Show language is decidable

A problem with a yes/no answer. Determining whether some potential solution to a question is actually a solution or not. E.g. "Is 43669" a prime number?".This is in contrast to a "search problem" which must find a solution from scratch, e.g. Nov 08, 2020 · Question: Are there languages decidable in linear time by RAM machines that have superlinear time complexity lower bounds for Multitape Turing machines? Background: I recently stumbled upon the dissertation "Computational complexity of multitape Turning machines and Random Access Machines" by Takumi Kasai (see here: https://repository.kulib ... Typically, show how to “Reduce” any string from language into a string for a language you know is decidable Convert the answer from the known decider into an answer for the desired decider For undecidability problems, form a contradiction Make sure you know what the language is (call it B 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.aaaa ACCEPT algorithm alphabet Answer applying automaton bbbb blank called computable Consider consists contain context free grammar Convert corresponding decidable defined derivation Design a Turing element enter enumerable equal equation equivalent Example Construct Example Design exists final finite automata function getting Give given ... To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5Show that ALL DFA is decidable. Proof #1: The following TM decides ALL DFA: S = "On input 〈A〉, where A is a DFA: 1. Construct a DFA B such that L(B) is the complement of L(A). 2. Use the TM T from Thm 4.4 (deciding EDFA) on input <B> 3. Accept if T accepts, reject if T rejects."To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5Warm-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 4.1) ACCEPT {( DFA ) | i i L(DFA)} string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w ...Language: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. Oct 26, 2020 · Show that the following language is decidable by finding the algorithm for the finite automaton. 2. How to show that this language is decidable. 2. I We show a problem decidable/undecidable by reducing it to another problem. One type of reduction: mapping reduction. De nition I Let A, B be languages over . A is mapping reducible to B, written A m B, if there is a computable function f : ! such that w 2A if and only if f (w) 2B. I Function f is called the reduction of A to B. De nitionShowing that a Language is Decidable/Recognizable. How do we show that a language is TD: Build a decider for it ; How do we show that a language is TR: Build a decider or recognizer for it ; How do we show that language L is TR and not TD: Build a Recognizer for L ; PROVE that there is no Decider for l ; We will see examples soonHow can you show that every CFL is a Decidable Language? How do you build a TM that decides a CFL A? since A is a CFL, there is a CFG G that generates A using G, build a TM M[G] that decides A M[G] = "on input w: 1. run the A[CFG] TM S on input 2. if S accepts, accept otherwise, reject" How are the 4 classes of languages related? Formulate this problem as a language and show that it is decidable. Consider a language L, there is a Turing Machine M which either accept or reject input string s. Also, because there is a decider that decides language L, it is Turing-decidable. languages • We defined a regular language to be one that is accepted by some DFA (or NFA). • We can prove that a language is regular by this definition if and only if it corresponds to some regular expression. Thus, – 1) Given a DFA or NFA, there is some regular expression to describe the language it accepts is not a Decidable Language • We cannot reuse the same reasoning used to show that EQ DFA is a decidable language since CFGs are not closed under complement and intersection • As it turns out, EQ CFG is not decidable! 10/10/19 Theory of Computation - Fall'19 Lorenzo De Stefani 16 Problem I. Decidable and Semi-decidable Languages (Score: _____ out of 20 points) Let Sigma be an alphabet. (20 points) Show that the set of decidable languages is closed under intersection.That is, show that if L1 and L2 are decidable languages, then L1 intersection L2 is a decidable language.Prove the following: a. Show that a language L is acceptable iff some enumerator E enumerates it. b. Show that a language L is decidable iff some enumerator E enumerates the language in L-ordering. Question: Prove the following: a. Show that a language L is acceptable iff some enumerator E enumerates it. b. Show that a language L is decidable ...Show that a language is decidable iff some enumerator enumerates the language in lexicographic orderHelpful? Please support me on Patreon: https://www.patre...T decides a language L if T recognizes L, and halts in all inputs. That is, a decider T is guaranteed to either accept, or reject, and never fall into