Show that alldfa is decidable Otherwise, A is called undecidable. Show that ALLDFA is decidable. We know that whether PDA has an empty language is decidable, and we may reduce the ques-tion of whether a given state q is useless to this question by making q the only accept state and then determining whether the resulting PDA has an empty language. Show that the problem of determining whether a CFG generates all strings in 1* is decidable. 3. 3) Let A = {⟨R, S⟩| R and S are regular expressions and L (R) ⊆ L (S)}. e. To show that a language is decidable, we need to create a Turing machine which will halt on any input string from the language's alphabet. In other words, show that $\ {\langle G\rangle | G$ is a CFG over {0, 1} and $1^ {*} \subseteq L (G)$} is a decidable language. Thus ALLDFA is decidable. Using CFGs in Chomsky normal form circumvents all these difficulties. 2 AllDFA Let AllDFA = fhDi j D is a DFA and L(D) = g. ) Consider the problem of determining whether a DFA and a regular expression are equivalent. called a reduction. Kolmogorov Complexity. Why study decidability? The following two key theorems show the usefulness of this new framework. Question: (20 points) Let language ALLDFA = { (A): A is a DFA and L (A) = 2*}. Step 1 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 Aug 10, 2023 · To show that INFINITEPDA is decidable, we need to prove that there exists an algorithm that can determine whether a given PDA recognizes an infinite language or not. I have exam right now! For your answer, give an explicit counterexample and briefly explain why it is a counterexample. Engineering Computer Science Computer Science questions and answers Problem 2 (Exercise 4. This fact is used for subset checking by leveraging the equivalence L (R) ? L (S) if and only if L (R) ? (complement L (S)) is empty. 400 Let L be the set of codes for TM's that never make a move left on any input. Submitted by David C. The amount of money to be spent. for emptiness testing for CFGs, we can show the problem is decidable ECFG = {<G> | G is a CFG and L(G) = Ø} Theorem: ECFG is a decidable language proof idea could use TM S that states whether a CFG generates a particular w to determine if L(G) = Ø, we could try generating all possible w’s, one by one, but infinite number of w’s The proof is given in the below: If $A$ is decidable, the enumerator operates by generating the strings in lexicographic order and testing each in turn for membership 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. Oct 26, 2020 · For sure you can effectively tests if a string codes a CFG (of course, you have to know how this code is built, but you can assume that the grammar is given in some "normal" form). Please help me fast as possible.  Show that the Problem 2 Let |):} is a DFA that accepts w R whenever it accepts {: w} Show that S is decidable. Assume to the contrary that EQCFG is decidable. In addition, give a short proof (2-3 lines are sufficient) if the statement is true, and give a counterexample otherwise. Oct 28, 2020 · I have following problem: INFPDA={ A |A is PDA and L(A)=infinite language} Prove that this is decidable problem. Theorem If A ≤ m B and B is decidable, then A is decidable.  Let L1 and L2 be decidable languages. If not Reductions A better option: to prove NEW is decidable, show how to transform it (e ectively) into a known decidable problem OLD so that solution to OLD can be used to solve NEW. Show that the set fMjM is a DFA not accepting any string with odd number of 1'sg is decidable. 10, page 183. We have described constructions which show that applying these operations to decidable languages result in decidable languages. Mar 19, 2017 · View Homework Help - Discrete Homework 4 from COT 4210 at University of Central Florida. Here’s the best way to solve it. Let the Turing Machine R decide membership in EQCFG. Show that S is decidable. In other words, show that { G | G is a CFG over {0,1} and 1* L (G)} is a decidable language. A problem is called partially decidable, semi-decidable, solvable, or provable if A is a recursively enumerable set. Therefore you need to simulate all possible execution paths. , Describe 3 languages that are not semi-decidable. We can reduce the problem (whether given state q is usel ss) by making q the only accepting state. ) Let ALLDFA = { (A) | A is a DFA and L (A) = L*} Show that ALLDFA is decidable. Exercise 9. Study with Quizlet and memorize flashcards containing terms like State 4 languages that are undecidable. Is it true that if A is a subset of B, and B is decidable, than A is guaranteed to be decidable? I believe it would be true because all the subsets of B should also be decidable making A decidable. iv) Let ETM { (M)| M is a TM and L (M) = 0 A decision problem A is called decidable or effectively solvable if the formalized set of A is a recursive set. . Machine M is a valid decider for ALLCFG since in step 1, we can easily pick a CFG that accept every string in finite time. Question: Show that the problem of determining whether a CFG generates all strings in 1* is decidable. Feb 7, 2025 · The set of all deterministic finite automata (DFAs) where the language accepted by the DFA is empty, denoted as alldfa hai a is a DFA and L(a) , can be shown to be decidable by constructing a Turing machine that can determine if a given DFA accepts an empty language. Since its a decider, it halts on all inputs. If a language is decidable, then there exists a decider M for it. , L (A) = Σ*). By converting the regular expressions to finite automata and applying operations of complementation and intersection, one can decide whether the resulting Nov 26, 2020 · Show that the following language is decidable by finding the algorithm for the finite automaton Ask Question Asked 4 years, 11 months ago Modified 4 years, 11 months ago Engineering Computer Science Computer Science questions and answers 1. Aug 29, 2016 · E (dfa) is a decidable language. Since M was a decider, so is M, and we To get a contradiction, let us assume that STM is decidable, and let S be a decider for STM. Transcribed Image Text: 1. Question 1 ALLDFA = { A |A is a DFA and L(A) = P∗}. Reduce ATM ≤m EQTM, and apply Fact on slide 5-34. We know this TM 3) Show that the following language is decidable ALLDFA = {< A>|A is a DFA and L (A) = {*} Acceptance Tests: ADFA = { (B, w)| B is a DFA that accepts input string w}. Hence, all operations yield decidable languages when applied to decidable languages. The algorithm M1 inputs hAi, where A is a DFA. Decidable languages are not closed under homomorphism Proof. Each symbol of the alphabet will contain a transition from 0 to 0 -. A Turing machine M decides ALLDFA as follows: M = On input⟨A⟩, where A is a DFA: Enumerate all possible strings w in Σ× (the Kleene closure of the alphabet Σ). Oct 25, 2023 · Solution For Show that the set of real numbers that are solutions of quadratic equations ax^ {2} + bx + c = 0 , where a, b and c are integers, is countable. , State an undecidable language whose description does not involve Turing machines. Run the decider on input C for the language E D F A The turning machine M is a decider and for all the Show that the problem of determining whether a CFG generates some string in 1* is decidable. 3) Show that ALLDFA is decidable. Accept if F accepts, reject if F rejects. If the language L of all yes instances to P is decidable then decision problem P is also decidable. Show that LEN_CFG is decidable. Sep. Because . May 2, 2023 · To show that the set is decidable, we need to prove that there exists a decision procedure which can determine whether a given deterministic finite automaton (DFA) accepts the language L(A) = Σ∗. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Let language ALLDFA= {<A> : A is a DFA and L (A) = *} 2. If there exist a Turing machine that accepts and halts on every input string of the language then that language is known as Decidable or Recursive. 1. Construct a DFA B such that L(B) is the complement of L(A). 4. Users with CSE logins are strongly encouraged to use CSENetID only. 3. All the decidable language is also Turing-Acceptable. Let ALLDFA-Mi Mis a DFA and L (M)-Σ*} . Aug 11, 2023 · To show that INFINITEDFA is decidable, we need to demonstrate the existence of an algorithm that can decide whether a given DFA recognizes an infinite language or not. This one is easy. iii) Let AECFG = { (G)| G is a CFG that generates &}. Problem 4. 