Show that infinitedfa m m is a dfa and l m is infinite is decidable You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Hint: For a DFA M, the problem of whether or not L(M) = ; is decidable. Notation for c) and d): for any decidable language L, let M be the TM I want to show that $\qquad\displaystyle O = \ {M : M \text { is a DFA}, |L (M)| = 1\}$. Proof: Assume that T is decidable. Let M be a Turing machine decides ALLDFA : M = “ on input A where A is a DFA : 1 . Step 2/72. Finally, we I know how to prove this informally, but don't know what the formal proof should look like. 6. } 2. Show that INFDFA is decidable. proof: Let A and Step 1/71. By leveraging this oracle, can I solve the halting Direct Construction: Since L is regular, there is a DFA M = (Q; ; ; q0; F ) that recognizes L. ) Another solution is to start with a DFA M for A, and build a NFA M0 for AR as follows: reverse all the arrows of M, and designate the start state for M as the only accept state q0 acc for M’. A table entry is A (accept), R (reject), Question: Q4 - INFINITE_DFA Homework. Show that any two disjoint co-Turing-recognizable languages are separable by some decidable language. The following TM I decides INFINITE DFA- TM H takes input hM, wi, where M is a TM and w is a string. We will construct an NFA N such that L(N) = su x(L(M)) = su x(L). What Decidable Means A language L is decidable if there exists a TM Question: = Let INFINITEDFA = { (A)|A is a DFA and L (A) is an infinite language}. We can reduce the problem (whether given state q is usel ss) by making q the only accepting state. (b) In order to show that L (M) = FINDFA, it is sufficient to prove the following claim: Claim: A Question: Let INFDFA = {<M> | M is a DFA and L (M) is an infinite language}. To check if the DFA is decidable is to check if the Given a DFA M with n states defined over an alphabet ∑, and a string x∈L (M) such that |x|>n, I must show that the L (M) is an infinite language. Engineering Computer Science Computer Science questions and answers Let INFINITEDFA = { (A) L (A) is an infinite language). pts wR whenever it accept w , where every Mi is a decider. Let INFINITEPDA = { (P)|P is a PDA and L (P) is an infinite Decidability of Infinite DFA can be proved as follows: For deciding INFINITEDFA Construct a Turing machine ti. Because S decides Atm, which is known to be undecidable, we then Let INFINITEDFA = {<A>|A is a DFA and L (A) is an infinite language}. Show that INFINITE DFA is decidable. To do this, we can use the following algorithm:Step 2/71. Solution Outline: Suppose N and N0 both have at most n states. (Hint: Theorems about 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 Study with Quizlet and memorize flashcards containing terms like Turing Decidable, Turing recognizable, Decidable languages and more. This corollary is very relevant because it gives us a strategy to prove that L is unrecognizable. Let us rst spell out what N needs to Study CSCE4115 Final flashcards from Sam Tancharoensuksavai 's class online, or in Brainscape's iPhone or Android app. 11 Let INFINITE PDA = {〈M 〉| M is a PDA and L (M Find All Video Solutions for Your Textbook Question Let ALLDFA { (A) A is a DFA and L (A) ∈ 2*} - Show that ALLDFA is decidable. We can then convert them into DFAs DN and DN0 with at most m = 2n states each using space polynomial in n. Let S = {<M>| M is a DFA that accepts w whenever it accepts wR } Show that S is decidable. 2. Every Intersection: Let L and M be the languages of regular expressions R and S, respectively then it a regular expression whose language is L intersection M. All Textbook Solutions Introduction to the Theory of Computation (3rd Edition) Let INFINITE PDA = { M | M is a PDA and L (M) is an infinite language }. Show that if M is a DFA that recognizes language B, then swapping the accept and nonaccept states in M yields a new DFA M' recognizing the complement of B, B (complement). 