# Show That Infinite Dfa Is Decidable

Show that INFINITEDFA is decidable. 1: Decidable Languages, pp. In E DFA and E CFG, the input string was an encoded DFA or grammar ; In EQ DFA, the input string was 2 encoded DFAs ; In A DFA, the input string for the TM has 2 parts: An encoded DFA M ; An input string w for the DFA; A DFA simulates the computation of M on D. 196) A REX. We want to show that the problem of test to see if two DFA's recognize the same language is decidable. In the proof, we constructa new DFA CfromAand B, where Caccepts. Then if the DFA accepts any string of length ≥ p,. The automaton works based on rules (a function) that tells it to transist from one state to another based on the current character of input. Example- (I) (Acceptance problem for DFA) Given a DFA does it accept a given. Since EQ DFAis decidable, there is an algorithm. $\begingroup$ Any language accepted by a DFA (i. DFA = { | M is a DFA and M accepts w }This language is decidable. Then we will prove something stronger: There are semi-decidable (r. Examples of using closed operations to show languages regular: September 30th: Prove ValidComp(M) is valid using homomorphism, O(n) notation, Decision Procedures, Prove Emptiness Problem, Membership Problem and Equivalence Problem in Regular Language are all decidable. NFA Acceptance Is Decidable Theorem: ANFA = {hN,wi | N is a NFA that accepts w} is a decidable language. Visit Stack Exchange. 5] Let INFINITEDFA = { |A is a DFA and L(A) is an inﬁnite language }. Prove that Halting problem is undecidable. Reducibility & Undecidable Problems Overview. Mark the start state of M. If A ≤ B and A is undecidable then so is B. Show that L is undecidable. Such a Turing Machine is called a decider. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Second, if both A and its complement are Turing-recognizable, show A is decidable. Moreover we obtain additional results concerning the case of finitary AFs. We show that L 1 \L 2 is decidable too. An in nite binary sequence is an unending sequence of 0s and 1s. Proof: In regular language donot matter which weuse like DFA, NFA or regular expression the actual problem is that the string w of the L is the decidable or not. Prove this fact by completing the following algorithm. A1 A2 A3 A4 A5 B NP C Note that the second step implies. Various transformations of CFL as explained and proved, to show that if null-production, unit productions, and useless productions are removed from a given CFG, its generating power remains the same. Show that S has an infinite, decidable subset. TM with a one way infinite tape. Otherwise, construct a DFA D for the language L(R) (refer to chapter 1 of Sipser for the algorithm that constructs an equivalent NFA for L(R) from R,. Show that the language A = f< D > jD a DFA that accepts some string of the formakbkjk 0g Solution 1. William Gasarch-U of MD Decidability of WS1S and S1S (An Exposition). ) Palindromes on a DFA. S 1 is over alphabet { a , b } and it is infinite. That is what makes them so weak. Question 6 (25): These questions use the definition of the language COMP C above. –So L c is Turing-recognizable, by Theorem 2. Consider set union as an example. Theorem 5 (Theorem 4. 1 is a decidable language ="On input , , where is a DFA and is a string: 1. To prove this, we first reduce the problem to a derivation problem for an infinite transition system, and then we show how to abstract this problem into a. Be able to show that a given language is finite-state. Do the Problem 4. Let T = "On input where A is a DFA i. Formally speaking, construct TM S: S = "On input : Construct DFA D_A equivalent to A; Construct DFA D_0 that accepts. This is true since the DFA can be simulated by a Turing machine (DFA is weaker than Turing machine). Examples – We will now consider few important Decidable problems:. Additionally, in your PDF submission, clearly label "TopHat Q4" and explain why this algorithm decides INFINITEDFA. 4, page 168) EDFA is a decidable language. So suppose EQ CFG is decidable and let Mbe the decider. This follows from Rice’s theorem as the property defining L is not trivial and it depends on the language accepted by the Turing machine. Decidability of Regular Languages - DFA We showed earlier that a TM can simulate a DFA Another way to look at this is: The acceptance problem for DFA is A DFA = fhB;wijB is a DFA that accepts wg A DFA is a TM-decidable language Note that this language deals with all possible DFAs and inputs w, not a speci c instance. Show that of Turing decidable languages is closed under concatenation. (To show the property of non-context-free. is reachable from from its start state. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Also use the fact that regular languages are closed under intersection. The DFA also has a *next-state function*, which specifies what it should do next, depending on what state it happens to be in and what input it then receives. It then constructs the DFA D_0 that accepts the language {0,1}*, we can then simulate the decider for EQ_DFA on. Also, suppose L is the infinite language {0 * 1 *} With this notation, Exercise 5. Covers E_DFA, Infinite_DFA, Three_DFA, and the pumping lemma for regular languages Correction: Step 4 should say Run THE DECIDER for E_DFA on If THE DECI. Construct a DFA for L. Such problems are termed as Turing Recognisable problems. To show that something is decidable, present an algorithm (pseudo-code plus English, or even entirely English is ne, because there will be no ambiguity). We show that EQ 2DIM DFA is undecidable. 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. In Section 14. