I am looking for a algorithm for string processing, I have searched for it but couldn't find a algorithm that meets my requirements. I will explain what the algorithm should do with an example.
There are two sets of word sets defined as shown below:
**Main_Words**: swimming, driving, playing
**Words_in_front**: I am, I enjoy, I love, I am going to go
The program will search through a huge set of words as soon it finds a word that is defined in Main_Words it will check the words in front of that Word to see if it has any matching words defined in Words_in_front.
i.e If the program encounters the word "Swimming" it has to check if the words in front of the word "Swimming" are one of these: I am, I enjoy, I love, I am going to go.
Are there any algorithms that can do this?
A straightforward way to do this would be to just do a linear scan through the text, always keeping track of the last N+1 words (or characters) you see, where N is the number of words (or characters) in the longest phrase contained in your words_in_front collection. When you have a "main word", you can just check whether the sequence of N words/characters before it ends with any of the prefixes you have.
This would be a bit faster if you transformed your words_in_front set into a nicer data structure, such as a hashmap (perhaps keyed by last letter in the phrase..) or a prefix/suffix tree of some sort, so you wouldn't have to do an .endsWith over every single member of the set of prefixes each time you have a matching "main word." As was stated in another answer, there is much room for optimization and a few other possible implementations, but there's a start.
Create a map/dictionary/hash/associative array (whatever is defined in your language) with key in Main_Words and Words_in_front are the linked list attached to the entry pointed by the key. Whenever you encounter a word matching a key, go to the table and see if in the attached list there are words that match what you have in front.
That's the basic idea, it can be optimized for both speed and space.
You should be able to build a regular expression along these lines:
I (am|enjoy|love|am going to go) (swimming|driving|playing)
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've been given a problem in my data structures class to find the solution to this problem. It's similar to an interview question. If someone could explain the thinking process or solution to the problem. Pseudocode can be used. So far i've been thinking to use tries to hold the dictionary and look up words that way for efficiency.
This is the problem:
Oh, no! You have just completed a lengthy document when you have an unfortunate Find/Replace mishap. You have accidentally removed all spaces, punctuation, and capitalization in the document. A sentence like "I reset the computer. It still didn't boot!" would become "iresetthecomputeritstilldidntboot". You figure that you can add back in the punctation and capitalization later, once you get the individual words properly separated. Most of the words will be in a dictionary, but some strings, like proper names, will not.
Given a dictionary (a list of words), design an algorithm to find the optimal way of "unconcatenating" a sequence of words. In this case, "optimal" is defined to be the parsing which minimizes the number of unrecognized sequences of characters.
For example, the string "jesslookedjustliketimherbrother" would be optimally parsed as "JESS looked just like TIM her brother". This parsing has seven unrecognized characters, which we have capitalized for clarity.
For each index, n, into the string, compute the cost C(n) of the optimal solution (ie: the number of unrecognised characters in the optimal parsing) starting at that index.
Then, the solution to your problem is C(0).
There's a recurrence relation for C. At each n, either you match a word of i characters, or you skip over character n, incurring a cost of 1, and then parse the rest optimally. You just need to find which of those choices incurs the lowest cost.
Let N be the length of the string, and let W(n) be a set containing the lengths of all words starting at index n in your string. Then:
C(N) = 0
C(n) = min({C(n+1) + 1} union {C(n+i) for i in W(n)})
This can be implemented using dynamic programming by constructing a table of C(n) starting from the end backwards.
If the length of the longest word in your dictionary is L, then the algorithm runs in O(NL) time in the worst case and can be implemented to use O(L) memory if you're careful.
You could use rolling hashes of different lengths to speed up the search.
You can try a partial pattern matcher for example aho-corasick algorithm. Basically it's a special space optimized version of a suffix tree.
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.
I have a dictionary which contains a big number of strings. Each string could have a range of 1 to 4 tokens (words). Example :
Dictionary :
The Shawshank Redemption
The Godfather \
Pulp Fiction
The Dark Knight
Fight Club
Now I have a paragraph and I need to figure out how many strings in the para are part of the dictionary.