an infinite loop. Languages recognized by a TM are called recognizable. Languages decided by a TM are called decidable.A language L is decidable if and only if L is CE and L is co-CE. Corollary The complement of HALT is not CE. Proof: we know that HALT is CE but not decidable if complement of HALT wereCE, then HALT is CE and co-CE hence decidable. Contradiction. Bottom line: For every \strictly semi-decidable language", its complement cannot be semi-decidable.A problem with a yes/no answer. Determining whether some potential solution to a question is actually a solution or not. E.g. "Is 43669" a prime number?".This is in contrast to a "search problem" which must find a solution from scratch, e.g. How can you show that every CFL is a Decidable Language? How do you build a TM that decides a CFL A? since A is a CFL, there is a CFG G that generates A using G, build a TM M[G] that decides A M[G] = "on input w: 1. run the A[CFG] TM S on input 2. if S accepts, accept otherwise, reject" How are the 4 classes of languages related? ii.Show that A is decidable. 2.(Decidable languages, part II) (a)(Persistent variable) A variable A in CFG G is persistent if it appears in every derivation of every string w in G. Given a CFG G and a variable A, consider the problem of testing whether A is persistent. Formulate this problem as a language and show that it is decidable. Show that a language is decidable iff some enumerator enumerates the language in lexicographic orderHelpful? Please support me on Patreon: https://www.patre...A regular tree language L is locally testable if membership of a tree in L depends only on the presence or absence of some fix set of neighborhoods in the tree. In this paper we show that it is decidable whether a regular tree language is locally testable. The decidability is shown for ranked trees and for unranked unordered trees. 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.Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,ii.Show that A is decidable. 2.(Decidable languages, part II) (a)(Persistent variable) A variable A in CFG G is persistent if it appears in every derivation of every string w in G. Given a CFG G and a variable A, consider the problem of testing whether A is persistent. Formulate this problem as a language and show that it is decidable. Jan 02, 2015 · Regular languages are generated by regular grammars, so a similar method of proof can be used to show that the language is decidable. Regular languages are also generated by FSA. Thus, it is sufficient to execute a graph traversal on the graph representation of the FSA from the input string, . Oct 26, 2020 · Show that the following language is decidable by finding the algorithm for the finite automaton. 2. How to show that this language is decidable. 2. aaaa ACCEPT algorithm alphabet Answer applying automaton bbbb blank called computable Consider consists contain context free grammar Convert corresponding decidable defined derivation Design a Turing element enter enumerable equal equation equivalent Example Construct Example Design exists final finite automata function getting Give given ... the obvious choice is an assumed decider for a given decidable language. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. To prove that a given language is Turing-recognizable: Construct an algorithm that accepts exactly those strings that are in the language.I We show a problem decidable/undecidable by reducing it to another problem. One type of reduction: mapping reduction. De nition I Let A, B be languages over . A is mapping reducible to B, written A m B, if there is a computable function f : ! such that w 2A if and only if f (w) 2B. I Function f is called the reduction of A to B. De nitionNov 08, 2020 · Question: Are there languages decidable in linear time by RAM machines that have superlinear time complexity lower bounds for Multitape Turing machines? Background: I recently stumbled upon the dissertation "Computational complexity of multitape Turning machines and Random Access Machines" by Takumi Kasai (see here: https://repository.kulib ... Show that if A and B are Turing-recognizable languages and A ( B, then machine(A) is not decidably separable from machine(B). 1This is an interesting property in the following situation: imagine two undecidable, disjoint languages A and B. If some language C decidable separates them, then C can help give | decidable! | hints for strings of A and B, Formulate this problem as a language and show that it is decidable. Consider a language L, there is a Turing Machine M which either accept or reject input string s. Also, because there is a decider that decides language L, it is Turing-decidable. A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 diagram −. 