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following. Those theorems though, always involve infinite quantities, one example is: Do this system of Diophantine equations have a finite or infinite number of solutions?". And thus we have derived a contradiction. Is BIGGER DFA decidable? Either prove that it is. This is true since the DFA can be simulated by a Turing machine (DFA is weaker than Turing machine). First, if A is decidable, show both A and its complement are Turing-recognizable. Solution for A low voltage supply from which one needs high currents must have very low internal resistance. Decidable languages Definition: A language L ⊆Σ* is decidable if there is a Turing Machine M which: 1. The language L is infinite, we do not know if the FA is finite or infinite. M =  On input 〈𝑫〉, where 𝑫 is a DFA: Use BFS to determine if an accepting state of D. In particular we show: (a) The set of recursive omega-words with decidable monadic second order theories is Sigma_3-complete. M simulates on w and if M reaches a final state, it accepts. Unanswered INFINITEDFA is Turing-decidable. for the following languages. Show that S has an infinite, decidable subset. Infinite time decidable equivalence relation theory Posted on September 25, 2011 by Joel David Hamkins S. Proof: In regular language donot matter which weuse like DFA, NFA or regular expression the actual problem is that the string w of the L is the decidable or not. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Show that BALDFA is decidable. is reachable from from its start state. 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. 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. In this model the transitions are labelled by letters or variables ranging over an infinite alphabet and guarded by conjunction of equalities and disequalities. 3 - A REX is a decidable language. 4 E DFA is a decidable language. Undecidable Languages. To show L is not regular, find an infinite set S of pair-wise distinguishable strings. Therefore, χ(P) is the collection of all. , have an algorithm) if the language L of all yes instances to P is decidable. 10) Ask Question Asked 2 years, 5 months ago. 195) A NFA = {| B is an NFA that accepts w} (p. We also show that if the input words are restricted to lasso. Show that a language is decidable if it is Turing-recognizable and co- recognizable 19. We will check for its strings and try to show that they are pairwise distinguishable with respect to L 1. Let’s construct a TM K. Solution We will create a Turing Machine M to decide INFINITEDFA. Covers E_DFA, Infinite_DFA, Three_DFA, and the pumping lemma for regular languages Correction: Step 4 should say Run THE DECIDER for E_DFA on If THE DECI. We want to show that the problem of test to see if two DFA's recognize the same language is decidable. Those theorems though, always involve infinite quantities, one example is: Do this system of Diophantine equations have a finite or infinite number of solutions?". If so, execute DFA M on input w and check M ends up in its final state. Let INFINITE PDA = {< A > |A is a PDA and L(A) is an inﬁnite language}. Department of Software Systems 167 OHJ-2306 Introduction to Theoretical Computer Science, Fall 2011 13. It remains to show that the set of TR languages is infinite-- this follows, for example, from the fact that every singleton language {w} is TR. Hamkins, “Infinite time decidable equivalence relation theory,” Notre Dame Journal of Formal Logic , vol. To show DFA for L needs at least k states, find a set S of k pair-wise distinguishable strings. 1 finite automata. TMs and Infinite Loops. We show that this problem is decidable, EXPSPACEcomplete. w A f(w) B so w A ⇒ M’ acceptsw. Formulate this problem as a language and show that it is decidable. Recall also that regular languages are closed under intersection. DFA = { | M is a DFA and M accepts w }This language is decidable. If we rst wanted to convert the NFA to a DFA, the total time would be doubly exponential From regex to FA’s We can build an. TM is not Turing-decidable. Total O(n) steps. It remains to show that the set of TR languages is infinite-- this follows, for example, from the fact that every singleton language {w} is TR. (We will never actually construct them, since they are going to be exponentially. CSC 473 Automata, Grammars and Languages 11/23/10 5 C SC 473 Automata, Grammars and Languages 13 TMs can Act as Recognizers or Transducers •Defn 5. (j) The complements of some Turing-recognisable languages are Turing-recognisable. [15 points] Solution:We construct a TM that decides PreﬁxFree REX as follows1. For all PDA's M there exists CFL G such that L(M) = L(G). In particular we show: (a) The set of recursive omega-words with decidable monadic second order theories is Sigma_3-complete. Unanswered INFINITEDFA is Turing-decidable. (To show the property of non-context-free. This follows from Rice’s theorem as the property defining L is not trivial and it depends on the language accepted by the Turing machine. The Borel reductions are replaced by the more general class of infinite time computable functions. A decidable language • To show that a language is decidable, we have to describe an algorithm that decides it ‣ Weʼll 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. Emptiness Problem for DFA’s The emptiness problem for DFA’s is to test whether the language recognized by a given DFA A is empty or not. Speaking informally, you can show this by constructing a Turing Machine constructs a DFA D_A equivalent to NFA A. We will check for its strings and try to show that they are pairwise distinguishable with respect to L 1. Show that for any infinite language L, L is decidable iff some enumerator TM enumerates L in lexicographic order. Show that CONTAIN PDA DFA is decidable. Show that ALL DFA is decidable. E 2DIM DFA is m-reducible to EQ 2DIM DFA in the same way that E TM is m-reducible to EQ TM, so if we show that E 2DIM DFA is undecidable, this will show that EQ. (2) Call a DFA "symmetric" iff for any string w, either it accepts both w and its reverse, or rejects both. if M 1 accepted then goto 4. 