Infinite language decidable. Build a Turing machine that will do the following.

Infinite language decidable Hope this helps! When using the pumping lemma, you typically assume a language is regular and conclude that what the pumping lemma says about regular languages - what we have said above - is true. • Theorem 2: If L is Turing-decidable then L is Turing-recognizable. 121 3 3 bronze badges $\endgroup$ Add a comment | 0 The problem of determining whether a recursively enumerable language is empty or infinite cannot be solved. So, S 1 and S 2 both are correct. A DFA • Can generate infinite strings and if not in the language, will never know it. An infinite decidable language that is a subset of $\overline { A_{TM}}$ Ask Question Asked 7 years, 1 month ago. TMs and Infinite Loops . Stack Exchange Network. Mark current state and delete the There are many infinite languages which are decidable, for example Σ∗ Σ ∗, or the language of all (binary representations of) primes. A set is considered decidable if there exists an algorithm that can determine membership within a finite amount of There's a famous theorem that every infinite Turing-recognizable language has an infinite decidable subset. 1(Decidable Language). Question: problem 4. Use the facts that regular languages are closed under complement and intersection, and that certain properties of DFA’s are decidable. If someone attempted to make a finite list of everything the language could represent, you could always legally create something one 'b' character longer than anything in his finite list. On the other hand, undecidable problems demonstrate the limits of computation, where Is the problem of Proper Subset of decidable languages decidable? Hot Network Questions Is sales tax determined by the state in which the SELLER is located, or the state in which the PURCHASER is located? $\begingroup$ @flashburn This works for the following reason: (1) Not every language is RE; (2) Every language can be expressed as an infinite union of languages containing a single word; (3) therefore, there are languages that are not RE that can be expressed as an infinite union of languages containing a single word. A finite subset of it is always trivially decidable. Consider all programs of the form f(x) = x + i-- definitely an infinite and decidable set. Claim 1. A PDA is defined by its states, input alphabet, stack alphabet, transitions, initial state, and But for strings not in the language (the first given machine cannot generate all the strings the second one can), our machine may halt and reject, or may never halt. Follow edited Jan 31, 2016 at 16:48. A table entry is A (accept), R (reject), But that's okay: all finite languages are decidable, anyway. Let M L be a TM accepting L, and let M L be a TM accepting L . Otherwise, the enumerator will never list $\langle M \rangle$, so it never However, it can only have countably many decidable languages, as before, so the number of decidable languages over $\{0,1\}$ must also be countably infinite; i. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it. Turing decidable means it halts in an accepting state if the input word is in the language, and halts in a rejecting state if the word is not in the language, Turing recognizable means it halts in an accepting state if the word is in the language, and in a rejecting state or fails to halt if the word Somewhere it explained as following: Statement (P1) is the membership problem of regular language , which is decidable problem . Question: Show that the following languages are decidable. True or false: If languages \(K\) and \(L\) are both undecidable, then their union is also undecidable. I basically said that there is a language S which is a sublanguage of L and it is a regular language. This subtle point is the reason for the star on the problem. There are any number of algorithms for that. INF_DFA = {{D))|D is a DFA and L(D) is infinite} E. Definition: For an undecidable language, there is no Turing Machine which accepts the language and makes a decision for every input string w. But this implies that Turing Machine that performs these checks ends up in an infinite loop, and thus never returns accepts nor rejects it's input. There's a stronger result that any infinite language has a subset that is not decidable. IfR accepts $\begingroup$ I think you have to (or someone has to) do some work to make clear what the result of an infinite number of steps is: how do we know what state the components of the machine are in? Also, can we travel an infinite distance along the tape? and if so, is it indexed by the ordinal numbers? otherwise, what do we do about programmes that would to do so? 