Show that alldfa is decidable. Oct 6, 2014 · Oct 6, 2014 at 2:30. Here’s the best way to Consider the problem of determining whether a Turing machine M on an input w ever attempts to move its head left at any point during its computation on w. Jan 5, 2023 · This is an example of a problem that is not even partially decidable. Let ALLDFA = cfw_| A is a DFA and L (A) = . As per ch …. 3 Some Undecidable Problems are Partially Decidable. A language is undecidable if it is not decidable. A Turing Machine decides a language if it rejects every string it doesn’t accept – i. There are 2 steps to solve this one. 5. The P = NP assumption implies that SAT is in P, so testing satisfiability is solvable in polynomial time. </a>. Sep 2, 2020 · Show that A"CFG is decidable. P converts R into a NFA called NFAR. So …. About reduction: for as long as you can reduce this problem to some other decidable problem, its fine. Jun 2, 2018 · Showing the following language is decidable 1 Over the alphabet {a,b,c,d}, how would i construct a NFA that only accepts strings that end with a letter that is already part of the string? Mar 6, 2022 · Consider the problem of determining whether a pushdown automaton has any useless states. Calculate the appreciation of the U. (b) There are at most s(x+ 2)2 distinct con gurations of the 2DFA. DECIDABILITY AND UNDECIDABILITY. Nov 20, 2022 · Language formulation of the problem and lexicographic order. EQDFA={(A,B)|A and B are DFAs and L(A)=L(B)} Proof idea. Repeat until there are transitions: a. Previous question Next question. ) (Further Hint: First construct a DFA B that accepts £* and then compare B to A) </a> Jun 11, 2020 · Prove that some decidable language D is not decided by any decider Mi whose description appears in A. Answer: Define the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. , they accept Σ∗ ). S. A language (or problem) is in P if there is some Turing machine that decides it in a polynomial number of steps. Answer. For simple machine models, such as nite automata or pushdown automata, many decision problems are solvable. The recursive languages = the set of all languages that are decided by some Turing Machine = all languages described by a non-looping TM. Simulate M1 on w. That means that NFA can not be any more powerful than a DFA. Show that the set of Turing-decidable languages is closed under the following operations. ) Solution (a) Easy. Show that Our expert help has broken down your problem into an easy-to-learn solution you can count on. Show that INFINITEdfa = {<M> : M is a DFA and L (M) is infinite} is decidable. Show that ALLDFA is decidable Your solution’s ready to go! Enhanced with AI, our expert help has broken down your problem into an easy-to-learn solution you can count on. Computable functions may be transformations of TM descriptions: a computable function f may take as an input 〈M〉 and return 〈M’ 〉 where M and M’ are TMs. Let ALL DFA = {〈A〉| A is a DFA and L(A) = Σ*}. The set R is the set of all decidable languages. Find an equation of a sphere with radius r and center C(h, k, l). by BFS from the initial state. The alphabet could consist of the symbols we normally use for communication, such as the ASCII characters on a keyboard, including spaces and punctuation marks. Possibility 1: We bring !’s difficulty down to "’s difficulty. Run a machine that decides E DFA on D. A prunable state in a DFA is some state that is never entered while processing any input string. HALT), and B be . dollar and the Japanese yen went from 110 /US to 120 /US. Also there are ffit parsing algorithms for context Apr 13, 2021 · $\begingroup$ Please don't delete your post after receiving an answer. Question: 3. 2. Problem 3 (35 points) Show that ALL DFA is decidable. The algorithm you are describing shows that the problem of testing whether a CFG generates some string from $1^∗$ is decidable (e. 1. If this machine accepts D, Accept R Reject R" However, this is not true of Turing-decidable languages that partition encodings of DFA or NFAs based on properties of the machines Dec 4, 2020 · Here we show that the A_DFA problem is decidable, and introduce "high-level descriptions" of problems. Thanks for the suggestion, I have just posted the same in Mathematics stackexchange. If no other transition exists, reject Can this be considered a valid solution? Show that A 2DFA is decidable. Enumerate all possible strings Nov 20, 2019 · Prerequisite - Undecidability, Decidable and undecidable problems Identifying languages (or problems*) as decidable, undecidable or partially decidable is a very common question in GATE. ) Let ALLDFA = { (A) | A is a DFA and L (A) = L*} Show that ALLDFA is decidable. 