23 suggests that A is decidable if and only if A ≤ m L. 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. This means that given any two decidable languages, and , applying these operators to the languages results in a set that is still decidable. ” Despite their falsity, they will be part of the Proof Gallery. Say that language Cseparates Aand Bif A Cand B C. There is a tape head that can read and write symbols and move around on the tape. To show that K is decidable, lets construct two Turing machines. Given hMi, which represents an automaton M, we can construct an automaton Nsuch that L(N) = L(M)R, that is, Naccepts a word wif and only if Maccepts wR. Now M is a TM with no oracle! Theorem: If A T B and B is decidable, then A is decidable Proof: Exactly the same proof as the one for mapping reductions!. Repeat until no new states get marked:. On the computational side, we show that several decision and construction problems which are known to be polynomial time solvable in finite afs are decidable in the context of the proposed formalism and we provide the relevant algorithms. Show that ALL DFA is decidable. TMs that always halt. To show DFA for L needs at least k states, find a set S of k pair-wise distinguishable strings. In the context of TMs and looping, it's useful to think about looping for all of our machines. DFA is decidable, where A DFA = fhB;wijB is a DFA and w 2L(B)g (Proof idea) IWe construct a Turing machine M to decide the problem. ), but itself not an r. 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. For example, a DFA can model a software that decides whether or not online user-input such as email addresses are valid. TM with a one way infinite tape. 1: Finite Automata, pp. Let INFINITEDFA = f A j A is a DFA and L(A) is an inﬁnite language g. 1 that accepts L 1 and a dfa M 2 that accepts L 2, construct a new dfa M’ with states and transition function resulting from a combination of the states and transition functions from M 1 and M. Otherwise, construct a DFA D for the language L(R) (refer to chapter 1 of Sipser for the algorithm that constructs an equivalent NFA for L(R) from R,. Decidable fields By James Ax and Simon Kochen 0. Show that every in nite semi-decidable language has an in nite decidable subset. Example Let L 1 = { a n b n | n is a positive integer } over alphabet { a , b } show its nonregular using Myhill-Nerode 1)Consider the set of strings S 1 = { a n | n is a positive integer }. Proof: Suppose that it is decidable. 1 Answer to Let INFINITEDFA = {(A)| A is a DFA and L(A) is an infinite language}. 1) To decide whether a particular DFA accept a given string nonaccepting state,. is decidable. Call (218) 693-2264 for more information. Show that EQNFA is in PSPACE. – Produce a Turing Machine M that decides whether or not its input w is in L or Lc. on input 𝐺𝐺,𝑤𝑤 2. Construct a DFA G that accepts strings containing an odd number of 1's. Rec-DFA is decidable. First, notice that a DFA which accepts in nitely many strings must accept arbitrarily long strings. 3, Linz makes connections between this hierarchy and the Chomsky hierarchy of languages. Consider a 2-tape TM M: M= "On input x: 1. Problem 27 Let EQ 2DIM DFA = fhA;BijAand B are 2DIM-DFA and L(A) = L(B)g. $\endgroup$ - MHS Oct 7 '14 at 4:34. Show that INFINITE DFA is decidable. DFA =fhB;wijB is a DFA that accepts the string wg B Therefore testing whether DFA B accepts w is the same as testing whether hB;wi2A DFA B Other computational problems are formulated in terms of testing membership in a language B To show that a computational problem is decidable is to show that the encoding of the problem is decidable 2 / 10. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. decidable Π " # classes. Next we show that the problem of checking whether there exists a matrix in the semigroup which has an infinite number of factorizations is decidable and NP-hard in SL $$(2,\mathbb {Z})$$. Since EQ DFAis decidable, there is an algorithm. Show that the set of all inﬁnite-length binary strings is uncountable. is reachable from from its start state. There is a tape head that can read and write symbols and move around on the tape. (a, 10) Prove that the language COMP NFA is Turing decidable. proof: Let MΣ∗ be a DFA that accepts Σ∗ (this can be easily constructed), then for every DFA A, A ∈ ALLDFA <=>< A,MΣ∗ >∈ EQDFA So, to decide whether A ∈ ALLDFA, we just need to decide whether < A,MΣ∗ >∈ EQDFA. (b) We characterise those sets P subset of Z that yield bi-infinite words (Z,<=,P) with decidable monadic second order theories. Let INFINITE PDA = fhMi j M is a PDA and L(M) is infiniteg. An amazing website. Take the DFA that accepts L and check if w is accepted. L 1 = {< M,w > | M is a. A single language is a set of strings over a finite alphabet and is therefor countable. Introduction In  we gave a complete set of elementary axioms for the valued field of p-adic numbers. DFA is decidable, where A DFA = fhB;wijB is a DFA and w 2L(B)g (Proof idea) IWe construct a Turing machine M to decide the problem. TMs and Infinite Loops. DFA = { | A is a DFA that recognizes Σ*}. IScan the input string repeatedly. We use \higher-order" automata constructions to extend these properties to monadic. ) Palindromes on a DFA. (20 points) Show that EQ DFA = fhA;BijA;B are DFAs, and L(A) = L(B)gis decidable by testing the two DFAs on all strings up to a certain size. To prove this, we first reduce the problem to a derivation problem for an infinite transition system, and then we show how to abstract this problem into a. (a, 10) Prove that the language COMP NFA is Turing decidable. We show that characteristic formulae for nite-state systems up to bisimulation-like equivalences (e. else M rejects 4. So, this is decidable. called co-semi-decidable if its complement is semi-decidable. Graphical representation of DFA. In several previous papers [1, 4, 17,19,11], we have developed a technique to show that many properties of automatic sequences are decidable. Simulate on input. copy x on the second tape 2. • An infinite binary sequence is an unending sequence of 0s and 1s. In fact there are six official ways to get into the special processing lane. Show that INFINITEDFA is decidable. Prove this fact by completing the following algorithm. There should be an equivalent DFA MD = (QD, ∑D, δD, q0, AD) such that L(MD) = L(MN). On input M to K, lets construct a NFA K’ that accepts the reverse of M. An amazing website. 