# Show That Alldfa Is Decidable

Show that, if P=NP. Let A = {hR,Si| R and S are regular expressions and L(R) L(S)}. Run M1 on w. Think back to what you learned about NP-completeness. Note: The next question is a programming question. Hint: In this case, all the states that are reachable from the start state are final states. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. Show that NP is closed under union and concatenation. Let LEN_CFG = { | G is a CFG, k elementof Z^noneg, and L (G) Intersection sigma^k notequalto (where sigma is the alphabet of G)}. Let build M0as follows: M0= \On input w: 1. If T rejects, “reject” 2. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. P is the class of all languages that are decidable by deterministic single-tape Turing machines running in polynomial time. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Let ALLDFA-{(A): A Is A DFA And L(A) = Σ*). Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Let Lpal2 be the language wwRvvR:w, v E E' a. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. 17 December cs701 paper CS701 Q#1suppose there is a language L we know that L is Turing. Show that the collection of decidable languages is closed under union. A Prunable State In A DFA Is Some State That Is Never Entered While Processing Any Input String. Show a Turing Machine accepts all DFAs. (Hint: Look at the proof for EDFA to get an idea. CSE 105 Sp04, Problem Set 3 Solutions 3 Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. Then, let M be a DFA that. 1 (a) a Yes, because M on input 0100 ends in an accept state. Show thatALLDFAis decidable. proof DFA defines same language as minimal DFA. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. We show that Sis decidable. Show That The Following. Show that SUB_DFA is decidable. 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, by giving an example, that part (a) is not necessarily true if you replace both A's by A. M 1 works as follows. (b) Show that L is decidable. cs701 fall 2016 mid term paper. Run M 1 on w. Full text of "Introduction To Theory Of Computation" See other formats. 2) Let INFINITE DFA = { | A is a DFA and L(A) contains an infinite number of strings}. >> Anonymous Wed Oct 19 17:44:05 2016 No. Give an example of a DFA Astallion that in in ALLDFA. The algorithm M 1 inputs hAi, where Ais a DFA. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. pdf - Free download as PDF File (. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. 1 (a) a Yes, because M on input 0100 ends in an accept state. Problem 3 (5 Points). Get best Help for Others questions and answers in computer-architecture Page-2877, step-by-step Solutions, 100% Plagiarism free Question Answers. cs701 fall 2016 mid term paper. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Show That LEN_CFG Is Decidable. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. , looping cannot happen. E(dfa) is a decidable language. Berkshire Hathaway AGM is still the Warren Buffett show ; US election: Trump’s man in Vegas. Let A(CFG = { | G is a CFG that generates (}. Show that ALL DFA is decidable. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Show that, if A is nonempty, A contains some string of length at most k. Show that SUB_DFA is decidable. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. To show PALDFA is decidable, we construct a decider D for PALDFA as follows (Let K be a TM that decides ECFG): D = \On input hMi, 1. DFA with an accepting state in the initial state. Else, accept. Knowing your stuff. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. DFA = fhA;Bi j Aand Bare DFAs and L(A) = L(B)g is decidable. (a) Show that L is not regular. Lecture 32/65: Decidability and Decidable Problems hhp3. Then it accepts if L(D) is empty, otherwise it rejects. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Show that SUB_DFA is decidable. Question 2. But we only show one here. (b) No, because M on input 011 ends in a non-accept state. It contains 8 KBytes and has a four-way set-associative organization and a block length of four 32-bit words. To show PALDFA is decidable, we construct a decider D for PALDFA as follows (Let K be a TM that decides ECFG): D = \On input hMi, 1. We show that Sis decidable. Show that AεCFG is decidable. } Using the procedure given in class, convert the regular expression aa∪bb into an NFA, and then a DFA, M 1. Give an example of a DFA Astallion that in in ALLDFA. There is a single “line valid bit” and three bits, B0, B1,. 28) Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. Consider the following algorithm for deciding whether a given non-empty string s of length n belongs to A∗: For ev-ery possible way of splitting s into non-empty substrings s =. Let T = “On input where A is a DFA i. By Theorem 4. CSCI 2670 Introduction to Theory of Computing October 13, 2005. (a) Show that L is not regular. Show that A is decidable. