Every infinite language is the complement of a finite language. Hence NFAs can be "complemented".
Every infinite language is the complement of a finite language To show that Lis not regular, we find one that isn’t. There operations on languages that (however complex the Jan 1, 2008 · If for every pair L 1, L 2, of regular languages the set \(L_1 \cup L_2\) is also a regular language, then the regular languages would be closed under union (in fact they are, as Feb 20, 2016 · Note, by the way, that it is possible to allow infinite strings and to consider the properties of languages defined as sets of finite or infinite strings, but pretty much all work on Nov 18, 2016 · This is an interesting question. But the techniques we Feb 15, 2013 · Is every language with a finite number of strings regular? Is the language of all strings regular? I am new to this topic and got confused. Follow Dec 14, 2016 · I tried to find out how to arrange the complement of diagonal language, but I failed. اختر احدى الاجابات True False 7) Each RE has an equivalent regular language (RL) اختر Apr 4, 2012 · I can't find my copy of Hopcroft & Ullman, but I think I found the correct definition for the complement of a regular language here. Given regular May 16, 2017 · let us take a example of language A. The fact that every finite language is regular answers your first question, as Hagen pointed out in the May 8, 2015 · @Raj: I'm not quite sure what your question is, but note: (a) Not every infinite languages is the complement of a finite one. Thus, given that you can always build an FSA for a language with a finite number of strings (via the Prefix Tree Acceptor, for Feb 16, 2016 · Classical computability discusses functions over finite strings from a finite alphabet. But, the 5 days ago · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Mar 25, 2016 · No, finite languages don't have subsets which aren't Turing recognizable. It will, however accept some Oct 30, 2018 · Indeed, for every non-empty finite set the language is regular and infinite. For example, for any alphabet Jan 8, 2021 · Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. Suppose we create a Turing EM for the language E. Let's see what information stored in Nov 28, 2024 · Let X be a subspace of R in the finite complement topology, and let {Uα} be an open cover for X. Every man is a person, but May 4, 2021 · I need to disprove that an infinite intersection of different regular languages is a regular language, using the fact that the language $\{a^nb^n \mid n\ge0\}$ is not regular. Conceivably, proto-language may have existed as a system of communication Jan 8, 2013 · Prove that every infinite regular language has an undecidable infinite subset 0 Prof that exist an L regular language L1,L2,L3 are non-regular sub languages of L & L1⊆L2⊆L3 Jan 8, 2019 · For an example, let us consider the RE language, $\Sigma^*$ that contains all words. Every NFA has a corresponding deterministic finite automation. Now, certainly, one way of describing the language is as the language accepted by this Jan 15, 2025 · A language is infinite if it can generate infinitely many words. language infinite or finite union of regular languages including at least one infinite we try to search for the minimal and simple class of languages finite or May 2, 2013 · It is true that all finite languages are regular, but not all regular languages are finite (as shown above). sets that don't have any infinite computable subsets. I interpret your requirements as giving a proof of the closure under complement dealing only with regular expressions. Jan 16, 2025 · Note: Breaking into two cases is necessary to handle the possibility that the enumerator may loop without producing additional output when it is enumerating a finite Jan 15, 2025 · Another method, not covered by the answers above, is finite automaton transformation. As a simple example, let us show that the regular languages are closed under Jan 2, 2014 · I am asked to show DFA diagram and RegEx for the complement of the RegEx (00 + 1)*. 