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I'm learning Haskell by writing an OSC musical sequencer to use it with SuperCollider. But because I'd like to make fairly complex stuff with it, it will work like a programming language where you can declare variables and define functions so you can write music in an algorithmic way. The grammar is unusual in that we're coding sequences and sometimes a bar will reference the last bar (something like "play that last chord again but a fifth above").
I don't feel satisfied with my own explanation, but that's the best I can without getting too technical.
Anyway, what I'm coding now is the parser for that language, stateless so far, but now I need some way to implement a growing list of the declared variables and alikes using a dictionary in the [("key","value")] fashion, so I can add new values as I go parsing bar by bar.
I know this involves monads, which I don't really understand yet, but I need something meaningful enough to start toying with them or else I find the raw theory a bit too raw.
So what would be a clean and simple way to start?
Thanks and sorry if the question was too long.
Edit on how the thing works:
we input a string to the main parsing function, say
"afunction(3) ; anotherone(1) + [3,2,1]"
we identify closures first, then kinds of chars (letters, nums, etc) and group them together, so we get a list like:
[("word","afunction"),("parenth","(3)"),("space"," "),("semicolon",";"),("space"," "),("word","anotherone"),("parenth","(1)"),("space"," "),("opadd","+"),("space"," "),("bracket","[3,2,1]")]
then we use a function that tags all those tuples with the indices of the original string they occupy, like:
[("word","afunction",(0,8)),("parenth","(3)",(9,11)),("space"," ",(12,13)) ...]
then cut it in a list of bars, which in my language are separated using a semicolon, and then in notes, using commas.
And now I'm at the stage where those functions should be executed sequentially, but because some of them are reading or modifying previously declared values, I need to keep track of that change. For example, let's say the function f(x) moves the pitch of the last note by x semitones, so
f(9), -- from an original base value of 0 (say that's an A440) we go to 9
f(-2), -- 9-2 = 7, so a fifth from A
f(-3); -- 9-2-3, a minor third down from the last value.
etc
But sometimes it can get a bit more complicated than that, don't make me explain how cause I could bore you to death.
Adding an item to a list
You can make a new list that contains one more item than an existing list with the : constructor.
("key", "value") : existing
Where existing is a list you've already made
Keeping track of changing state
You can keep track of changing state between functions by passing the state from each function to the next. This is all the State monad is doing. State s a is a value of type a that depends on (and changes) a state s.
{- ┌---- type of the state
v v-- type of the value -}
data State s a = State { runState :: s -> (a, s) }
{- ^ ^ ^ ^
a function ---|--┘ | |
that takes a state ---┘ | |
and returns | |
a value that depends on the state ---┘ |
and a new state ------┘ -}
The bind operation >>= for State takes a value that depends on (and changes) the state and a function to compute another value that depends on (and changes) the state and combines them to make a new value that depends on (and changes) the state.
m >>= k = State $ \s ->
let ~(a, s') = runState m s
in runState (k a) s'
I'm writing a Haskell program that reads a wordlist of the English language and a rectangular grid of letters such as:
I T O L
I H W S
N H I S
K T S I
and then finds a Hamiltonian path through the grid from the top-left corner that spells out a sequence of English words, such as:
--> $ runghc unpacking.hs < 4x4grid.txt
I THINK THIS IS SLOW
(If there are multiple solutions, it can just print any one it finds and stop looking.)
The naïve, strict approach is to generate a full path and then try to split it up into words. However, assuming that I'm doing this (and currently I am forcing myself to -- see below) I'm spending a lot of time finding paths like:
IINHHTOL...
IINHHTOW...
IINHHWOL...
These are obviously never going to turn out to be words, looking at the first few letters ("IINH" can't be split into words, and no English word contains "NHH".) So, say, in the above grid, I don't want to look at the many[1] paths that begin with IINHH.