24 on page 88 in the textbook. Rao, CSE 322 4 The Chomsky Hierarchy – Then & Now… CFLs Decidable T. Lecture 22:(4/14) Turing reducibility: if A ≤ B and B is decidable then so is A. DFA = { | M is a DFA and L(M) = Ø } Theorem. DFA = { 𝑫 | 𝑫. Run the DFA on all strings over the alphabet of length [n, 2n). Next we show that the problem of checking whether there exists a matrix in the semigroup which has an infinite number of factorizations is decidable and NP-hard in SL $$(2,\mathbb {Z})$$. We show that there are plenty of infinite sets that satisfy the omniscience principle, in a minimalistic setting for constructive mathematics that is compatible with classical mathematics. is reachable from from its start state. Otherwise, construct a DFA D for the language L(R) (refer to chapter 1 of Sipser for the algorithm that constructs an equivalent NFA for L(R) from R,. One difference from DFA or CFG is that the input for the Turing machine is actually on the tape: i nitially the tape contains only the input string and is blank everywhere. up vote 6 down vote favorite 4. $\begingroup$ Any language accepted by a DFA (i. for the following languages. – Suppose that L is Turing-decidable. Recall also that regular languages are closed under intersection. 9) Prove that INFINITEDFA = { | A is a DFA and L(A) is an infinite language} is decidable. Decidable language-A decision problem P is said to be decidable (i. DFA :First check if the input is well-formed and indeed encodes a DFA M. Show that Breduces to A(i. If A ≤ B and A is undecidable then so is B. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. Then if the DFA accepts any string of length ≥ p,. Decidable Problems for Regular Languages: DFAs Emptiness problem for DFAs Theorem 4 4: E is a decidable language E DFA ={| A is a DFA and L(A) =0/} Theorem 4. Theorem: A NFA is a decidable language. Theorem 4 (Theorem 4. A DFA is defined as an abstract mathematical concept, but due to the deterministic nature of a DFA, it is implementable in hardware and software for solving various specific problems. Formal language & automata theory 1. Many basic aspects of the classical theory remain intact, with the added bonus that it becomes sensible to study some special equivalence relations whose complexity is beyond Borel or even analytic. 10) Ask Question Asked 2 years, 5 months ago. Second, if both A and its complement are Turing-recognizable, show A is decidable. To prove the claim we simply need to give an algorithm A that on input hM;wi determines whether M recognizes its input. ; design S to. On input where M is a DFA: 1. 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. Construction: the following TM MA decides A. • Define a new DFA M' = (Q, Σ, δ, q 0, Q-F) • This has the same transition function δ as M, but for any string x ∈ Σ* it accepts x if and only if M rejects x • Thus L(M') is the complement of L • Because there is a DFA for it, we conclude that the complement of L is regular The complement of any regular. T = “On inpput , where A is a DFA: 1. show how to construct the 5-tuple for. Decidable Problems for Regular Languages: DFAs Emptiness problem for DFAs Theorem 4 4: E is a decidable language E DFA ={| A is a DFA and L(A) =0/} Theorem 4. If not, stop and reject. We don’t like TM’s that sometimes loop. Indeed, we show that these classes form a strictly increasing in nite hierarchy. Show that Aε CFG is decidable. On a single move depending on the state of finite control and symbol scanned by each of tape heads ,the machine can change state print a new symbol on each cells scanned by tape head, move each of its tape head independently one cell to the left or right or remain. We like deciders. Also assume that a Turing machine can construct DFA based on closures of regular languages and perform conversions among DFA, NFA, and regular expression. Covers E_DFA, Infinite_DFA, Three_DFA, and the pumping lemma for regular languages Correction: Step 4 should say Run THE DECIDER for E_DFA on If THE DECI. (c) Let A be any undecidable language (e. When we convert NFA N to DFA D and run the Turning machine M for A DFA on input (D,w). Show that INFINITEDFA is decidable. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. Active 2 years, 5 months ago. (Hint: You may want to use Problem #3 from the first homework. We use \higher-order" automata constructions to extend these properties to monadic. Proof sketch: The procedure we gave for translating an NFA to an equivalent DFA was mechanistic and terminating, so a halting Turing machine can do that job. DFA = { 𝑫 | 𝑫. Construct a DFA for L. (Hint: Use the fact that it is possible to contruct a DFA that recognizes the regular language sigma*555sigma*. What is the difference between these three characters?. Question 1 (10): True or false with justification: The language INF CFG is Turing decidable. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following. (infinite strings are not used in any formal language) Language: A language is a collection of sentences of finite length all constructed from a finite alphabet of symbols. Theory of Computation Lectures delivered by Michael Sipser Notes by Holden Lee Fall 2012, MIT Last updated Tue. M =  On input 〈𝑫〉, where 𝑫 is a DFA: Use BFS to determine if an accepting state of D. How do we show that a language is TD: Build a decider for it. Proof: The following TM M decides E. ) • Next show that a known NP-Complete language B can be reduced to C in polynomial time; i. Also, suppose L is the infinite language {0 * 1 *} With this notation, Exercise 5. This is the first in a series of “false proofs. It follows from the following claim. Then if the DFA accepts any string of length ≥ p,. Your instructor prefers the name Turing-decidable for these languages. Some more decidable problems are − Does DFA accept the empty language? Is L 1 ∩ L 2 = ∅ for regular sets? Note − If a language L is decidable, then its complement L' is also decidable. Last time we talked about decidable problems relating to finite automata and context-free grammars. •Note: since all finite languages are recursive, (they’re regular in fact) any decision problem. DFA is decidable, where A DFA = fhB;wijB is a DFA and w 2L(B)g (Proof idea) IWe construct a Turing machine M to decide the problem. We start with the formal definition of an nfa, which is a 5-tuple, and add two things to it: is a finite set of symbols called the stack alphabet, and z is the stack start symbol. DFA = { | M is a DFA and L(M) = Ø } Theorem. is reachable from from its start state. On input where M is a DFA: 1. What are the various models of Turing Machine? 17. Show that INFINITE PDA is decidable. If not, then reject. called co-semi-decidable if its complement is semi-decidable. DFA =fhB;wijB is a DFA that accepts the string wg B Therefore testing whether DFA B accepts w is the same as testing whether hB;wi2A DFA B Other computational problems are formulated in terms of testing membership in a language B To show that a computational problem is decidable is to show that the encoding of the problem is decidable 2 / 10. Then, there exists a decider TM D such that L(D) = L. *** c) Prove that the language in part (b) is decidable. 3/16/2020 CS332 - Theory of Computation 22. T = “On inpput , where A is a DFA: 1. If not, then reject. DFA = { 𝑫 | 𝑫. The following TM decides 𝐿𝐿. We can also de ne a DFA for f g: it has a single state that is both the start state and an accept state, and it does not allow any transitions. If so, execute DFA M on input w and check M ends up in its final state. School of EECS, WSU. on input 𝐺𝐺,𝑤𝑤 2. It then constructs the DFA D_0 that accepts the language {0,1}*, we can then simulate the decider for EQ_DFA on. is decidable. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Proof: The following TM M decides E. Equivalence of deterministic and nondeterministic machines. This is the first in a series of “false proofs. Express this problem as a language and show that it is decidable. E(dfa) is a decidable language. IKeep track of the current DFA state and position of w. (b) Known characterisations of the $\omega$-words with decidable monadic second order theories are transfered to the corresponding question for bi-infinite words. 2 - A NFA is a decidable language. let n = number of states of that DFA. Consider all possible cases of pairs of strings in S. Show that the following language is decidable: INFINITE DFA = fhAi jA is a DFA and L(A) is an in nite languageg Sipser Problem 4. It remains to show that the set of TR languages is infinite-- this follows, for example, from the fact that every singleton language {w} is TR. Quiz review sections. Nondeterministic FSMs. 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. Ideally we must use a bulletted style to facilitate reading/grading. It then constructs the DFA D_0 that accepts the language {0,1}*, we can then simulate the decider for EQ_DFA on. That is, all words in the language are accepted by the TM. Theorem: If A m. We introduce an analogue of the theory of Borel equivalence relations in which we study equivalence relations that are decidable by an infinite time Turing machine. Let INFINITE PDA = {< A > |A is a PDA and L(A) is an inﬁnite language}. To show that K is decidable, lets construct two Turing machines. 12/11/2012 Contents Lecture 1 Thu. We show that the resulting classes of timed omega-languages are robust, and we explain their relationship. Show that PreﬁxFree REX is decidable. A decidable language • To show that a language is decidable, we have to describe an algorithm that decides it ‣ Weʼll 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. Visit Stack Exchange. By drawing the FA diagram there are a finite number of states which means this minimal DFA is finite. 196) A REX. T = “On inpput , where A is a DFA: 1. October 2nd. It ‘decides’the language L. Repeat until no new states are marked: 3. S 1 is over alphabet { a , b } and it is infinite. DFA is decidable, where A DFA = fhB;wijB is a DFA and w 2L(B)g (Proof idea) IWe construct a Turing machine M to decide the problem. Some more decidable problems are − Does DFA accept the empty language? Is L 1 ∩ L 2 = ∅ for regular sets? Note − If a language L is decidable, then its complement L' is also decidable. Halts on every input x∈Σ*. The halting problem can be used to show that other problems are undecidable. Run the decider for 𝐴𝐴. Those theorems though, always involve infinite quantities, one example is: Do this system of Diophantine equations have a finite or infinite number of solutions?". Solution: a): not Turing decidable. Recall that the proof of Theorem 4. The fundamental tool is the following: Theorem 1. 24 on page 88 in the textbook. Show that ALL DFA is decidable. Describe an algorithm for using the descriptions of A and B to produce a DFA, C, that recognizes the language L M, that is each string in L that is not also in M. Hint: consider the DFA that recognizes the reverse language. After constructing K’ convert K’ to DFA K ' DFA. We show that EQ CFG is undecidable by showing that if it would be decidable, then so would ALL CFG, which is not true (Theorem 5. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. 4 E DFA is a decidable language. Example 2 | Prove language is decidable 1. Proof: The following TM M decides E. That is, all words in the language are accepted by the TM. Dfa problems with solutions pdf. Every decidable language is Turing-Acceptable. (h) The latest D-Wave quantum computer is deterministic. We will show there is no onto function from the set of all Turing Machines to the set of all languages over {0,1}. We introduce an analogue of the theory of Borel equivalence relations in which we study equivalence relations that are decidable by an infinite time Turing machine. Notice the inclusion is proper. Mark the start state of M. It follows that the refinement problem and the /spl forall/CTL* model-checking problem are decidable for 2D rectangular automata. You must show that the following algorithm accepts if and only if L(D) is infinite. Additionally, in your PDF submission, clearly label "TopHat Q4" and explain why this algorithm decides INFINITEDFA. Show that INFINITEdfa = { : M is a DFA and L(M) is infinite} is decidable Let p be the number of states in the DFA. Now show that FDFA M) : M is a DFA and L(M) is finite} is decidable, by using part 3a. A Deterministic Finite Automaton (DFA) is a Turing machine where the tape head always moves to the right at every step (and thus there is no use in. ) — {(1Vf)1 is a DFA that accepts some palindrome}. 4 - E DFA is a decidable language. Last time we talked about decidable problems relating to finite automata and context-free grammars. First, note that for every string in [ ], is in exactly one equivalence class, namely [ ]. A language L is decidable if and only if both L and the complement of L are TM-recognizable. (c) If C is decidable, then the complement of D is decidable. (a, 10) Prove that the language COMP NFA is Turing decidable. (g) The complement of every Turing-recognisable language is Turing-recognisable. No TMs exist “Undecidable” problems “Decidable” problems. M is clearly a decider since steps 1, 3, and 4 will not create an infinite loop and step 2 calls a decider. In a Deterministic Finite Machine, your input is a word, such as abbaaabb. Solution: The set of strings of odd length is a regular language. In fact there are six official ways to get into the special processing lane. Show that L is Turing-decidable. You must show that the following algorithm accepts if and only if L(D) is infinite. Theorem: ADFA is a decidable language. So we can see that the languages de ning the \base-cases" of the regular languages are all decidable by DFAs. (2) If L and Lc are Turing-recognizable, L is decidable. Pf: Observe that Because of closure properties, there is an algorithm to construct a DFA C from A, B that accepts Use Theorem 4. We will check for its strings and try to show that they are pairwise distinguishable with respect to L 1. Fauzan has 2 jobs listed on their profile. Proof of (1): L is decidable, so that L is Turing-recognizable (why?). , χ(P), which has entries χi ∈ {0, 1}, indicating whether a string in Σ ∗ is being included in P at that corresponding location ∀i, with infinitely many such i’s. From FA to regex We need to compute n. 23 Let — {0/1>1 M is a DFA that accepts some string containing an equal number of Os and Is}. Let INFINITE PDA = fhMi j M is a PDA and L(M) is infiniteg. First, note that for every string in [ ], is in exactly one equivalence class, namely [ ]. 17 Prove that EQ DFA is decidable by testing the two DFAs on all strings up to a certain size. EQ={, B|areDFAndL()=} DFA. let n = number of states of that DFA. If not, then reject. Property of Decidable Language Theorem: Let Lc denote the complement of L. Mark start state of A. Problem3 : Check for a context-free language is decidable. Show that A is Turing decidable. DFA: Write. Covers E_DFA, Infinite_DFA, Three_DFA, and the pumping lemma for regular languages Correction: Step 4 should say Run THE DECIDER for E_DFA on If THE DECI. Express this problem as a language and show that it is decidable. 1 that accepts L 1 and a dfa M 2 that accepts L 2, construct a new dfa M’ with states and transition function resulting from a combination of the states and transition functions from M 1 and M. Hint: Suppose M never moves left, but. Examples of using closed operations to show languages regular: September 30th: Prove ValidComp(M) is valid using homomorphism, O(n) notation, Decision Procedures, Prove Emptiness Problem, Membership Problem and Equivalence Problem in Regular Language are all decidable. The problem is in general undecidable, although for particular examples you might find it decidable. Example: 011111… is an infinite string which has infinite length. On a single move depending on the state of finite control and symbol scanned by each of tape heads ,the machine can change state print a new symbol on each cells scanned by tape head, move each of its tape head independently one cell to the left or right or remain. 