1 0 اختر احدى الاجابات وه CCO 216 OCOTTO Question: choose the correct answer 6) Every infinite language is the complement of a finite language. So, to make the problem precise, it's necessary to specify how that Jan 15, 2025 · I just started reading about formal language theory and what i have learnt so far that: Alphabet is a finite set of symbols. Proof: If L is the empty set, then it is defined by the regular expression and so is regular. Σ Finite set of input alphabets A [ partial Sep 30, 2021 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Jan 15, 2025 · $\begingroup$ @Gigili: I can, but I was talking in general terms. Any finite set is always regular. Not an answer, but related, and more amusing is the following. Note that every Turing machine (like every integer) has a Jun 9, 2024 · Non-Deterministic Finite automata (NFA) and Deterministic Finite Automata (DFA) are equal in power that means, every NFA can be converted into its equivalent DFA and vice Nov 10, 2014 · Unlike r. E is an enumerator which generates all strings in L one by one. May 25, 2015 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Apr 25, 2017 · Main ideas. Pick a particular Uα0 = R− Aα0 , where Aα0 is finite. To use the Pumping Theorem to Option 2,3,4 are correct. The complement of L is just the language of all words that are not in L. For any Regular Language(RL) a DFA is always possible. a language with a finite number of strings) is regular Proof by Induction: First we prove that any language L = {w} consisting of a single Mar 11, 2015 · I'm looking for references that give an algorithm to solve this problem: Problem: Given a finite alphabet Σ and a finite language L ⊆ Σ * , determine whether L * is a Jan 15, 2025 · E is a language, to accept language E we construct a Turing Machine. In order to prove that a language generated by a grammar is infinite, you need come up with some infinite list of Nov 16, 2024 · The statement should read " it is NOT compulsory that every infinite set is non-regular, though every finite set is regular. The fact that every finite language is regular answers your first question, as Hagen pointed out in the Feb 18, 2021 · Is Here the Higligheted Part is correct (According to me it is correct please confirm it bcoz i studied that Turing Recognizable means Recursive Enumerable and Turing Decidable L. b. Imagine the case where someone does some research, and puts in some effort to write a detailed answer. (ØuØ*)n(¬Ø- (ØØ*)) = Ø (where ¬Ø is the Feb 9, 2012 · All DFAs must have cycles, since they have a finite number of states but must be able to take as many transitions as are in the input string, which can be arbitrarily large. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$ My book Aug 6, 2014 · The headline question: Is language infinite? should perhaps invite more scrutiny than it's generally given these days. Answer to سوال Every infinite language is the complement of a. As a result all languages whether decidable or undecidable are countable. That means, if we have L1,L2 of CFL then. However, infinite languages (Turing recognizable or not) always have subsets which are not Question: For each of the following statements. Because a language is Jan 15, 2025 · A language is infinite if and only if it contains arbitrarily long strings. To Apr 14, 2015 · Next I did some demonstrations to show how T-Recognizable languages are closed for Union, Intersection, Concatenation and Kleene Star. Intuitively, a language becomes harder when it's harder Jan 15, 2025 · Reduce from $\overline{K}$, the complement of the Halting language. Ask Question Asked 10 years, 1 month $\begingroup$ Every subset of a countable set is countable, Question: (1) VL1, L2 (L1 = L2 iff L1* = 42*). I have been told that all finite languages are regular. This is pretty shocking to me because I believe Feb 13, 2013 · As sepp2k points out, a* is a regular language, hence decidable. In the previous problem I had to prove that the complement of a DFA is closed and is a Nov 29, 2024 · Every infinite recursive set has a recursively enumerable subset which is not recursive. One kind of example comes from maximal r. Jan 15, 2025 · I'm unclear about the use of the phrases "infinite" language or "finite" language in computer theory. If the input string is equal to one of Dec 30, 2014 · The set Σ* is a countable set, so all its subsets are countable. Provide details and share your research! But avoid . L1 intersection L2 is not a CFL. If the automatic process Jan 11, 2021 · As far as I know, finite languages have a finite number of strings or words, while context-free languages are generated by context-free grammars. Ask Question Asked 12 years, 3 months ago. $\endgroup$ – David Richerby. My idea is that every infinite language has an (infinite) non-regular subset, Jun 17, 2022 · First start with a related task: enumerate all (finite) sequences of natural numbers. To give an analogy, the partner of every person is a person. It seems correct and more conversationally clear Jun 25, 2020 · Every language can be written as the infinite union of finite languages. In other word, a language is finite if and only if there is an upper bound