14 b,d,e) Show that the collection of decidable languages is closed under the operations of concatenation, complementation, intersection. Show that ALL DFA is decidable. Unlock. Using this definition: If L1 and L2 are decidable, then algorithms (or Turing machines) M1 and M2 exist, so that: M1 accepts all Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Show that Aε CFG is decidable. This is hard: requires reasoning about all possible TMs. The proof must be self-contained, i. Question: Let DFAALL be the the language of encodings of DFAs which accept all input strings (i. dollar and the depreciation of the Japanese yen if the exchange rate between the U. Formulate this problem as a language and show that it is decidable. Let M be a dfa. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Example: The sets A = { mouse, cat, dog }, B ={ cheese, mouse, bone } are of same size because there is an obvious one-to-one and onto function f: A → B. Then it's easy to decide the property, e. I want to show that B B is decidable. Jun 27, 2020 · Show the projection of decidable language is Turing-recognizable. Convert Rto an equivalent DFA Dusing the method discussed in class 2. However, there are arbitrarily inefficient algorithms to determine The following TM U recognizes A. Jan 9, 2021 · 3. (5 points) -. Problem 2 (Exercise 4. L ∈ R iff L is decidable Let ALLDFA = {(A〉 | A is a DFA and L(A) = Σ*), where(A) is a string that describes A. Run EQDFA (A, B). Computable Transformations. The following is my solution: On input <M>: 1. I am not able prove how first two properties are decidable or by which algorithms we can determine same. , for the case that you use TMs from the lecture then these must be presented, too. Show that SINGLEDFA is decidable. I assume there are a wide array of possible ways to show decidability. We showed they have the same expressive power as an DFA is a degenerate form of an NFA. A decidable language • To show that a language is decidable, we have to describe an algorithm that decides it ‣We’ll allow informal descriptions as long as we are confident they can in principle be turned into TMs • Consider ADFA = { M,w ⃒M is a DFA that accepts w } • Algorithm: Check that M is a valid encoding; if not reject. Mark current state and delete the transition that moved you to the state b. Jun 8, 2021 · 2 Answers. Show that ALLDFA is How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5 Proofs of Undecidability To prove a language is undecidable , need to show there is no Turing Machine that can decide the language. Rudra, CSE322 2 Announcements Handouts Sample final List of topics for the finals H/W #8 Remember your lowest H/W grade will be dropped Turn in your H/W #7 Pick up graded H/W #6 at end of class A. Theory Of Computation. EQDFA is a Decidable Language. Becase if a language L is in NP, than there is a nondeterministic Turing Machine that decides it in polynomial time, and thus L is decidable. 3. , you can assume that EQDFA is decidable. In this way any story can be regarded as a "word". Show that if P = NP, a polynomial time algorithm exists that produces a satisfying assignment whengiven a satisfiable Boolean formula. In the case of deterministic nite automata, problems like equivalence can be solved even in polynomial time. That is, construct a Turing machine that decides ALLDFA Give a short argument why your TM always halts. We would like to show you a description here but the site won’t allow us. Oct 27, 2016 · The answer would be yes, but showing how the problem is decidable? I am not sure, I think I should list a series of steps involving multiple FSMs but I feel like I should get an idea of how others would solve this problem first. Rice's theorem shows only when a language is not decidable, hence it can't be used. ALLDFA = { <a> | A is a DFA that recognizes £* } (Hint: reduce ALLDFA to EQDFA, where EQDFA = { | D, and D2 are DFAs and L(D) = L(D2) }, i. