# Show That Infinite Dfa Is Decidable

, Problem 4. We show that this problem is decidable, EXPSPACEcomplete. Show that for any infinite language L, L is decidable iff some enumerator TM enumerates L in lexicographic order. $\begingroup$ Any language accepted by a DFA (i. Explain the halting problem is un-decidable. Decidable Problem. • 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. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. In general, these are not stably infinite, and the Nelson-Oppen scheme cannot be used to integrate them into SMT solvers. The data shows the average number of days of rainfall in Rangoon for each month of the year. Formulate this problem as a language and show that it. Equivalence decidable in poly time over fields. Show that every infinite recognizable set has an infinite decidable subset. This is surprising considering that with the strict (non-reflexive) navigation relations the satisfiability problem is undecidable. We show that the state-. Decidable. Proof: Consider the following Turing machine F: F = "On input , Determine the number k of states of A. Construct a DFA G that accepts strings containing an odd number of 1’s. Give an example in the spirit of the recursion theorem of a program in a real programming. We need to show that a TM can test to see if a DFA accepts any. In a finite language no string is pumpable. It is also useful to be familiar with some of the example languages we have seen, like A DFA, E DFA, EQ DFA, A CFG, E CFG, EQ CFG, A TM, etc. If infinite words are in your scope, you can generalize DFA (with parity condition) to the so-called Good-for-Games automata (GFG), that still have polynomial containment. 1 is a decidable language ="On input , , where is a DFA and is a string: 1. Assg 9 - Solution sketches with study points added. RE December 2, 2015 10 / 23. Finally, we give an application to the theory of games. This is a divergent series because the absolute value of r is greater than 1. We show that various aspects of k-automatic sequences -- such as having an unbordered factor of length n -- are both decidable and effectively enumerable. Show that if a language L and its complement L’ are both recursively enumerable then L is recursive. Since EF is a very weak fragment of the modal µ-calculus, model checking with EF is decidable for many more classes of infinite-state systems. page 161: An alternative to Definition 4. COT 4210 Homework #4: Decidability Due Date: Wednesday July 13, 2011 (in class) 1) Let ALL DFA = { | A is a DFA that recognizes Σ *}. 13 Some other decidable problems • ANFA = { M,w ⃒M is an NFA that accepts w} ‣By direct simulation, or by reduction to ADFA. Show that S is decid-able. Show that INFINITEDFA is decidable. An in nite binary sequence is an unending sequence of 0s and 1s. A Turngi Machine. $\begingroup$ Any language accepted by a DFA (i. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. If not, then reject. Show that INFDFA is decidable. (20 points) Show that the set of decidable languages is closed under intersection. Abstract We present new results on a constraint satisfaction problem arising from the inference of resource types in automatic amortized analysis for object-oriented programs by Rodriguez and Hofmann. The complete algorithm is now as follows: "On input hRi:. First, notice that a DFA which accepts in nitely many strings must accept arbitrarily long strings. IScan the input string repeatedly. Formulate this problem as a language and show that it. Decidability Informally: •Decidability of problem X -whether the problem has an algorithm which solves it (decides it), i. Speaking informally, you can show this by constructing a Turing Machine constructs a DFA D_A equivalent to NFA A. We use the same idea as in Question 1, by constructing a DFA that recognizes ˙ 1 ˙ n. (B) Prove that L 5 can not be recognized by a 4-state DFA. is a decidable language DFA DFA EQABAB LALB EQ == Theorem Proof. (a, 10) Prove that the language COMP NFA is Turing decidable. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Construct a PDA for L. Idea: Let p be the number of states in the DFA. A NFA is GFG if there is a strategy $\sigma:A^*\times Q\times A\to \Delta$, that given the prefix read so far and the current state and letter, chooses a transition to go to. The first-order theory of arbitrary boolean algebras with a sequence of distinguished ideals is decidable. whether there is an TM which: 1. Let Mbe a Turing machine that decides this language. ” EQ DFA = { | M. 1) To decide whether a particular DFA accept a given string nonaccepting state,. Run the universal U on. CSE396, Spring 2018 Final Exam May 15, 2018 One notes binder allowed, no Internet, closed neighbors, 170 minutes. Convert A into an equivalent NFA A´ by using the procedure for this conversion given in Theorem 1. You should also comment on why the provided algorithm is decidable. If no accept state is marked, accept; otherwise reject. 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. RL The accepting problem for DFAs of testing if a particular DFA accepts a given string can be expressed as the following language: ADFA = {|B is a DFA that accepts w. Let S = {hMi | M is a DFA that accepts wR whenever it accepts w}. If the simulation ends in an accept state,. -So L c is Turing-recognizable, by Theorem 2. Nakayama 1. Regular language iv. 17 Prove that EQ DFA is decidable by testing the two DFAs on all strings up to a certain size. Show that ALLDFA is. Not all decidable are context-free. The TM always halts. RE is a decidable language. A Turing-recognizable language L is the one that has a Turing-m. Express this problem as a language and show that it is decidable. boolean algebras with quantification over ideals, is decidable. 4 E DFA is a decidable language. Donate to arXiv. (20 points) (Sipser 4. Let's construct a Turing machine M to decide the ALL DFA problem. (15%) [Sipser 4. 