So one of the exercises in my class was to perform the Boyer-Moore Algorithm with the pattern abc and the string aabcbcbabcabcabc and abababababababab. I was also meant to note the number of comparisons.
Which I had done using the extended bad character rule, however I was told today that one needs to use the strong good suffix rule once a match is found in the string. However, I'm a bit confused on how the strong good suffix rule would be used here as the pattern length is only 3.
When there is a full match, as the pattern abc has no border, when it is encountered in the string would I simply just shift by 3 symbols?
Thank You!
Related
I'm writing a program that makes word declension for Polish language. In this language stems can vary in some cases (because of palatalization or mobile/fleeting e and other effects).
For example, we have word "karzeł" and it is basic dictionary form of word. It's stem is also 'karzeł'. But genitive form of this word is "karła" and stem is "karł". We can see here that 'e' dissapeared and 'rz' changes to 'r'.
Another example:
'uzda' -> stem 'uzd'
'uździe' -> stem 'uździ'
Alternation: 'zd' -> 'ździ'
I'd like to store in dictionary only basic form of stem ('karzeł' and 'uzd') and when I'll put in my program stem 'karł' or 'uździ' it will find proper basic stems. Alternations takes place only at the end of stem and contains maximum 4 letters of it.
Is there any algorithms that could do that? Levensthein distance treats all letters equally so if I type word 'barzeł' then the distance to stem 'karzeł' will be less than to stem 'karł'.
I thought also about neural networks but I'm not sure how to encode words (give each stem variation different id?).
Another idea is to write algorith which makes something like reversed alternation and creates set of possible stems and try to find them in dictionary.
I would like to highlight that I only want store basic form of stem and everything else makes on the fly.
First of all, I remember seeing a number of projects on Polish morphology around. So I would look at them first, before starting one of your own.
Regarding Levenshtein, as Pierre correctly noted in the comment, the distance function can be customized. And it should be. Let me put it this way: think of Levenshtein not as an algorithm of and in itself, but as a solution to a specific error model. First he suggests a model which says that when you are typing a word every letter can be either dropped or replaced by another one due to some random process (fingers not pressing the right keys). Then, his algorithm is just a generator of maximum likelihood solutions under this model. The more errors you allow, the smaller is the probability of this sequence of errors actually happening, the bigger is the score.
You (implicitly) state a very different hypothesis, though. That Polish stems may have certain flexibility at the end (some linguistic process that you do not fully understand within this framework). Then, when you strip your suffix (or something that looks like one), there are three options:
1) there is a chance that what you have here is just a different form of a stem you have stored in your dictionary, or
2) it is a completely different stem, or
3) you've stripped your suffix improperly and what you have is not stem at all.
You can heuristically estimate these probabilities by looking at how many letters in the beginning of the supposed stem match some dictionary entries, for example (how to find these entries is a related but different question). And then you can pick the guess that is the most plausible according to your metric/heuristic.
Now, note that you can use any algorithm to find the candidates in the dictionary. Including the Levenshtein algorithm - as long as you are reasonably sure that the right ones will be picked up. But obviously you are better off writing your own dictionary search algorithm that follows your own metric or emulates it. For example, by giving the biggest/prohibitive cost to the change of letters in the beginning of the word and reducing it as you go towards the end.
I'm wondering whether there is a way to generate the most specific regular expression (if such a thing exists) that matches a given string. Here's an illustration of what I want the method to do:
str = "(17 + 31)"
find_pattern(str)
# => /^\(\d+ \+ \d+\)$/ (or something more specific)
My intuition was to use Regex.new to accumulate the desired pattern by looping through str and checking for known patterns like \d, \s, and so on. I suspect there is an easy way for doing this.
This is in essence an algorithm compression problem. The simplest way to match a list of known strings is to use Regexp.union factory method, but that just tries each string in turn, it does not do anything "clever":
combined_rx = Regexp.union( "(17 + 31)", "(17 + 45)" )
=> /\(17\ \+\ 31\)|\(17\ \+\ 45\)/
This can still be useful to construct multi-stage validators, without you needing to write loops to check them all.
However, a generic pattern matcher that could figure out what you mean to match from examples is not really possible. There are too many ways in which you could consider strings to be similar or not. The closest I could think of would be genetic programming where you supply a large list of should match/should not match strings and the code guesses at the best regex by constructing random Regexp objects (a challenge in itself) and seeing how accurately they match and don't match your examples. The best matchers could be combined and mutated and tried again until you got 100% accuracy. This might be a fun project, but ultimately much more effort for most purposes than writing the regular expressions yourself from a description of the problem.
If your problem is heavily constrained - e.g. any example integer could always be replaced by \d+, any example space by \s+ etc, then you could work through the string replacing "matchable units", in fact using the same regular expressions checked in turn. E.g. if you match \A\d+ then consume the match from the string, and add \d+ to your regex. Then take the remainder of the string and look for next matching pattern. Working this way will have its limitations (you must know the full set of patterns you want to match in advance, and all examples would have to be unambiguous). However, it is more tractable than a genetic program.
I have 1,000,000 strings that I want to categorize. The way I do this is to bucket it if it contains a set of words or phrases. The set of words is about 10,000. Ideally I would be able to support regular expressions, but I am focused on making it run fast right now. Example phrases:
ford, porsche, mazda...
I really dont want to match each word against the strings one by one, so I decided to use regular expressions. Unfortunately, I am running into a regular expression issue:
Regexp.new("(a)"*253)
=> /(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)...