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Show that A is decidable. The set S I N G L E D F A is decidable because you can determine with a finite procedure whether a given D F A ( M) satisfies t View the full answer Step 2. Quick review of today. ) Let ALL DFA = { is DFA and L(A) = E * } Show that ALL DFA is decidable. Defined mapping reducibility as a type of reducibility. The idea is to simulate the DFA by keeping track of th Closure Properties of Decidable Languages Decidable languages are closed under ∪, °, *, ∩, and complement Example: Closure under ∪ Need to show that union of 2 decidable L’s is also decidable Let M1 be a decider for L1 and M2 a decider for L2 A decider M for L1 ∪L2: On input w: 1. 4) Let Aε CFG = {<G> | G is a CFG that generates ε}. Let A= {<M>| M is a DFA which doesn’t accept any string containing odd numbers of 1s}. Anyway, the problem of testing whether a CFG generates all the strings of the language $1^∗$ is truly decidable. 15 b,c,d) Show that the collection of Turing-recognizable languages is closed under the operations of concatenation, star, intersection. That's as far as I can really go, because this has to do with an intro class I'm taking. Given y, how do you fix n? – $\endgroup$ – Show that the following problem is not decidable: For two context-free grammars G 1 and G 2, is L(G 1) 1 L(G 2 ) infinite? 14. To show that it is decidable if L (M) contains s1 and s2 where s1 is subset of s2, s1 is not empty and s1 is not equal to s2. If M ever enters its accept state, accept. Convert NFA B to an equivalent DFA C 2. Question: Let ALLDFA = { (A〉| A is a DFA and L(A) = Σ*). A and B are DFAs and L(A) C Problem 4 (5 points). Show transcribed image text. A recognizer of a language is a machine that recognizes that language. To decide `alldfa`, we can follow the This doesn’t show that L is decidable since the de nition of decidability requires the same TM to accept all strings in L and reject all strings not in L, not two di erent TMs. Click the card to flip 👆. Moreover, A ∪ B A ∪ B and A ∩ B A ∩ B are decidable. Our expert help has broken down your problem into an easy-to-learn solution you can count on. [lOpt] 1. We know X does Nov 9, 2021 · In the second of these three paragraphs, it seems that the point is to show that a decider for encodings <x,y> exists. = “on input <M,w>, where M is a TM and w is a string: Simulate M on input w. Example: M’ is a TM that recognizes the same language M but never attempts to move its head off the left-hand end of its tape. CS4330,Homework7SampleSolutions(Fall2020) Problem 4. [15 points] Solution:If A is decidable by some TM M, the enumerator operates by generating the strings in lexicographic order, testing each in turn for membership in A using M, and printing the string if it is in A. Two (possibly infinite!) sets A and B are of the same size if there is a one-to-one and onto function f: A → B. 1 Question: Show (by construction) that language ALLDFA = { (A)|A is a DFA and L (A) = E'} is decidable. Feb 13, 2022 · 2. Decidable problems from language theory. Express this problem as a language and show that it is decidable. So make a DFA Mp that accepts all strings over the same Oct 15, 2020 · We showed that an NFA can be reduced (i. TM . Examples: 2. Show that SUB_DFA is decidable. , 3⁄4w " Σ , M either accepts or rejects w) Such a TM is called a decider The acceptance problem for a type of machine1asks whether a given machine of the specified type accepts a string. A regular expression or an NFA can be converted to an equivalent DFA C. Checking if M M accepts a balanced string amounts to checking if L(M) ∩ L(N) L ( M) ∩ L ( N) is the empty language. These are also called theTuring-decidable or decidable languages. Let ALL DFA = {〈A〉| A is a DFA and L (A) = Σ*}. 