13 Some other decidable problems • ANFA = { M,w ⃒M is an NFA that accepts w} ‣By direct simulation, or by reduction to ADFA. A NFA is GFG if there is a strategy $\sigma:A^*\times Q\times A\to \Delta$, that given the prefix read so far and the current state and letter, chooses a transition to go to. Murawski2, and C. Note that we are assuming that halts is decidable (it always halts, returning true or false). That is, all words in the language are accepted by the TM. In Inductive Inference [5, 7, 20] the basic model of learning is as follows. 3 and 4 only. (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. Prove that the class of decidable languages is closed under the operations of union, concatenation, star, complementation and intersection. 4 Warm-Up: Some Decidable Languages Show that the following languages are decidable by describing (at a high level) an algorithm that deci des them (see more in Sipser 4. Problems about context-free languages may or may not be decidable. Homework 8Solutions 1. • HALTTM = {< M , w > | M is a TM and M halts on input w }. Visit Stack Exchange. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. Proof Idea: Specify a TM M that decides A DFA. 1) To decide whether a particular DFA accept a given string nonaccepting state,. Assignment 7 Sipser 4. Equivalently, if A is undecidable and reducible to B, B is undecidable. To our knowledge, NNCT is the first extension of Petri nets, in the class of nets with an infinite set of token types, that has primitive recursive coverability. Let f be a mapping reduction from A to B To decide A, we build a machine M’ M’(w): 1. 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). Prerequisite - Turing Machine A problem is said to be Decidable if we can always construct a corresponding algorithm that can answer the problem correctly. Recalling these examples is also useful for proving other languages (like INFINITE DFA) are or are not decidable or computably enumerable. 4 Decidable and Undecidable Languages 37/38-7 Complementary, My Dear Watson co-RE is the set of languages whose complements are RE. Show that ag-TMs can recognize any Turing-recognizable language. by iterated application of the latter morphism (cf. (A counterexample suffices). Show that INFDFA is decidable. This is directly from the textbook (page 168) and we went over it in class. Then, there exists a decider TM D such that L(D) = L. We'll say L is semi-decidable if it's semi-decidable+ or semi-decidable- Dec = Recursive. Show that A is decidable. Instance: CFL L, and a string x. (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. First, notice that a DFA which accepts in nitely many strings must accept arbitrarily long strings. We focus, in this work, on the limitation of Atl * to allow only one temporal goal for each strategic assertion and study the fragment of Sl with the same restriction. 9 in the textbook). Proof: Suppose that it is decidable. Repeat until no new variables get marked: a) Mark any variable A where G has a rule A-->U 1,. Mark the start state of A 2. These (W)SnS-interpretations show that the undecidable system Fsub possesses consistent decidable extensions, i. The data shows the average number of days of rainfall in Rangoon for each month of the year. Proof: The following TM M decides E. I'll tell you the ingredients you need for the proof, but you'll have to figure out the recipe: - regular language. Repeat until no new variables get marked: a) Mark any variable A where G has a rule A-->U 1,. E(dfa) is a decidable language. let D be the DFA and R the RE, they are equal if L(R)=L(D). M =  On input 〈𝑫〉, where 𝑫 is a DFA: Use BFS to determine if an accepting state of D. Since we can encode a DFA D as a string hDi, the acceptance problem is equivalent to testing for membership in the language ADFA = {hD,wi | D is a DFA that accepts w} The acceptance problem is called decidable if the corresponding language is (Turing-)decidable. Show that B is uncountable using a proof by diagonalization. 2009 • The TM can work as follows on input ¢A²: 1. Idea: Let p be the number of states in the DFA. CFG = { (G, w) | G is a CFG that generates string w } So we can define the following languages: Theorem: A. The full definition of 2DIM-DFA can be found in Sipser's "Introduction to the Theory of Computation" (5. Show that every infinite Turing-recognizable language has an infinite decidable subset. The following problem is decidable: given a DFA. Automata Theory: The study of abstract machines and automata, including the computational. 1 Answer to Let INFINITEDFA = {(A)| A is a DFA and L(A) is an infinite language}. This is directly from the textbook (page 168) and we went over it in class. You need to show that, given a Context Free Grammar G, one can check whether the language L(G) recognized by G is included in a*b*. 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). View Homework Help - Discrete Homework 4 from COT 4210 at University of Central Florida. Each logical system comes with both a syntactic component, which among other things determines the notion of provability, and a semantic component, which determines the notion of logical validity. Method II: Suppose on the contrary that L is regular. CSE396, Spring 2018 Final Exam May 15, 2018 One notes binder allowed, no Internet, closed neighbors, 170 minutes. We consider three qualitative decision problems: (i) the positive decision problem asks whether there is a word that is accepted with positive probability; (ii) the almost decision problem asks whether there is a word that is accepted with probability 1; and (iii) the limit decision. We show that 2D rectangular automata, which model discrete reactive systems with continuous environments, define effectively presented infinite graphs with finite similarity relations. Show that A is decidable. The set of decidable problems $\mathbb{D}$ is countable so $\lvert \mathbb{D} \rvert = \lvert \mathbb{N} \rvert$ and this led me to the foll Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and. Proof: The