2011 • The TM can work as follows on input ¢A²: 1. (15 points) Show that the set of all in nite binary sequences is uncountable. (Hint: Theorems about CFLs are helpful here. (Exercise 4. Strings in a Turing-decidable language will either be accepted or rejected by their associated Turing machine. A1 A2 A3 A4 A5 B NP C Note that the second step implies. Develop a DFA for the language. The size of a language is the number of strings in the language. This loop can contain just a single state, in the case of a self-loop. Problem 27 Let EQ 2DIM DFA = fhA;BijAand B are 2DIM-DFA and L(A) = L(B)g. Let INFINITE PDA = {< A > |A is a PDA and L(A) is an inﬁnite language}. Visit Stack Exchange. Furthermore, M will accept those DFA’s whose language is the same as B’s language – i. 18) Let Aand Bbe two disjoint languages. Publisher's on-line resources. • This might seem surprising---it seems simpler than the general acceptance problem, since it involves just one particular string. This means that given any two decidable languages, and , applying these operators to the languages results in a set that is still decidable. Then B is decidable, and A is a subset of B. We also show that if the input words are restricted to lasso. We show that Sis decidable. Moreover we obtain additional results concerning the case of finitary AFs. This banner text can have markup. Show that A ε CFG is decidable. T = “On inpput , where A is a DFA: 1. In this problem we chose the two approaches e the first approach lik to simulate NFA and other approach on w that the is. The machines never go into an infinite loop. 1 finite automata. Be able to prove that a language is not finite-state. 9 in Sipser, with solution included. Exercise 9. (15 points) Show that the set of all in nite binary sequences is uncountable. Example 2 | Prove language is decidable 1. $\begingroup$ Any language accepted by a DFA (i. A decision problem P is decidable if the language L of all yes instances to P is decidable. In nite languages. Consider the following language L 1. BREX is the decidable languages. The Turning machine is similar to DFA but with an infinite tape serving as unlimited memory. Hint: Suppose M never moves left, but. 4 Let A ε CFG = { | G is a CFG that generates ε}. "Turing recognizable" vs. A first example of an omniscient set is the one-point compactification of the natural numbers , also known as the generic convergent sequence. A Turing machine U to decide the language A works as follows U = \On input hMiwhere M is a DFA, 1. It remains to show that the set of TR languages is infinite-- this follows, for example, from the fact that every singleton language {w} is TR. Show that Aε CFG is decidable. Prove (you may use the Church-Turing thesis) that. This is directly from the textbook (page 168) and we went over it in class. s (contʼd) •Theorem 4. 4, page 168) EDFA is a decidable language. Graphical representation of DFA. On input 𝑤𝑤: 1. From the lecture, we know the complement of L 1 is decidable, so does L 1. To show DFA for L needs at least k states, find a set S of k pair-wise distinguishable strings. We will show there is no onto function from the set of all Turing Machines to the set of all languages over {0,1}. You can be excused from setting an online appointment to apply for your passport at the Department of Foreign Affairs (DFA). With an example explain the universal Turing machine 18. Show that. The new year may bring the need for you to apply for a passport - whether it's a new one, up for renewal, or regrettably, lost. Show that Aε CFG is decidable. Show that there is a language E such that both E and Ec (the complement of E) have non-empty intersection with every in nite semi-decidable set. [15 points] Solution:We construct a TM that decides PreﬁxFree REX as follows1. The notions of ANFA and AREX are similarly de-ﬁned. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following. A language L is decidable if and only if both L and the complement of L are TM-recognizable. Let INFINITE PDA = fhMi j M is a PDA and L(M) is infiniteg. HALT), and B be. $\endgroup$ - MHS Oct 7 '14 at 4:34. Totality Problem: A function (or program) F is said to be total if F(x) is defined for all x (or similarly, if F(x) halts for all x). In fact there are six official ways to get into the special processing lane. Proof: In regular language donot matter which weuse like DFA, NFA or regular expression the actual problem is that the string w of the L is the decidable or not. •Note: since all finite languages are recursive, (they’re regular in fact) any decision problem. 5 - EQ DFA is a decidable language. Prove by construction. Show that the following language is decidable: INFINITE DFA = fhAi jA is a DFA and L(A) is an in nite languageg Sipser Problem 4. 1 and L 2. Examples of using closed operations to show languages regular: September 30th: Prove ValidComp(M) is valid using homomorphism, O(n) notation, Decision Procedures, Prove Emptiness Problem, Membership Problem and Equivalence Problem in Regular Language are all decidable. Accept if not. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following. On input R, reject if R is not a valid regular expression. DFA •can we build one TM that will work for all DFAs? •is there an algorithmic way to solve this problem? CS 310 –Fall 2016 Pacific University Theorem •A DFA is decidable –given we can decide if A DFA or A DFA •Proof Idea: –Use a TM, M, to simulate B with input w –Keep track of current state and current. Theory of Computation Lectures delivered by Michael Sipser Notes by Holden Lee Fall 2012, MIT Last updated Tue. 5, page 169) EQDFA is a decidable language. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. M =  On input 〈𝑫〉, where 𝑫 is a DFA: Use BFS to determine if an accepting state of D. accept the same language Emptiness testing problem : Decide whether the language of an automaton is empty Can be applied to DFA, NFA, REX, PDA, CFG, TM, Informatik Theorie II (A) WS2009/10 acs-07: Decidability 4 4. The halting TM/program P for A. – Produce a Turing Machine M that decides whether or not its input w is in L or Lc. In order to do so, we would apply the characteristic sequence method onto P; i. Since NPSPACE = PSPACE, we are done. Let's quickly go through some other decidability questions. Formally speaking, construct TM S: S = "On input : Construct DFA D_A equivalent to A; Construct DFA D_0 that accepts. Prove that the class of decidable languages is closed under the operations of union, concatenation, star, complementation and intersection. (b) L 2 = fhN;wi: Nis a NFA and w2L(N)g Decidable. If it contains such a string, pumping up shows infinitely many other strings to be in the langauge. Simulate on input. Consider the DFA M with start state q0 and set of accept states F. 1 Answer to Let INFINITEDFA = {(A)| A is a DFA and L(A) is an infinite language}. We want to show that the problem of test to see if two DFA's recognize the same language is decidable. The set of decidable languages is closed under the following operations: set union, set intersection, set complementation, string concatenation, and Kleene closure. Such an algorithm is easily seen at a high level: A ﬂrst checks whether the input is legitimate and if not it rejects. 