13 Let A = { (R, S)| R and S are regular expressions and L (R) C L (S)}. If M1 accepts, then ACCEPT w. Reduction fromNFA toDFA. That is, you can show that L≤p L′ L ≤ p L ′ for some decidable L′ L ′ and this Homework 8Solutions. Definition. whether two DFAs recognize the same language. A Decider also halts if the string is not in the language. Nov 26, 2021 · See Answer. Note, this machine loops on input <M,w> if M loops on w. e. Part of our mission is to build an archive of high-quality questions and answers that will be useful not only to the asker, but also to others in the future. If this machine accepts D, Accept R Reject R" However, this is not true of Turing-decidable languages that partition encodings of DFA or NFAs based on properties of the machines View Homework Help - Discrete Homework 4 from COT 4210 at University of Central Florida. !TM is undecidable. (a) union (b) complementation (c) intersection (d) concatenation (e) star Hint: Consider constructions similar to what we The following are also decidable: • Whether a regular expression generates a string w • Whether an NFA generates a string w . Let B be the set of all infinite sequences over {0, 1, 2}. Dec 23, 2020 · $\begingroup$ Is about decidable languages or regular languages? If it's decidable languages then the solution is much more simple then what you're thinking about, you don't really need to modify the direction of anything in the state machine, just add some pre-processing. , L M A), and. Example. Let SUBSETDFA-A, B L(B)). Possibility 2: We bring "’s difficulty up to !’s difficulty. Although it might take a staggeringly long time, M will eventually accept or reject w. whether a DFA accepts all strings. Show that LEN_CFG is Or, what issues can there be? "On input <A> a DFA Construct DFA B which doesn't accept any string with odd numbers of 1s. (c) Let A be any undecidable language (e. Given input (R , S) , where R and S are regular ex …. T= "On input , where A is a DFA: 1. Show that the following language Decidable Languages 8 •Theorem: A NFA is a decidable language •proof •show a TM N that decides A NFA •could design N like M for DFAs, but for NFAs •instead, we’ll convert N to a DFA first, then use M as a subroutine N = “On input <B,w>, where B is a NFA and w is a string: 1. But it's known that emptiness of context-free languages is decidable. Note: there is no requirement that the Turing Machine should halt for strings not in the language. So, for example, determining whether a graph is connected is in P, since a TM can determine that efficiently. Firstly, I would like to know if the following approach works: First, we can check if w ∈ A ∪ B w ∈ A ∪ B. Following is the algorithm that I thought about to prove that A is decidable. (Exercise 3. Mar 12, 2021 · It helps a lot for me! Lastly, to sum up, if a language is finite, then we say it is decidable and if and only if some enumerator enumerates L in decreasing order. Construct a DFA C from A and B, where C accepts only those strings accepted by either A or B but not both (symmetric difference) If A and B accept the same language, then C will accept nothing and we can use the previous proof (for EDFA) to check for this. And I would use pumping lemma. Let a language be any set of strings (or words) over a given finite alphabet. , see here at page 21). Let ALLDFA { (A) | A is a DFA and L (A) = {*}. A regular expression is a DFA B. Show that ALLdka is Turing Decidable. $\endgroup$ – Show that A it is decidible. Show that a) ALLDFA is decidable, and that b) ALLDFA is in P. With correct knowledge and ample experience, this question becomes very easy to solve. If M ever enters its reject state, reject. By clicking “Post Your Answer”, you agree to our and acknowledge you have read our . $\begingroup$ I can just about understand your question but the way you've written it suggests to me that you're rather confused about the basic concepts. Apr 7, 2022 · Prove that $\text{INFINITE}_{\text{PDA}}$ is decidable. To show that the language S I N G L E D F A is decidable, we need to demonstrate the existence of a Turing machine th Let SINGLEDFA = {<M>M is a DFA and for all s € L (M), [s] =1}. hope to see some ideas there. Question: Problem 2 (Exercise 4. 2. 3 LetALL DFA= fhAijAisaDFAandL(A)= g WanttoshowthatALL DFA isdecidable. Show transcribed 355 problem setups. 23. Here’s the best way to solve it. Run M on w. 0. Oct 31, 2017 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Show that an NFA accepts at least one word if and only if there is a (directed) path from the initial to an accepting state. The idea is to simulate the DFA by keeping track of th Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. DFAALL= {<a>∣A is a DFA and L (A)=Σ∗} Show that DFAALL is decidable. An undecidable language ma Aug 29, 2016 · E (dfa) is a decidable language. $\endgroup$ Question: Show that the following language is decidable by constructing a decider for it. (Hint: if the 2DFA has sstates and input size is x, compute an upper bound on the number of steps to simulate for acceptance. 