following TM M decides A. First, notice that a DFA which accepts in nitely many strings must accept arbitrarily long strings. Give an example in the spirit of the recursion theorem of a program in a real programming. Explain the halting problem is un-decidable. Let k be the number of states of A. Aug Sept 13 Apr May 26 MONTH Jan Feb Mar Jun Jul Oct Nov Dec 28 DAYS OF RAIN 2 17 28 22 Calculate for this rainfall data the A. Express this problem as a language and show that it is decidable. IEvery nite language is decidable: For example, by a TM that has all the strings in the language \hard-coded" into it IWe just saw some example algorithms all of which terminate in a nite number of steps, and output yes or no (accept or reject. Repetition might lead to infinite computations although the DFA accepts some string. Proof: Suppose that it is decidable. The following machine decides it: T := "On input M, where M is a DFA, 1. 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). I'll tell you the ingredients you need for the proof, but you'll have to figure out the recipe: - regular language. (Hint: Theorems about CFLs are helpful here. 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. Show that if Lis Turing decidable, the ˜ L(x) computable. 13 Let A = { (R, R and S are regular expressions and L(R) C L(S)}. X is Turing decidable if and only if both X and its complement X-bar are Turing recognizable. Simulate M1 on w. edu for assistance. Solutions to Assignment 8 November 23, 2000 Exercise 1 (30 pts) Please do Exercise 4. If T accepts, accept. Show that S is decid-able. com - id: 3e23d0-NWM3N. (A) Show that L 5 can be recognized by an 5-state DFA. Recalling these examples is also useful for proving other languages (like INFINITE DFA) are or are not decidable or computably enumerable. 20 is decidable. These are some decidable problems-Problem1 : There is a DFA M and input string w decide if, M accepts w. Show that AGREE DFA is decidable by testing the two DFAs on all strings up to a certain length. Let S = {hMi | M is a DFA that accepts wR whenever it accepts w}. Assume there is a procedure TOTAL. We show that if either the memoizing assumption or the early-assumes assumption is relaxed, the verification problem becomes undecidable, arguing for the necessity of these assumptions. So, this is decidable. Show that INFINITEDFA is decidable. (15%) [Sipser 4. Proof: The following TM M decides A. Prove that INFINITEDFA is decidable. Express this problem as a language and show that it is decidable. We can write an infinite table for every pair. Note that standard (qubit-based) quantum computers are PSPACE-reducible, thus they are still a Turing-complete model of computation (i. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following. Let L 1, L 2, and L 3 be languages de ned over the alphabet = fa;bg, where L 1 consists of all possible strings over except the strings w 1;w 2;:::;w 100; i. always halts 2. (b)[5 marks] Show that the set of decidable languages is closed under concatenation. Kleene’s theorem v. 22,208 results. 1 Usage of the PCP solver in Asg7; 7. A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. A useless state in a pushdown automaton is never entered on any input string. Definition 2. Decidable First-Order Transition Logics for PAProcesses Article in Information and Computation 203(1):75-113 · July 2000 with 16 Reads How we measure 'reads'. First, it's easy to tell whether the DFA accepts an infinite language: check whether there is a cycle that is reachable from the start state and from which the accepting state is reachable. Solution: From last problem (sipser 3. This alternative point of view may make it easier to see why the satisfaction problem is decidable. DFA on input hCi. Show that S is decid-able. Show that A is Turing decidable. Hence INFINITE DFA is Turing decidable (recursive). Decidability regular CF decidable Turing recognizable • D is a DFA that accepts w • N is an NFA that accepts w • D is a DFA that accepts a non-empty language • A, B are DFAs and L(A) = L(B) • C is a CFL that accepts w • C is a CFL that accepts a non-empty language • But, A, B are CFLs and L(A) = L(B) not decidable • When the model of comp increases in power. Consider the language A = {hMi|M is a DFA which doesn’t accept any string with an odd number of 1’s}. We show that characteristic formulae for nite-state systems up to bisimulation-like equivalences (e. Let B be the set of all inﬁnite sequences over {0,1}. Department of Software Systems 167 OHJ-2306 Introduction to Theoretical Computer Science, Fall 2011 13. A Turing-recognizable language L is the one that has a Turing-m. If $L$ is decidable, then we can just generate strings in lexicographic order, and test if each is in $L$ , thus generating an enumerator that prints in standard string order. edu) Notes: Recall that the symbol denotes reducibility. It doesn't take in infinitely many DFAs or strings - it just takes in a single DFA and a single string at a time. Note that the number of states p in the DFA is the p of the Pumping Lemma applied to the regular language L(A). If not, then reject. Show that the union of two recognizable sets is. Undecidable Languages. If the simulation ends in an accept state,. True; A DFA has infinite number of states. that these machines do not have the usual ability to move the head one symbol left. [40 points] Solution Outline: One method to decide the language is to convert the DFA into an equivalent regular expression. 3 Lecture 17: Proving Undecidability 13 Acceptance Language A TM= { < M, w> | M is a TM description and M accepts input w} We proved ATM is undecidable last class. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. 