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 4.1) ACCEPT {( DFA ) | i i L(DFA)} string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w ...Equivalent language: AH = { <M,w> | TM M halts on input w} Need to show AH is undecidable We know ATM = {<M,w> | TM M accepts w} is undecidable Show ATM is reducible to AH (Theorem 5.1 in text) Suppose AH is decidable there's a decider MH for AH Then, we can construct a decider DTM for ATM: On input <M,w>, run MH on <M,w>.Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,Show that ALL DFA is decidable. Proof #1: The following TM decides ALL DFA: S = "On input 〈A〉, where A is a DFA: 1. Construct a DFA B such that L(B) is the complement of L(A). 2. Use the TM T from Thm 4.4 (deciding EDFA) on input <B> 3. Accept if T accepts, reject if T rejects."Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,I We show a problem decidable/undecidable by reducing it to another problem. One type of reduction: mapping reduction. De nition I Let A, B be languages over . A is mapping reducible to B, written A m B, if there is a computable function f : ! such that w 2A if and only if f (w) 2B. I Function f is called the reduction of A to B. De nitionA language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 diagram −. A regular tree language L is locally testable if membership of a tree in L depends only on the presence or absence of some fix set of neighborhoods in the tree. In this paper we show that it is decidable whether a regular tree language is locally testable. The decidability is shown for ranked trees and for unranked unordered trees. A regular tree language L is locally testable if membership of a tree in L depends only on the presence or absence of some fix set of neighborhoods in the tree. In this paper we show that it is decidable whether a regular tree language is locally testable. The decidability is shown for ranked trees and for unranked unordered trees. Language: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. is not a Decidable Language • We cannot reuse the same reasoning used to show that EQ DFA is a decidable language since CFGs are not closed under complement and intersection • As it turns out, EQ CFG is not decidable! 10/10/19 Theory of Computation - Fall'19 Lorenzo De Stefani 16 ii.Show that A is decidable. 2.(Decidable languages, part II) (a)(Persistent variable) A variable A in CFG G is persistent if it appears in every derivation of every string w in G. Given a CFG G and a variable A, consider the problem of testing whether A is persistent. Formulate this problem as a language and show that it is decidable. Show that a language is decidable iff some enumerator enumerates the language in lexicographic orderHelpful? Please support me on Patreon: https://www.patre...Show that ALL DFA is decidable. Proof #1: The following TM decides ALL DFA: S = "On input 〈A〉, where A is a DFA: 1. Construct a DFA B such that L(B) is the complement of L(A). 2. Use the TM T from Thm 4.4 (deciding EDFA) on input <B> 3. Accept if T accepts, reject if T rejects."Language: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. Oct 26, 2020 · Show that the following language is decidable by finding the algorithm for the finite automaton. 2. How to show that this language is decidable. 2. To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5Prove the following: a. Show that a language L is acceptable iff some enumerator E enumerates it. b. Show that a language L is decidable iff some enumerator E enumerates the language in L-ordering. Question: Prove the following: a. Show that a language L is acceptable iff some enumerator E enumerates it. b. Show that a language L is decidable ...Nov 08, 2020 · Question: Are there languages decidable in linear time by RAM machines that have superlinear time complexity lower bounds for Multitape Turing machines? Background: I recently stumbled upon the dissertation "Computational complexity of multitape Turning machines and Random Access Machines" by Takumi Kasai (see here: https://repository.kulib ... 1. (Problem 3.15) Show that the collection of decidable languages is closed under (a) Union: (in the textbook). (b) Concatenation: Let K,L be decidable languages. The concatenation of languages K and L is the language KL = {xy|x ∈ K and y ∈ L}. Since K and L are decidable languages, it follows that there exist turing machines M K and MLanguage: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. A problem with a yes/no answer. Determining whether some potential solution to a question is actually a solution or not. E.g. "Is 43669" a prime number?".This is in contrast to a "search problem" which must find a solution from scratch, e.g. $\begingroup$ Any language accepted by a DFA (i.e. a regular language) is decidable. This is true since the DFA can be simulated by a Turing machine (DFA is weaker than Turing machine). $\endgroup$ -Jan 02, 2015 · Regular languages are generated by regular grammars, so a similar method of proof can be used