7 Let B be the set of all infinite sequences over {0,1}. Nov 11, 2021 · We could reject strings whose length-n prefix is the length of the shortest string on which the recognizer loops. Engineering Computer Science Computer Science questions and answers Let S = { (M)| M is a DFA that accepts w^R whenever it accepts w}. Aug 3, 2023 · Identifying languages (or problems*) as decidable, undecidable or partially decidable is a very common question in GATE. (Sipser, Problem 3. P is a class of language which are decidable in polynomial time on DFA. Prove that L1∪L2 is also decidable. This algorithm takes as input a PDA M and checks whether the language recognized by M is infinite or not. A and B are DFAs and L (A) C Problem 4 (5 points). Hint: For a DFA M, the problem of whether or not L(M) = ; is decidable. ) 3. Using the Cartesian Product construction, create a DFA C that accepts the language L (R)⊆L (S). Question: Let ALLDFA= { A ∣A is a DFA and L (A)=Σ∗}. Show that s (n) ≤ (6n + 2)3n . Express this problem as a language and show that this language is decidable Say that language C separates A and B if A ⊆ C and B ⊆ C. 3 Let ALLDFA = { A | A is a DFA and L (A) = Σ ∗ } Show that ALLDFA is decidable. Express this problem as a language and show that it is decidable. Show transcribed image text Here’s the best way to solve it. Show that B is uncountable using a proof by diagonalization. The real problem is with the second part: your "algorithm" do not decide if the recognized language is empty, and this because it is actually a semi-algorithm, as it does not halt if the language is empty (you can We have proven that is decidable, and it is TM possible to construct a PDA from a given CFG. 1), and briefly explain why your TM decides the language. Solution For Show that the problem of determining whether a CFG generates all strings in 1^ { \ast } is decidable. \item [4. Show that SUBSETDFA is decidable. Sep 2, 2021 · Find All Video Solutions for Your Textbook Question Let ALLDFA { (A) A is a DFA and L (A) ∈ 2*} - Show that ALLDFA is decidable. Problem 6. Step 1 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 M = all Turing machines Observation: M is countable. In other words, show that { G ∣ G is a CFG over {0, 1} and 1 ∗ ⊆ L (G)} is a decidable language. Show that L is decidable. Sep 26, 2012 · To show that NP is a subset of EXPTIME, observe that one can just do a brute-force search for all possible y. Continue to help good content that is interesting, well-researched, and useful, rise to the top! Since EQDFA is decidable, the machine M always terminates and returns the output ACCEPT/REJECT. OG- { (G) 1 G is a grammar that generates €). We construct a T … View the full answer Problem 2 (5 points). Oct 21, 2016 · I must show that the following problem is decidable: Given $ \\Sigma = \\{a,b\\}$ and $\\alpha$ a regular expression, is it true that the language defined by $\\alpha$ contains all the odd-length strin Solution for sider the problem of deter equiva 4. Engineering Computer Science Computer Science questions and answers Let ALLDFA = { <A> | A is a DFA and L (A) = Σ* }. ii) Let ALLDFA = { (A)| A is a DFA and L (A) = "}. Oct 6, 2019 · Let BALDFA = {<M>∣ M is a DFA that accepts some string containing an equal number of 0's and 1's } B A L D F A = {<M>∣ M is a DFA that accepts some string containing an equal number of 0's and 1's } Show that BALDFA B A L D F A is decidable. State whether each of the following statements is true. to prove NEW is undecidable, show how to transform a known undecidable problem OLD into NEW so that solution to NEW could be used to solve OLD. Produce a DFA 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∗`. 3 Let ALLDFA = { (A)| A is a DFA and L (A) = *}. m. [nb 1] 1. Show that ALLDFA is decidable. VIDEO ANSWER: Show that the problem of determining whether a CFG generates all strings in 1^* is decidable. Show that A is decidable. 2. Mar 13, 2018 · However, the algorithm you have given is indeed enough to show that the language you have described in the question is decidable. What's reputation and how do I get it? Instead, you can save this post to reference later. One option would be to construct a machine much like the one we built for EDFA: M1 = \On input hDi: (20 points) Let language ALLDFA= {<A> : A is a DFA and L (A) = 2*}. Travis Montey COT 4210 Homework #4 1. For your answer, give the high-level description of a Turing machine that decides your language using Sipser notation (see Section 4. Jun 2, 2018 · Show that the Post Correspondence Problem (PCP) is decidable over the unary alphabet ? = {0}. 