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 Conclusion. OK, but what expressions will you allow? $\{a^nb or decidable, or even context-free languages. As an attempt to solve the problem, I could only think of a proof by construction. Classic examples (think Halting problem) are semi-decidable but not decidable. Here I would like to address the following question. Hope this helps! Problem 5 (3 Points) Show that every infinite Turing-recognizable language has an infinite decidable subset. Since both sets of decidable languages are as numerous as the natural numbers, there must be as many decidable languages over $\{0\}$ as over $\{0 The question whether decidable languages evolve into infinite language is vacuous since infinite generation is the null hypothesis for a generative procedure. Thus language defined by {ϵ}∪{01}∪{0011} Note that we sort of waved our hands there when we "split up" stuff not in L1 or L2 into two infinite languages, one co-Turing-recognizable and the other co-Turing-recognizable. . So it is an infinite language, right? That is, "infinite set" means "infinite language", right? Clearly, a* is a regular language. Here, • Decidable Languages • decidable problems concerning regular languages • An infinite binary sequence is an unending sequence of 0s and 1s. Improve this question. Contradiction The question is about searching for an element in an infinite stream S of ordered elements: a very natural algorithmic question. This means that our Turing Machine is Recognizable, but it is not decidable. This makes the decision problem simple: A few answers has addressed the confusion about the length of a word being infinite. It seems like it's safe to assume that in any infinite language, there must be exist a proper subset of that language which is recognizable but not decidable. I'll present an example of a decidable language, followed by a general result about decidable languages. A problem is considered decidable if there exists an algorithm that can always determine the correct answer in a finite amount of time. gnasher729 gnasher729. Given an infinite decidable language $L$, then if $S \subset L$ such that $L \setminus S$ is finite, then $S$ must be decidable. I did get some info on list of undecidable problems How to come up with a solution of finite or infinite language using context free grammar? 0. ) We can prove stuff using the fact that every language has this form, but not the kind of stuff we know about regexes. The proof goes by reduction to the problem of decidability, which is known to be unfeasible for recursively enumerable languages. You need to reason by cases. 11 Let INFINITEPDA = {?M?| M is a PDA and L(M) is an infinite language}. Construct a Turing The set $\{L' : L' \subseteq L\}$ is an uncountable set of languages, and since there is only countably many decidable languages, it has to contain an undecidable language. A language is Turing-decidable (or decidable) if some Turing machine decides it; Aka Recursive Language ; The language is infinite iff its grammar can generate an infinite number of words, Is it decidable whether a given context free grammar generates an infinite number of strings? 1. Because each Turing machine can recognize a single language and there are more languages than Turing The task is to demonstrate that the language INFINITE PDA={ M ∣M is a PDA and L(M) is an infinite language} is decidable. The program is not run to determine its recognizability. " I demonstrated this by contradiction. There are finitely many possible strings of at most n n symbols. If the machine has no accepting states, then it cannot possibly accept any string given to it. Diagonalization is a reasonable way of proving this. (e) {(A) A is a PDA and L(M) is an infinite language} (d) {(RS) R and S are regular expressions and L(R) CL(S)} Show transcribed image text. L = { <M> | M is a DFA that accepts infinitely many strings } In other words, the computational problem of determining whether a given DFA accepts an infinite language or not is decidable. We will modify E to only enumerate strings in lexicographic order. $\begingroup$ Don't know about hangs, that is an unnecessary concept. An example of this is Hilbert's 10th problem, which is recognizable and undecidable, but we showed in the last problem set that the Here prove that A is decidable. A better way to phrase my question is: Is there a language of any number of 1s over $\Sigma=\{0,1\}$ that you can make unrecognizable?. This Prove that $\text{INFINITE}_{\text{PDA}}$ is decidable. 4. In other words, decidable languages are those for which there exists an effective method to determine whether a given string is a member of the language or not. We can do this by simulating the PDA on all possible inputs of increasing length. A language is not decidable if there exists no machine that decides it. This means, in particular, that every infinite language is countable, even though not all infinite languages are recursively enumerable. ) How is that possible? I mean, how would we build a Turing machine that would accept (or reject) a possibly infinite string of 0's? I also thought that maybe we could create an enumerator that would create all words from $0^*$ with increasing length, but I am not sure if we can. You can try all input strings up to the pumping length for the resulting DFA (strings that don't cause the DFA to enter any state more than once). HINT: think about the fact that any computably enumerable set which we can enumerate "in order" is computable. There are 3 steps to solve this one. If the simulation halted within that number of steps, then M