Regexp.new("(a)"*254)
RegexpError: regular expression too big: /(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)...
where a would be one of my words or phrases. Right now, I am planning on running 10,000 / 253 matches. I read that the length of the regex heavily impacts performance, but my regex match is really simple and the regexp is created very quickly. I would like to get around the limitation somehow, or use a better solution if anyone has any ideas. Thanks.
You might consider other mechanisms for recognizing 10k words.
Trie: Sometimes called a prefix tree, it is often used by spell checkers for doing word lookups. See Trie on wikipedia
DFA (deterministic finite automata): A DFA is often created by the lexer in a compiler for recognizing the tokens of the language. A DFA runs very quickly. Simple regexes are often compiled into DFAs. See DFA on wikipedia
Im learning about automata. Can you please help me understand how automata with Kleene closure works? Let's say I have letters a,b,c and I need to find text that ends with Kleene star - like ab*bac - how will it work?
The question seems to be more about how an automaton would handle Kleene closure than what Kleene closure means.
With a simple regular expression, e.g., abc, it's pretty straightforward to design an automaton to recognize it. Each state essentially tells you where you are in the expression so far. State 0 means it's seen nothing yet. State 1 means it's seen a. State 2 means it's seen ab. Etc.
The difficulty with Kleene closure is that a pattern like ab*bc introduces ambiguity. Once the automaton has seen the a and is then faced with a b, it doesn't know whether that b is part of the b* or the literal b that follows it, and it won't know until it reads more symbols--maybe many more.
The simplistic answer is that the automaton simply has a state that literally means it doesn't know yet which path was taken.
In simple cases, you can build this automaton directly. In general cases, you usually build something called a non-deterministic finite automaton. You can either simulate the NDFA, or--if performance is critical--you can apply an algorithm that converts the NDFA to a deterministic one. The algorithm essentially generates all the ambiguous states for you.
The Kleene star('*') means you can have as many occurrences of the character as you want (0 or more).
a* will match any number of a's.
(ab)* will match any number of the string "ab"
If you are trying to match an actual asterisk in an expression, the way you would write it depends entirely on the syntax of the regex you are working with. For the general case, the backwards slash \ is used as an escape character:
\* will match an asterisk.
For recognizing a pattern at the end, use concatenation:
(a U b)*c* will match any string that contains 0 or more 'c's at the end, preceded by any number of a's or b's.
For matching text that ends with a Kleene star, again, you can have 0 or more occurrences of the string:
ab(c)* - Possible matches: ab, abc abcc, abccc, etc.
a(bc)* - Possible matches: a, abc, abcbc, abcbcbc, etc.
Your expression ab*bac in English would read something like:
a followed by 0 or more b followed by bac
strings that would evaluate as a match to the regular expression if used for search
abac
abbbbbbbbbbac
abbac
strings that would not match
abaca //added extra literal
bac //missing leading a
As stated in the previous answer actually searching for a * would require an escape character which is implementation specific and would require knowledge of your language/library of choice.
Arising out of this question, I'm looking for an elegant (ruby) way to compute the word signature suggested in this answer.
The idea suggested is to sort the letters in the word, and also run length encode repeated letters. So, for example "mississippi" first becomes "iiiimppssss", and then could be further shortened by encoding as "4impp4s".
I'm relatively new to ruby and though I could hack something together, I'm sure this is a one liner for somebody with more experience of ruby. I'd be interested to see people's approaches and improve my ruby knowledge.
edit: to clarify, performance of computing the signature doesn't much matter for my application. I'm looking to compute the signature so I can store it with each word in a large database of words (450K words), then query for words which have the same signature (i.e. all anagrams of a given word, that are actual english words). Hence the focus on space. The 'elegant' part is just to satisfy my curiosity.
The fastest way to create a sorted list of the letters is this:
"mississippi".unpack("c*").sort.pack("c*")
It is quite a bit faster than split('') and join(). For comparison it is also best to pack the array back together into a String, so you dont have to compare arrays.
I'm not much of a Ruby person either, but as I noted on the other comment this seems to work for the algorithm described.
s = "mississippi"
s.split('').sort.join.gsub(/(.)\1{2,}/) { |s| s.length.to_s + s[0,1] }
Of course, you'll want to make sure the word is lowercase, doesn't contain numbers, etc.
As requested, I'll try to explain the code. Please forgive me if I don't get all of the Ruby or reg ex terminology correct, but here goes.
I think the split/sort/join part is pretty straightforward. The interesting part for me starts at the call to gsub. This will replace a substring that matches the regular expression with the return value from the block that follows it. The reg ex finds any character and creates a backreference. That's the "(.)" part. Then, we continue the matching process using the backreference "\1" that evaluates to whatever character was found by the first part of the match. We want that character to be found a minimum of two more times for a total minimum number of occurrences of three. This is done using the quantifier "{2,}".
If a match is found, the matching substring is then passed to the next block of code as an argument thanks to the "|s|" part. Finally, we use the string equivalent of the matching substring's length and append to it whatever character makes up that substring (they should all be the same) and return the concatenated value. The returned value replaces the original matching substring. The whole process continues until nothing is left to match since it's a global substitution on the original string.
I apologize if that's confusing. As is often the case, it's easier for me to visualize the solution than to explain it clearly.
I don't see an elegant solution. You could use the split message to get the characters into an array, but then once you've sorted the list I don't see a nice linear-time concatenate primitive to get back to a string. I'm surprised.
Incidentally, run-length encoding is almost certainly a waste of time. I'd have to see some very impressive measurements before I'd think it worth considering. If you avoid run-length encoding, you can anagrammatize any string, not just a string of letters. And if you know you have only letters and are trying to save space, you can pack them 5 bits to a letter.
---Irma Vep
EDIT: the other poster found join which I missed. Nice.