to show that the language is decidable. Regular languages are also generated by FSA. Thus, it is sufficient to execute a graph traversal on the graph representation of the FSA from the input string, . Prove the following: a. Show that a language L is acceptable iff some enumerator E enumerates it. b. Show that a language L is decidable iff some enumerator E enumerates the language in L-ordering. Question: Prove the following: a. Show that a language L is acceptable iff some enumerator E enumerates it. b. Show that a language L is decidable ...Show the collection of decidable languages is closed under the operation of 1)Complementation 2)Intersection 3)Concatenation* Hint: You can construct non-deterministic Turing machines! A non-deterministic machine is a "decider" if every path for any input eventually halts with accept or reject.Oct 26, 2020 · Show that the following language is decidable by finding the algorithm for the finite automaton. 2. How to show that this language is decidable. 2. provided. When proving closure of the class of decidable languages under a given operation the obvious choice is an assumed decider for a given decidable language. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. To prove that a given language is Turing-recognizable: To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5Typically, show how to “Reduce” any string from language into a string for a language you know is decidable Convert the answer from the known decider into an answer for the desired decider For undecidability problems, form a contradiction Make sure you know what the language is (call it B A language 'L' is decidable if it is a recursive language. All decidable languages are recursive languages and vice-versa. Recursively enumerable language(RE) - A language 'L' is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in ...A language 'L' is decidable if it is a recursive language. All decidable languages are recursive languages and vice-versa. Recursively enumerable language(RE) - A language 'L' is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in ...Show that a language is decidable iff some enumerator enumerates the language in lexicographic orderHelpful? Please support me on Patreon: https://www.patre...Some language is not decidable. Because there are more languages than TMs. We will show some specific language 0. TM. is not decidable. Check-in 8.1 . Hilbert’s 1. st . question asked if there is a set of intermediate size between ℕand ℝ. Gödel and Cohen showed that we cannot answer this question by using the standard axioms of mathematics. languages • We defined a regular language to be one that is accepted by some DFA (or NFA). • We can prove that a language is regular by this definition if and only if it corresponds to some regular expression. Thus, – 1) Given a DFA or NFA, there is some regular expression to describe the language it accepts If A is decidable, the enumerator operates by generating the strings in lexicographic order and testing each in turn for membership in A using the decider. Those strings which are found to be in A are printed. If A is enumerable in lexicographic order, we consider two cases. If A is finite, it is decidable because all finite languages are decidable. Definition: A language is called semi-decidable (or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. In case the string does not belong to the language, the algorithm either rejects it or runs forever. Clearly, any decidable language is recognizable.Showing that a Language is Decidable/Recognizable. How do we show that a language is TD: Build a decider for it ; How do we show that a language is TR: Build a decider or recognizer for it ; How do we show that language L is TR and not TD: Build a Recognizer for L ; PROVE that there is no Decider for l ; We will see examples soonSome language is not decidable. Because there are more languages than TMs. We will show some specific language 0. TM. is not decidable. Check-in 8.1 . Hilbert’s 1. st . question asked if there is a set of intermediate size between ℕand ℝ. Gödel and Cohen showed that we cannot answer this question by using the standard axioms of mathematics. To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 diagram −. Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 diagram −. a. Show that for any infinite language L, L is decidable iff some enumerator TM enumerates L in lexicographic order. Proof. (Æ) Let L be a decidable language. Then, there exists a decider TM D such that L(D) = L. We can use D to construct an enumerator E for L as follows: E = "Ignore the input. 