regular-language

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

问题 Typically in our work we use regular expressions in capture or match operations. However, regular expressions can be used - manually at least - to generate legal sentences that match the regular expression. Of course, some regular expressions can match infinitely long sentences, e.g., the expression .+ . I have a problem that could be solved by using a regular expression sentence generating algorithm. In pseudocode, it would operate something like this: re = generate("foo(bar|baz)?", max

问题 I have a large collection of regular expression that when matched call a particular http handler. Some of the older regex's are unreachable (e.g. a.c* ⊃ abc* ) and I'd like to prune them. Is there a library that given two regex's will tell me if the second is subset of the first? I wasn't sure this was decidable at first (it smelled like the halting problem by a different name). But it turns out it's decidable. 回答1: Trying to find the complexity of this problem lead me to this paper. The

问题 I'm studying for my computing languages test, and there's one idea I'm having problems wrapping my head around. I understood that regular grammars are simpler and cannot contain ambiguity, but can't do a lot of tasks that are required for programming languages. I also understood that context-free grammars allow ambiguity, but allow for some things necessary for programming languages (like palindromes). What I'm having trouble with is understanding how I can derive all of the above by knowing

问题 How do you say δ: Q × Σ → Q in English? Describing what × and → mean would also help. 回答1: δ is like a mathematical function called the transition function . Something like. z = f(x, y) A function in mathematical defines mapping of elements in one set to another set. In function set of input arguments are called Domain of a function and output is the rage. [ANSWER] In expression "δ:Q×Σ → Q" , × means Cartesian product (that is a set), and → is a mapping . "δ:Q×Σ → Q" says δ is a transition

问题 What is the direct and easy approach to draw minimal DFA , that accepts the same language as of given Regular Expression(RE) . I know it can be done by: Regex ---to----► NFA ---to-----► DFA ---to-----► minimized DFA But is there any shortcut way? like for (a+b)*ab 回答1: Regular Expression to DFA Although there is NO algorithmic shortcut to draw DFA from a Regular Expression(RE) but a shortcut technique is possible by analysis not by derivation, it can save your time to draw a minimized dfa.

问题 I know a n b n for n > 0 is not regular by the pumping lemma but I would imagine a*b* to be regular since both a,b don't have to be the same length. Is there a proof for it being regular or not? 回答1: Answer to your question: imagine a*b* to be regular, Is there a proof for it being regular or not? No need to imagine, expression a*b* is called regular expression (re), and regular expressions are possible only for regular languages. If a language is not a regular then regular expression is also

问题 Lets say you have a language L and you want to determine if it is context free. Context free languages intersected with regular languages are context free. Is that enough to prove that L is context free? Meaning, L intersect P = T Where P is a regular language and T is context free. Does this imply that L is context free? 回答1: No, your statement is not true. Consider the following counter-example: L = {0 n 1 n 2 n | n > 0}, P = T = Ø . Clearly we have L ∩ P = L ∩ Ø = Ø = T , and Ø is both

问题 Maybe I have missed something, but what are wrong with this regular expresion? var str = "lorem ipsum 12345 dolor"; var x = /\d+/.exec(str); var y = /\d*/.exec(str); console.log(x); // will print 12345 console.log(y); // will print "" but why ? Can you please explain why /\d*/.exec(str); returns an empty string instead of "12345". * means zero or more number of matches. 回答1: \d* matches zero or more digits in a row. When you run exec on a regex, it starts at the beginning of the input and

问题 I know that to prove that a language is non-regular one can use the pumping-lemma. I think I understand how it works, but when it comes to showing that a context-free-grammar is (or isn't regular) I'm having big problems. Here is an example of a CFG that I can't understand how to show is regular (or non-regular): i) S → NP VP ii) NP → DET N iii) VP → TV NP iv) N → N N v) N → A N vi) NP → Mary |John vii) DET → a |the |her |his viii) TV → bought |loves |misses ix) N → bike |jersey |mountain