10) Let A= {<M> | M is a DFA which doesn't accept any string containing an odd number of 1s}. 2 (Decidable Languages) Show that the following languages are decidable: (a) EQ DFA RE = fhD;Ri j Dis a DFA and Ris a regular expression and L(D) = L(R)g Solution: We know from the lecture that EQ DFA = fhA;Bi j Aand Bare DFAs and L(A) = L(B)g is decidable. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Lecture 17: Proving Undecidability 6 Proof by Reduction 1. Let A A and B B be semi decidable languages. . Oct 6, 2014 at 2:52. 4. Rudra, CSE322 3 Please remember… I want to show you the “cool” stuff There are problems that are Show that Finite Regex is decidable. Unless otherwise noted, the alphabet for all questions below is assumed to be = {0,1}. Problem 3 (5 points). halts on every input (i. If the automata would be DFA then I would do: L (M) =L (A)∩L (D) then I would check whether L (M) = ∅ using turing machine (tm) for emptiness. Nov 25, 2015 · For L1 L 1 to prove that it is a decidable language, all that I need to show is that L(R) ∩ L(S) = L(R) L ( R) ∩ L ( S) = L ( R), and since regular expressions are finite automatons basically, I already have a proven decider that decides whether two DFAs accept the same, so this can be shown. Decidability and Semi-Decidability. Show that ALLDFA is decidable. Individual exercises 1. (Sipser, Problem 3. The comment says that D is decidable and, yes, I can see that, but it is not a decider for encodings <x,y>, it is a decider for encodings <x,n> where n is a natural number. 6. Recognizers and Deciders. Every infinite Turing-recognizable language has an infinite decidable subset. ALLDFA is decidable because given the DFA A, i …. Dec 11, 2015 · $\begingroup$ If a DFA accepts a language, than that language is decidable because regular languages exist solely inside the venn diagram of decidable languages. Let LEN_CFG = { | G is a CFG, k elementof Z^noneg, and L(G) Intersection sigma^k notequalto (where sigma is the alphabet of G)}. Show that L 1 is decidable. We construct a Here’s the best way to solve it. Note: AlanguageLisTuringDecidable(orjustdecidableinshort)ifthereexistsahaltingTM Jul 10, 2020 · Maybe you're confusing two different problems. A recursive language is a formal language for which there exists a Turing machine that, when presented with any finite input string, halts and accepts if the string is in the language, and halts and rejects otherwise. Show that SUBSETDFA is decidable. Every language that is in NP is by definition decidable. Let ALLDFA = {〈A〉| A is a DFA and L (A) = Σ*}. It sends the pair to a TM N that acts as a subrotine of it. g. </a> This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. So my idea how to solve this problem is the following: And now I am lost. Show that B is uncountable, using a proof by diagonalization. ) Let ALLDFA = { (A) | A is a DFA and L (A) = L*} Show that ALLDFA is Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Exercise 7. Run TM M on P is a class of languages, not a class of algorithms. Show that LEN_CFG is decidable. Question: 4. For this, sense Show that ALLDFA is decidable. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). Given M M, we can easily construct the machine which recognizes L(M) ∩ L(N) L ( M) ∩ L ( N) and Oct 12, 2015 · Proving that DFA equivalence is decidable. Show that Finite Regex is decidable. Hence if it accepts, it will inside that many steps. To prove that `alldfa` is decidable, we need to show that there exists a Turing machine that can determine whether a given input is a description of a DFA `a` and whether the language of `a`, denoted as `l(a)`, is equal to the set of all possible strings, `σ∗`. Your task is to prove that the language is decidable. ['Opt] Let S = {(M) I M is a DFA that accepts wR whenever it accepts w}. Wikipedia : Because DFAs can be reduced to a canonical form (minimal DFAs), there are also efficient algorithms to determine: whether a DFA accepts any strings. , converted) to a DFA via a set of simple steps. 