1.1. (Problem 3.15) Show that the collection of decidable languages is closed under (a) Union: (in the textbook). (b) Concatenation: Let K,L be decidable languages. The concatenation of languages K and L is the language KL = {xy|x ∈ K and y ∈ L}. Since K and L are decidable languages, it follows that there exist turing machines M K and MWarm-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 4.1) ACCEPT {( DFA ) | i i L(DFA)} string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w ...Some language is not decidable. Because there are more languages than TMs. We will show some specific language 0. TM. is not decidable. Check-in 8.1 . Hilbert’s 1. st . question asked if there is a set of intermediate size between ℕand ℝ. Gödel and Cohen showed that we cannot answer this question by using the standard axioms of mathematics. aaaa ACCEPT algorithm alphabet Answer applying automaton bbbb blank called computable Consider consists contain context free grammar Convert corresponding decidable defined derivation Design a Turing element enter enumerable equal equation equivalent Example Construct Example Design exists final finite automata function getting Give given ... A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w.Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 ...A language 'L' is decidable if it is a recursive language. All decidable languages are recursive languages and vice-versa. Recursively enumerable language(RE) - A language 'L' is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in ...Definition: A language is called semi-decidable (or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. In case the string does not belong to the language, the algorithm either rejects it or runs forever. Clearly, any decidable language is recognizable.$\begingroup$ Any language accepted by a DFA (i.e. a regular language) is decidable. This is true since the DFA can be simulated by a Turing machine (DFA is weaker than Turing machine). $\endgroup$ -1. Show that a language is decidable iff some enumerator enumerates the language in lexico-graphic order. 2. Let INFINITEPDA = {hMi | M is a PDA and L(M) is an infinite language}. Show that INFINITEPDA is decidable. 3. Let A = {hRi | R is a regular expression describing a language containing at least one string How can you show that every CFL is a Decidable Language? How do you build a TM that decides a CFL A? since A is a CFL, there is a CFG G that generates A using G, build a TM M[G] that decides A M[G] = "on input w: 1. run the A[CFG] TM S on input 2. if S accepts, accept otherwise, reject" How are the 4 classes of languages related? If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 52. (Sipser, Problem 3.18) Show that a language is decidable iff some enumerator enumerates the language in lexicographic order. [15 points] Solution:If A is decidable by some TM M, the enumerator operates by generating the strings in lexicographic order, testing each in turn for membership in A using M, and printing the string if it is in A. 1Show that C CFG is decidable. 6. Let A and B be two disjoint languages. Say that language C separates A and B if A ⊆ C and B ⊆ C. Show that any two disjoint co-Turing-recognizable languages are separable by some decidable language. 7. Given a natural number n, define f(n) to be n/2 if n is even and 3n+1 if n is odd. The (3n+1) T decides a language L if T recognizes L, and halts in all inputs. That is, a decider T is guaranteed to either accept, or reject, and never fall into an infinite loop. Languages recognized by a TM are called recognizable. Languages decided by a TM are called decidable.Formulate this problem as a language and show that it is decidable. Consider a language L, there is a Turing Machine M which either accept or reject input string s. Also, because there is a decider that decides language L, it is Turing-decidable. is not a Decidable Language • We cannot reuse the same reasoning used to show that EQ DFA is a decidable language since CFGs are not closed under complement and intersection • As it turns out, EQ CFG is not decidable! 10/10/19 Theory of Computation - Fall'19 Lorenzo De Stefani 16 a. Show that for any infinite language L, L is decidable iff some enumerator TM enumerates L in lexicographic order. Proof. (Æ) Let L be a decidable language. Then, there exists a decider TM D such that L(D) = L. We can use D to construct an enumerator E for L as follows: E = "Ignore the input. 1.Language: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. $\begingroup$ Any language accepted by a DFA (i.e. a regular language) is decidable. This is true since the DFA can be simulated by a Turing machine (DFA is weaker than Turing machine). $\endgroup$ -If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. Show that if A and B are Turing-recognizable languages and A ( B, then machine(A) is not decidably separable from machine(B). 