Now, my functions look like this:
paths :: Coord -> Coord -> [[Coord]]
paths (w, h) (1, 1) = [[(1, 1), (1, 2), ... (x, y)], ...]
lexes :: Set String -> String -> [[String]]
lexes englishWordset "ITHINKTHISWILLWORK" = [["I", "THINK", "THIS", ...], ...]
paths just finds all the paths worth considering on a (w, h) grid. lexes finds all the ways to chop a phrase up, and is defined as:
lexes language [] = [[]]
lexes language phrase = let
splits = tail $ zip (inits phrase) (tails phrase)
in concat [map (w:) (lexes language p') | (w, p') <- splits,
w `S.member` language]
Given "SAMPLESTRING", it looks at "S", then "SA", then "SAM"... as soon as it finds a valid word, it recurses and tries to "lex" the rest of the string. (First it will recurse on "PLESTRING" and try to make phrases with "SAM", but find no way to chop "plestring" up into words, and fail; then it will find ["SAMPLE", "STRING"].)
Of course, for an invalid string above, any hope of being "lazy" is lost by following this approach: in the example from earlier we need to still search beyond a ridiculous phrase like "ITOLSHINHISIST", because maybe "ITOLSHINHISISTK" (one letter longer) might form a valid single word.
I feel like somehow I could use laziness here to improve performance throughout the entire program: if the first few characters of phrase aren't a prefix of any word, we can bail out entirely, stop evaluating the rest of phrase, and thus the rest of the path.[2] Does this make sense at all? Is there some tree-like data structure that will help me check not for set membership, but set "prefix-ness", thereby making checking validity lazier?
[1] Obviously, for a 4x4 grid there are very few of these, but this argument is about the general case: for bigger grids I could skip hundreds of thousands of paths the moment I see they start with "JX".
[2] phrase is just map (grid M.!) path for some Map Coord Char grid read from the input file.
In a program I need to efficiently answer queries of the following form:
Given a set of strings A and a query string q return all s ∈ A such that q is a subsequence of s
For example, given A = {"abcdef", "aaaaaa", "ddca"} and q = "acd" exactly "abcdef" should be returned.
The following is what I have considered considered so far:
For each possible character, make a sorted list of all string/locations where it appears. For querying interleave the lists of the involved characters, and scan through it looking for matches within string boundaries.
This would probably be more efficient for words instead of characters, since the limited number of different characters will make the return lists very dense.
For each n-prefix q might have, store the list of all matching strings. n might realistically be close to 3. For query strings longer than that we brute force the initial list.
This might speed things up a bit, but one could easily imagine some n-subsequences being present close to all strings in A, which means worst case is the same as just brute forcing the entire set.
Do you know of any data structures, algorithms or preprocessing tricks which might be helpful for performing the above task efficiently for large As? (My ss will be around 100 characters)
Update: Some people have suggested using LCS to check if q is a subsequence of s. I just want to remind that this can be done using a simple function such as:
def isSub(q,s):
i, j = 0, 0
while i != len(q) and j != len(s):
if q[i] == s[j]:
i += 1
j += 1
else:
j += 1
return i == len(q)
Update 2: I've been asked to give more details on the nature of q, A and its elements. While I'd prefer something that works as generally as possible, I assume A will have length around 10^6 and will need to support insertion. The elements s will be shorter with an average length of 64. The queries q will only be 1 to 20 characters and be used for a live search, so the query "ab" will be sent just before the query "abc". Again, I'd much prefer the solution to use the above as little as possible.
Update 3: It has occurred to me, that a data-structure with O(n^{1-epsilon}) lookups, would allow you to solve OVP / disprove the SETH conjecture. That is probably the reason for our suffering. The only options are then to disprove the conjecture, use approximation, or take advantage of the dataset. I imagine quadlets and tries would do the last in different settings.
It could done by building an automaton. You can start with NFA (nondeterministic finite automaton which is like an indeterministic directed graph) which allows edges labeled with an epsilon character, which means that during processing you can jump from one node to another without consuming any character. I'll try to reduce your A. Let's say you A is:
A = {'ab, 'bc'}
If you build NFA for ab string you should get something like this:
+--(1)--+
e | a| |e
(S)--+--(2)--+--(F)
| b| |
+--(3)--+
Above drawing is not the best looking automaton. But there are a few points to consider:
S state is the starting state and F is the ending state.