1) string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w Oct 23, 2019 · Pushdown automata are nondeterministic. This Turing machine can simulate the operation of the DFA on all possible inputs and determine if it ever reaches an accepting Question: Let ALLDFA { (A) | A is a DFA and L (A) = {*}. 4 Let AεCFG = { G | G is a CFG that generates ε} Show that AεCFG is decidable. , a that is not a symbol in G. Moreover, EQCFG is decidable so step 2 and 3 will terminate in finite time. 02, 2021 11:05 p. We will show some specific language 0TM is not decidable. Construct a decider R for ALLDFA 4. Show that ALLDFA is decidable 째- 빼@ prn c 8 0 Show transcribed image text Here’s the best way to solve it. 5. 2) Let B be the set of all infinite sequences over {0,1}. Show that S Engineering Computer Science Computer Science questions and answers Let ALLDFA = { <A> | A is a DFA and L (A) = Σ* }. Example 1 Equivalence of DFA and REX: Consider the problem of testing whether a DFA and a regular expression are equivalent. Accept if T accepts, reject if T rejects. . , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Turing-recognizable 3. It seems that would produce a decidable subset, but how do we know that the subset so defined in infinite? Mar 23, 2021 · Let A = { | M is a DFA which doesn't accept any string containing an odd number of 1s}. Jan 15, 2015 · Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, 1. A prunable state in a DFA is some state that is never entered while processing any input string. By Theorem 4. Show that SUB_DFA is decidable. We will perform a breadth- rst search of the possible states of our Turing machine, until we have either discovered every reachable state, or found a reachable state on which it moves left. Run M on A, W Question: Let ALLDFA = { (A〉 | A is a DFA and L (A) = Σ*), where (A) is a string that describes A. Your UW NetID may not give you expected permissions. g. (Hint: Show that such an automaton accepts all strings if and only if it accepts all strings of length one. Homework Review Exercise 4. 31 We are given a CFG G, so we can construct a grammar G’ that has all the rules in G. Let SUB_DFA = { (A, B) | A, B are DFAs, and L (A) L (B)}. Show that AECFG is decidable. Show that ALLDFA is Nov 30, 2023 · To show that ALLDFA is decidable, the given input represents a DFA language that accepts all possible strings (i. It shows that some languages are not decidable or even Turing-recognizable, for the reason that there are uncountably many languages yet only countably many Turing machines. 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. How much is a tee in the range? 4. You also need to recall the closure properties of regular language and consider what will be the input to that subroutine. Mark the start state of A. Show that AεCFG is decidable. Create a DFA W that has an initial state of q0 only. Solutions to Practice Midterm 1 1. Now, we will build a Turing Machine S that decides membership in ALLCFG as follows (assume an alphabet of {0, 1} for simplicity): Problem 15 Show that the problem of determining whether a CFG generates all strings in 1 ∗ is decidable. Upvoting indicates when questions and answers are useful. The Turing machine can solve this q Apr 7, 2022 · You'll need to complete a few actions and gain 15 reputation points before being able to upvote. 2]Consider the problem of determining whether a DFA and a regular expression are equivalent. period. Since each Turing machine can recognize a single language and there are more languages than Turing machines, some languages are not recognized by any Turing machine. 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 = { accepts w } M,w ⃒M is a DFA that Algorithm: Check that M is a valid encoding; if not reject. 