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. A triangle in an undirected graph is a 3-clique. 57149995 >>57149874 I'm making a prop logic theorem prover. Since EQ DFAis decidable, there is an algorithm. Prove that ALLDFA is decidable. Show that ALLDFA is decidable. Then, let M be a DFA that. (b) Show that L is decidable. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Show that SUB_DFA is decidable. (Hint: Look at the proof for EDFA to get an idea. Show that INFINITE DFA is decidable. Businessman with a ‘golden touch’ who gave the presidential candidate his place in Sin City. (Proof idea) IWe construct a Turing machine S to decide the problem. Knowing your stuff. CSCI 2670 Introduction to Theory of Computing October 13, 2005. pdf - Free download as PDF File (. Submit to the decider for EQ DFA 3. Let CONNECTED={IG is a connected undirected graph}. Keeping a good spirit, one of optimism and openness. Give a PDA that accepts Lpal2 b. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Homework Solution - Set 7 Due: Friday 10/24/08 1. Please attach your answer to this in the file Spiral. Convert P0 into an equivalent CFG G 4. Give an example of a DFA Astallion that in in ALLDFA. Format your answers in the following style: if you think that the decidable languages are closed under union, show how to write def deciderUnion(w) assuming that deciderL 1 (w) and deciderL 2 (w) exist; if you think that the decidable languages are not closed. A triangle in an undirected graph is a 3-clique. •Recall that decidable languages are languages that can be decided by TM (that means, the corresponding TM will accept or reject correctly, never loops) •In this lecture, we investigate some decidable languages that are related to DFA, NFA, and CFG –Testing Acceptance, Emptiness, or Equality •Also, we show how TM can simulate CFG Objectives. To show PALDFA is decidable, we construct a decider D for PALDFA as follows (Let K be a TM that decides ECFG): D = \On input hMi, 1. INTRODUCTION TO THE THEORY OF COMPUTATION. Let ALLDFA-{(A): A Is A DFA And L(A) = Σ*). Show that the collection of decidable languages is closed under union. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Observe that indeed we cannot copy the proof that EQ DFA is decidable as the the CFL's are not closed under comple-ment, while the regular languages are. In general, people believe P is a proper subclass of NP In general, it is difficult to find a lower time bound for a given problem NP-completeness Polynomial time reducibility: A <=P B B is NP-complete if B is in NP A <=P B for all A in NP Cook-Levin Theorem: SAT is NP-complete More NP-Complete Problems 3SAT HAMPATH CLIQUE SUBSET-SUM Final Exam. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Convert P0 into an equivalent CFG G 4. Show that NP is closed under union and concatenation. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. (Families Affected by Fetal Alcohol Spectrum Disorder). 64 Let N be an NFA with k states that recognizes some language A. Since EQ DFA is decidable, we assume that is an algorithm M (i. Show That The Following. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Slideshow 3925035 by eliza. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. Show that decidable languages are closed under: (a) union (b) concatenation (c) Kleene star (d) complement (e) intersection Answer: For all of these answers, let L, L1 , and L2 be decidable languages, and M , M1 , and M2 be the TM's that decide them. A “good old boy” who can get along with others. (b) Show that A TM is Turing-recognizable. Let A(CFG = { | G is a CFG that generates (}. Show that the set of incompressible strings is un decidable. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Showing up. Transcription. Lecture 32/65: Decidability and Decidable Problems hhp3. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. Consider the following algorithm for deciding whether a given non-empty string s of length n belongs to A∗: For ev-ery possible way of splitting s into non-empty substrings s =. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Thelanguageweareconcernedwithis INFINITEDFA =fhAi: A isaDFAandL(A)isaninﬂnitelanguageg: Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. Homework 8Solutions 1. A language L belongs to P iff there is a constant k and a decider M running in time O(nk) such that L = L(M). Sipser provides an algorithm for ALL NFA that runs in nondeterministic space O(n), so ALL. 