halts. It will, however accept some strings that are not in language 1 (I think), so its language is not a subset of the original recognized language. Prove that Turing Machine ever writes a blank symbol over a non blank symbol is undecidable. Call it G0. Decidable • A language L is Turing recognizable if some Turing machine recognizes it. Yu Gu Yu Gu. Here are the steps to prove this: Input Representation: We take a PDA M as input. The following is my solution: On input <M>: 1. Thus, by Kleene's Theorem it cannot be a regular language. Is the difference between an unrecognizable language and a finite language There are certainly examples for all "levels": It is easy to give an infinite set of encodings that is decidable. It may accept some and/or reject some strings. The set R is the set of all decidable languages. This problem is indeed decidable, although in a somewhat sneaky way. We say a language L is decidable (L ∈L D(TM)) if there exists a Turing machine M such that • w ∈L ⇐⇒M accepts w • w ̸∈L ⇐⇒M rejects w Definition 1. I also that if a DFA has a finite language with size greater than 50, all words must be finite so I can iterate through the set of all words over the alphabet in lenlex order, using that a whether a particular word is accepted by a DFA is decidable, and reject Then A and B become Turing Decidable, therefore C also becomes turing decidable. To show that INFINITE a = {A ∣ A is a DFA and L (A) is an infinite language} is decidable, we need to find an algorithm that confirms whether a given Deterministic Finite Automaton (DFA) accepts an infinite language. Next, we need to check if the language of the PDA is infinite. True or false: If a language \(L\) is undecidable, then \(L\) is infinite. the language of words that only contain the letter "a". The idea is to appeal to the pumping lemma for regular languages REX is decidable Theorem The language A REX r–R,w‰¶R is a regular expression that generates the string wx is decidable. Context free grammar assistance. I have this question for a homework. A language is decidable if there exists a machine that decides it. A decision procedure (algorithm) must take a finite input so how would you give an infinite language as input? Maybe you want to use expressions such as $\{a^nb^n\mid n\ge 0\}$. Otherwise, M deliberately goes into an infinite I know being a decidable language doesn't necessarily mean that every subset is of it is decidable, but does every undecidable language contain a decidable subset? formal-languages; Does there exist a undecidable infinite language with only a Summary:: Show that every infinite Turning-recognizable language has an infinite decidable subset Sipser's Theory of Computation, third edition, chapter three contains and exercise that asks us to demonstrate this. And for a lot of languages, we don't know about their Warm-Up: Some Decidable Languages Show that the following languages are decidable by describing (at a high level) an algorithm that deci des them infinite 2D table whose rows are Turing machines and whose columns are input strings. decidable We know that a language may be semi-decidable but not decidable. answer for problem 1 Problem 1 Show INFINITEPDA, from Problem 4. The idea, as described in the answer you provided, is that a string with length between $k$ and $2k-1$ can be pumped up repeatedly, to Show that INFINITEPDA is decidable. formal-languages; undecidability; uncountability; Share. Do a breadth-first search of the grammar rules of G0 In this chapter, we formally prove that almost all languages are undecidable using the countability and uncountability concepts from a previous chapter. Then there would exist a machine M B that recognized (but did not decide) B. Convert G to Chomsky Normal Form. A language is Turing-decidable (or decidable) if some Turing machine decides it; Aka Recursive Language ; The question whether decidable languages evolve into infinite language is vacuous since infinite generation is the null hypothesis for a generative procedure. This language contains all natural numbers, or; This language contains no natural numbers, or; This language contains all natural numbers greater than some natural number n. every infinite language (regular or not) has an undecidable subset. Prove that $\text{INFINITE}_{\text{PDA}}$ is decidable. with infinite tape but in finite time)? This is, are there words, w, in a decidable language for which cannot be determined a bound f(|w|)? Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. Explanation of Decidability. For an input $v$ If $D$ accepts one of these strings, $L(D)$ is infinite. Question: Is the following language decidable? Prove your answer: INFINITEDFA = (<A>|A is a DFA and L(A) is an infinite language) Is the following language decidable? Prove your answer: INFINITEDFA = (<A>|A is a DFA and L(A) is an infinite language) There are 2 TMs and Infinite Loops . Stack Exchange network consists of 183 Q&A communities including Stack Overflow, You now ask whether the following language is decidable: Each of these machines recognizes a different language: the language of terminating programs; the language of programs terminate and don't produce any input; the language of terminating programs that print a finite prefix of the decimal representation of $\pi$; the language of programs that run for an even number of clock cycles; the empty language. – Can we use the information on the length of w? To demonstrate that the set INFINITEPDA is decidable, we must show that there exists an algorithm to determine whether a given Pushdown Automaton (PDA) accepts an infinite language. If you insist on the classical Turing Machine model, it can be done that way too, though less perspicaciously. The standard proof of this result works by constructing an enumerator for the Turing-recognizable language, then including the first enumerated string in the decidable language, then the first string that comes after it lexicographically, then the first string that In Michael Sipser's Introduction to the Theory of Computation, he states: "some languages are not decidable or even Turing recognizable, for the reason that there are uncountably many languages yet only countably many Turing machines. Showing that the language is decidable is the same thing as showing that the computational problem of testing acceptance is decidable 4 . Therefore the language is NOT Turing Recognizable and hence NOT RE. I have been given the following problem and was wondering if my solution is correct (taken from the textbook exercise in the book Introduction to the Theory of Computation by Martin Sipser): Given $$\text{INFINITE}_{\text{PDA}} =\{\langle M\rangle \mid M\text{ is a PDA }\text{and L(M) is infinite}\}$$. , as numerous as the natural numbers. 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 By definition, machine M accepts a string s if it reaches an accepting state when given input s. Does there exist a undecidable infinite language with only a finite undecidable subset? 1. In contrast to that, the situation for pushdown automata is much more "Demonstrates that the non-regular language L={0^n 1^n : n natural} had no infinite regular sublanguages. However, the If M accepts w, then this language is infinite and so the LBA will accept infinitely many inputs. On input < M >: 1. Follow Theorem $26D$ makes no real use of the finiteness of the language. You switched accounts on another tab or window. Step 1: Understanding Infinite Pushdown Automata (PDA) Languages Robb T. But that set could, in fact, be empty. Infinite languages cannot be enumerated in reverse order. You signed in with another tab or window. 1) infinite 2D table whose rows are Turing machines and whose columns are input strings. However, in some sense "most" languages $I$ is infinite, because for any index $i$ one can always find words longer than $w_i$ in $L$, since $L$ is infinite. If M does not accept w, then this language is empty. 1. The set of decidable languages is infinite. Footnote: You can find a proof of the standard theorem above in many places; Now, given an infinite string over this set of symbols it is only semi-decidable if the string belongs to this language. You got this! Solution. Show that INFINITEDFA is decidable. We could clearly construct a decider for Bby running M A TM on hM B;wi. g. 2(Recognizable Language). If the graph has a cycle, the language is infinite. i'm intending to do it by having the player describe their strategy in a decidable language, which is then incorporated into a larger program in that decidable . Question: Let INFINITE PDA = {<M>|M is a PDA and L(M) is an infinite language}. Show that INFINITE DFA is decidable. Viewed 2k times An undecidable language is necessarily infinite. Some infinite languages are decidable, some are not. (Hint: Go over the proof of the pumping lemma for context free languages and see what it implies if the language is also prefix-closed). Then, language (B-A) transforms to an undecidable, as its equivalent to the language which is empty. Without loss of generality, let the alphabet be $\Sigma=\{0,1\}$. The language L is thus not semi-decidable. Warm-Up: Some Decidable Languages Show that the following languages are decidable by describing (at a high level) an algorithm that decides them (see more in Sipser 4. Then, we have "natural" input encodings (grammars, Turing machines There are languages L that are not semi-decidable (and therefore undecidable as well). 7k 35 35 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 (2) All languages are countable; none are uncountable. Suppose regular expressions were allowed to be infinite. Commented Nov 20, 2013 at 17:23. dar_bur dar Does there exist a undecidable infinite language with only a finite undecidable subset? 0. Question: If a TM is given a string, what can happen to the computation? Answer: The machine can either ; Definition: Turing Decidable Language . 