1This is an interesting property in the following situation: imagine two undecidable, disjoint languages A and B. If some language C decidable separates them, then C can help give | decidable! | hints for strings of A and B, Language: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. Showing that a Language is Decidable/Recognizable. How do we show that a language is TD: Build a decider for it ; How do we show that a language is TR: Build a decider or recognizer for it ; How do we show that language L is TR and not TD: Build a Recognizer for L ; PROVE that there is no Decider for l ; We will see examples soonEquivalent language: AH = { <M,w> | TM M halts on input w} Need to show AH is undecidable We know ATM = {<M,w> | TM M accepts w} is undecidable Show ATM is reducible to AH (Theorem 5.1 in text) Suppose AH is decidable there's a decider MH for AH Then, we can construct a decider DTM for ATM: On input <M,w>, run MH on <M,w>.1 Answer1. Show activity on this post. By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not in the language. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it.Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). Run M on w. Although it might take a staggeringly long time, M will eventually accept or reject w. The set R is the set of all decidable languages. L ∈ R iff L is decidableNov 08, 2020 · Question: Are there languages decidable in linear time by RAM machines that have superlinear time complexity lower bounds for Multitape Turing machines? Background: I recently stumbled upon the dissertation "Computational complexity of multitape Turning machines and Random Access Machines" by Takumi Kasai (see here: https://repository.kulib ... the obvious choice is an assumed decider for a given decidable language. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. To prove that a given language is Turing-recognizable: Construct an algorithm that accepts exactly those strings that are in the language.How do I show this language {<C,A,B> | C,A,B are DFAs, L(C) contains the shuffle of L(A) and L(B)} is decidable ? I believe if I can construct automatas for A and B, then I can get an automata that contains the shuffle of them. I am also thinking about using emptiness testing but I have not made any progress yet.is not a Decidable Language • We cannot reuse the same reasoning used to show that EQ DFA is a decidable language since CFGs are not closed under complement and intersection • As it turns out, EQ CFG is not decidable! 10/10/19 Theory of Computation - Fall'19 Lorenzo De Stefani 16 To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5A regular tree language L is locally testable if membership of a tree in L depends only on the presence or absence of some fix set of neighborhoods in the tree. In this paper we show that it is decidable whether a regular tree language is locally testable. The decidability is shown for ranked trees and for unranked unordered trees. Typically, show how to “Reduce” any string from language into a string for a language you know is decidable Convert the answer from the known decider into an answer for the desired decider For undecidability problems, form a contradiction Make sure you know what the language is (call it B If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5Show that ALL DFA is decidable. Proof #1: The following TM decides ALL DFA: S = "On input 〈A〉, where A is a DFA: 1. Construct a DFA B such that L(B) is the complement of L(A). 2. Use the TM T from Thm 4.4 (deciding EDFA) on input <B> 3. Accept if T accepts, reject if T rejects."To prove a language is decidable, we can show how to construct a TM that decides it. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. Lecture 17: Proving Undecidability 4 Proofs of Un decidability How can you prove a language is un decidable ? Lecture 17: Proving Undecidability 5If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. ii.Show that A is decidable. 2.(Decidable languages, part II) (a)(Persistent variable) A variable A in CFG G is persistent if it appears in every derivation of every string w in G. Given a CFG G and a variable A, consider the problem of testing whether A is persistent. Formulate this problem as a language and show that it is decidable. A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 diagram −. A language 'L' is decidable if it is a recursive language. All decidable languages are recursive languages and vice-versa. Recursively enumerable language(RE) - A language 'L' is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in ...How can you show that every CFL is a Decidable Language? How do you build a TM that decides a CFL A? since A is a CFL, there is a CFG G that generates A using G, build a TM M[G] that decides A M[G] = "on input w: 1. run the A[CFG] TM S on input 2. if S accepts, accept otherwise, reject" How are the 4 classes of languages related? A problem with a yes/no answer. Determining whether some potential solution to a question is actually a solution or not. E.g. "Is 43669" a prime number?".This is in contrast to a "search problem" which must find a solution from scratch, e.g. the obvious choice is an assumed decider for a given decidable language. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. To prove that a given language is Turing-recognizable: Construct an algorithm that accepts exactly those strings that are in the language.Language is Turing recognizable if some Turing machine recognizes it •Also called "recursively enumerable" Machine that halts on all inputs is a decider. A decider that recognizes language L is said to decide language L Language is Turing decidable, or just decidable, if some Turing machine decides it 2 Example non-halting machinea. Show that for any infinite language L, L is decidable iff some enumerator TM enumerates L in lexicographic order. Proof. (Æ) Let L be a decidable language. Then, there exists a decider TM D such that L(D) = L. We can use D to construct an enumerator E for L as follows: E = "Ignore the input. 1.Language: Share on Facebook Twitter. Get our app ... We show that this is indeed the case for all the cases mentioned and in fact many more. a. Show that for any infinite language L, L is decidable iff some enumerator TM enumerates L in lexicographic order. Proof. (Æ) Let L be a decidable language. Then, there exists a decider TM D such that L(D) = L. We can use D to construct an enumerator E for L as follows: E = "Ignore the input. 1.Showing that a Language is Decidable/Recognizable. How do we show that a language is TD: Build a decider for it ; How do we show that a language is TR: Build a decider or recognizer for it ; How do we show that language L is TR and not TD: Build a Recognizer for L ; PROVE that there is no Decider for l ; We will see examples soonA language 'L' is decidable if it is a recursive language. All decidable languages are recursive languages and vice-versa. Recursively enumerable language(RE) - A language 'L' is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in ...A language L is decidable if and only if L is CE and L is co-CE. Corollary The complement of HALT is not CE. Proof: we know that HALT is CE but not decidable if complement of HALT wereCE, then HALT is CE and co-CE hence decidable. Contradiction. Bottom line: For every \strictly semi-decidable language", its complement cannot be semi-decidable.If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,Formulate this problem as a language and show that it is decidable. Consider a language L, there is a Turing Machine M which either accept or reject input string s. Also, because there is a decider that decides language L, it is Turing-decidable. If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. Oct 03, 2021 · Recursively enumerable languages are the formal languages that can be decide-able, ( fully or partially). According to the Chomsky hierarchy of formal languages, we can see the recursively enumerable languages as type 0 languages. some examples of recursively enumerable languages are; Recursive languages. Regular languages. Jan 02, 2015 · Regular languages are generated by regular grammars, so a similar method of proof can be used to show that the language is decidable. Regular languages are also generated by FSA. Thus, it is sufficient to execute a graph traversal on the graph representation of the FSA from the input string, . aaaa ACCEPT algorithm alphabet Answer applying automaton bbbb blank called computable Consider consists contain context free grammar Convert corresponding decidable defined derivation Design a Turing element enter enumerable equal equation equivalent Example Construct Example Design exists final finite automata function getting Give given ... Definition: A language is called semi-decidable (or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. In case the string does not belong to the language, the algorithm either rejects it or runs forever. Clearly, any decidable language is recognizable.Formulate this problem as a language and show that it is decidable. Consider a language L, there is a Turing Machine M which either accept or reject input string s. Also, because there is a decider that decides language L, it is Turing-decidable. provided. When proving closure of the class of decidable languages under a given operation the obvious choice is an assumed decider for a given decidable language. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. To prove that a given language is Turing-recognizable: 1 Answer1. Show activity on this post. By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not in the language. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it.A E, the language A is Turing-recognizable.(i.e. A E is recursively enumerable.) 