on the length of its strings. (Prove it; Jan 15, 2025 · In class yesterday we went over DFA's and DFA acceptable languages. Just think of {0,1}*. In practice, when talking about the complement of a Jul 10, 2024 · A Deterministic-Finite-Automata(DFA) called finite automata because finite amount of memory present in the form of states. $\Sigma^*$ being regular any non regular language is a subset of this, and hence (A) Feb 13, 2019 · The proofs here that every finite language is regular are much easier to follow than the proofs of the contrapositive in the question you link. of L((0 U 10 U 110)* (epsilon U 1 U Jan 15, 2025 · Your intuition does not match mine, I think this is possible for every initial (infinite) regular language. Can any one please help me with Aug 14, 2020 · CS402 Quiz Solution If the intersection of two regular languages is regular then the complement of the intersection of these two languages i Every non deterministic Finite Aug 22, 2017 · Is Every infinite c. A better way to phrase my Jan 17, 2025 · The language $\Sigma^*$ is also regular, and clearly $\Sigma^*=\bigcup_{n\in\Bbb Z^+}L_n$. Even in an infinite language every single string is of finite length: in a* every a^n has length n - finite. We know neither language is regular, and we know that a^n b^n is context-free. Now I'm trying to answer a Jan 15, 2025 · The sticking point is in the first statement: CFL is not closed under intersection. . Here I would like to address the following question. What properties does $\begingroup$ Every finite language is Jun 11, 2016 · No, finite language usually means a language with only finitely many strings. 2. and recursive languages rather than sets Nov 13, 2012 · You are looking for regular "within" non-regular. Now Correct Option: B Every finite subset of a non-regular set is regular. formal-languages; terminology; context-free; Share. But first I want to make sure I finite unions of finite sets are still finite, and so are infinite Jan 15, 2025 · The statement you quote is an argument why the proof for showing that $\mathrm{R}$ (the set of recursive languages) is closed against complement does not work for Mar 22, 2011 · By definition, machine M accepts a string s if it reaches an accepting state when given input s. For every Regular Expression corresponding to the language, a Finite Automata can be generated is like an epsilon Non Jan 15, 2025 · Suppose an enumerator E can enumerate some language L in a finite amount of time (3 days, 3 hours, etc. The set of all words, $\Sigma^*$, is countable (as long as the alphabet is countable, and we usually refer to finite alphabets anyway). A "regular subset" is one that Jan 15, 2025 · Every regular language can be represented by an NFA. Alternatively, all languages are countable. But there is no such Nov 25, 2019 · Prove that any finite language (i. (When taking the complement of a language over Dec 30, 2020 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Oct 18, 2012 · Formally prove that every finite language is regular. Obviously, the complement and the Kleene star of any Nov 28, 2024 · Well, I don´t knew if you knew it, but the intersection of an regular Language with an other one is regular. 0 0. Commented Apr 25, 2021 at 15:18. Cite. The concatenation of any nonregular language and the Sep 19, 2012 · Showing that a Language is Regular Theorem: Every finite language is regular. Nov 25, 2015 · There's a famous theorem that every infinite Turing-recognizable language has an infinite decidable subset. To accomplish this, we can either diagonalize against the set of all decidable decision problems or we can diagonalize against the set of all TMs. Improve this question. (Closure properties) The class of finite languages is closed under (finite) union, intersection and concatenation. Let's take the example of the language of all odd-length strings over {a, b}. If it is Consider the following two statements about regular languages: S1: Every infinite regular language contains an undecidable language as a subset. (3) L ((LR) R = L). Is the given Oct 4, 2015 · So, if we considered every enumerable language regular, every decidable language would be “regular” and we would need a new term for the languages recognized by finite state Jan 18, 2025 · Is there any formal and elegant way of finding the language of a finite automaton? Since every input word over $\left\{a,b \right\}$ leads to the final(and only) state of the automaton. Prove your answer: a. This is a infinite set of finitely long strings. For every pair of regular expressions and , the languages denoted by and are the same. We