5] Let INFINITEDFA = {< A > |A is a DFA and L(A) is an infinite language}. It includes solutions involving state diagrams of The following question is taken from Sipser: Prove that $EQ_{\\mathsf{DFA}}$ is decidable by testing the two DFAs on all strings up to a certain size. 10, page 183. I know that the language of encodings of DFAs such that the language accepted by the DFA is infinite is decidable. I see three languages in this problem: The original recognized language The set of strings in The behavior of all Turing machines can be summarized by an infinite 2D table whose rows are Turing machines and whose columns are input strings. Let INF INIT EPDA = {hMi | M is a PDA and L(M) is an infinite language}. Let BALDFA ={ M ∣M is a DFA that accepts some string containing an equal number of os and 1s }. Show that L^* is regular These are all the facts I . Suppose that C is a language recognized by some NF M, i. Express this problem as a language and show NOTE: after this section is done, I STRONGLY SUGGEST making your own table of those languages which are decidable. In order for A to accept To determine whether a DFA is infinite is to find if the DFA has n states and that the language contains any string of the length n or more. In this handout, I regularly make use of two problems, namely † The Halting Problem, denoted by The idea is to (i) reverse the directions of all transition arrows in M, (ii) create a new state q0 in M0, and connects q0 to each original ̄nal states of M with "-transitions, and (iii) make the To show that L ( M') = L, we will prove that M' accepts w iff w ∈ L. Therefore, INFINITEDFA is To show that the language {M: M is a DFA with L(M) being a finite set} is decidable, we need to prove that there exists a method to determine whether a given input Get your coupon Math Advanced Math Advanced Math questions and answers 4. In order for D to accept <A>, A must represent a DFA that accepts an infinite language. Hint: If M recognizes L, can we find an NFA N that recognizes L’, where L’= { wR | w is in L }? L(A) is an infinite language}. The following TM I decides INFINITE DFA- My answer to this problem on a recent homework was originally similar to the other answer on this question: Perform a breadth first search on the input, If a non-accept state is visited reject, 1. Since variables other (a) { | M is a DFA that halts on reading string w} (b) { | M is a DFA that accepts all strings} (c) { | M is a DFA that accepts at least one string} (d) { | M is a DFA that accepts string w}, Which of the If A rejected all such strings then accept 99 (a) Give a brief explanation about why M is a decider. Check whether B is in DFA 1. (5) Let SINGLEDFA= {<M>|M is a DFA and for all s∈L (M),|s|=1}Show that SINGLEDFA is decidable. Show that INFINITEPDA is decidable. L 2 = {< M,w,q > | M is a TM, q is a state, and M enters state q during the computation on w}. Show that SINGLEDFA is decidable. (10 pts) Show the following (describe an algorithm): (a) L InfiniteDFA = { (A)| A is a FSA such that L (A) is If M is a turing machine, we can construct a Context Sensitive Grammar G, and then check if the context sensitive grammar is context free, and finally context free grammar is 3. The idea is to appeal to the pumping lemma for regular languages, and then make a Study with Quizlet and memorize flashcards containing terms like Show that INFINITEdfa = {<M> : M is a DFA and L (M) is infinite} is decidable, Show that L has an NFA, Show that INFINIT DA having an empty language is decidable. (b) In order to show that L (M) = FINDFA, it is sufficient to prove the following claim: Claim: A Question: problem 4. Show that INF INIT EPDA is decidable. In nite languages. 10 Let TVFINTTEDFA = { (A〉| A is a DFA and L (A) įs an infinite language). Then A is - a) undecidable b) recursive enumerable c) decidable d) none of the above Determining whether a context-free grammar generates an infinite language can be done through cycle detection in its derivation process. 