5. Problem 4, page 131 Show that the a language is decidable if and only if there is some enumerator that enumerates the language in lexicographic order. Solution: Since we are proving an "if and only if" statement, we divide the proof into two parts. One part$\begingroup$ Any language accepted by a DFA (i.e. a regular language) is decidable. This is true since the DFA can be simulated by a Turing machine (DFA is weaker than Turing machine). $\endgroup$ -Show the collection of decidable languages is closed under the operation of 1)Complementation 2)Intersection 3)Concatenation* Hint: You can construct non-deterministic Turing machines! A non-deterministic machine is a "decider" if every path for any input eventually halts with accept or reject.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.Show the collection of decidable languages is closed under the operation of 1)Complementation 2)Intersection 3)Concatenation* Hint: You can construct non-deterministic Turing machines! A non-deterministic machine is a "decider" if every path for any input eventually halts with accept or reject.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 4.1) ACCEPT {( DFA ) | i i L(DFA)} string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w • DFA = {(<DFA>, w) | w is in L(DFA)} languages • We defined a regular language to be one that is accepted by some DFA (or NFA). • We can prove that a language is regular by this definition if and only if it corresponds to some regular expression. Thus, – 1) Given a DFA or NFA, there is some regular expression to describe the language it accepts A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w.Every decidable language is Turing-Acceptable. A decision problem P is decidable if the language L of all yes instances to P is 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 ...Showing that a Language is Decidable/Recognizable. How do we show that a language is TD: Build a decider for it ; How do we show that a language is TR: Build a decider or recognizer for it ; How do we show that language L is TR and not TD: Build a Recognizer for L ; PROVE that there is no Decider for l ; We will see examples soonFormulate this problem as a language and show that it is decidable. Consider a language L, there is a Turing Machine M which either accept or reject input string s. Also, because there is a decider that decides language L, it is Turing-decidable. Nov 08, 2020 · Question: Are there languages decidable in linear time by RAM machines that have superlinear time complexity lower bounds for Multitape Turing machines? Background: I recently stumbled upon the dissertation "Computational complexity of multitape Turning machines and Random Access Machines" by Takumi Kasai (see here: https://repository.kulib ... 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 4.1) ACCEPT {( DFA ) | i i L(DFA)} string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w ...Equivalent language: AH = { <M,w> | TM M halts on input w} Need to show AH is undecidable We know ATM = {<M,w> | TM M accepts w} is undecidable Show ATM is reducible to AH (Theorem 5.1 in text) Suppose AH is decidable there's a decider MH for AH Then, we can construct a decider DTM for ATM: On input <M,w>, run MH on <M,w>.Show that the following language is decidable by finding the algorithm for the finite automaton. 2. How to show that this language is decidable. 2. Why is it enough to test only the words up to length n ( number of state) in the given algorithm for a decidability problem. Hot Network QuestionsShow language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,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.Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. - I can take both recognizers and run them in parellel, simulating a step on each one, eventually one will accept,How do I show this language {<C,A,B> | C,A,B are DFAs, L(C) contains the shuffle of L(A) and L(B)} is decidable ? I believe if I can construct automatas for A and B, then I can get an automata that contains the shuffle of them. I am also thinking about using emptiness testing but I have not made any progress yet.If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Simply, Decidable ---- always halt Recognizable ---- halt or loop Let INFINITEDFA={<A> A is a DFA and L(A) is an infinite language} . Show that INFINITEDFA is decidable. Showing that a Language is Decidable/Recognizable. How do we show that a language is TD: Build a decider for it ; How do we show that a language is TR: Build a decider or recognizer for it ; How do we show that language L is TR and not TD: Build a Recognizer for L ; PROVE that there is no Decider for l ; We will see examples soonA language L is decidable if and only if L is CE and L is co-CE. Corollary The complement of HALT is not CE. Proof: we know that HALT is CE but not decidable if complement of HALT wereCE, then HALT is CE and co-CE hence decidable. Contradiction. Bottom line: For every \strictly semi-decidable language", its complement cannot be semi-decidable. catholic schools curriculumferraillage poteau 8x8umthakathi storieswhere to buy lvl beams near me X_1