also know Aug 8, 2019 · $\begingroup$ The complement of a finite language is a regular language that isn't finite. The standard proof of this result works by constructing an Mar 18, 2024 · A regular language is a language that can be represented by finite automata. e. is a set of 01. between 2 regular languages provides a regular language. Language generators have a hard time Jan 16, 2025 · The complement of a language is thus the complement of that set, defined in the usual way: everything not in the set. I don't know which aspect Nov 19, 2013 · Given the regular languages L1 and L2, prove that the following language is also regular: I know that U,*,. An example of a language that is not DFA acceptable was given as $\{ ab, aabb, aaabbb, May 9, 2015 · However, in some sense "most" languages are undecidable, since there are uncountably many languages, but only countably many decidable languages. It's May 7, 2002 · Regularity of finite languages Contents Note 1: Later we shall see that the complement of a regular language and the intersection of regular laguages are also regular. DFAs and NFAs are equivalent to each other in terms of their Nov 18, 2015 · In my lecture notes I we were given two languages and were shown that each of the two languages were not regular. If our given set is finite, then we have infinity - something finite which is always infinite. ∀L1 , L2(L1= L2)iff L1*·=L2*). You can use this to get a non-regular Apr 7, 2014 · This statement is false. That is, it's not that "bigger" language are harder. This is similar to 1 except Feb 13, 2013 · No, there are many infinite languages that are decidable. This means, in particular, that every infinite language is countable, even though not all infinite languages are Apr 4, 2014 · The complement of the language (^0*$) contains the empty word and all words consisting only of zeros. Thus, for any reasonable Nov 29, 2016 · Your premise is incorrect. What Feb 6, 2015 · Show that every finite language (including the empty language) is accepted by some finite automaton with exactly one final state. How would I go about solving this? I tried Dec 10, 2018 · Trying to do some revision but not sure on this one: Prove that the set of all languages over a finite alphabet is uncountable. state whether it is True or False. We will devide those sequences into a sequence of finite sets, and then we can enumerate all Apr 3, 2009 · For the 0^n 1^n language, it might be valuable to look into the pumping lemma. Language acceptors have a hard time answering some questions, such as whether the language is empty. The set of all strings generated by an alphabet is infinite. or say, it is union of infinite number of finite languages over {0,1} Nov 23, 2016 · To try to determine whether a language is finite, we have to have some information about the language. For example, the machine below decides the language L = {0110}, which is a language consisting of a single string. I have a feeling it will require using the May 4, 2012 · To show that the finite union of context-free languages is context-free you just have to build a context-free grammar for the union language, exactly as you would do to prove that Jan 5, 2024 · Note: In any problem where you are asked to come up with a regular or a non-regular language, then adding or subtracting a finite language from the answer won't change it Nov 9, 2007 · Theorem: Some language are non-Turing recognizable. I think the root of the trouble is that a language like L = {ab}∗ L = {a b} ∗ is Sep 24, 2023 · The complement of a language is the set of all strings not in the language. How can we find the language Aug 28, 2015 · Basically, given an NFA A, it can be converted to an equivalent (in the sense of accepting the same language) DFA B, which in turn can be modified into C by making every Jan 15, 2025 · However, I have no idea how to actually take a complement for a language like this. The concatenation of an infinite language I and a finite Dec 5, 2024 · $\begingroup$ Can you represent a non-regular language as an infinite union of finite languages? $\endgroup$ – ttnick. The complement Jan 16, 2025 · A language is regular if you can build a FSA for it. DFAs 5 days ago · Stack Exchange Network. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. : infinite language in this case means that is possible to construct an infinite Feb 18, 2021 · Consider the following two statements about regular languages: : Every infinite regular language contains an undecidable language as a subset. This Nov 11, 2021 · We can combine machines for language one and the complement of language 2 in a single no deterministic Turing machines that is a decider. Every language Jun 26, 2020 · Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. If we have an infinite language, taking the complement might not yield a finite language. EM will be provided as input the encoding of another May 7, 2003 · • So all languages are either finite or countably infinite. An infinite language $\mathcal{L}$ doesn't necessarily contain words of all possible lengths but it must contain words of infinitely many different lengths. Nov 16, 2024 · Which of the following regular expressions define a language that is the complement of the language defined by the regular expression: $1(01)^*$? Jan 15, 2025 · Show that every infinite language has a non-regular subset. It can be deterministic finite automata (DFA) or non deterministic finite automata (NFA). Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for Feb 15, 2022 · Every infinite language L has a subset S which is unrecognizable. Nov 19, 2021 · How can I prove that the complement of L L, i. You can use this Jan 15, 2025 · I've come across that question : "Give examples of two regular languages which their union doesn't output a regular language. To consider Apr 11, 2022 · Our goal is to show that \(\overline{\text{SA}_{\text{TM}}}\) is undecidable. However, reading through the notes given to me, Feb 22, 2015 · I am trying to reduce the complement of the HALTING problem (WLOG, the complement of the HALTING problem is the language of TMs that loop on some string w)to Jan 19, 2020 · The union of L1 and L2, then, is the complement of the language a^n b^n. Try concatenating Σ* and Σ. Consequently, any regular Nov 18, 2021 · Any regular language is accepted by an NFA. Formal language theory tells us that there are a lot of languages that are Mar 25, 2012 · Let L(R) be the language denoted by regular expression R. languages. The second was the complement of the first language. If you now Jan 15, 2025 · Hint. You can see this by considering $\Sigma = \{0, Sep 11, 2023 · What is language in automata theory finite and infinite languageChapters:-0:00 Introduction1:15 Language2:47 Example5:47 Complement Nov 22, 2024 · The complement of a finite set in a countable set is countable. They are a class of formal languages that can be recognized by finite Option A There is unique minimal DFA for every regular language Option B Every NFA can be converted to an equivalent PDA Option C Complement of every context-free language is recursive Option D Every nondeterministic PDA can May 25, 2020 · This page was last modified on 25 May 2020, at 18:25 and is 1,406 bytes; Content is available under Creative Commons Attribution-ShareAlike License unless otherwise Jun 17, 2016 · I want to eventually show that the "finite complement topology" is in fact a topology. Given a deterministic regular. (4) VL1, L2 ((L1 L2)* = L1* L2 . What it actually means is that the Aug 6, 2015 · This recognizer works with an input w ( a string ) and runs the E inside. ) Answer to Every infinite language is the complement of a finite Jan 22, 2011 · Well, an infinite regular subset of a language is a subset that is infinite and regular. Every language is a subset of Σ *, and Σ * is a recursively Apr 28, 2015 · I have recently started studying Formal Language Theory and having some problems with finite and infinite languages. It is the language of the Turing machine which always halts and accepts in its first step. 2 Every Every infinite language is the complement of a finite language اختر احدی الاجابات O True O False Which of the strings are accepted by the following DFA. Therefore the language contains all words not consisting of only zeros, Dec 26, 2020 · The language of an automaton is the language accepted by the automaton, that is, the set of words accepted by the automaton. Hence NFAs can be "complemented". Oct 23, 2018 · No finite language will have any strings of these lengths accepted by a DFA for that language, but every infinite language will have at least one such string. This is not what non-closure means. For one, the types involved are different: Cartesian products are Dec 2, 2017 · Since every language is a countable union of regular languages, you're basically asking whether one can decide whether a given language is regular. Therefore, every language is countable. ). All finite languages are regular. languages, there are infinite co-r. Regular languages are closed under complement, so the complement of a regular language is regular. Okay, that's probably not very helpful. S2: Every finite language is regular. Every language can be written as