11 Let is a PDA and L (M) is an infinite language Namely, we show that if E LBA were decidable, A TM would also be. If an accept state is ever marked, then the string that is obtained by going from the start state to that accept state must be in the language L(A) L (A), and hence it is not empty. Solution We know that ALLCFG is undecidable, so we can use this as a starting point. To prove a language is decidable, we can show how to construct a TM that decides it. The idea is to appeal to the pumping lemma for regular languages May 15, 2017 · How do I show this language {<C,A,B> | C,A,B are DFAs, L(C) contains the shuffle of L(A) and L(B)} is decidable ? I believe if I can construct automatas for A and B, then I can get an automata that contains the shuffle of them. In this case, showing that ALL_DFA is decidable means there exists a Turing machine that can determine whether a given DFA accepts all possible strings over its alphabet. That means the set of decidable languages is closed under these operations. Show that ALL_DFA is decidable. is a TM} ⊆ Σ∗ . ” Construct a DFA B such that L(B) = ∑*. 4. Solution For Let ALLDFA = { A ∣A is a DFA and L (A)=Σ∗}. I am also thinking about using emptiness testing but I have not made any progress yet. We need to show that the language ALLDFA is decidable, meaning there exists an algorithm that can determine whether a given DFA A accepts all possible strings in Σ*. Here we show that the problem of checking whether a DFA's language is infinite is decidable. A language L is decidable if there exists a Turing machine T such that L = L (T). 6(b), pg. The problem is that since PDAs support $\epsilon$ -moves, there could potentially be infinitely many execution paths, and it's not clear whether there's an a priori bound on an accepting computation. So my idea how to solve this problem is the following: k = number of states of Show that DQAs are equivalent in power to Turing machines: that is, any given language L is decidable by a DQA if and only if it’s decidable by a Turing machine. { (G) G is a CG that generates… Assume that $!%&TM is decidable and show that !TM is decidable (false!). and more. Then also its complement would be decidable, as well as the intersection of its complement with the language of all algorithms for recognizing languages. Therefore if it is easy to check if the input is in the . Question: Let ALLDFA = { (A)| A is a DFA and L (A) = {*}. 3 in the textbook. 1^* \subse Aug 22, 2020 · Show that the problem of determining whether a CFG generates all strings in 1 ∗ is decidable. Proposition 4. Show that any two disjoint co-Turing-recognizable languages are separable by some decidable language. (c) Show that ALLDFA is decidable. Intersection: similar to concatenation, except you just run both deciders on the input string and answer yes if both answered yes, no otherwise. 4) Let S = {⟨M⟩| M is a DFA that accepts w^R whenever it accepts w}. 4 (EDFA is a decidable language) as a subroutine. The language L is decidable in the case when a decider is there to decide the language L. That L ∈ LD(T M) ⇐⇒ L ∈ LD(T M). Let B be the set of all infinite sequences over {0, 1, 2}. In that case, I start warying, and try to define an algorithm deciding the language. , you can assume that EQDFA is decidable. Check if hM, wi is a valid encoding of a TM M and string w. Get your coupon Engineering Computer Science Computer Science questions and answers Theory Of Computation 3. Let ALLDFA- { (A): A is a DFA and L (A) = Σ*). We can show that ALLDFA is decidable by design a turing machine for it: M: on input A , where A is a DFA Find step-by-step Computer science solutions and your answer to the following textbook question: Show that the problem of determining whether a CFG generates all strings in $1^ {*}$ is decidable. Otherwise, we know that any string in the language is of length at most (excluding), otherwise it could be pumped forever and the ( ) would be infinite, which we already know is not true. 4, we know that EDFA is decidable and therefore has a decider S. 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. Decidability Problem Problem Set Set Six Six is is due due in in the the box box up up front front if if using using a a late late period.  Show that B is uncountableusing a proof by diagonalization.  Show that AεCFG is decidable. 