If you are at F state it means your string qualifies as a subsequence.
The rule of propagating within an autmaton is that you can consume e (epsilon) to jump forward, therefore you can be at more then one state at each point in time. This is called e closure.
Now if given b, starting at state S I can jump one epsilon, reach 2, and consume b and reach 3. Now given end string I consume epsilon and reach F, thus b qualifies as a sub-sequence of ab. So does a or ab you can try yourself using above automata.
The good thing about NFA is that they have one start state and one final state. Two NFA could be easily connected using epsilons. There are various algorithms that could help you to convert NFA to DFA. DFA is a directed graph which can follow precise path given a character -- in particular, it is always in exactly one state at any point in time. (For any NFA, there is a corresponding DFA whose states correspond to sets of states in the NFA.)
So, for A = {'ab, 'bc'}, we would need to build NFA for ab then NFA for bc then join the two NFAs and build the DFA of the entire big NFA.
EDIT
NFA of subsequence of abc would be a?b?c?, so you can build your NFA as:
Now, consider the input acd. To query if ab is subsequence of {'abc', 'acd'}, you can use this NFA: (a?b?c?)|(a?c?d). Once you have NFA you can convert it to DFA where each state will contain whether it is a subsequence of abc or acd or maybe both.
I used link below to make NFA graphic from regular expression:
http://hackingoff.com/images/re2nfa/2013-08-04_21-56-03_-0700-nfa.svg
EDIT 2
You're right! In case if you've 10,000 unique characters in the A. By unique I mean A is something like this: {'abc', 'def'} i.e. intersection of each element of A is empty set. Then your DFA would be worst case in terms of states i.e. 2^10000. But I'm not sure when would that be possible given that there can never be 10,000 unique characters. Even if you have 10,000 characters in A still there will be repetitions and that might reduce states alot since e-closure might eventually merge. I cannot really estimate how much it might reduce. But even having 10 million states, you will only consume less then 10 mb worth of space to construct a DFA. You can even use NFA and find e-closures at run-time but that would add to run-time complexity. You can search different papers on how large regex are converted to DFAs.
EDIT 3
For regex (a?b?c?)|(e?d?a?)|(a?b?m?)
If you convert above NFA to DFA you get:
It actually lot less states then NFA.
Reference:
http://hackingoff.com/compilers/regular-expression-to-nfa-dfa
EDIT 4
After fiddling with that website more. I found that worst case would be something like this A = {'aaaa', 'bbbbb', 'cccc' ....}. But even in this case states are lesser than NFA states.
Tests
There have been four main proposals in this thread:
Shivam Kalra suggested creating an automaton based on all the strings in A. This approach has been tried slightly in the literature, normally under the name "Directed Acyclic Subsequence Graph" (DASG).
J Random Hacker suggested extending my 'prefix list' idea to all 'n choose 3' triplets in the query string, and merging them all using a heap.
In the note "Efficient Subsequence Search in Databases" Rohit Jain, Mukesh K. Mohania and Sunil Prabhakar suggest using a Trie structure with some optimizations and recursively search the tree for the query. They also have a suggestion similar to the triplet idea.
Finally there is the 'naive' approach, which wanghq suggested optimizing by storing an index for each element of A.
To get a better idea of what's worth putting continued effort into, I have implemented the above four approaches in Python and benchmarked them on two sets of data. The implementations could all be made a couple of magnitudes faster with a well done implementation in C or Java; and I haven't included the optimizations suggested for the 'trie' and 'naive' versions.
Test 1
A consists of random paths from my filesystem. q are 100 random [a-z] strings of average length 7. As the alphabet is large (and Python is slow) I was only able to use duplets for method 3.
Construction times in seconds as a function of A size:
Query times in seconds as a function of A size:
Test 2
A consists of randomly sampled [a-b] strings of length 20. q are 100 random [a-b] strings of average length 7. As the alphabet is small we can use quadlets for method 3.