1 Lecture 22 (3/2/16) covered material: A CFG and E CFG are decidable, introduction to undecidability, diagonalization reading: finish chapter 4. The alphabet could consist of the symbols we normally use for communication, such as the ASCII characters on a keyboard, including spaces and punctuation marks. We will check for its strings and try to show that they are pairwise distinguishable with respect to L 1. Let INFINITE PDA ={<M>|M is a PDA and L(M) is an infinite language} Show that INFINITE PDA is decidable. a DFA and an RE are 'equivalent' if they have the same language. IEvery nite language is decidable: For example, by a TM that has all the strings in the language \hard-coded" into it IWe just saw some example algorithms all of which terminate in a nite number of steps, and output yes or no (accept or reject. There is a tape head that can read and write symbols and move around on the tape. Show that the following languages are decidable: 1. A TM can check hBiis valid encoding of DFA (using any reasonable convention), then use tape. tex Find file Copy path ryandougherty Added Table of Contents, changed document format 3008db2 Jul 12, 2016. 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. Let L = fhMijM is a DFA and for every string w, if M accepts w, then M also accepts wRg. I'll tell you the ingredients you need for the proof, but you'll have to figure out the recipe: - regular language. DFA;REX m EQ DFA By Theorem 5. (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. Enter an Infinite Loop (ie the computation never ends) Looping and DFA, NFA, PDA. ” EQ DFA = { | M. Construct a DFA M such that L(M) = L(A) ∩ L(D). 67) Lemma 1. CFG = { (G, w) | G is a CFG that generates string w } So we can define the following languages: Theorem: A. recognizes ∅. Lecture 5 (scribe: Agnishom Chattopadhyay) 29/1/2019: Lecture 6: Reduce(Transpose(Reduce(Transpose(A)))) gives the minimal WA for the function (over a field). Let L 17 = fhMi: M is a Turing machine that accepts " and at least one other stringg. Alternative Solution 2. It is easy to see that such TM decides the language L[L0. 18) : If we have an enumerator which enumerates a language in lexicographic order, then we can make a decider using it. Visit Stack Exchange. Graphical representation of DFA. IKeep track of the current DFA state and position of w. Definition 2. Let k be the number of states of A 2. The size of a language is the number of strings in the language. We show how to add transparent memoization to stream predicates: without changing their code, we can cause our predicates to apply themselves to partially available data and yield the residual. I = "On input 〈A〉, where A is a DFA: 1. Test all smaller DFA’s for equivalence with A. Assg 9 - Solution sketches with study points added. I'll present an example of a decidable language, followed by a general result about decidable languages. Simulate on input. 22,208 results. Consider the sets {0}, {01}, {0011}, etc. Show that ALLDFA is decidable. Show how to construct a DFA that accepts L n. 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. Finally, we give an application to the theory of games. 1, begin chapter 4. Since EF is a very weak fragment of the modal µ-calculus, model checking with EF is decidable for many more classes of infinite-state systems. Let A and B be Turing-recognizable languages such that A S B =. CS3231 : Tutorial - 7 Rahul Jain 18-Oct-2010 Q1 : Let In nite DFA = fhAi: A is a DFA and L(A) is an in nite languageg. There may be other problems in the theory that are undecidable, in which case no solution exists. (In each transition, we can read. Decidability Let a language be any set of strings (or words) over a given finite alphabet. boolean algebras with quantification over ideals, is decidable. Morphic words have been intensively studied in both of these contexts. Assume for contradiction that there exists a four state DFA M. , Fsub is not essentially undecidable (Tarski, 1949). This alternative point of view may make it easier to see why the satisfaction problem is decidable. L is said to beTuring-decidable(Recursiveor simply decidable) if there exists a TM M which decides L. (2) If L and Lc are Turing-recognizable, L is decidable. Donate to arXiv. There is a tape head that can read and write symbols and move around on the tape. 10 Show that A is decidable, where A = { | M is a DFA which does not accept any string containing an odd number of 1's}. com - id: 3e23d0-NWM3N. How to show that the set of universal sentences with infinite models is a decidable set? \ \phi$is a universal sentence with an infinite model$ \}$is a. boolean algebras with quantification over ideals, is decidable. •It is sufficient to show the set E of encoding of TMs is countable (as each TM DFA, E CFG, EQ DFA are decidable languages •TM is more powerful than CFG •A TM is undecidable •The complement of A. Consequently, this class is a natural target for proving a decidable synthesis result. 110 is decidable by constructing a decider for it. Consider the following language: C = {hM,wi|M is a Turing machine and never halts. DEGREES OF INDISCERNIBLES IN DECIDABLE MODELS BY H. Let k be the number of states of A. But such a proof does not imply that the theory itself is complete or decidable. Method 2: Since EQ DFA is decidable, we assume that is an algorithm M (i. A useless state in a pushdown automaton is never entered on any input string. (b)[5 marks] Show that the set of decidable languages is closed under concatenation. -Asol L, c is Turing-decidable, by Theorem 3. (16) decidable problem. DFA Introduction; DFA Examples. We need to show that a TM can test to see if a DFA accepts any. Decidable and recognizable languages 1. We construct a. The notion of tiling means that we have an infinite set of square tiles of the same size, but only finitely many are different from each other. • If a language is both semi-decidable and co-semi-decidable, then it is decidable.