regular-language

问题 This is a theoretical Computer Science question (Computation Theory). I know that RegExps can take a very long time to calculate. However, from Theory of Computation we know that matching with a Regular Expression can be done extremely fast in a few clock cycles. If RegExps are equivalent to Finite Automata, why RegExps have (or require) a timeout method? Using a DFA, the computation time for matching can be exteremely fast. By RegExps I mean the Regular Expressions pattern matching classes

问题 String S= "multiply 3 add add 3 3 1" I want to get two string arrays The one is {"multiply", "add", "add"} Another out is {"3","3","3",1} How can I get it? I tried to use String operators[] = s.split("[0-9]+"); String operands[] =s.split("(?:add|multiply)"); But, it doesn't work. 回答1: You should use Matcher instead of split : import java.util.regex.Matcher; import java.util.regex.Pattern; ... List<String> operators = new ArrayList<String>(); Matcher m = Pattern.compile("add|multiply").matcher

问题 We all know that (a + b)* is a regular language for containing only symbols a and b . But (a + b)* is a string of infinite length and it is regular as we can build a finite automata, so it should be finite. Can anyone please explain this? 回答1: Finite automaton can be constructed for any regular language, and regular language can be a finite or an infinite set. Of-course there are infinite sets those are not regular sets. Check the Venn diagram below: Notes : 1. every finite set is a regular

问题 I was looking at the question posed in this stackoverflow link (Regular expression for odd number of a's) for which it is asked to find the regular expression for strings that have odd number of a over Σ = {a,b} . The answer given by the top comment which works is b*(ab*ab*)*ab* . I am quite confused - a was placed just before the last b* , does this ordering actually matter? Why can't it be b*a(ab*ab*)*b* instead (where a is placed after the first b* ), or any other permutation of it?

问题 According to my textbook, the complement of L1 = A* - L1 is a regular language as long as L1 is a regular language. Doesn't A* also include Context Free languages, Context Sensitive languages, and Recursively Enumerable languages? A*-L1 would include all of them too, wouldn't it? How can it be regular then? Under the representation of a Finite State Machine I understand why the complement is still a regular language. However, I can't understand the theory behind it. Also, A* - L1 = A*

问题 I apologize in advance for my primitive english; i will try my best to avoid grammatical errors and such. Two weeks ago i decided to freshen my knowledge of Scheme (and its enlightnings) whilst implementing some math material i got between hands, specifically, Regular Languages from a course on Automata theory and Computation in which i am enrolled. So far, i've been representing alphabets as lists of symbols instead of lists of chars because i want to have letters of variable size. lists of

问题 I apologize in advance for my primitive english; i will try my best to avoid grammatical errors and such. Two weeks ago i decided to freshen my knowledge of Scheme (and its enlightnings) whilst implementing some math material i got between hands, specifically, Regular Languages from a course on Automata theory and Computation in which i am enrolled. So far, i've been representing alphabets as lists of symbols instead of lists of chars because i want to have letters of variable size. lists of

问题 As we know, given a regular grammar, we have algorithm to get its regular expression. But if the given grammar is context-free grammar (but it only generates regular language), like S->aAb A->bB B->cB|d Is there any existing algorithm that can get the regular expression in general? Thanks！ 回答1: In the most general sense, there is no solution. The problem of determining whether a CFG is regular is undecidable (Greibach Theorem, last 3 pages of http://www.cis.upenn.edu/~jean/gbooks/PCPh04.pdf )

问题 My regExp do not get a correct result, I can not find where is mistake. let ori_str = "abcXabcYabcZ" // there are 3 Capital character let pattern = "[A-Z]" let regular = try!NSRegularExpression(pattern: pattern, options: .caseInsensitive) let results = regular.matches(in: ori_str, options: .reportProgress , range: NSMakeRange(0, ori_str.characters.count)) print("results have: \(results.count) count") // log 'result have: 12 count' why 12 without 3 ??? I want to get the Capital Character, then

问题 I understand the reason and the proof why {a^n b^n | n >= 0} is NOT regular. Why is {a^nb^n | n >= 0} not regular? The solution of one of my exercises is: {a^n a^n | n >= 0} is regular. How can I prove this thesis? 回答1: Yes, Language {a n a n | n >= 0} is a regular language . To proof that certain language is regular, you can draw its dfa/regular expression. And you can drive do for this language as follows: Because " a n a n for n >= 0 " is same as " a 2n for n >=0", and that is "set of all