the words themselves. And it is easy to implement.
Worst-case cost will be the size of the alphabet (26?) raised to the power of the number of wildcards. But unless your query begins with N wildcards, a simple left-to-right search will work well in practice. I'd suggest forbidding a query to begin with too many wildcards, or else create multiple dawgs, e.g., dawg for mirror image, dawg for rotated left three characters, and so on.
Matching an arbitrary sequence of wildcards, e.g., ______ is always going to be expensive because there are combinatorially many solutions. The dawg will enumerate all solutions very quickly.
I would first test the regex solution and see whether it is fast enough - you might be surprised! :-)
However if that wasn't good enough I would probably use a prefix tree for this.
The basic structure is a tree where:
The nodes at the top level are all the possible first letters (i.e. probably 26 nodes from a-z assuming you are using a full dictionary...).
The next level down contains all the possible second letters for each given first letter
And so on until you reach an "end of word" marker for each word
Testing whether a given string with wildcards is contained in your dictionary is then just a simple recursive algorithm where you either have a direct match for each character position, or in the case of the wildcard you check each of the possible branches.
In the worst case (all wildcards but only one word with the right number of letters right at the end of the dictionary), you would traverse the entire tree but this is still only O(n) in the size of the dictionary so no worse than a full regex scan. In most cases it would take very few operations to either find a match or confirm that no such match exists since large branches of the search tree are "pruned" with each successive letter.
No matter which algorithm you choose, you have a tradeoff between speed and memory consumption.
If you can afford ~ O(N*L) memory (where N is the size of your dictionary and L is the average length of a word), you can try this very fast algorithm. For simplicity, will assume latin alphabet with 26 letters and MAX_LEN as the max length of word.
Create a 2D array of sets of integers, set<int> table[26][MAX_LEN].
For each word in you dictionary, add the word index to the sets in the positions corresponding to each of the letters of the word. For example, if "orange" is the 12345-th word in the dictionary, you add 12345 to the sets corresponding to [o][0], [r][1], [a][2], [n][3], [g][4], [e][5].
Then, to retrieve words corresponding to "or..ge", you find the intersection of the sets at [o][0], [r][1], [g][4], [e][5].
You can try a string-matrix:
0,1: A
1,5: APPLE
2,5: AXELS
3,5: EAGLE
4,5: HELLO
5,5: WORLD
6,6: ORANGE
7,8: LONGWORD
8,13:SUPERLONGWORD
Let's call this a ragged index-matrix, to spare some memory. Order it on length, and then on alphabetical order. To address a character I use the notation x,y:z: x is the index, y is the length of the entry, z is the position. The length of your string is f and g is the number of entries in the dictionary.
Create list m, which contains potential match indexes x.
Iterate on z from 0 to f.
Is it a wildcard and not the latest character of the search string?
Continue loop (all match).
Is m empty?
Search through all x from 0 to g for y that matches length. !!A!!
Does the z character matches with search string at that z? Save x in m.
Is m empty? Break loop (no match).
Is m not empty?
Search through all elements of m. !!B!!
Does not match with search? Remove from m.
Is m empty? Break loop (no match).
A wildcard will always pass the "Match with search string?". And m is equally ordered as the matrix.
!!A!!: Binary search on length of the search string. O(log n)
!!B!!: Binary search on alphabetical ordering. O(log n)
The reason for using a string-matrix is that you already store the length of each string (because it makes it search faster), but it also gives you the length of each entry (assuming other constant fields), such that you can easily find the next entry in the matrix, for fast iterating. Ordering the matrix isn't a problem: since this has only be done once the dictionary updates, and not during search-time.
If you are allowed to ignore case, which I assume, then make all the words in your dictionary and all the search terms the same case before anything else. Upper or lower case makes no difference. If you have some words that are case sensitive and others that are not, break the words into two groups and search each separately.
You are only matching words, so you can break the dictionary into an array of strings. Since you are only doing an exact match against a known length, break the word array into a separate array for each word length. So byLength[3] is the array off all words with length 3. Each word array should be sorted.
Now you have an array of words and a word with potential wild cards to find. Depending on wether and where the wildcards are, there are a few approaches.
If the search term has no wild cards, then do a binary search in your sorted array. You could do a hash at this point, which would be faster but not much. If the vast majority of your search terms have no wildcards, then consider a hash table or an associative array keyed by hash.
If the search term has wildcards after some literal characters, then do a binary search in the sorted array to find an upper and lower bound, then do a linear search in that bound. If the wildcards are all trailing then finding a non empty range is sufficient.
If the search term starts with wild cards, then the sorted array is no help and you would need to do a linear search unless you keep a copy of the array sorted by backwards strings. If you make such an array, then choose it any time there are more trailing than leading literals. If you do not allow leading wildcards then there is no need.
If the search term both starts and ends with wildcards, then you are stuck with a linear search within the words with equal length.