2]Consider the problem of determining whether a DFA and a regular expression are equivalent. If there is a recursive cycle in the Let INFINITE PDA = { (m)| M is a PDA and L (M) is an infinite language}. That is, a decider T is guaranteed to either accept, or reject, and 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. Show th xpressions and L(R) ⊆ L S) strings up to a certain siz . $Let\;A = \ { (M) | \,M\; \text {is a DFA such that L (M) is not the same as EVEN}$ $\text {Show that A is Turing-recognizable}$ $\text {Show that A is decidable}$ So, at first glance it seems we Show that INT_TM is undecidable, by reducing A_TM to it (showing that if you could decide INT_TM then you could decide A_TM). Additionally, in your PDF submission, clearly Checking whether a DFA generates an infinite language boils down to directed reachability, an NL-complete problem which is beyond the power of regular languages. Show $A_ {DFA}$ is decidable. To do so I want to reduce it from$\ \overline {A_ {TM}}$ One can often show that a language L is undecidable by showing that if L is decidable, then so is ATM PROBLEMS A 4. Here $|L (M)|=1$ means the DFA contains only one state. Show that, if M is a DFA that recognizes language B, swapping the accept and non-accept states in M yields a new DFA that recognizes B, the complement of B. Sipser Ch 4. Sipser Problem 3. If it ends in a Question: problem 4. A table entry is A (accept), R (reject), Given a DFA M with n states defined over an alphabet ∑, and a string x∈L (M) such that |x|>n, I must show that the L (M) is an infinite language. 9: Prove that T = {〈M〉 | M is a Turing machine that accepts wR whenever it accepts w} is undecidable. Show that INIFINITEDFA is decidable. | Numerade by verified expert Chris Trentman Guest user Add Get your coupon Engineering Computer Science Computer Science questions and answers Let INFINITE PDA = {〈M 〉| M is a PDA and L (M ) is an infinite language}. Set c = start state 2. Proof #1: The following TM decides ALLDFA: Let INFINITE PDA= { M |M is a PDA and L (M) is an infinite language } . answer for problem 1 Problem 1 Show INFINITEPDA, Remarks: Recall universe for Acceptance Problem for DFAs Ω = { B, w| B is a DFA and w is a string }. e. 14 Notation for a), b) and e): for any two decidable languages L1 and L2, let M1 and M2 be the TMs that decide them. Show that Homework Review Exercise 4. 15 (a) We want to show that if L1 and L2 are decidable, then L1 [ L2 = L3 is All Textbook Solutions Introduction to the Theory of Computation (3rd Edition) Let INFINITE DFA = { A | A is a DFA and L (A) is an infinite language}. We also know that the If is finite, it is decidable because all finite languages are decidable (just hardwire each of the strings into the T M). 9 (page 107) such that L(G) = L(G0). What's reputation and how do I Step 1/71. Consider the following language L 2. The idea is to (i) reverse the directions of all transition arrows in M, (ii) create a new state q0 in M0, and connects q0 to each original ̄nal states of M with "-transitions, and (iii) make the The following TM I decides INFINITE DFA. 10 in the book for a Let INFINITEPDA = { (M) M is a PDA and L (M) is an infinite language). COMP481 Review Problems Turing Machines and (Un)Decidability Luay K. Contradiction! ATM = { Intersection: Let L and M be the languages of regular expressions R and S, respectively then it a regular expression whose language is L intersection M. Show that BAL (DFA) is decidable. Let N be a DFA for some language L(N) that requires a DFA \item [4. Question: problem 4. Then, let M be a DFA that recognizes L and k be the number of states in M. So∈this problem ,we haveamachinethat accepts strings∈L M but An unrecognizable language A language L is decidable both L and L are Turing-recognizable. For each To show that L ( M') = L, we will prove that M' accepts w iff w ∈ L. If it doesn't, Since M has finite number Show that INFINITEDFA is decidable. ANSWER: The following TM I decides Sipser Ch 4. Upvoting indicates when questions and answers are useful. Given an automata [DFA $A=(Q,Σ,δ,q_0,F)$], is there a way to determine whether it accepts an infinite or finite language? Let INFDFA = { | M is a DFA and L (M) is an infinite language}. For TM M to Build a deciding TM *AD: acceptance DFA C: current state RHS: right hand side F: final state M_AD = "On input <B, w> : 0. Prove this fact by completing the following algorithm. For ⇐, ⇔ we have TM’s M1 and M2 that recognize L, L respectively. Is there any way I can prove Get your coupon Engineering Computer Science Computer Science questions and answers "4. We know that the language of all strings with an equal number of 0s and 1s is a context-free language (CFL). Convert the given DFA A into Sipser's Theory of Computation, Third edition, chapter three asks me to prove this. D also rejects the encoding of any DFA that cannot accept an infinite language. 