1 answer ALL of It ALLways ALL Over the Place 10 points Using a reduction show that the language ALLDFA = (D) D is a DFA and L (D) = Sigma is decidable (ie for a given alphabet E D accepts all strings composed of that alphabet) You may use any problem we have shown to be decidable in lecture for a reduction Answered step-by-step 1 answer Mar 7, 2023 · The collection of decidable languages is closed under union, concatenation, star, complementation, and intersection. Let SUBSETDFA-A, B L (B)). Proof Here is a decider for A: On input x, where is a potential member of A: Compute f (x) Run the decider for B on f (x) Since A ≤ m B, we have that f (x) ∈ B if and only if x ∈ A. To show that ATM reduces to STM, we will now use the decider S as a subroutine to build a TM A that decides ATM, as follows: A = “On input hM, wi, where M is a TM and w is a string: 0. (Hint: use the known fact that the class of regular languages is closed under complement) 1. Show that it is decidable, given a pushdown automaton M with one state, whether L(M) = Σ*. Let M be a Turing machine decides allDFA: M = "on input A where A is a DFA". Hard-wired into M is the specification hBi for a DFA B that recognizes §¤; this is easy to build by making the start state an accept state with self-loops corresponding to each symbol in §. A TM = {〈M, w〉 | M is a TM and M accepts w} Note that Given a TM M and a string w , simulating M on input w cannot be done on an LBA (since M may use arbitrarily large portion of its tape, while LBA is limited in tape usage). In other words, show that \left\ {\langle G\rangle \mid G\right. Lets start with some definitions:- Decidable language -A decision problem P is said to be decidable (i. We now construct another algorithm M1 to decide language ALLDFA. Let INT_TM = { (M_1, M_2): M For the second claim, suppose that M would be decidable. Other Decidable Problems We can use decidable problems or construction algorithms to show new problems are decidable We know EDFA (checking if the language of a DFA is empty) is decidable (Th. Because there are more languages than TMs. 3 Let ALLDFA = { (A)| A is a DFA and L (A) = Σ*}. Also, we want to replace all A’s on the right hand side with a new symbol, e. The decidability of INFINITEPDA has been proven, which means that there exists such an algorithm. (12 points) Hint: You can use the TM T that we constructed in Theorem 4. 1 A First Theorem First we show that the decidable languages are closed under complement. neither? Assuming the Church-Turing thesis, these are fundamental properties of languages and algorithms. Show that AllDFA is decidable. Let INFINITEPDA = fhMi j M is a PDA and L(M) is an infinite languageg: Show that INFINITEPDA is decidable. Let ALLDFA = cfw_| A is a DFA and L (A) = .  Prove that the language is a CFG that generates εlon. Turing-decidable 2. Let ALLDFA {(A) A is a DFA and L(A) ∈ 2*} - Show that ALLDFA is decidable. Solution:Since Σ = { 1 }, there is only a difference in the number of 1s on the top of each domino compared to the bottom. Let TM , decide $!%&TM. T= "On input , where A is a DFA: 1. “On input B, w , Decidable Languages we now turn to examples of languages that are decidable by algorithms 3. Show that B is uncountable, using a proof by diagonalization. 4) Can show NDFA = { <D> | D is a DFA and L(D) is nonempty} is decidable. (Hint: use the known fact that the class of regular languages is closed under complement). In other words, show that {G|G is a CFG over {0, 1} and 1∗∩L (G) !=∅} is a decidable language. ) (Further Hint: First construct a DFA B that accepts £* and then compare B to A) </a> Question: Theory Of Computation 3. In other words, show that $\left\ {\langle G\rangle \mid G\right. Let's consider the Turing Machine M described below: When the regular expressions R and S are given to M, it Constructs DFAs X and Y equivalent to the regular expression R and S respectively. May 31, 2019 · Enhanced Document Preview: 1. To show that U is decidable we design Turing machine which accepts only strings in U. Jul 10, 2020 · The algorithm you are describing shows that the problem of testing whether a CFG generates some string from $1^∗$ is decidable (e. (25 point) Show that the following two problems are decidable: AllDFA (M) | M is a DFA and L (M) -2*. Corollary 2: Some language is not decidable. Math Advanced Math Advanced Math questions and answers 1. M1 works as follows: First, M1 generates a DFA B that accepts language . 