I've tried my best reading most of the literature on this, and still haven't understood anything about how the failure function used in KMP algorithm is constructed. I've been referring mostly to http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=stringSearching tutorial which most of the people consider excellent. However, I still have not understood it. I'd be thankful if you could take the pain of giving me a simpler and easy to understand explanation on it.
The failure function actually tells us this: if you matched X characters of a string, what is the longest suffix of such string such that it's also a prefix of a search string.
You are asking how it's built, the approach is quite straightforward.
If you add a new character at the end of a string, that is you are building f[x], and if it matches with character at position f[x-1], then f[x] is simply f[x-1]+1.
In the other cases where it doesn't match, you try to find smaller and smaller suffixes and check if they match.
For example, you have a word "accadaccac" for which you are building a failure function and you just added the letter 'c'. Let's say you are building a failure function for the last letter, the letter 'c'.
First you check the failure function of the previous letter, its failure function was 4 because you can match suffix "acca" with the prefix "acca", and now you add the letter 'c', it doesn't match with the letter 'd' succeeding prefix "acca".
So you backtrack, to the last good suffix. You are now searching for a suffix of "acca" which is also a prefix of "accadaccac", but is smaller than "acca". The answer to that question is f[length("acca")-1], or f[3], which is f[3] = 1, because suffix of length 1 (just the letter 'a') is also a prefix of a search string.
And now you can try if the 'c' matches with the character on the position 1, and voila, it matches, so now you know f[9] = f[f[8]-1]+1 = 2.
I hope this will help you. Good luck! :)
http://www.oneous.com/Tutorial-Content.php?id=24
U can use the learning resources in this website for understanding the KMP Algorithm and the failure function. Also try to take the code and do some runs on it for an example string by hand. However, the best way to understand its working would be to code it yourself on some variations of the basic algorithm. I suggest u start with NHAY and PERIOD on SPOJ.
Related
In this problem we've to split a string into meaningful words. We're given a dictionary to see If the word exists or not.
I"ve seen some other approaches here at How to split a string into words. Ex: "stringintowords" -> "String Into Words"?.
I thought of a different approach and was wondering If it would work or not.
Example- itlookslikeasentence
Algorithm
Each letter of the string corresponds to a node in a DAG.
Initialize a bool array to False.
At each node we have a choice- If the addition of the present letter to the previous subarray still produces a valid word then add it, if it does not then we will begin a new word from that letter and set bool[previous_node]=True indicating that a word ended there. In the above example bool[1] would be set to true.
This is something similar to the maximum subarray sum problem.
Would this algorithm work?
No, it wouldn't. You solution takes the longest possible word at every step, which doesn't always work.
Here is counterexample:
Let's assume that the given string is aturtle. Your algorithm will take a. Then it will take t as at is valid word. atu is not a word, so it'll split the input: at + urtle. However, there is no way to split urtle into a sequence of valid English words. The right answer would be a + turtle.
One of the possible correct solutions uses dynamic programming. We can define a function f such that f(i) = true iff it's possible to split the first i characters of the input into a valid sequence of words. Initially, f(0) = true and the rest of the values are false. There is a transition from f(l) to f(r) if s[l + 1, r] is a valid word for all valid l and r.
P.S. Other types of greedy algorithms would not work here either. For instance, if you take the shortest word instead of the longest one, it fails to work on, for instance, the input atnight: there is no way to split tnight after the a is stripped off, but at + night is clearly a valid answer.
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 take a input from the user and its a string with a certain substring which repeats itself all through the string. I need to output the substring or its length AKA period.
Say
S1 = AAAA // substring is A
S2 = ABAB // Substring is AB
S3 = ABCAB // Substring is ABC
S4 = EFIEFI // Substring is EFI
I could start with a Single char and check if it is same as its next character if it is not, I could do it with two characters then with three and so on. This would be a O(N^2) algo. I was wondering if there is a more elegant solution to this.