So an array of arrays of strings. Each array of strings is sorted, and contains strings of equal length. Optionally duplicate the whole structure with the sorting based on backwards strings for the case of leading wildcards.
The overall space is one or two pointers per word, plus the words. You should be able to store all the words in a single buffer if your language permits. Of course, if your language does not permit, grep is probably faster anyway. For a million words, that is 4-16MB for the arrays and similar for the actual words.
For a search term with no wildcards, performance would be very good. With wildcards, there will occasionally be linear searches across large groups of words. With the breakdown by length and a single leading character, you should never need to search more than a few percent of the total dictionary even in the worst case. Comparing only whole words of known length will always be faster than generic string matching.
Try to build a Generalized Suffix Tree if the dictionary will be matched by sequence of queries. There is linear time algorithm that can be used to build such tree (Ukkonen Suffix Tree Construction).
You can easily match (it's O(k), where k is the size of the query) each query by traversing from the root node, and use the wildcard character to match any character like typical pattern finding in suffix tree.
I have a lot of compound strings that are a combination of two or three English words.
e.g. "Spicejet" is a combination of the words "spice" and "jet"
I need to separate these individual English words from such compound strings. My dictionary is going to consist of around 100000 words.
What would be the most efficient by which I can separate individual English words from such compound strings.
I'm not sure how much time or frequency you have to do this (is it a one-time operation? daily? weekly?) but you're obviously going to want a quick, weighted dictionary lookup.
You'll also want to have a conflict resolution mechanism, perhaps a side-queue to manually resolve conflicts on tuples that have multiple possible meanings.
I would look into Tries. Using one you can efficiently find (and weight) your prefixes, which are precisely what you will be looking for.
You'll have to build the Tries yourself from a good dictionary source, and weight the nodes on full words to provide yourself a good quality mechanism for reference.
Just brainstorming here, but if you know your dataset consists primarily of duplets or triplets, you could probably get away with multiple Trie lookups, for example looking up 'Spic' and then 'ejet' and then finding that both results have a low score, abandon into 'Spice' and 'Jet', where both Tries would yield a good combined result between the two.
Also I would consider utilizing frequency analysis on the most common prefixes up to an arbitrary or dynamic limit, e.g. filtering 'the' or 'un' or 'in' and weighting those accordingly.
Sounds like a fun problem, good luck!
If the aim is to find the "the largest possible break up for the input" as you replied, then the algorithm could be fairly straightforward if you use some graph theory. You take the compound word and make a graph with a vertex before and after every letter. You'll have a vertex for each index in the string and one past the end. Next you find all legal words in your dictionary that are substrings of the compound word. Then, for each legal substring, add an edge with weight 1 to the graph connecting the vertex before the first letter in the substring with the vertex after the last letter in the substring. Finally, use a shortest path algorithm to find the path with fewest edges between the first and the last vertex.
The pseudo code is something like this:
parseWords(compoundWord)
# Make the graph
graph = makeGraph()
N = compoundWord.length
for index = 0 to N
graph.addVertex(i)
# Add the edges for each word
for index = 0 to N - 1
for length = 1 to min(N - index, MAX_WORD_LENGTH)
potentialWord = compoundWord.substr(index, length)
if dictionary.isElement(potentialWord)
graph.addEdge(index, index + length, 1)
# Now find a list of edges which define the shortest path
edges = graph.shortestPath(0, N)
# Change these edges back into words.
result = makeList()
for e in edges
result.add(compoundWord.substr(e.start, e.stop - e.start + 1))
return result
I, obviously, haven't tested this pseudo-code, and there may be some off-by-one indexing errors, and there isn't any bug-checking, but the basic idea is there. I did something similar to this in school and it worked pretty well. The edge creation loops are O(M * N), where N is the length of the compound word, and M is the maximum word length in your dictionary or N (whichever is smaller). The shortest path algorithm's runtime will depend on which algorithm you pick. Dijkstra's comes most readily to mind. I think its runtime is O(N^2 * log(N)), since the max edges possible is N^2.
You can use any shortest path algorithm. There are several shortest path algorithms which have their various strengths and weaknesses, but I'm guessing that for your case the difference will not be too significant. If, instead of trying to find the fewest possible words to break up the compound, you wanted to find the most possible, then you give the edges negative weights and try to find the shortest path with an algorithm that allows negative weights.
And how will you decide how to divide things? Look around the web and you'll find examples of URLs that turned out to have other meanings.
Assuming you didn't have the capitals to go on, what would you do with these (Ones that come to mind at present, I know there are more.):
PenIsland
KidsExchange
TherapistFinder
The last one is particularly problematic because the troublesome part is two words run together but is not a compound word, the meaning completely changes when you break it.
So, given a word, is it a compound word, composed of two other English words? You could have some sort of lookup table for all such compound words, but if you just examine the candidates and try to match against English words, you will get false positives.