0. $\begingroup$ Was thinking if this simple argument is valid: "keep generating strings infinitely on given alphabet and make Turing Machine dovetail on these strings. , acceptance problem for DFAs of testing whether a DFA accepts a given string •A DFA = {<B,w> | B is a DFA that accepts w} •the Is the decision problem "Does a given context free grammar generate an infinite number of strings" decidable? {n+2}$, and check if the resulting grammar is non-empty. Therefore, an algorithm exists that can decide if the language of a DFA is infinite. See the Encyclopedia of Mathematics for more on recognizable and undecidable languages (specifically The existence of a non-deciding TM for a language A does not imply that the language is not Turing-decidable. The question stems from the fact that you can determine whether a regular language is empty by using a Turing machine to count the states n in the given FSM. For example, language equivalence and inclusion are decidable (see [9]), and for many subclasses of the regular languages it is decidable whether a given automaton accepts a lan-guage inside this subclass (see [19] for some results of this kind). The negative answer to decidable = non-contracting grammar? suggests the following question: Is there a decidable language that can be recognized only by a space unrestricted Turing Machine (i. This can be shown by reducing the halting problem to L: For the halting problem instance (N, y), create a new machine M for the L problem. The following language L is decidable. " input n Run Pe on 1n for n steps if Pe accept any number then enter an infinite loop and start accepting infinite number if pe does not accept any number then accept 1" Now if i can define program P like above then asking the question whether Language accepted by P belong to Linf will tell me whether Language accepted by Pe is null or not. L(M) is a set of strings. First, construct a deterministic finite automaton M3 whose accepted language is the intersection of L1 and L2 by using the Cartesian Product Machine construction - an algorithm which produces the desired machine. When you generate all strings from length 0 to n, if the machine accepts any of them then the language is non-empty. Closure properties for TM. decidable problem: has both counting (bijection with $\mathbb N $) and membership algorithm (TM halts for both member and non member strings ); semidecidable problem: has counting algorithm and TM halts for member strings; countable problem: has only counting algorithm; Now I have come across undecidable language set which does not have Find an infinite recognizable not decidable subset of an infinite decidable language. Similarly: (1) Not all mammals are people; 2 Recall: Recognizable vs. Engineering; Computer Science; Computer Science questions and answers; Problem 2 Show that an infinite language is decidable if and only if some enumerator enumerates the language in the standard string order (shortest strings first and alphabetically sorted among equal length). Also by definition, L(M), referred to as the "Language accepted by M", is the set of all strings accepted by M. I could not understand the rejection case in the last line in the second paragraph. Prove that determining if a PDA has an infinite language is decidable. Question: Prove that the following language is decidable. $\begingroup$ If valid words in a language define an infinite loop, that does not make it unrecognizable. Since a language Lis defined to be a subset of , the set of all Recall the definition of decidable and recognizable languages. But that doesn't say anything about being decidable or not. An example of a countably infinite language over the alphabet would be ${a}^*$ But do uncountably infinite languages even exist? Each item (string) Like countable languages, we can use Turing machines to define what it means for one of A language is decidable if there exists a Turing machine that accepts all strings in the language and rejects all strings not in the language. No infinite loops. This makes the property trivial. All regular languages are decidable. Otherwise not. In this question, we need to determine which of the given statements are decidable. I don't think diagonalization would be particularly useful to prove the existence of a non-regular subset. We don't have to check words with length > n because It says that a cyclic path is a necessary feature for a DFA that accepts an infinite language. Show transcribed image text. As we need to check infinite number of strings thus the language is NOT semi decidable (NOT RE) For L’ – We can apply non-monotonic property as the Lyes subset Lno (phi subset sigma*). The argument that increasing cardinality leads to infinite language is incoherent since it essentially conflates concepts of performance with notions of competence. Follow asked Nov 29, 2016 at 20:04. algorithmic properties. We also present (with proofs) several explicit examples of undecidable languages. • Obviously. Then, you provide a counterexample - a string of length greater than the pumping length p (more on what p is in a moment) - and conclude the assumption is false, that is, that the I came across this problem: Show that in every infinite computably enumerable set, there exists an infinite decidable set. Deciding whether the language of a TM has a trivial property is always decidable: if it's a property all languages have, then answer yes; if it's a property no languages have, answer no. The algorithm that you give, however, doesn't work, for two reasons: When the