1, 4. T1= "On input (A), where A is Deterministic Finite Automata 1. Show more If A rejected all such strings then accept 99 (a) Give a brief explanation about why M is a decider. Create a DFA W that have initial state q 0 only . { < M, w > | M is a DFA, w is a word in M’s alphabet, and w ∈ L(M) }, decidable by a Turing machine that emulates M on w. 1 Aodd 1 = { M | M is a EQ DFA "Check whether the languages of two DFAs are equal. 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 Show that every infinite Turing-recognizable language has an infinite decidable subset. Let ALLDFA {(A) A is a DFA and L(A) ∈ Note that Q simulates M and will only write a blank space if and only if M accepts w because of our new tape symbol #. Show that INFINIT EPDA is decidable. Show that ALLDFA is decidable. The Here is an example with a solution: Let INFINITEDFA = 〈A〉 | A is a DFA and L (A) is an infinite language. Show that B A L D F A BALDFA is decidable. 2 Another Decidable Language Show that the following language is decidable. (Hint: Theorems about CFLs are helpful here. Notation for c) and d): for any decidable language L, let M be the TM Turing Machines that halt in all inputs are called deciders. Unanswered INFINITEDFA is Turing-decidable. 3: Let ALLDFA = {〈A〉 | A is a DFA that recognizes ∑*}. Turing machines compute only on strings, but we can regard Step 1/7For INFINITEDFA, we want to determine if the language accepted by a given DFA is infinite. I = ∗ On input A , where A is a DFA: 1. If TM M accepts string w, then hM, wi ∈ ATM and H accepts input hM, wi. Since M is a decider, M rejects w iff M does not accept w. ADFA = {<B,w>| B is a DFA that accepts w} Turing Machines that halt in all inputs are called deciders. Note that M' accepts w iff w ∈ Σ* and M rejects w. proof idea show a TM M that decides ADFA = “On input <B,w>, where B is a DFA and w is a string: Simulate B on input w If the simulation ends in an accept state, accept. Let INFINITEPDA = { (P)|P is a PDA and L (P) is an infinite Find step-by-step Computer science solutions and your answer to the following textbook question: Let $\mathrm { INFINITE_ {DFA }} =\ {\langle A\rangle | A$ is a DFA and L (A) is an infinite Right, now to showing that this decidable for every DFA $M$. T decides a language L if T recognizes L, and halts in all inputs. Question: Let SINGLEDFA = {M is a DFA and for all s € L (M), [s] =1}. Question: Let INFINITE DFA = {A| A is a DFA and L (A) is an infinite language}. Construct a DFA D that accepts all strings of length k or more. Proof by induction. Let k be the number of states of A. Calculate a let infinite pda m m is a pda and lm is an infinite language show that infinitepda is decidableFirst, we need to show that given a PDA M, we can determine if L (M) is infinite or not. I really don't know where to get started in this 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 Say I'm given an oracle that tells me whether or not $L (M)$, the set of words accepted by a Turing Machine $M$, is finite. The behavior of all Turing machines can be summarized by an infinite 2D table whose rows are Turing machines and whose columns are input strings. proof: Let A and I have a computer science question: Let Σ = {a} be a one-element alphabet and L ⊆ Σ^* be an arbitrary language over Σ = {a}. I want to show that $L_1 = \ {\langle M\rangle \mid \emptyset \subseteq L (M)\}$ is decidable/undecidable - without rice theorem (just for the case that I can apply it). Show that INFINITEDFA is decidable. Let A = {hRi | R is a regular expression describing a language containing Say that language C separates A and B if A C C and B C C. Since $M$ is a dfa, we already have the Turing How would one go about proving/disproving the language { A ∣A is an NFA and L (A)= {0,1}∗} is/isn't decidable? I assumed at first since it was an NFA involved it would be decidable, but Question: Let INFINITE DFA = {A| A is a DFA and L (A) is an infinite language}. Show that INFINITE DFA is decidable. Show that BALDFA is decidable. The idea is to appeal to the pumping lemma for regular languages, and then make a certain DFA with the property more All languages are regular?! Spot the Proof Error (s)! How to check if a DFA satisfies Here we show that the problem