Construction times in seconds as a function of A size:
Query times in seconds as a function of A size:
Conclusions
The double logarithmic plot is a bit hard to read, but from the data we can draw the following conclusions:
Automatons are very fast at querying (constant time), however they are impossible to create and store for |A| >= 256. It might be possible that a closer analysis could yield a better time/memory balance, or some tricks applicable for the remaining methods.
The dup-/trip-/quadlet method is about twice as fast as my trie implementation and four times as fast as the 'naive' implementation. I used only a linear amount of lists for the merge, instead of n^3 as suggested by j_random_hacker. It might be possible to tune the method better, but in general it was disappointing.
My trie implementation consistently does better than the naive approach by around a factor of two. By incorporating more preprocessing (like "where are the next 'c's in this subtree") or perhaps merging it with the triplet method, this seems like todays winner.
If you can do with a magnitude less performance, the naive method does comparatively just fine for very little cost.
As you point out, it might be that all strings in A contain q as a subsequence, in which case you can't hope to do better than O(|A|). (That said, you might still be able to do better than the time taken to run LCS on (q, A[i]) for each string i in A, but I won't focus on that here.)
TTBOMK there are no magic, fast ways to answer this question (in the way that suffix trees are the magic, fast way to answer the corresponding question involving substrings instead of subsequences). Nevertheless if you expect the set of answers for most queries to be small on average then it's worth looking at ways to speed up these queries (the ones yielding small-size answers).
I suggest filtering based on a generalisation of your heuristic (2): if some database sequence A[i] contains q as a subsequence, then it must also contain every subsequence of q. (The reverse direction is not true unfortunately!) So for some small k, e.g. 3 as you suggest, you can preprocess by building an array of lists telling you, for every length-k string s, the list of database sequences containing s as a subsequence. I.e. c[s] will contain a list of the ID numbers of database sequences containing s as a subsequence. Keep each list in numeric order to enable fast intersections later.
Now the basic idea (which we'll improve in a moment) for each query q is: Find all k-sized subsequences of q, look up each in the array of lists c[], and intersect these lists to find the set of sequences in A that might possibly contain q as a subsequence. Then for each possible sequence A[i] in this (hopefully small) intersection, perform an O(n^2) LCS calculation with q to see whether it really does contain q.
A few observations:
The intersection of 2 sorted lists of size m and n can be found in O(m+n) time. To find the intersection of r lists, perform r-1 pairwise intersections in any order. Since taking intersections can only produce sets that are smaller or of the same size, time can be saved by intersecting the smallest pair of lists first, then the next smallest pair (this will necessarily include the result of the first operation), and so on. In particular: sort lists in increasing size order, then always intersect the next list with the "current" intersection.
It is actually faster to find the intersection a different way, by adding the first element (sequence number) of each of the r lists into a heap data structure, then repeatedly pulling out the minimum value and replenishing the heap with the next value from the list that the most recent minimum came from. This will produce a list of sequence numbers in nondecreasing order; any value that appears fewer than r times in a row can be discarded, since it cannot be a member of all r sets.
If a k-string s has only a few sequences in c[s], then it is in some sense discriminating. For most datasets, not all k-strings will be equally discriminating, and this can be used to our advantage. After preprocessing, consider throwing away all lists having more than some fixed number (or some fixed fraction of the total) of sequences, for 3 reasons:
They take a lot of space to store
They take a lot of time to intersect during query processing
Intersecting them will usually not shrink the overall intersection much
It is not necessary to consider every k-subsequence of q. Although this will produce the smallest intersection, it involves merging (|q| choose k) lists, and it might well be possible to produce an intersection that is nearly as small using just a fraction of these k-subsequences. E.g. you could limit yourself to trying all (or a few) k-substrings of q. As a further filter, consider just those k-subsequences whose sequence lists in c[s] are below some value. (Note: if your threshold is the same for every query, you might as well delete all such lists from the database instead, since this will have the same effect, and saves space.)
One thought;
if q tends to be short, maybe reducing A and q to a set will help?
So for the example, derive to { (a,b,c,d,e,f), (a), (a,c,d) }. Looking up possible candidates for any q should be faster than the original problem (that's a guess actually, not sure how exactly. maybe sort them and "group" similar ones in bloom filters?), then use bruteforce to weed out false positives.