input is not in a language, you never find out. Improve this answer. So I'm stuck. Accept if F accepts; reject if F rejects. Here we show that the problem of checking whether a DFA's language is infinite is decidable. INFPDA={ A |A is PDA and L(A)=infinite language} Prove that this is decidable problem. Finding –L s iTuring-decidable if there is some TM that decides L. If L is an infinite language, then the number of subsets of L is uncountably infinite (since the power set 2^inf is uncountably infinite). So my idea how to solve this problem is the following: k = number of states of A, create finite automata D, which accepts all words which have length=k and more; Create context-free grammar G based on A; And now I am lost. Modified 5 years, 1 month ago. However, if the PDA accepts inputs of all lengths, then we know the language is infinite. Repeat until there are transitions: a. 11, is recognizable. Show that INFINIT EPDA is decidable. In other words, if A TM was decidable, then every Turing-recognizable language would also be decidable. Whether the intersection of two regular languages is infinite Since the universe of strings over any finite alphabet is countable, every language can be mapped to a subset of the natural numbers. If elements in S have an upper bound, then S is finite, and hence it is decidable since every finite set is decidable. Share. A language is prefix-closed if the prefix of any string in the language is also in the language. Follow answered Sep 4, 2019 at 4:19. This is because every sentence can only use finitely many symbols. If every subset of L were a regular language, then there would only be countably many subsets of For instance the language of lists of integers in increasing orders is not finite, but it is easy to design an algorithm testing whether a list is sorted in increasing order, so this language is decidable. Every infinite language L has a subset S which is undecidable. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Here’s how to approach this question. Viewed 1k times 2 $\begingroup$ What is an example of } Some languages are Turing-decidable A Turing Machine will halt on all inputs (either accepting or rejecting). Show that INFINITEPDA is decidable. For example, take the regular language a*. If it is, the original grammar generated an infinite language, and otherwise - a finite language. Commented Nov 2, 2015 at 22:40 $\begingroup$ However, one could enumerate the language of M so it will always accept, meaning that it has an infinite language, a contradiction. (It is also decidable, but that is harder to show, do it if you like. $\endgroup$ – Davislor. Conclusion – Both L and L’ are NOT RE. $\begingroup$ Finite language are decidable, period, end of story. (You can get a) INFINITE DFA is decidable. I N F I N I T E P D A = M | M i s a P D A a n d L M i s a n i n f i n i t e l a n g u a g e. This question appears to be off-topic because it is about computability theory prove all finite languages are regular. However, since the set of all languages is uncountably infinite and the set of all Turing machines is countably infinite, there are more languages than Turing machines. Infinite u decidable languages. One trivial example is the language {n € N | a^n} , i. Languages recognized by a TM are called recognizable. I can't think of a turing machine that would decide this. $\endgroup$ Proof that whether a Such the set of Turing-decidable languages is an infinite subset of the set of Turing-recognizable languages, the set of Turing-decidable languages is also countably infinite. Show that every infinite prefix-closed context free language contains an infinite regular subset. Step 1. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). We need to find a string end with final state of DFA or not ? No problem here . You signed out in another tab or window. This is a good exercise. But, at any point in time, when TM has rejected some number of strings, we will never know if it will accept any string in future, making it impossible to say its language is Let INFINITE DFA = {〈A〉| A is a DFA and L(A) is an infinite language}. e. The language accepted by a DFA M with n states is infinite if and only if M accepts a string of length k, where n≤k<2n. Therefore, if a TM could decide whether the LBA had an infinite language, it could decide whether M accepts w, contradicting that this is impossible. Answer to Is the following language decidable or. The question whether decidable languages evolve into infinite language is vacuous since infinite generation is the null hypothesis for a generative procedure. Modified 7 years, 1 month ago. How to build a DFA that recognizes a language. Since {0, 1}* is theoretically infinite does mean a turing machine may never halt thus the language is undecidable? If so A and B are DFAs and L(A)=L(B)} (we know that EQ_{DFA} is a decidable language). So you just have to take a Recursively enumerable language wich is not decidable and map it into a subset of $\{1\}^*$. Cite. (A word of general caution, though it doesn't apply here. The Halting Problem - Undecidable Languages Robb T. On input x, M simulates (N, y) for