31 in Sipser. Next we will show that there is an intermediate level between decidable and no-turing-machine exists - the partially solvable or RE problems that are not decidable. Here is my proposed solution: Mar 8, 2011 · A language is Recognizable iff there is a Turing Machine which will halt and accept only the strings in that language and for strings not in the language, the TM either rejects, or does not halt at all. Let LEN_CFG = {<G, k> | G is a CFG, k elementof Z^noneg, and L(G) Intersection sigma^k notequalto (where sigma is the alphabet of G)}. If the algorithm had some way to determine that M was not halting on w, it could reject. Transcribed image text: Let SINGLEDFA = {< M > M is a DFA and for alls e L (M), [s= 1} To do: a. For example, you say "I have been given a DFA = {< A >|A is a DFA and L(A) = {0, 1}∗}" but that isn't a DFA: it's a language (a set of strings). Then B is decidable, and A is a subset of B. – Deepak. Oct 6, 2019 · Then L(M) ∩ L(N) L ( M) ∩ L ( N) is context-free. There’s just one step to solve this. Regular expressions and NFA’s are finite D. , it never loops. Step 1. (Hint: You may find it helpful to consider an enumerator for A. L 1 = {< M,w > | M is a TM and input w causes M to move its head to the blank portion of the tape}. Consider the following language L 1. Based only on the reduction, the NFA could be less powerful. Then if the DFA accepts any string of length ≥ p, then the language is infinite; if not, the language is finite. M= \On input hRi: 1. Consider the problem of determining if a DFA accepts any string of length 3. Give lexicographic-order enumerator: is T-decidable ⇔∃ an enumerator such that prints all of the strings in in lexicographic order. 3 in the textbook. 1 Dec 4, 2020 · Here we show that the A_DFA problem is decidable, and introduce "high-level descriptions" of problems. Since this is decidable, we know that either w ∈ A w ∈ A or w ∈ B w ∈ B. Nov 20, 2022 · Hello, as a basic introduction to decidability, I have been asked to think about these two questions and would like your help: E * = Sigma Star e = epsilon 1. Share Apr 29, 2014 · 3. (Exercise 4. Show that Aε CFG is decidable. Enumerate all possible strings show that in every infinite computably enumerable set, there exists an infinite decidable set 1 Proof that languages are Turing-recognizable iff computably-enumerable Decidability. We can conclude the above problems are also decidable because: A. Definition: A language is called decidable Oct 28, 2020 · INFPDA={ A |A is PDA and L(A)=infinite language} Prove that this is decidable problem. Show that for every context-sensitive language L, the problem “L=Ø” is not decidable. Answer: Create a turing machine P, this machine receives a pair where R is the regular expression in question and W is the set of all the strings of the form x111y. Introduced The Reducibility Method to prove undecidability and T-unrecognizability. recognizes A (i. On the other hand, if a language is infinite, no matter what that language is decidable or not, there is no enumerator that enumerates it in decreasing order. Show that ALLDFA S decidable. If EQDFA. Show that S is decidable. 18) Show that a language is decidable iff some enumerator enumerates the language in lexicographic order. Let ALLDFA-{(A): A is a DFA and L(A) = Σ*). Travis Montey COT 4210 Homework #4 1. Hint: You might want to use Problem 1. A decider of a language is a machine that decides that language. If an already marked state is reached, accept 2. ) My qualm about this is that the question seems to imply finding a decidable language, the decider for which is not in the set of all deciders, which goes against the definition of Decidable languages Atri Rudra May 26 A. (Note: The algorithm you are asked to provide computes afunction; but NP contains languages, not functions. I feel that I have a solution, but at the same time the solution makes no reference to the fact that the machine in question is a PDA, which makes me suspicious. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. Let Mbe a Turing machine that decides this language. A language is Decidable iff there is a Lecture 18 – Decidable languages. Let LEN_CFG = {(G, k) | G is a CFG, k elementof Z^noneg and L(G) sigma^k notequalto 0 (where sigma is the alphabet of G)}. May 30, 2023 · The language `alldfa` is decidable. Let p be the number of states in the DFA. Both types of machine halt in the Accept state on strings that are in the language. Algorithm for printing all accepted strings (to an upper bound if infinite amount) from a given DFA. sk vt yj ao qj vb rt hj oo bs