If A strings are lengthy, you could make the characters unique based on their occurence, so that would be {(a1,b1,c1,d1,e1,f1),(a1,a2,a3,a4,a5,a6),(a1,c1,d1,d2)}. This is fine, because if you search for "ddca" you only want to match the second d to a second d. The size of your alphabet would go up (bad for bloom or bitmap style operations) and would be different ever time you get new A's, but the amount of false positives would go down.
First let me make sure my understanding/abstraction is correct. The following two requirements should be met:
if A is a subsequence of B, then all characters in A should appear in B.
for those characters in B, their positions should be in an ascending order.
Note that, a char in A might appear more than once in B.
To solve 1), a map/set can be used. The key is the character in string B, and the value doesn't matter.
To solve 2), we need to maintain the position of each characters. Since a character might appear more than once, the position should be a collection.
So the structure is like:
Map<Character, List<Integer>)
e.g.
abcdefab
a: [0, 6]
b: [1, 7]
c: [2]
d: [3]
e: [4]
f: [5]
Once we have the structure, how to know if the characters are in the right order as they are in string A? If B is acd, we should check the a at position 0 (but not 6), c at position 2 and d at position 3.
The strategy here is to choose the position that's after and close to the previous chosen position. TreeSet is a good candidate for this operation.
public E higher(E e)
Returns the least element in this set strictly greater than the given element, or null if there is no such element.
The runtime complexity is O(s * (n1 + n2)*log(m))).
s: number of strings in the set
n1: number of chars in string (B)
n2: number of chars in query string (A)
m: number of duplicates in string (B), e.g. there are 5 a.
Below is the implementation with some test data.
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.TreeSet;
public class SubsequenceStr {
public static void main(String[] args) {
String[] testSet = new String[] {
"abcdefgh", //right one
"adcefgh", //has all chars, but not the right order
"bcdefh", //missing one char
"", //empty
"acdh",//exact match
"acd",
"acdehacdeh"
};
List<String> subseqenceStrs = subsequenceStrs(testSet, "acdh");
for (String str : subseqenceStrs) {
System.out.println(str);
}
//duplicates in query
subseqenceStrs = subsequenceStrs(testSet, "aa");
for (String str : subseqenceStrs) {
System.out.println(str);
}
subseqenceStrs = subsequenceStrs(testSet, "aaa");
for (String str : subseqenceStrs) {
System.out.println(str);
}
}
public static List<String> subsequenceStrs(String[] strSet, String q) {
System.out.println("find strings whose subsequence string is " + q);
List<String> results = new ArrayList<String>();
for (String str : strSet) {
char[] chars = str.toCharArray();
Map<Character, TreeSet<Integer>> charPositions = new HashMap<Character, TreeSet<Integer>>();
for (int i = 0; i < chars.length; i++) {
TreeSet<Integer> positions = charPositions.get(chars[i]);
if (positions == null) {
positions = new TreeSet<Integer>();
charPositions.put(chars[i], positions);
}
positions.add(i);
}
char[] qChars = q.toCharArray();
int lowestPosition = -1;
boolean isSubsequence = false;
for (int i = 0; i < qChars.length; i++) {
TreeSet<Integer> positions = charPositions.get(qChars[i]);
if (positions == null || positions.size() == 0) {
break;
} else {
Integer position = positions.higher(lowestPosition);
if (position == null) {
break;
} else {
lowestPosition = position;
if (i == qChars.length - 1) {
isSubsequence = true;
}
}
}
}
if (isSubsequence) {
results.add(str);
}
}
return results;
}
}
Output:
find strings whose subsequence string is acdh
abcdefgh
acdh
acdehacdeh
find strings whose subsequence string is aa
acdehacdeh
find strings whose subsequence string is aaa
As always, I might be totally wrong :)
You might want to have a look into the Book Algorithms on Strings and Sequences by Dan Gusfield. As it turns out part of it is available on the internet. You might also want to read Gusfield's Introduction to Suffix Trees. As it turns out this book covers many approaches for you kind of question. It is considered one of the standard publications in this field.