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)}. Problem 5. In otherwords, show that {G|Gis a CFG over {0,1}and 1∗⊂L (G)} is a decidable language. Let ALLDFA = {⟨A⟩| A is a DFA and L (A) = Σ*}. [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. However, there are some exceptions such that we want to delete rules that include A on the left hand side. The algorithm works as Let M = all Turing machines Observation: M is countable. (5 points) - R ¶ B is a DFA, w " L B , and w " L B x is decidable R be the decider for ADFA. Question: ALL of It, ALLways, ALL Over the Place [ 10 points]Recall that to show a language is decidable, you can show a halting reduction to a known decidable problem. The questions seems simple so I designed the following TM D that decides whether t Solution We will create a Turing Machine M to decide ALLDFA. Proof. Show that ALLDFA is decidable. , have an algorithm) if the language L of all yes instances to P is decidable ANS):- To show that ALLDFA is decidable, we need to demonstrate an algorithm that can determine wheth 1) Let AεCFG = {⟨G⟩| G is a CFG that generates ε}. Since $M$ is a dfa, we already have the Turing Machine and just need to show that the dfa halts on every input. GG is a C Show that Ace Database System Concepts 7th Edition ISBN: 9780078022159 SOLUTION- ALLDFA = { (D| D is a DFA and L (D) = Σ*). 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. Let ALL_DFA = { (A) | A is a DFA and L (A) = sigma* (where sigma is the alphabet of A)}. Anyway, the problem of testing whether a CFG generates all the strings of the language $1^∗$ is truly decidable. (5) Let SINGLEDFA = {< M > M is a DFA and for all s E L (M),\s] =1} Show that SINGLEDFA is decidable. ” DA having an empty language is decidable. 23. 18) Show that a language is decidable iff some enumerator enumerates the language in lexicographic order. Construct a new Turing machine M which is just M but we swapped its accept and reject state. Mar 23, 2021 · Let A = { | M is a DFA which doesn't accept any string containing an odd number of 1s}. 4 Let is a CFG that generates ε. isaCFGo 2. Let s (n) be the number of possible n-state, single-tape Turing machines over the 3-symbol alphabet {0, 1, #}. On input <M> 5. If this doesn't look easy, it sometimes help to split up the work in both a recognizing, and a co-recognizing algorithms. 5 Show that the Post Correspondence Problem is decidable over the unary alphabet Σ = { 1 }. Therefore it is sufficient to check this finite number of Question: Show that the following language is decidable by constructing a decider for it. (Exercise 4. Aug 24, 2020 · For regular languages, the emptiness problem (determining whether a given language is empty) is decidable. , see here at page 21). Step: 1 Unlock To show that ALLDFA is decidable we need to show that there exists an View full answer We consider the questions: Which languages are 1. This is proven by constructing Turing machines for each operation that effectively decides the resulting languages. Nov 19, 2015 · QED. On input hAi, M will feed hA; Bi as input to T MEQDFA, a TM that decides whether the languages of two DFA’s are equal. But this intersection is exactly L, the language shown to be unrecognizable, and thus surely undecidable, in the previous proof. We will show a decidable language L and a homomorphism h such that h(L) is undecidable Let L = fxy j x 2 f0; 1g ; y 2 fa; bg ; x = hM; wi; and y encodes an integer n such that the TM M on input w will halt in n steps g L is decidable: can simply simulate M on input w Problem 4. To show that a language A is decidable if and only if A ≤m PRIMES (the set of all prime numbers), we need to prove two implications: If A is decidable, then A ≤m PRIMES. Problem 3 (5 points). is a CFG over \ {0,1\} and \left. And every EXPTIME langauge is obviously decidable and recognizable. Which of the following problems about context-free grammars are solvable, and which are undecidable? Explain your answers carefully. 2) Consider the problem of testing whether a DFA and a regular expression are equivalent. Question: Paragraph Styles 3. With correct knowledge and ample experience, this question becomes very easy to solve. 3: Let ALLDFA = {〈A〉 | A is a DFA that recognizes ∑*}. Since EQDFA is decidable, we assume that is an algorithm M (i. Show that a) ALLDFA is decidable, and that b) ALLDFA is in P. Apr 25, 2017 · Usually, when studying a language, I assume that it is decidable, unless it shows some form of reference to the way Turing Machine work. vwikurd xmp xeroim plitt wsxl idick jkwwwng mmqvrhb ldnb kafd oydjgy lilheq ipdqrhl qrcwct fejyt