3, Linz makes connections between this hierarchy and the Chomsky hierarchy of languages. Show that S has an infinite, decidable subset. s (contʼd) •Theorem 4. • Proof attempt: – Try a reduction---show if you could decide Acc01 TM then you could decide general acceptance problem Acc TM. IScan the input string repeatedly. In particular we show: (a) The set of recursive $\omega$-words with decidable monadic second order theories is $\Sigma_3$-complete. 4 with to test whether If that algorithm accepts, then accept; otherwise, reject. Let’s another TM E that checks if two DFA’s are equal or not. Proof sketch: The procedure we gave for translating an NFA to an equivalent DFA was mechanistic and terminating, so a halting Turing machine can do that job. There is a tape head that can read and write symbols and move around on the tape. The machines never go into an infinite loop. E 2DIM DFA is m-reducible to EQ 2DIM DFA in the same way that E TM is m-reducible to EQ TM, so if we show that E 2DIM DFA is undecidable, this will show that EQ. Recall also that regular languages are closed under intersection. n a (ω) mod 2 = 1 can also be written as n a (ω) ≅ 1 (mod 2). Express the problem of determining whether a DFA and a regular expression are equivalent, using a language of strings accepted by a Turing machine. , Problem 4. if M 1 accepted then goto 4. (15 points) Show that the set of all in nite binary sequences is uncountable. Let INFINITEDFA = fhAijA is a DFA and L(A) is an inﬁnite language g:Show that INFINITEDFA is decidable. Example- (I) (Acceptance problem for DFA) Given a DFA does it accept a given. The language accepted by a DFA is the set of all strings accepted by it. DFA = {|A is a DFA and L(A) = }. Hence, we want to show that P is uncountable. Simulate on input. 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. Overlapping type families can cause soundness problems, so you can't have them. Solution for A low voltage supply from which one needs high currents must have very low internal resistance. INFINITE DFA = fhAi : Ais a DFA and L(A) is in niteg Show that this language is decidable. If a language is decidable, then there is an enumerator for it. Proof: Let 𝐺𝐺be a CFG generating 𝐿𝐿. Enter an Infinite Loop (ie the computation never ends) Looping and DFA, NFA, PDA. (1) If L is decidable, both L and Lc are Turing-recognizable. Also, L is decidable implies Lc is decidable (why?). if M 1 accepted then goto 4. In the theory of computation, a branch of theoretical computer science, a deterministic finite automaton (DFA)—also known as deterministic finite acceptor (DFA), deterministic finite-state machine (DFSM), or deterministic finite-state automaton (DFSA)—is a finite-state machine that accepts or rejects a given string of symbols, by running through a state sequence uniquely determined by the. Example: 011111… is an infinite string which has infinite length. This idea can be used to study determinacy of semi-decidable sets, i. Prove this fact by completing the following algorithm. Consider the language A = {hMi|M is a DFA which doesn’t accept any string with an odd number of 1’s}. We can use D to construct an enumerator E for L as follows: E = "Ignore the input. Show how to construct a DFAfor LE,given a DFA for L. On input M to K, lets construct a NFA K’ that accepts the reverse of M. Repeat until no new states get marked:. is decidable. (Hint: You may want to use Problem #3 from the first homework. NFA Acceptance Is Decidable Theorem: ANFA = {hN,wi | N is a NFA that accepts w} is a decidable language. Show that the language A = f< D > jD a DFA that accepts some string of the formakbkjk 0g Solution 1. The set of decidable languages is closed under the following operations: set union, set intersection, set complementation, string concatenation, and Kleene closure. Let S be an infinite, Turing-recognizable language. 16 Prove that EQ DFA is decidable by testing the two DFAs on all strings up to a certain length. In this problem we chose the two approaches e the first approach lik to simulate NFA and other approach on w that the is. Show that L is decidable. Recall that the proof of Theorem 4. Notice the inclusion is proper. 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. To show DFA for L needs at least k states, find a set S of k pair-wise distinguishable strings. q 0 a 1 b b 2 a a,b L = a*b* Q: What is the complement of L?. DFA DECIDABILITY THEOREM We need the following easy theorem: Theorem: The following problem is decidable: given a DFA determine if it accepts ANY strings. Unanswered INFINITEDFA is Turing-decidable. First, note that for every string in [ ], is in exactly one equivalence class, namely [ ]. A= fhMijMis a DFA which doesn’t accept any string containing an odd number of 1sg: Show that Ais decidable. Call (218) 693-2264 for more information. Theory of Computation Lectures delivered by Michael Sipser Notes by Holden Lee Fall 2012, MIT Last updated Tue. (b) We characterise those sets P subset of Z that yield bi-infinite words (Z,<=,P) with decidable monadic second order theories.