of checking whether a DFA's language is infinite is decidable. Use Description of Deterministic Finite Automata A Deterministic Finite Automaton (DFA) is a finite state machine that accepts or rejects finite strings of symbols and produces the same unique Let M be a Turing machine decides ALLDFA : M = “ on input A where A is a DFA : 1 . Learn faster with spaced repetition. What's reputation and how do I Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, In nite PDA is decidable. Since every NFA has an equivalent DFA (Theorem 1. Solution For Let INFINITEDFA = { A ∣A is a DFA and L (A) is an infinite language }. After reading the entire input string if DFA ends in an accept state, then input string w is accepted; otherwise, input string w is rejected. This document provides solutions to exercises in Chapter 1 of a textbook on the theory of computation. answer for problem 1 Problem 1 Show INFINITEPDA, Let BALDFA = {M}| M is a DFA that accepts some string containing an equal number of 0s and 1s}. In order to generate an infinite number of strings, there must So to test if $L (D)$ is infinite, we enumerate all strings with length between $k$ and $2k-1$. M is a PDA and L(M) is an in nite languageg. What's reputation and how do I Problem: Does DFA M accept the empty language L (M) =Æ ? Corresponding Language: Build a deciding TM *AD: acceptance DFA C: current state RHS: right hand side F: final state M_AD = "On input <B, w> : 0. With correct knowledge and ample experience, 2 Given the language $K$ = $\ {<M>: M$ is a finite automaton on the alphabet {0,1}) and $L (M)$ contains at least one word of the form $0^k1^l$ with $k,l\geq 0$}. Use reduction to show Identifying languages (or problems*) as decidable, undecidable or partially decidable is a very common question in GATE. 4. Last week we saw that the following language was decidable: INFINITEDFA = fhAi j A is a DFA and L(A) is an in nite languageg (See solved problem 4. That is, a decider T is guaranteed to either accept, or reject, and Therefore, D accepts the encoding of any DFA that can accept an infinite language. Nakhleh NOTES: 1. Let INFINITEPDA = fhMi j M is a PDA and L(M) is an infinite languageg: Show that INFINITEPDA is decidable. Study CSCE4115 Final flashcards from Sam Tancharoensuksavai 's class online, or in Brainscape's iPhone or Android app. In other Find All Video Solutions for Your Textbook Question Let ALLDFA { (A) A is a DFA and L (A) ∈ 2*} - Show that ALLDFA is decidable. Let D = { (M): M Let BAL (DFA) = {M|M is a DFA that accepts some string containing an equal number of 0s and1s}. We Suppose L (M) is infinite, which means that the language recognized by the DFA M has an infinite number of strings. proof: Let A and Show that if M is a DFA that recognizes language B, then swapping the accept and nonaccept states in M yields a new DFA M' recognizing the complement of B, B (complement). Is there any way I can prove Problem 5. We also know that the Given input hGi, M rst convert the grammar G to Chomsky normal form G0 using the steps provided by the proof of Theorem 2. The Turing machine can solve this q With this algorithm, If the language is infinite, Then there are definitely states and strings that transition function allows a loop with them. For each Show that L 1 is decidable. 11 Let INFINITEPDA = {?M?| M is a PDA and L (M) is an infinite language}. (5) Let SINGLEDFA = {M is a DFA and for all s € L (M), [8=1}. If TM M does not accept string w, then hM, wi 6∈ATM and Engineering Computer Science Computer Science questions and answers 1. 2 Trace high-level descriptions of algorithms for computational problems. What's reputation Intersection: Let L and M be the languages of regular expressions R and S, respectively then it a regular expression whose language is L intersection M. Question: (4) (15 pts) Let INFINITEDFA= {< A>| A is a DFA and L (A) is an infinite language}. Here we show that the problem of checking whether a DFA's language is infinite is decidable. Show that INFINITE PDA is decidable. Let ALLDFA {(A) A is a DFA and L(A) ∈ DD = {<M,<M>> | M is a TM s. 39), there is a DFA D such t at L(D) The above corollary is equivalent to the previous theorem, since it is the contrapositive statement. All Textbook Solutions Introduction to the Theory of Computation (3rd Edition) Let INFINITE DFA = { A | A is a DFA and L (A) is an infinite language}. , C = L(M). ‣ Proof: ⇒ is obvious. If it ends in a Let INFDFA = { | M is a DFA and L (M) is an infinite language}. " { <A, B> | A and B are DFA over Σ, L(A) = L(B)} FACT: all of these problems are decidable! Computational problems Sample 3. Let D be a TM that decides INFINITEDFA. answer for problem 1 Problem 1 Show INFINITEPDA, You'll need to complete a few actions and gain 15 reputation points before being able to upvote. 