Get a fast longest common subsequence algorithm implementation. Actually it suffices to determine the length of the LCS. Notice that Gusman's book has very good algorithms and also points to more sources for such algorithms.
Return all s ∈ A with length(LCS(s,q)) == length(q)
Xapian docs talk about a query constructor that takes a term position parameter, to be used in phrase searches:
Quote:
This constructor actually takes a couple of extra parameters, which
may be used to specify positional and frequency information for terms
in the query:
Xapian::Query(const string & tname_,
Xapian::termcount wqf_ = 1,
Xapian::termpos term_pos_ = 0)
The term_pos represents the position of the term in the query. Again,
this isn't useful for a single term query by itself, but is used for
phrase searching, passage retrieval, and other operations which
require knowledge of the order of terms in the query (such as
returning the set of matching terms in a given document in the same
order as they occur in the query). If such operations are not
required, the default value of 0 may be used.
And in the reference, we have:
Xapian::Query::Query ( const std::string & tname_,
Xapian::termcount wqf_ = 1,
Xapian::termpos pos_ = 0
)
A query consisting of a single term.
And:
typedef unsigned termpos
A term position within a document or query.
So, say I want to build a query for the phrase: "foo bar baz", how do I go about it?!
Does term_pos_ provide relative position values, ie define the order of terms within the document:
(I'm using here the python bindings API, as I'm more familiar with it)
q = xapian.Query(xapian.Query.OP_AND, [xapian.Query("foo", wqf, 1),xapian.Query("bar", wqf,2),xapian.Query("baz", wqf,3)] )
And just for the sake of testing, suppose we did:
q = xapian.Query(xapian.Query.OP_AND, [xapian.Query("foo", wqf, 3),xapian.Query("bar", wqf, 4),xapian.Query("baz", wqf, 5)] )
So this would give the same results as the previous example?!
And suppose we have:
q = xapian.Query(xapian.Query.OP_AND, [xapian.Query("foo", wqf, 2),xapian.Query("bar", wqf, 4),xapian.Query("baz", wqf, 5)] )
So now this would match where documents have "foo" "bar" separated with one term, followed by "baz" ??
Is it as such, or is it that this parameter is referring to absolute positions of the indexed terms?!
Edit:
And how is OP_PHRASE related to this? I find some online samples using OP_PHRASE as such:
q = xapian.Query(xapian.Query.OP_PHRASE, term_list)
This makes obvious sense, but then what is the role of the said term_pos_ constructor in phrase searches - is it a more surgical way of doing things!?
int pos = 1;
std::list<Xapian::Query> subs;
subs.push_back(Xapian::Query("foo", 1, pos++));
subs.push_back(Xapian::Query("bar", 1, pos++));
querylist.push_back(Xapian::Query(Xapian::Query::OP_PHRASE, subs.begin(), subs.end()));
The Problem: A large static list of strings is provided. A pattern string comprised of data and wildcard elements (* and ?). The idea is to return all the strings that match the pattern - simple enough.
Current Solution: I'm currently using a linear approach of scanning the large list and globbing each entry against the pattern.
My Question: Are there any suitable data structures that I can store the large list into such that the search's complexity is less than O(n)?
Perhaps something akin to a suffix-trie? I've also considered using bi- and tri-grams in a hashtable, but the logic required in evaluating a match based on a merge of the list of words returned and the pattern is a nightmare, furthermore I'm not convinced its the correct approach.
I agree that a suffix trie is a good idea to try, except that the sheer size of your dataset might make it's construction use up just as much time as its usage would save. Theyre best if youve got to query them multiple times to amortize the construction cost. Perhaps a few hundred queries.
Also note that this is a good excuse for parallelism. Cut the list in two and give it to two different processors and have your job done twice as fast.
you could build a regular trie and add wildcard edges. then your complexity would be O(n) where n is the length of the pattern. You would have to replace runs of ** with * in the pattern first (also an O(n) operation).