Edit: looks as if I am going to have to go to provide some examples. Words I was thinking of include:
accustomednesses != accustomed + nesses
adulthoods != adult + hoods
agreeabilities != agree + abilities
willingest != will + ingest
windlasses != wind + lasses
withstanding != with + standing
yourselves != yours + elves
zoomorphic != zoom + orphic
ambassadorships != ambassador + ships
allotropes != allot + ropes
Here is some python code to try out to make the point. Get yourself a dictionary on disk and have a go:
from __future__ import with_statement
def opendict(dictionary=r"g:\words\words(3).txt"):
with open(dictionary, "r") as f:
return set(line.strip() for line in f)
if __name__ == '__main__':
s = opendict()
for word in sorted(s):
if len(word) >= 10:
for i in range(4, len(word)-4):
left, right = word[:i], word[i:]
if (left in s) and (right in s):
if right not in ('nesses', ):
print word, left, right
It sounds to me like you want to store you dictionary in a Trie or a DAWG data structure.
A Trie already stores words as compound words. So "spicejet" would be stored as "spicejet" where the * denotes the end of a word. All you'd have to do is look up the compound word in the dictionary and keep track of how many end-of-word terminators you hit. From there you would then have to try each substring (in this example, we don't yet know if "jet" is a word, so we'd have to look that up).
It occurs to me that there are a relatively small number of substrings (minimum length 2) from any reasonable compound word. For example for "spicejet" I get:
'sp', 'pi', 'ic', 'ce', 'ej', 'je', 'et',
'spi', 'pic', 'ice', 'cej', 'eje', 'jet',
'spic', 'pice', 'icej', 'ceje', 'ejet',
'spice', 'picej', 'iceje', 'cejet',
'spicej', 'piceje', 'icejet',
'spiceje' 'picejet'
... 26 substrings.
So, find a function to generate all those (slide across your string using strides of 2, 3, 4 ... (len(yourstring) - 1) and then simply check each of those in a set or hash table.
A similar question was asked recently: Word-separating algorithm. If you wanted to limit the number of splits, you would keep track of the number of splits in each of the tuples (so instead of a pair, a triple).
Word existence could be done with a trie, or more simply with a set (i.e. a hash table). Given a suitable function, you could do:
# python-ish pseudocode
def splitword(word):
# word is a character array indexed from 0..n-1
for i from 1 to n-1:
head = word[:i] # first i characters
tail = word[i:] # everything else
if is_word(head):
if i == n-1:
return [head] # this was the only valid word; return it as a 1-element list
else:
rest = splitword(tail)
if rest != []: # check whether we successfully split the tail into words
return [head] + rest
return [] # No successful split found, and 'word' is not a word.
Basically, just try the different break points to see if we can make words. The recursion means it will backtrack until a successful split is found.
Of course, this may not find the splits you want. You could modify this to return all possible splits (instead of merely the first found), then do some kind of weighted sum, perhaps, to prefer common words over uncommon words.
This can be a very difficult problem and there is no simple general solution (there may be heuristics that work for small subsets).
We face exactly this problem in chemistry where names are composed by concatenation of morphemes. An example is:
ethylmethylketone
where the morphemes are:
ethyl methyl and ketone
We tackle this through automata and maximum entropy and the code is available on Sourceforge
http://www.sf.net/projects/oscar3-chem
but be warned that it will take some work.
We sometimes encounter ambiguity and are still finding a good way of reporting it.
To distinguish between penIsland and penisLand would require domain-specific heuristics. The likely interpretation will depend on the corpus being used - no linguistic problem is independent from the domain or domains being analysed.
As another example the string
weeknight
can be parsed as
wee knight
or
week night
Both are "right" in that they obey the form "adj-noun" or "noun-noun". Both make "sense" and which is chosen will depend on the domain of usage. In a fantasy game the first is more probable and in commerce the latter. If you have problems of this sort then it will be useful to have a corpus of agreed usage which has been annotated by experts (technically a "Gold Standard" in Natural Language Processing).
I would use the following algorithm.
Start with the sorted list of words
to split, and a sorted list of
declined words (dictionary).
Create a result list of objects
which should store: remaining word
and list of matched words.
Fill the result list with the words
to split as remaining words.
Walk through the result array and
the dictionary concurrently --
always increasing the least of the
two, in a manner similar to the
merge algorithm. In this way you can
compare all the possible matching
pairs in one pass.
Any time you find a match, i.e. a
split words word that starts with a
dictionary word, replace the
matching dictionary word and the
remaining part in the result list.
You have to take into account
possible multiples.
Any time the remaining part is empty,
you found a final result.
Any time you don't find a match on
the "left side", in other words,
every time you increment the result
pointer because of no match, delete
the corresponding result item. This
word has no matches and can't be
split.
Once you get to the bottom of the
lists, you will have a list of
partial results. Repeat the loop
until this is empty -- go to point 4.