You can do this in linear time and constant additional space by inductively computing the period of each prefix of the string. I can't recall the details (there are several things to get right), but you can find them in Section 13.6 of "Text algorithms" by Crochemore and Rytter under function Per(x).
Let me assume that the length of the string n is at least twice greater than the period p.
Algorithm
Let m = 1, and S the whole string
Take m = m*2
Find the next occurrence of the substring S[:m]
Let k be the start of the next occurrence
Check if S[:k] is the period
if not go to 2.
Example
Suppose we have a string
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
For each power m of 2 we find repetitions of first 2^m characters. Then we extend this sequence to it's second occurrence. Let's start with 2^1 so CD.
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
CDCD CDCD CDCD CDCD CD
We don't extend CD since the next occurrence is just after that. However CD is not the substring we are looking for so let's take the next power: 2^2 = 4 and substring CDCD.
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
CDCD CDCD
Now let's extend our string to the first repetition. We get
CDCDFBF
we check if this is periodic. It is not so we go further. We try 2^3 = 8, so CDCDFBFC
CDCDFBFCDCDFDFCDCDFBFCDCDFDFCDC
CDCDFBFC CDCDFBFC
we try to extend and we get
CDCDFBFCDCDFDF
and this indeed is our period.
I expect this to work in O(n log n) with some KMP-like algorithm for checking where a given string appears. Note that some edge cases still should be worked out here.
Intuitively this should work, but my intuition failed once on this problem already so please correct me if I'm wrong. I will try to figure out a proof.
A very nice problem though.
You can build a suffix tree for the entire string in linear time (suffix tree is easy to look up online), and then recursively compute and store the number of suffix tree leaves (occurences of the suffix prefix) N(v) below each internal node v of the suffix tree. Also recursively compute and store the length of each suffix prefix L(v) at each node of the tree. Then, at an internal node v in the tree, the suffix prefix encoded at v is a repeating subsequence that generates your string if N(v) equals the total length of the string divided by L(v).
We can actually optimise the time complexity by creating a Z Array. We can create Z array in O(n) time and O(n) space. Now, lets say if there is string
S1 = abababab
For this the z array would like
z[]={8,0,6,0,4,0,2,0};
In order to calcutate the period we can iterate over the z array and
use the condition, where i+z[i]=S1.length. Then, that i would be the period.
Well if every character in the input string is part of the repeating substring, then all you have to do is store first character and compare it with rest of the string's characters one by one. If you find a match, string until to matched one is your repeating string.
I too have been looking for the time-space-optimal solution to this problem. The accepted answer by tmyklebu essentially seems to be it, but I would like to offer some explanation of what it's actually about and some further findings.
First, this question by me proposes a seemingly promising but incorrect solution, with notes on why it's incorrect: Is this algorithm correct for finding period of a string?
In general, the problem "find the period" is equivalent to "find the pattern within itself" (in some sense, "strstr(x+1,x)"), but with no constraints matching past its end. This means that you can find the period by taking any left-to-right string matching algorith, and applying it to itself, considering a partial match that hits the end of the haystack/text as a match, and the time and space requirements are the same as those of whatever string matching algorithm you use.
The approach cited in tmyklebu's answer is essentially applying this principle to String Matching on Ordered Alphabets, also explained here. Another time-space-optimal solution should be possible using the GS algorithm.
The fairly well-known and simple Two Way algorithm (also explained here) unfortunately is not a solution because it's not left-to-right. In particular, the advancement after a mismatch in the left factor depends on the right factor having been a match, and the impossibility of another match misaligned with the right factor modulo the right factor's period. When searching for the pattern within itself and disregarding anything past the end, we can't conclude anything about how soon the next right-factor match could occur (part or all of the right factor may have shifted past the end of the pattern), and therefore a shift that preserves linear time cannot be made.
Of course, if working space is available, a number of other algorithms may be used. KMP is linear-time with O(n) space, and it may be possible to adapt it to something still reasonably efficient with only logarithmic space.
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)