$ Show that $ALL_{DFA}$ is decidable. •Repeat until the last input symbol has been read. Is L(G)=∑*? M is a Turing machine. Decidability Let a language be any set of strings (or words) over a given finite alphabet. (15 points) Show that the set of all in nite binary sequences is uncountable. Moreover, we show that the join levels and the subsequent intersection levels do not coincide. Show that INFINITE DFA is decidable. Separation logic (SL) has gained widespread popularity because of its ability to succinctly express complex invariants of a program’s heap configurations. Sofya Raskhodnikova; based on slides by Nick Hopper. •Read an input symbol and move to the indicated state. S 1 is over alphabet { a , b } and it is infinite. $\begingroup$ DFA accepting a string of length greater than the number of states seems like a sufficient condition to prove that the DFA accept an infinite language. Show that INFINITE PDA is decidable. L(D) is infinite. How can we compare sizes of infinite sets? Definition: Two sets have the same size if there is a correspondence (bijection) between them. You may assume that the language EMPTY DFA which is de ned similarly, but with condition L(A) = ; is decidable. There is some novelty in the presentation. 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. edu for assistance. Let L be a recognizable language over the alphabet {0,1} such that L contains exactly one string of length n for every n. The exam has seven problems and totals 240 pts. Accept otherwise. Construct a DFA D that accepts all strings of length k or more 3Construct a DFA M such that L(M)=L(A) intersect L(D) 4Test L(M. Decidability of a logical system. For example, the set of odd-length strings — L={0,1,000,001,010,011,100,101,110,111,…} is a language over the alphabet set {0,1}. A language is a subset of strings over some alphabet. Is L(G)=∑*? M is a Turing machine. A classic use of where the pumping lemma helps in a. Thus, the language of E' is an infinite decidable subset of A. That is, all words in the language are accepted by the TM. E(dfa) is a decidable language. Finally, to show the semi-linearity of the range of the QL defined by a monolithic expression E with d atoms, we (1) put E in normal form ϕ (W 1, …, W d), (2) construct the unambiguous WA over Z d obtained by the product of the d atoms of E, (3) apply Lemma 3 on this product to show that its range is semi-linear, and combine it with ϕ to. The halting problem can be used to show that other problems are undecidable. INFINITE DFA = fhDijDis a DFA such that L(D) is in niteg: 2. Q2 : Let A = fhMi: M is a DFA which doesn't accept any string containing an odd number of 1sg. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following. [email protected] Consequently, this class is a natural target for proving a decidable synthesis result. (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. 19) Let S = fhMijM is a DFA that accepts wR whenever it accepts wg. (16) decidable problem. You may assume that the language EMPTY DFA which is de ned similarly, but with condition L(A) = ; is decidable. The Turning machine is similar to DFA but with an infinite tape serving as unlimited memory. 2) Let INFINITE DFA = { | A is a DFA and L(A) contains an infinite number of strings}. com - id: 3e23d0-NWM3N. Is L(G)=∑*? M is a Turing machine. •Proof:⇐ -Gvein M 1 recognizing L, and M 2 recognizing Lc. Then if the DFA accepts any string of length p, then the language is in nite; if not, the language is nite. Repeat until no new states get marked:. The following language L is decidable. • AREX = { R,w ⃒R is a regular expression that generates w} ‣By reduction to ANFA. Using the fully abstract game se-mantics of RML, our algorithm reduces the problem to the language equivalence of visibly pushdown automata. Formally speaking, construct TM S: S = "On input : Construct DFA D_A equivalent to A; Construct DFA D_0 that accepts. a DFA and an RE are 'equivalent' if they have the same language. Each logical system comes with both a syntactic component, which among other things determines the notion of provability, and a semantic component, which determines the notion of logical validity. 9 To understand the algorithm for this one, there is some setup that is required. In 1964, Ginsburg and Hibbard showed in their JACM paper Solvability of Machine Mappings of Regular Sets to Regular Sets that on input regular languages L and R it is decidable whether there is a generalized sequential machine (gsm) which maps L onto R. Prove that Regular Sets are NOT closed under infinite union. CS3231 : Tutorial - 7 Rahul Jain 18-Oct-2010 Q1 : Let In nite DFA = fhAi: A is a DFA and L(A) is an in nite languageg. Express this problem as a language and show that it is decidable. Take the DFA that accepts Check if there is a walk with cycle from the initial state to a final state Question: Answer: Costas Busch - LSU * DFA is infinite DFA is finite Costas Busch - LSU * Given regular languages and how can we check if ?. Show that ALL DFA is decidable, i. (10 points) Show that INFINITE DFA is a decidable language. 13 Some other decidable problems • ANFA = { M,w ⃒M is an NFA that accepts w} ‣By direct simulation, or by reduction to ADFA. How can we compare sizes of infinite sets? Definition: Two sets have the same size if there is a correspondence (bijection) between them. 2 CFLs decidable; TMs having CF languages not! 3 Dealing with IFF; 4 Notes on Cardinality and Infinities; 5 RE sets being closed under Kleene-star; 6 TM Questions on Asg7; 7 The Post Correspondence Problem (PCP) 7. If infinite words are in your scope, you can generalize DFA (with parity condition) to the so-called Good-for-Games automata (GFG), that still have polynomial containment. If X is decidable, so is X-bar (by exchanging the accepting and rejecting states, as we did for DFA’s). 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. Graphical representation of DFA. RL The accepting problem for DFAs of testing if a particular DFA accepts a given string can be expressed as the following language: ADFA = {|B is a DFA that accepts w. Decidable. Do the Problem 4. Show that A ε CFG is decidable. Show that the union of two recognizable sets is. Notepad/Notepad ++ editor 3. 