Example, when the para below :
The Shawshank Redemption considered the greatest movie ever made according to the IMDB Top 250.For at least the year or two that I have occasionally been checking in on the IMDB Top 250 The Shawshank Redemption has been
battling The Godfather for the top spot.
is run against the dictionary, I should be getting the ones in bold as the ones that are part of the dictionary.
How can I do this with the least dictionary calls.
Thanks
You might be better off using a Trie. A Trie is better suited to finding partial matches (i.e. as you search through the text of a paragraph) that are potentially what you're looking for, as opposed to making a bunch of calls to a dictionary that will mostly fail.
The reason why I think a Trie (or some variation) is appropriate is because it's built to do exactly what you're trying to do:
If you use this (or some modification that has the tokenized words at each node instead of a letter), this would be the most efficient (at least that I know of) in terms of storage and retrieval; Storage because instead of storing the word "The" a couple thousand times in each Dict entry that has that word in the title (as is the case with movie titles), it would be stored once in one of the nodes right under the root. The next word, "Shawshank" would be in a child node, and then "redemption" would be in the next, with a total of 3 lookups; then you would move to the next phrase. If it fails, i.e. the phrase is only "The Shawshank Looper", then you fail after the same 3 lookups, and you move to the failed word, Looper (which as it happens, would also be a child node under the root, and you get a hit. This solution works assuming you're reading a paragraph without mashup movie names).
Using a hash table, you're going to have to split all the words, check the first word, and then while there's no match, keep appending words and checking if THAT phrase is in the dictionary, until you get a hit, or you reach the end of the paragraph. So if you hit a paragraph with no movie titles, you would have as many lookups as there are words in the paragraph.
This is not a complete answer, more like an extended-comment.
In literature it's called "multi-pattern matching problem". Since you mentioned that the set of patterns has millions of elements, Trie based solutions will most probably perform poorly.
As far as I know, in practice traditional string search is used with a lot of heuristics. DNA search, antivirus detection, etc. all of these fields need fast and reliable pattern matching, so there should be decent amount of research done.
I can imagine how Rabin-Karp with rolling-hash functions and some filters (Bloom filter) can be used in order to speed up the process. For example, instead of actually matching the substrings, you could first filter (e.g. with weak-hashes) and then actually verify, thus reducing number of verifications needed. Plus this should reduce the work done with the original dictionary itself, as you would store it's hashes, or other filters.
In Python:
import re
movies={1:'The Shawshank Redemption', 2:'The Godfather', 3:'Pretty Woman', 4:'Pulp Fiction'}
text = 'The Shawshank Redemption considered the greatest movie ever made according to the IMDB Top 250.For at least the year or two that I have occasionally been checking in on the IMDB Top 250 The Shawshank Redemption has been battling The Godfather for the top spot.'
repl_str ='(?P<title>' + '|'.join(['(?:%s)' %movie for movie in movies.values()]) + ')'
result = re.sub(repl_str, '<b>\g<title></b>',text)
Basically it consists of forming up a big substitution instruction string out of your dict values.
I don't know whether regex and sub have a limitation in the size of the substitution instructions you give them though. You might want to check.
lai
I've list of words. Number of words is around 1 million.
I've strings coming at runtime, I've to check which word from the list is present in string and return that word (need not to return all words occurring in sentence, returning first one also suffice the requirement).
One solution is checking all words one by one in string but it's inefficient.
Can someone please point out any efficient method of doing it?
Use the Knuth-Morris-Pratt algorithm. Although a million words is not all that much. You can also convert your text body into a Trie structure and then use that to check your search list against. There is a special kind of Trie called a Suffix Tree used especially for full text searching.
Put your word list in a tree or hash table.
Unless your word's list is ordered (or inserted in a efficient data structure like an ordered binary tree) to perform a binary search, the solution you are proposing is the most efficient one.