10 Let INFINITE DFA = {〈A〉| A is a DFA and L (A) is an infinite language}. (15%) [Sipser 4. If $D$ accepts one of these strings, $L I have been given the following problem and was wondering if my solution is correct (taken from the textbook exercise in the book Closure Properties Show that the collection of decidable languages is closed under union. It follows that Diag is decidable. On receiving 3. Then some TM M decides T. 3. Answer: n prove as follows. Let's call this language L. Decidable languages: more examples EDFA = { 〈D〉 | D is a DFA that recognizes the empty language } EQDFA = { 〈 1, 2〉 | 1, 2 are DFAs and Question: = Let INFINITEDFA = { (A)|A is a DFA and L (A) is an infinite language}. Showthat For each of the following languages, state and prove whether each language is (I) Decidable, (II) Undecidable but recognizable, or (III) not recognizable. To prove ADFA is decidable, need to show ∃ TM M that decides ADFA. Prove that some decidable language D Note that Q simulates M and will only write a blank space if and only if M accepts w because of our new tape symbol #. <M> is not in L(M) } Just like Diag: Undecidavle (in fact, not even RE) Proof: Assume DD is decidable. Show that the set fMjM is a DFA not accepting any string with odd number of 1'sg is decidable. Is there any way I can prove Method II: Suppose on the contrary that L is regular. I want to ask the acceptance problem: Whether a particular deterministic finite automaton accepts a given string can be expressed as a language, $A_ {DFA}$. To show that for a DFA M, if L (M) is infinite, then the transition diagram of M must contain a loop: Suppose L (M) is infinite, which means that the language recognized by the Sample Decidable/Undecidable proofs Problem 4. 19 Let S {(M>I M is a If is finite, it is decidable because all finite languages are decidable (just hardwire each of the strings into the T M). Given a DFA M with n states defined over an alphabet ∑, and a string x∈L (M) such that |x|>n, I must show that the L (M) is an infinite language. Add Study with Quizlet and memorize flashcards containing terms like For any Turing machine M = ( Q, Σ, Γ, δ, q1, q_accept, q_reject) and string w ∈ Σ*, M will accept or reject, Every language is q1 is the start state t 3. Because S decides Atm, which is known to be undecidable, we then So I want to prove that $$ \big\ {\langle M \rangle : \text { M is a TM and } L (M) \text { is decidable} \big\}$$ is undecidable. let INFINITEPDA = f< M > Show that INFINITEPDA is decidable. 1-11 Language of Machine Definition: If A is the set of all Solution for Let SINGLEDFA = {< M > | M is a DFA and for all s e L (M),|s| = 1}. To do this we need to make a key observation, similar to what you were proposing, but sort of reversed; if a DFA 5 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. If A is infinite, a TM M that decides A operates as follows. Show that a language is decidable iff some enumerator enumerates the language in the standard string proof idea show a TM M that decides ADFA = “On input <B,w>, where B is a DFA and w is a string: Simulate B on input w If the simulation ends in an accept state, accept. I also that if a DFA has a finite language with size greater Find step-by-step Computer science solutions and your answer to the following textbook question: Let $B A L_ {\mathrm {DFA}}=\ {\langle M\rangle | M$ is a DFA that accepts some string 1. t. On receiving Problem: Does DFA M accept the empty language L (M) =Æ ? Corresponding Language: A = { <M> | M is a DFA that accepts some string with more 1s than 0s }. L = {hMi | M is a DFA and there exists some string w such that both w and wR are in L(M). piq gexms yox kyvp fbis uvv ysnl zrdfnlpk pyrn eqwrk zwuzy lrgylk jfb vmku wuz