18]Show that a language is decidable iff some enumerator enumerates the language in the standard string order. a decidable). G generates initiely many strings, not all strings}. 2 Consider the problem of determining whether a DFA and a regular expression are equivalent. INFINITE DFA is decidable. Practice Problems for Final Exam CS 341: Foundations of Computer Science II Prof. A language is decidable if some TM decides it (chapter 3). Aug Sept 13 Apr May 26 MONTH Jan Feb Mar Jun Jul Oct Nov Dec 28 DAYS OF RAIN 2 17 28 22 Calculate for this rainfall data the A. Show that, unlike stacks which can only recognize context-free languages, a queue automaton can simulate a Turing machine. Theorem 4 (Theorem 4. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. We show that this problem is decidable, EXPSPACEcomplete. We say an infinite set A have the same size as N, if there exists a one-to-one correspondence f: N A. Show that the union of two recognizable sets is. Automata Theory: The study of abstract machines and automata, including the computational. Show that S is decidable. The idea is to attempt to fill the upper right quadrant of a coordinate plane with these tiles. A Turngi Machine. 2) Let INFINITE DFA = { | A is a DFA and L(A) contains an infinite number of strings}. You need to show that, given a Context Free Grammar G, one can check whether the language L(G) recognized by G is included in a*b*. Let INFINITE DFA = { | A is a DFA and L(A) is an infinite language}. D2/ using the Cartesian-product construction from class and check whether an accept state is reachable from the start state. A useless state in a pushdown automaton is never entered on any input string. (see related Question 4. The machines never go into an infinite loop. (Example 2. 4,5 Review Algorithms created in proofs from prior chapters (p. If $L$ is decidable, then we can just generate strings in lexicographic order, and test if each is in $L$ , thus generating an enumerator that prints in standard string order. Give an example in the spirit of the recursion theorem of a program in a real programming. Since EF is a very weak fragment of the modal µ-calculus, model checking with EF is decidable for many more classes of infinite-state systems. Prove that INFINITEDFA = { | A is a DFA and L(A) is an infinite language} is decidable. Tree unfoldings Caucal’s Hierarchies Higher Order Pushdown Graphs Decidable MSO: probing the boundaries Logic and Computation in Finitely Presentable Inﬁnite Structures Lecture 6: The Caucal (Pushdown) Hierarchy and MSO Valentin Goranko and Sasha Rubin ESSLLI 2006, Malaga, August 2006. Please write in the exam books only|if you need more paper you may ask. (15 points) Show that the set of all in nite binary sequences is uncountable. 18) : If we have an enumerator which enumerates a language in lexicographic order, then we can make a decider using it. (Example 2. Formulate this problem as a language and show that it is decidable. 9 in Sipser, with solution included. Compute f(w) 2. Proof: The following TM M decides E. 13) A Turing machine with stay put instead of left is similar to an ordinary Turing machine, but the transition function has the form δ : Q×T → Q×T ×{R,S} At each point the machine can move its head right or let it stay in the same position. 20 Show that = ∑*} show that is All DFA decidable (page#29 and 30 handouts, solutn manual #38) 13. The alphabet could consist of the symbols we normally use for communication, such as the ASCII characters on a keyboard, including spaces and punctuation marks. Construct DFA A to accept Σ ∗111Σ. 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. Decidable First-Order Transition Logics for PAProcesses Article in Information and Computation 203(1):75-113 · July 2000 with 16 Reads How we measure 'reads'. (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. The squares along the boundary of the rectangle contain the symbol # and the internal. Every inconsistent theory is decidable, as every formula in the signature of the theory will be a logical consequence of, and thus a member of, the theory. is decidable. In the first part of the proof, we assume that B A!= ∅. Finally, we examine the DFA to see if there is a path from the start state to an accept state that begins with 11 and ends with 0. 22 (page 208) and the fact that EQ DFA is decidable, EQ DFA;REX is also decidable. • AREX = { R,w ⃒R is a regular expression that generates w} ‣By reduction to ANFA. Calculate a size that works. We want to show that L(A) is infinite. The first-order theory of arbitrary boolean algebras with a sequence of distinguished ideals is decidable. Let $ALL_{DFA} = \{ \langle{ A }\rangle \mid A \text{ is a DFA and}\: L(A) = \Sigma^{\ast}\}. Finally, we examine the DFA to see if there is a path from the start state to an accept state that begins with 11 and ends with 0. [6 marks] Let ALL DFA = fhAi j A is a DFA and L(A) = g. (Example 2. Simulate on input. Disjunctive ASP with functions: Decidable queries and effective computation* - Volume 10 Issue 4-6 - MARIO ALVIANO, WOLFGANG FABER, NICOLA LEONE. This is directly from the textbook (page 168) and we went over it in class. So, this is decidable. The machines never go into an infinite loop. 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. Demonstrate that this problem is decidable. A TM for L = { w#w DFA is decidable Proof: A DFA is a special case of a TM. M =  On input 〈𝑫〉, where 𝑫 is a DFA: Use BFS to determine if an accepting state of D. NFA is decidable Proof idea 2: •Design transducer that converts NFA to equivalent DFA •Decidable: finite number of steps: breadth first search of DFA states based on transitions from start state and from resulting states; at most 2k states if |Q NFA |=k •Ran A DFA on the equivalent DFA we just defined above –decidable as proven in. Express this problem as a language and show that it is decidable. Consider the language A = {hMi|M is a DFA which doesn’t accept any string with an odd number of 1’s}. Construct a DFA M such that L(M) = L(A) ∩ L(D). Repeat until no new variables get marked: a) Mark any variable A where G has a rule A-->U 1,. Automata Theory: The study of abstract machines and automata, including the computational. These constraints are essentially linear inequalities between infinite lists of nonnegative rational numbers which are added and compared pointwise. Noticing some elementary results about the cardinalities of the models of these theories, we show that they can nevertheless be combined with almost any other decidable theory. 