If the list of words were I am an ox then the trie would look a bit like this:
(I ($ [I])
a (m ($ [am])
n ($ [an])
? ($ [am an])
* ($ [am an]))
o (x ($ [ox])
? ($ [ox])
* ($ [ox]))
? ($ [I]
m ($ [am])
n ($ [an])
x ($ [ox])
? ($ [am an ox])
* ($ [I am an ox]
m ($ [am]) ...)
* ($ [I am an ox]
I ...
...
And here is a sample python program:
import sys
def addWord(root, word):
add(root, word, word, '')
def add(root, word, tail, prev):
if tail == '':
addLeaf(root, word)
else:
head = tail[0]
tail2 = tail[1:]
add(addEdge(root, head), word, tail2, head)
add(addEdge(root, '?'), word, tail2, head)
if prev != '*':
for l in range(len(tail)+1):
add(addEdge(root, '*'), word, tail[l:], '*')
def addEdge(root, char):
if not root.has_key(char):
root[char] = {}
return root[char]
def addLeaf(root, word):
if not root.has_key('$'):
root['$'] = []
leaf = root['$']
if word not in leaf:
leaf.append(word)
def findWord(root, pattern):
prev = ''
for p in pattern:
if p == '*' and prev == '*':
continue
prev = p
if not root.has_key(p):
return []
root = root[p]
if not root.has_key('$'):
return []
return root['$']
def run():
print("Enter words, one per line terminate with a . on a line")
root = {}
while 1:
line = sys.stdin.readline()[:-1]
if line == '.': break
addWord(root, line)
print(repr(root))
print("Now enter search patterns. Do not use multiple sequential '*'s")
while 1:
line = sys.stdin.readline()[:-1]
if line == '.': break
print(findWord(root, line))
run()
If you don't care about memory and you can afford to pre-process the list, create a sorted array of every suffix, pointing to the original word, e.g., for ['hello', 'world'], store this:
[('d' , 'world'),
('ello' , 'hello'),
('hello', 'hello'),
('ld' , 'world'),
('llo' , 'hello'),
('lo' , 'hello'),
('o' , 'hello'),
('orld' , 'world'),
('rld' , 'world'),
('world', 'world')]
Use this array to build sets of candidate matches using pieces of the pattern.
For instance, if the pattern is *or*, find the candidate match ('orld' , 'world') using a binary chop on the substring or, then confirm the match using a normal globbing approach.
If the wildcard is more complex, e.g., h*o, built sets of candidates for h and o and find their intersection before the final linear glob.
You say you're currently doing linear search. Does this give you any data on the most frequently performed query patterns? e.g. is blah* much more common than bl?h (which i'd assume it was) among your current users?
With that kind of prior knowledge you can focus your indexing efforts on the commonly used cases and get them down to O(1), rather than trying to solve the much more difficult, and yet much less worthwhile, problem of making every possible query equally fast.
You can achieve a simple speedup by keeping counts of the characters in your strings. A string with no bs or a single b can never match the query abba*, so there is no point in testing it. This works much better on whole words, if your strings are made of those, since there are many more words than characters; plus, there are plenty of libraries that can build the indexes for you. On the other hand, it is very similar to the n-gram approach you mentioned.
If you do not use a library that does it for you, you can optimize queries by looking up the most globally infrequent characters (or words, or n-grams) first in your indexes. This allows you to discard more non-matching strings up front.
In general, all speedups will be based on the idea of discarding things that cannot possibly match. What and how much to index depends on your data. For example, if the typical pattern length is near to the string length, you can simply check to see if the string is long enough to hold the pattern.
There are plenty of good algorithms for multi-string search. Google "Navarro string search" and you'll see a good analysis of multi-string options. A number of algorithsm are extremely good for "normal" cases (search strings that are fairly long: Wu-Manber; search strings with characters that are modestly rare in the text to be searched: parallel Horspool). Aho-Corasick is an algorithm that guarantees a (tiny) bounded amount of work per input character, no matter how the input text is tuned to create worst behaviour in the search. For programs like Snort, that's really important, in the face of denial-of-service attacks. If you are interested in how a really efficient Aho-Corasick search can be implemented, take a look at ACISM - an Aho-Corasick Interleaved State Matrix.