length(x) steps. Run M on w. Note that if a language of any number of 1s is unrecognizable, then it is Let M1 and M2 be minimal deterministic finite automata whose accepted languages are L1 and L2, respectively. Can you say a language that isn't recognizable as undecidable? If given a language like We can combine machines for language one and the complement of language 2 in a single no deterministic Turing machines that is a decider. That is, a decider T is guaranteed to either accept, or reject, and never fall into an infinite loop. Definition 1. Since there are uncountably many languages (because By closure property of regular languages, regular language is not closed under infinite union so is the above . Explanation: for statements S 2 consider the following explanation. A table entry is A (accept), R (reject), The set Σ* is a countable set, so all its subsets are countable. First, consider the set of all languages. I. Is the Language of all encodings of Turing Machine that at least halts on one input and outputs 0 I know that. This is true since given a decider of Why is a subset of a undecidable language decidable? Ask Question Asked 5 years, 3 months ago. Follow answered Jun 4, 2021 at 14:11. Question: Let INFINITEPDA = {(M) M is a PDA and L(M) is an infinite language). } Some languages are Turing-recognizable, but not decidable. Reload to refresh your session. Consider the problem of determining whether a DFA and a regular expression are equivalent. • But the other direction does not hold---there are languages that are Turing-recognizable but not Turing-decidable. I'm confused on the definition of undecidable languages. Languages decided by a TM are called decidable. In conclusion, decidable and undecidable problems highlight the boundaries of what computers can and cannot solve. No, there are many infinite languages that are decidable. (Hint: Use the equivalences between recognizers and enumerators, and deciders and ordered enumerators). The set $\{w_i: i\in I\}$ is also decidable. Is the problem of deciding whether a Context free grammar generates exactly K strings is decidable? 4. A Turing Machine recognizes the language, but it will loop infinitely on some inputs } Some languages are not Turing-recognizable Just think about that the universal set of input is a decidable language, but there are infinite subsets of it are undecidable Share. If none of these are accepted by the DFA, then the DFA accepts no strings and the language is empty. That is, there is no TM which will tell you if an infinite string contains no $1$ s, but a TM can tell you if a string does contain a $1$ . L ∈ R iff L is decidable But because there are an infinite number of words, this means that we would have to pass an input to M an infinite number of times. If a language L and its complement L are both semi-decidable, then L is decidable. Koether Homework Review Universal Turing Machines The Acceptance Problem for Turing Machines INFINITEPDA = fhMijM is a PDA and L(M) is an infinite languageg: Show that INFINITEPDA is decidable. Here’s the best way to solve it. We say a L language is recognizable (L ∈ L Every infinite computably enumerable set has an infinite computable subset. c. Let INFINITEPDA = {(P)|P is a PDA and L(P) is an infinite language}. i'm trying to make a game in which the player faces an infinite (finitely specified) series of enemies and has to specify a strategy that provably defeats all of them (ie defeats enemy n in finite time for all n). Try focusing on one step at a time. , one can pursue an approach similar to that used in showing the decidability of EDPA; another is to consider other formalisms for describing the language INF_DFA. Iff you end up with a DFA that has a single non-accepting state, the language is empty. Let E be an enumerator for it. Decidability is a property that refers to whether a problem can be solved by an algorithm. 31. Languages (1) and (2) are, respectively, {0, 1}* and the empty language, both of which are decidable (so there are TMs that always halt that accept those languages). (The lists we can generate by computer are the decidable languages. If the PDA accepts an input of length n, but not an input of length n+1, then we know the language is finite. Since the possible Regular expressions for S are 0*, 1*, (1+0)* and (0o1)*. Otherwise, $L(D)$ is finite. I know that the language of encodings of DFAs such that the language accepted by the DFA is infinite is decidable. Decidable problems have solutions that can always be found by an algorithm, making them predictable and useful in everyday computing tasks. Skip to main content. WewanttobuildaTMM thatdecidesA REX. Question: Let INFINITEPDA = {< M >M is a PDA and L(M) is an infinite language). Solution. C = A-B = A, is Turing recognizable. It is the set {0, a, aa, aaa, } which is clearly an infinite set (0 = the empty string). Every infinite language L has a subset S which is unrecognizable. Express this problem as a language and show that it is decidable. Case 2: Intersection of A and B is either of them (Lets say A-B = A). ‣Some strings not in L may cause the TM to loop ‣Turing recognizable = recursively enumerable (RE) • A language L is Turing decidable if some Turing machine decides it ‣To decide is to return a definitive answer; the TM must halt on all inputs A language is a collection of strings--assuming we have an infinite number of words then concatenating this collection (keeping things simple, normally you need to at least be able to tell where one word ends and another begins) and viewing the language as the resulting single big string, if the Kolmogorov complexity of this string is infinite can the language still be These languages can be decided by an algorithm that always halts and correctly determines whether a given input string belongs to the language or not. Build a Turing machine that will do the following. 