9 in the textbook). 10 Show that A is decidable, where A = { | M is a DFA which does not accept any string containing an odd number of 1’s}. (a)[5 marks] Show that the set of decidable languages is closed under complement. (15 points) Show that the set of all in nite binary sequences is uncountable. 3] Let ALLDFA = { |A is a DFA that recognizes Σ∗}. 5, and so determining if L(A) = Sigma* is decidable as well. The following machine decides it: T := “On input M, where M is a DFA, 1. DFA = { ¢A²| A is a DFA and L(A) = Ø} Theorem 4. That is, show that if L1 and L2 are decidable languages, then L1 intersection L2 is a decidable language. Show that every infinite Turing-recognizable language has an infinite decidable subset. Convert P0 into an equivalent. decidable languages are not closed under complementation, but closed under intersection and union. 194) A DFA = {| B is a DFA that accepts w}. a DFA which can look at the front symbol of a queue; based on that symbol and its ﬁnite state, it is allowed to change its state, pop the front symbol, and/or push a symbol onto the back of the queue. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. boolean algebras with quantification over ideals, is decidable. Murawski2, and C. The Emptiness Problem asks, given some probability 0 ≤ λ ≤ 1, whether there exists a word accepted with probability greater than λ , and the Isolation Problem asks whether there. Show that Ln is recursively enumerable. Now suppose there is a program called. 1) To decide whether a particular DFA accept a given string nonaccepting state,. Speaking informally, you can show this by constructing a Turing Machine constructs a DFA D_A equivalent to NFA A. Decidable languages We can express diﬀerent computational problems as languages. Consequently, this class is a natural target for proving a decidable synthesis result. Construct the DFA for the symmetric difference language (closure property) 2. Show that ag-TMs can recognize any Turing-recognizable language. We show that the state-. Show that every infinite recognizable set has an infinite decidable subset. For a decidable language, for each input string,. is reachable from from its start state. How to show that the set of universal sentences with infinite models is a decidable set? \ \phi$ is a universal sentence with an infinite model$\}$ is a. Given a language L, let mid(L) be the following language fxj9y;z2 such that yxz2Lg: Show that if Lis Turing acceptable, then mid(L) is Turing acceptable. In the context of TMs and looping, it's useful to think about looping for all of our machines. COMP NFA = {(N, N'): L(N) = Σ * ∖ L(N')}. Short answers: (a) De ne the following terms and concepts: i. This alternative point of view may make it easier to see why the satisfaction problem is decidable. We need to show that a TM can test to see if a DFA accepts any. Theorem: A NFA is a decidable language. Assignment 7 Sipser 4. Given a language L, let mid(L) be the following language fxj9y;z2 such that yxz2Lg: Show that if Lis Turing acceptable, then mid(L) is Turing acceptable. 17) I show a reduction to the decidability of a problem which is known to be undecidable and hence prove the undecidability of the original language. out, and paths; 10. Mark the start state of A 2. Theorem: ADFA is a decidable language. We extend this result to the other join levels of the Trotter-Weil hierarchy. Let INFINITE PDA = fhMi j M is a PDA and L(M) is infiniteg. **** (6) The Odd One Out (From Michael Sipser, Introduction to the Theory of Computation, 2nd ed. Construct DFA B such that L(B) = L(A) T L(R). Using the subset construction, constructs DFA MR such that L(MR) = L(NR) 4. Test L(M) = ∅ using the E DFA decider T from. Prove that is Turing-recognizable if and only if there is a decidable language. Test L(M) = ∅ using the E DFA decider T from. Let INFINITE PDA = fhMi j M is a PDA and L(M) is infiniteg. Then given a string a Turing machine that accept the language starts the computation. decidable if there is an algorithm to - Check if the DFA can reach an accept state from the is stuck in an infinite loop. 2 Consider the problem of determining whether a DFA and a regular expression are. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. The intersection of. Consider the language A = {hMi|M is a DFA which doesn’t accept any string with an odd number of 1’s}. A context-free grammar is in Chomsky normal form if every rule is of the form: L(G) and |w| > 0, then any derivation of w in G has length 2|w| - 1 Proof (by induction on |w|):. But such a proof does not imply that the theory itself is complete or decidable. The complete algorithm is now as follows: "On input hRi:. Chapter 4: Decidability Decidable Languages: • Acceptance problem expressed as languages for regular expressions: o ADFA = {< B, w > | B is a DFA that accepts input string w} o The problem of testing whether a DFA B accepts an input w is the same as the problem of testing whether is a member of the language ADFA. Full text of "Introduction To Theory Of Computation" See other formats. Let INFINITE DFA = { | A is a DFA and L(A) is an infinite language}. Show that INFINITE PDA is decidable. • AREX = { R,w ⃒R is a regular expression that generates w} ‣By reduction to ANFA. Show that B is uncountable using a proof by diagonalization. accept string w ? (Decidable). Decidability regular CF decidable Turing recognizable • D is a DFA that accepts w • N is an NFA that accepts w • D is a DFA that accepts a non-empty language • A, B are DFAs and L(A) = L(B) • C is a CFL that accepts w • C is a CFL that accepts a non-empty language • But, A, B are CFLs and L(A) = L(B) not decidable • When the model of comp increases in power. Regular matching problems have a long history. Finally, we give an application to the theory of games. CFG is decidable Proof Idea: Transform G into Chomsky Normal Form. If T accepts, reject. In particular we show: (a) The set of recursive $\omega$-words with decidable monadic second order theories is $\Sigma_3$-complete. Almeida and Azevedo have shown that the join of R-trivial and L-trivial finite monoids is decidable; this is the first non-trivial join level of the Trotter-Weil hierarchy. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. A language L is called decidable (or recursive) if ∃ a. Show your work|this may help for partial credit. Question: Let INFDFA = { | M is a DFA and L(M) is an infinite language}. , quantum computers cannot be real computers or Zeno machines, and they do not break the Church. We can write an infinite table for every pair (M, w) all possible. Thus the question of membership in the language is decidable. 2 shows that not all Turing-recognizable are decidable. 4 JHU-CTY Theory of Computation (TCOM) Lancaster 2007 ~ Instructors Kayla Jacobs & Adam Groce. If there is a one-to-one map f: A --> B, we say that B is at least as big as A. Finally, we give an application to the theory of games. (10 points) Show that INFINITE DFA is a decidable language. If not, then reject. , subdivided as shown. Equivalently, if A is undecidable and reducible to B, B is undecidable. To find the solution of this problem, we can easily. Show that A is Turing decidable. 10 Show that A is decidable, where A = { | M is a DFA which does not accept any string containing an odd number of 1's}. So, this is decidable. Show that INFDFA is decidable. 20 is decidable. Assg 9 - Solution sketches with study points added. 67) Lemma 1. decidable if there is an algorithm to - Check if the DFA can reach an accept state from the is stuck in an infinite loop. The complete algorithm is now as follows: “On input hRi:. Question 6 (25): These questions use the definition of the language COMP C above. Since EF is a very weak fragment of the modal µ-calculus, model checking with EF is decidable for many more classes of infinite-state systems. There is a tape head that can read and write symbols and move around on the tape. Let q1 be the state that M transitions to on input 1, and let F0 be the set of states that lead to F on an input of 1. Let PALDFA = fhMi j M is a DFA that accepts some palindromeg. We show that if either the memoizing assumption or the early-assumes assumption is relaxed, the verification problem becomes undecidable, arguing for the necessity of these assumptions. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. (refer to sipser page 172) 3. 10INFINITEDFA = fhAijis a DFA and L(A) is an in nite languageg. Theorem 4 (Theorem 4. The language accepted by a DFA is the set of all strings accepted by it. $\begingroup$ Any language accepted by a DFA (i. The Turning machine is similar to DFA but with an infinite tape serving as unlimited memory. At any point in time, if the Turing machine is running, there is no way of telling whether it is in an infinite loop or along the way to a solution and it needs more time. Construct a DFA M such that L(M) = L(A) ∩ L(D). Prove that is Turing-recognizable if and only if there is a decidable language. Problem 2: Recognizibility is linked with the idea of unbounded existential search. Almeida and Azevedo have shown that the join of R-trivial and L-trivial finite monoids is decidable; this is the first non-trivial join level of the Trotter-Weil hierarchy. This is directly from the textbook (page 168) and we went over it in class. Busch - LSU 8 We can actually show: Decidable Turing-Acceptable HALT TM. Solution: From last problem (sipser 3. Every decidable language is Turing-Acceptable. Decidable and Semi-Decidable Languages Decidable A language L is Decidable if for every string w, there is a Turing Machine M that correctly Decides whether w∈L • M Halts and Accepts if w∈L • M Halts and Rejects if w∉L Semi-Decidable A language L is Semi-Decidable if for every string w, there is a Turing Machine M that Semi-Decides whether w∈L • M Halts and Accepts if w∈L. Thus, Lc is Turing-recognizable. Run the universal U on. Decidable and Semi-decidable Languages (Score: _____ out of 20 points) Let Sigma be an alphabet. Then contains some string of length at most. We will show there. e, or semi-decidable) if ∃ a Turing machine M such that L(M) = L. Introduction to the Theory of Computation Due Wednesday, June 25 (at the nal exam) Exercise 1. IEvery nite language is decidable: For example, by a TM that has all the strings in the language \hard-coded" into it IWe just saw some example algorithms all of which terminate in a nite number of steps, and output yes or no (accept or reject. 5] Let INFINITEDFA = { |A is a DFA and L(A) is an inﬁnite language }. • Encoding languages and Turing machines as binary strings. Let ALLDFA = cfw_| A is a DFA and L (A) =. Show that INFINITE DFA is decidable. Since we can encode a DFA D as a string hDi, the acceptance problem is equivalent to testing for membership in the language ADFA = {hD,wi | D is a DFA that accepts w} The acceptance problem is called decidable if the corresponding language is (Turing-)decidable. (Æ) Let L be a decidable language. 18 Show that a language is decidable i some enumerator enumerates the language in lexico-graphic order. It then constructs the DFA D_0 that accepts the language {0,1}*, we can then simulate the decider for EQ_DFA on. Understanding this should enable us to identify decidable fragments of Sl. Some decidable languages: cfl, regular… Some decidable decision problems: Membership of CFL L. Nakhleh ([email protected]
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