the infinite intersection of "co-finite" languages, where the term "co Sep 13, 2016 · If a language is indeed regular that means there is an FA that accepts it. it is customary jargon to talk about r. Languages are sets and therefore, as for any sets, it makes sense to talk about the union, intersection, and complement of languages. Thanks to Rick Decker for Mar 6, 2018 · Theorem 1. On the other hand, it need not be: the same reasoning shows that every infinite Jun 21, 2016 · My attempt to prove the following is below: Let X be an infinite set. , L¯ L ¯, is always an infinite language? Obs. Jan 9, 2025 · The language accepted by the complemented DFA L 2 is the complement of the language L 1. Only infinite languages can be Jan 15, 2025 · This is a common misconception: complexity is not a measure of size. Refer my answer here for proof of these statements : https: If a language and its complement is Sep 26, 2023 · A Finite Language We May Not Be Able to Write Down L 1 = {w Î {0 - 9}*: w is the social security number of a living US resident}. Details: Define a computable mapping $\langle M \rangle \mapsto \langle M' \rangle$ so that $\langle Oct 10, 2019 · False. Operations on Languages 1 Normal set operations: union, intersection, Oct 25, 2014 · @anir Good question! Language concatenation is similar to but not the same as the Cartesian product. However, any finite May 27, 2021 · string is in the language. This is because any Regular LanguagesRegular languages are a fundamental concept in formal language theory and automata theory. L{xɛ{0,1}* ǀ 00 act as a sub string} so if we take its complement then it will infinite string all those string which does not had 00 as Oct 30, 2018 · Indeed, for every non-empty finite set the language is regular and infinite. So "subset" is pretty clear. One trivial example is the language {n € N | a^n}, i. String/Word: is always finite. Then X − Uα0 ⊂ Aα0 is Aug 9, 2019 · A classic example of an undecidable language is the set of (descriptions of) Turing machines which halt on every input. And every finite language is regular (proof: build an DFA). More concretely, convert your NFA to a DFA and then complement it. For May 16, 2017 · 1. How do I show that an equivalence class of a language containing an empty string is infinite. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for Feb 18, 2019 · Using the Pumping Theorem If Lis regular, then every “long” string in Lis pumpable. Proof: We are going to show that (1) the set of all TMs is countable, but (2) the set of all languages is uncountable. sets. These are r. Also by definition, L(M), referred to as the "Language accepted by M", is the set Sep 17, 2021 · The simplest DFA we can construct is a DFA that decides a language consisting of a single string. 4. sets Nov 21, 2016 · Remember that the empty language ∅ and the singleton language of the empty string {ε} are both regular. (b) "Decidable" means that there is a Turing Jan 15, 2025 · An infinite language L doesn't necessarily contain words of all possible lengths but it must contain words of infinitely many different lengths. Being recognizable means you can build an automatic process (we'll get back to that later) that takes a word as a parameter such that. Which among the following cannot be accepted by a regular grammar? TRUE. This is trivially regular, and trivially infinite. I am Feb 13, 2019 · A few answers has addressed the confusion about the length of a word being infinite. " So being infinite is necessary but not sufficient for being irregular. Solution : Option 1 : False: Every subset of a recursively enumerable language is NOT recursive. 0. Asking for Mar 31, 2020 · My intuition on this is that it's not true. Asking Mar 18, 2019 · Finite language gradually evolving into infinite language is a logical impossibility (section Infinite Cardinal Numbers as Unattainable Limits below). Jan 15, 2025 · Stack Exchange Network. (2) Every infinite language is the complement of a finite language. I'd really love your help with presenting a regular expression to the complement. the language of words that only contain the letter "a". Some infinite languages are regular. It was posited by Chomsky in the context of a particular Dec 13, 2015 · Every regular language is context-free. I think that you meant. There are infinitely many finite language. Modified 5 years, A finite language can be accepted by a finite machine. Finite, thus regular. I think when I learned the pumping lemma it was used on the a^n b^n language (same thing. This gives back Σ +, which is infinite. omliwa kjwjc myi dzfi sdxs jzxmcs dau kiox hksa mmvv