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 Are you asking for a class of languages, contained in CSL but containing CFL, for finiteness is decidable? $\endgroup$ Therefore, determining whether a given context-sensitive grammar generates a finite or infinite language is undecidable. Related. I don't know how to do this; I have certain ideas. However, a recognizable language may or may not be decidable. 1 Show that single-tape TMs that cannot write on the portion of the tape containing the input string recognizes only regular languages. Koether Homework Review Universal Turing Machines The Acceptance 1 Semi-decidable vs. If the complement's language is empty, then the original DFA recognizes an infinite language. Taking the contrapositive of this statement gives you that all undecidable (non And every finite language is decidable. By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not in the language. Its property of being recognized means the program containing an infinite loop can be confirmed as a valid program in a finite amount of time and always accepts. Why is a subset of a undecidable language decidable? 1. However, if both L and its complement L are semi-decidable then L must in fact be decidable. Therefore, there are infinitely many infinite languages for which no TM exists. ;) $\endgroup$ – Raphael. Mark current state and It is not finite. Really the only way complexity can enter the picture is if the language we're looking at is itself of high complexity, in which case the set of sentences in that language itself (not even thinking about the set of sentences in that As a result, we end up showing that the language is decidable but we do not (and cannot) algorithmically construct the decider for the language from the enumerator for the language. Decidable Languages 4 •decidable problems concerning regular languages •computational problems concerning finite automata •represent computational problems through languages •already have a framework and terminology •e. LetR betheTMthatdecidesA NFA andbuild M “Oninput–R,w‰, 1 ConverttheregularexpressionR toanequivalentNFAN usingthestandard construction 2 RunR on–N,w‰. By In a finite language there will be a maximal length of any string in the language -- call it n n. Commented Aug 10, 2019 at 8:56 The language INFINITEDFA is decidable because we can construct the complement of a given DFA and check if its language is empty. Proof. TM was decidable, but that some other undecidable language Bwas Turing-recognizable. • The classes of Turing-recognizable and Turing-decidable languages are different. And it is an infinite language. Every infinite recognizable language has an infinite decidable subset. Question: = Let INFINITEDFA = {(A)|A is a DFA and L(A) is an infinite language}. Your language L is indeed undecidable. We could modify the recognizer to reject strings whose prefix matches the shortest strings on If you can show that S goes to something and itself ex: S->bS then your language is infinite. Theorem. So, no, not all non-recursive languages are infinite – twalberg. • We can show that B is uncountable by using a proof by diagonalization similar to the one we used to show that R is uncountable. So is $0^*$ a decidable language? And if so, why? Can you guys list the problems that you are aware that are decidable for Context free Language and for Deterministic Context free languages. the correct answer is option 4. To prove the result, we simply observe that the set of all languages is uncountable whereas the set of semi-decidable languages is countable. " Share. Show that of Turing decidable languages is closed under concatenation. Although it might take a staggeringly long time, M will eventually accept or reject w. Is the following language with a finite bound Turing Recognizable? 1. Are languages with an infinite amount of only finite strings always decidable? Given that it is a regular expression it is also a decidable language, however I don't quite understand how a Turing Machine can actually decide on such a language. Which is something that the accepted answer implies (if you have a think about why the theorem actually works), but never actually states. Solution:Idea: Let L be any infinite recognizable language. Show that INFINITE PDA is decidable. $\endgroup$ – BHK. Follow T decides a language L if T recognizes L, and halts in all inputs. fpfisjs adtjf pmleae wmv mnzbk xxkg eupxer heri bcii bjaiss