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. Show that ALLDFA is in P. CSCI670CSCI670IntroductiontoTheoryofIntroductiontoTheoryofComputingComputingComputingComputingOctober13005AgendaAgenda•Yesterday-Decidabilityandregularlanguages. Show That SUB_DFA Is Decidable. Show that ALLDFA is decidable. Since EQ DFAis decidable, there is an algorithm. Show that the collection of decidable languages is closed under union. Prove that ECFG is a decidable language. decidable, a decider for A NTM could be to decide A TM. Let A(CFG = { | G is a CFG that generates (}. Show thatALLDFAis decidable. Question 2. Clearly, hMi2Sif and only if L(M) = L(N). P is the class of all languages such that if w 2P then there is a deterministic single-tape Turing machine which accepts the string w in polynomial time. Let T = “On input where A is a DFA i. Let CONNECTED={IG is a connected undirected graph}. cs701 fall 2016 mid term paper. Answer: The universal TM U recognizes A TM, where U is deﬁned as follows: U = "On input hM,wi, where M is a TM and w is a string: 1. } Using the procedure given in class, convert the regular expression aa∪bb into an NFA, and then a DFA, M 1. Show that EQ CFG is undecidable. Show that LEN_CFG is decidable. Hint: First mark each variable that can yield ( by one derivation, then mark those variables that can yield ( by more derivations. Lecture 32/65: Decidability and Decidable Problems hhp3. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. Show that INFINITE DFA is decidable. Else, accept. By Theorem 4. Full text of "Introduction To Theory Of Computation" See other formats. We now construct another algorithm M 1 to decide language ALL DFA. Method II: Suppose on the contrary that L is regular. 4 on input , where T decides E DFA iii. Prove that ALLDFA is decidable. October 13, 2005. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. CS701 ALL Current Mid Term Papers Fall 2016 And Past Mid Term Papers at One Place from 17 December 2016 to 29 December 2016. txt) or read online for free. Show that AεCFG is decidable. Think back to what you learned about NP-completeness. Lecture 32/65: Decidability and Decidable Problems hhp3. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. Then it accepts if L(D) is empty, otherwise it rejects. ) * (3) All the King’s Horses, and All the King’s Men… Let ALLDFA = { | A is a DFA and L(A) = Σ*} Describe in English what the language ALLDFA consists of. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. 15 of Sipser book) Q16. CS701 - Current Mid Term Papers Dated: 04-07-2015. Answer: Deﬁne the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. 10 Marks (PROBLEM 5. M 1 works as follows. Else, accept. Let Lpal2 be the language wwRvvR:w, v E E' a. If T accepts, “accept”. But we only show one here. proof DFA defines same language as minimal DFA. Show that for any two languages A and B a language J exists, where A≤T J and B≤T J. Consider the following algorithm for deciding whether a given non-empty string s of length n belongs to A∗: For ev-ery possible way of splitting s into non-empty substrings s =. Run M1 on w. Show that A(CFG is decidable. Let CONNECTED={IG is a connected undirected graph}. Show that ALLDFA is decidable. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. Notice that this claim follows from Exercise 1. The algorithm M 1 inputs hAi, where Ais a DFA. Assg 9 - Solution sketches with study points added. A triangle in an undirected graph is a 3-clique. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Yesterday Decidability and regular languages Today Putting things in perspective More on decidability and regular languages Decidability and context free grammars. Announcements. Clearly, hMi2Sif and only if L(M) = L(N). Businessman with a ‘golden touch’ who gave the presidential candidate his place in Sin City. Show that INFINITEDFA is decidable. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Then it accepts if L(D) is empty, otherwise it rejects. Show that, by giving an example, that part (a) is not necessarily true if you replace both A's by A. 3 Let ALLDFA = { | A is a DFA that recognizes (*}. Convert P0 into an equivalent CFG G 4. The algorithm M 1 inputs hAi, where Ais a DFA. 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. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. There is a single "line valid bit" and three bits, B0, B1,. 15 of Sipser book) Q16. CS701 mid term paper shared by student. Show that LEN_CFG is decidable. 24 on page 88 in the textbook. The cache is organized into 128 sets. Posted by Irfan Khan MSCS on December 18, 2016 at 4:46pm in CS701 - Theory of Computation Mid Term Paper and Final Term Paper; Back to CS701 - Theory of Computation Mid Term Paper and Final Term Paper Discussions. Show that ALLDFA is decidable. >> Anonymous Wed Oct 19 17:44:05 2016 No. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. Show That The Following. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. Convert P0 into an equivalent CFG G 4. 64 Let N be an NFA with k states that recognizes some language A. Loading Unsubscribe from hhp3? Show more Show less. We construct a. 1 (a) a Yes, because M on input 0100 ends in an accept state. 4 Let AεCFG = { G | G is a CFG that generates ε}. (b) Show that A TM is Turing-recognizable. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. 3) Consider counting the ordered pairs of integers in the Cartesian plane. give the pseudocode for function def deciderSUB DFA (G) and discussion of correctness. Run M1 on w. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Show that NP is closed under union and concatenation. It contains 8 KBytes and has a four-way set-associative organization and a block length of four 32-bit words. CS701 mid term paper shared by student. 24 on page 88 in the textbook. pdf), Text File (. Knowing your stuff. Show that EQ CFG is undecidable. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. ) * (3) All the King's Horses, and All the King's Men… Let ALLDFA = { | A is a DFA and L(A) = Σ*} Describe in English what the language ALLDFA consists of. Slideshow 3925035 by eliza. Run M 1 on w. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Show that NP is closed under union and concatenation. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Show that the set of incompressible strings is un decidable. Show that SUB_DFA is decidable. 1 (a) a Yes, because M on input 0100 ends in an accept state. 4 Let AεCFG = { G | G is a CFG that generates ε}. Thelanguageweareconcernedwithis INFINITEDFA =fhAi: A isaDFAandL(A)isaninﬂnitelanguageg: Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. Videos recorded Spring 2014 for CSE355 at Arizona State University. (c) No, because the input is not in correct form: the second component of the input is missing. We show that EQ CFG is undecidable by showing that if it would be decidable, then so would ALL CFG, which is not true (Theorem 5. - 1796261. But we only show one here. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. Chapter 3: The Physical Science of the Environment Searching for life elsewhere Looking for life o Probes sent to space o Mars: new focus Viking ½ Evidence of water encouraging because all organisms require water to. Notice that this claim follows from Exercise 1. Announcements. Sipser provides an algorithm for ALL NFA that runs in nondeterministic space O(n), so ALL. proof DFA defines same language as minimal DFA. Let Cbe the DFA obtained by interchanging accepting and rejecting states of B. Submit to the decider for EQ DFA 3. Create a DFA B such that L(B) = Σ * 2. Since EQ DFA is decidable, we assume that is an algorithm M (i. 5 (EQUIVdfa is decidable), the condition L(A)=L(S) is decidable, and so determining if A belongs to ALLdfa is decidable too. decidable, a decider for A NTM could be to decide A TM. Using the procedure given in class, convert the regular expression. We show that Sis decidable. Loading Unsubscribe from hhp3? Show more Show less. Show That The Following. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. 17 December cs701 paper CS701 Q#1suppose there is a language L we know that L is Turing. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. Show that the SPCP is decidable. Last week Variants of Turing machines Definition of algorithm This week Chapter 4 Decidable and undecidable languages The halting problem. 24 on page 88 in the textbook. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. Slideshow 3925035 by eliza. Use the closure properties of regular languages and context-free languages to show that Lne is not regular 9. Show that ALLDFA is decidable. Show that, if A is nonempty, A contains some string of length at most k. 28) Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. Convert P0 into an equivalent CFG G 4. Show that ALLDFA is decidable. Show that INFINITE DFA is decidable. Show thatALLDFAis decidable. Chapter 3: The Physical Science of the Environment Searching for life elsewhere Looking for life o Probes sent to space o Mars: new focus Viking ½ Evidence of water encouraging because all organisms require water to. (Proof idea) IWe construct a Turing machine S to decide the problem. import Data. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. txt) or read online for free. - 1796261. Show that ALLDFA is in P. CS701 mid term paper shared by student. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Prove that ECFG is a decidable language. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Show that NP is closed under union and concatenation. We show that Sis decidable. Show that ALL DFA is decidable. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Show that ALLDFA is in P. Show that, if A is nonempty, A contains some string of length at most k. txt) or read online for free. (a) Show that L is not regular. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. CSCI 2670 Introduction to Theory of Computing October 13, 2005. Show That ALL_DFA Is Decidable. Knowing your stuff. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. Show that INFINITE DFA is decidable. Show that INFINITEDFA is decidable. CSCI 2670 Introduction to Theory of Computing. b Let A be a decidable language and let D be a polytime decider for it. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. 3) Answer: Consider the following Turing machine M = " on input , where A is a DFA 1. Then, let M be a DFA that. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. ) a) Give an example of a language that is NOT context free that can be accepted by a 2-PDA. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. P is the class of all languages that are decidable by deterministic multi-tape Turing machines running in polynomial time. October 13, 2005. List data Prop = Atom String | Not Prop | Imp Prop Prop der. import Data. Show that ALLDFA is decidable. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. a Show that P is closed under complement and concatenation. com - id: 56456d-ZGMxZ. If L(G) is empty, reject. Convert P0 into an equivalent CFG G 4. Being there. P is the class of all languages such that if w 2P then there is a deterministic single-tape Turing machine which accepts the string w in polynomial time. Show thatALLDFAis decidable. In general, people believe P is a proper subclass of NP In general, it is difficult to find a lower time bound for a given problem NP-completeness Polynomial time reducibility: A <=P B B is NP-complete if B is in NP A <=P B for all A in NP Cook-Levin Theorem: SAT is NP-complete More NP-Complete Problems 3SAT HAMPATH CLIQUE SUBSET-SUM Final Exam. Given hMi, which represents an automaton M, we can construct an automaton Nsuch that L(N) = L(M)R, that is, Naccepts a word wif and only if Maccepts wR. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. Show That LEN_CFG Is Decidable. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Lecture 32/65: Decidability and Decidable Problems hhp3. If T rejects, “reject” 2. But we only show one here. (c) No, because the input is not in correct form: the second component of the input is missing. proof DFA defines same language as minimal DFA. Idea: If G is in CNF, then it takes at most 2n 1 steps to generate w. } Using the procedure given in class, convert the regular expression aa∪bb into an NFA, and then a DFA, M 1. 4 on input , where T decides E DFA iii. Show that INFINITE DFA is decidable. pdf using cvssubmit. " Note that U only recognizes A TM and does not decide A TM Because when we run M on w,. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. If T rejects, “reject” 2. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. •Recall that decidable languages are languages that can be decided by TM (that means, the corresponding TM will accept or reject correctly, never loops) •In this lecture, we investigate some decidable languages that are related to DFA, NFA, and CFG -Testing Acceptance, Emptiness, or Equality •Also, we show how TM can simulate CFG Objectives. 10 Marks (PROBLEM 5. We show that Sis decidable. Show that ALLDFA is in P. A “good old boy” who can get along with others. DFA with an accepting state in the initial state. Loading Autoplay When autoplay is enabled,. Notice that this claim follows from Exercise 1. 64 Let N be an NFA with k states that recognizes some language A. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Since EQ DFA is decidable, we assume that is an algorithm M (i. 57149995 >>57149874 I'm making a prop logic theorem prover. A language L belongs to P iff L 2TIME(2n). Construct a PDA P such that L(P) = fw j w is a palindromeg 2. Using the procedure given in class, convert the regular expression. Run M on w. Show that the SPCP is decidable. 4 Let AεCFG = { G | G is a CFG that generates ε}. (a) Show that L is not regular. Homework due next Tuesday (10/26) Slideshow 3839749 by aislin. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. 5 (EQUIVdfa is decidable), the condition L(A)=L(S) is decidable, and so determining if A belongs to ALLdfa is decidable too. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Show thatALLDFAis decidable. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. - 1796261. Give 5 languages that are in the class P. Show that ALLDFA is in P. We show that Sis decidable. Clearly, hMi2Sif and only if L(M) = L(N). INTRODUCTION TO THE THEORY OF COMPUTATION. Show that ALLDFA is decidable. Prove that ALLDFA is decidable. Let LEN_CFG = { | G is a CFG, k elementof Z^noneg, and L (G) Intersection sigma^k notequalto (where sigma is the alphabet of G)}. Show that, if A is nonempty, A contains some string of length at most k. The algorithm M 1 inputs hAi, where Ais a DFA. A language L belongs to P iff there is a constant k and a decider M running in time O(nk) such that L = L(M). P is the class of all languages such that if w 2P then there is a deterministic single-tape Turing machine which accepts the string w in polynomial time. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Consider the following algorithm for deciding whether a given non-empty string s of length n belongs to A∗: For ev-ery possible way of splitting s into non-empty substrings s =. CS701 ALL Current Mid Term Papers Fall 2016 And Past Mid Term Papers at One Place from 17 December 2016 to 29 December 2016. Loading Unsubscribe from hhp3? Show more Show less. Else, accept. We construct a. >> Anonymous Wed Oct 19 17:44:05 2016 No. Irish: ·sea Proverb: Bainfidh an fharraige a cuid féin amach; beidh a cuid féin ag an bhfarraige. Keeping a good spirit, one of optimism and openness. Homework Solution - Set 7 Due: Friday 10/24/08 1. CSCI670CSCI670IntroductiontoTheoryofIntroductiontoTheoryofComputingComputingComputingComputingOctober13005AgendaAgenda•Yesterday-Decidabilityandregularlanguages. To show PALDFA is decidable, we construct a decider D for PALDFA as follows (Let K be a TM that decides ECFG): D = \On input hMi, 1. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. But we only show one here. 3) Consider counting the ordered pairs of integers in the Cartesian plane. FAFASD seeks to spread information, awareness, and hope for families impacted by fetal alcohol spectrum disorder. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. CSE 105 Sp04, Problem Set 3 Solutions 3 Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. E(dfa) is a decidable language. Mega Collection of CS701 Mid Term Exam Papers Spring 2015 Dated 04-07-2015 to 09-04-2015 of MSCS of Virtual University of Pakistan (VU) at mscsvu. Show that the collection of decidable languages is closed under union. Answer: Deﬁne the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. Submit to the decider for EQ DFA 3. Show that ALLDFA is decidable. 3) Answer: Consider the following Turing machine M = “ on input , where A is a DFA 1. If M accepts w, accept; if M rejects w, reject. Show that INFINITE DFA is decidable. Posted by + M. Show that, by giving an example, that part (a) is not necessarily true if you replace both A's by A. A “good old boy” who can get along with others. Show that NP is closed under union and concatenation. Let SUBSETDFA-A, B L(B)). Express this problem as a language and show that it is decidable. Notice that this claim follows from Exercise 1. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. Show that ALLDFA is decidable. Let Cbe the DFA obtained by interchanging accepting and rejecting states of B. The algorithm M 1 inputs hAi, where Ais a DFA. Answer: The universal TM U recognizes A TM, where U is deﬁned as follows: U = "On input hM,wi, where M is a TM and w is a string: 1. 2 (Decidable Languages) Show that the following languages are decidable: (a) EQ. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. Think back to what you learned about NP-completeness. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. 2 Consider the problem of determining whether a DFA and a regular expression are. P is the class of all languages that are decidable by deterministic multi-tape Turing machines running in polynomial time. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. Let A = {hR,Si| R and S are regular expressions and L(R) L(S)}. Show that ALLDFA is decidable. Theory also is relevant to you because it shows you a new, simpler, and more elegant side of computers, which we normally consider to be complicated machines. Show thatALLDFAis decidable. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. Say that a variable A in CFL G is usable if it appears in some derivation of some string w G. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. A language L belongs to P iff there is a constant k and a decider M running in time O(nk) such that L = L(M). 57149995 >>57149874 I'm making a prop logic theorem prover. CSCI 2670 Introduction to Theory of Computing. decidable, a decider for A NTM could be to decide A TM. , there is a Turing machine M such that M halts and accepts on any input w ∈ A, and M halts and rejects on input input w ∈ A; i. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Gaining skills and applying them well. Show that LEN_CFG is decidable. (c) No, because the input is not in correct form: the second component of the input is missing. Show that A(CFG is decidable. Sipser Problem 3. DFA is decidable. One-to-one function f: T N. A language L belongs to P iff there is a constant k and a decider M running in time O(nk) such that L = L(M). , there is a Turing machine M such that M halts and accepts on any input w ∈ A, and M halts and rejects on input input w ∈ A; i. CSCI 2670 Introduction to Theory of Computing. Answer: Deﬁne the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. By Theorem 4. Videos recorded Spring 2014 for CSE355 at Arizona State University. If L(G) is empty, reject. Show that the following language is decidable (ie. Formulate this problem as a language and show that it is decidable. Show That SUBSETDFA Is Decidable. Gaining skills and applying them well. Using the procedure given in class, convert the regular expression. A “good old boy” who can get along with others. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. Let CONNECTED={IG is a connected undirected graph}. A language L belongs to P iff L 2TIME(2n). Show that, if P=NP. Posted by + M. Loading Autoplay When autoplay is enabled,. Show that SUB_DFA is decidable. Give a PDA that accepts Lpal2 b. CS701 mid term paper shared by student. (Hint: Look at the proof for EDFA to get an idea. Create DFA that accept language where number of 0's is even and after every 1 goes 0. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Prove that ECFG is a decidable language. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. Show that ALL DFA is decidable. Lecture 32/65: Decidability and Decidable Problems hhp3. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. Show That LEN_CFG Is Decidable. Show that A(CFG is decidable. M 1 works as follows. We show that EQ CFG is undecidable by showing that if it would be decidable, then so would ALL CFG, which is not true (Theorem 5. We now construct another algorithm M 1 to decide language ALL DFA. 10 Marks (PROBLEM 5. Show that ALLDFA is decidable. FAFASD seeks to spread information, awareness, and hope for families impacted by fetal alcohol spectrum disorder. Last week Variants of Turing machines Definition of algorithm This week Chapter 4 Decidable and undecidable languages The halting problem. Let CONNECTED={IG is a connected undirected graph}. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. Lecture 32/65: Decidability and Decidable Problems hhp3. Show that T is countable. Show that, if A is nonempty, A contains some string of length at most k. Give 5 languages that are in the class P. A triangle in an undirected graph is a 3-clique. 3) Consider counting the ordered pairs of integers in the Cartesian plane. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. Let Lpal2 be the language wwRvvR:w, v E E' a. For MSCS, Mega Collection of Solved and Unsolved Past and Current Mid and Final Term Exam Papers, Academic Research Term Papers, Assignments, Guidance, and all about MSCS. (Proof idea) IWe construct a Turing machine S to decide the problem. Show that LEN_CFG is decidable. A language L belongs to P iff L 2TIME(2n). 4 on input , where T decides E DFA iii. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Answer: The universal TM U recognizes A TM, where U is deﬁned as follows: U = "On input hM,wi, where M is a TM and w is a string: 1. The Intel 80486 has an on-chip,unified cache. This decider will accept a DFA that does not accept all inputs if, for example, there is no transition for one of the characters in σ. Full text of "Introduction To Theory Of Computation" See other formats. Give an example of a DFA Astallion that in in ALLDFA. CS701 mid term paper shared by student. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. ) a) Give an example of a language that is NOT context free that can be accepted by a 2-PDA. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Show that ETM, the complement of. Convert P0 into an equivalent CFG G 4. Show that A(CFG is decidable. M 1 works as follows. Let build M0as follows: M0= \On input w: 1. (a) Show that L is not regular. It will not accept inputs with a 1, but there is no reject state that will cause this DFA to be rejected by the decider. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. 3) Consider counting the ordered pairs of integers in the Cartesian plane. Show that for any two languages A and B a language J exists, where A≤T J and B≤T J. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible. Show that ALLDFA is decidable. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. Method II: Suppose on the contrary that L is regular. Let ALLDFA-{(A): A Is A DFA And L(A) = Σ*). Given hMi, which represents an automaton M, we can construct an automaton Nsuch that L(N) = L(M)R, that is, Naccepts a word wif and only if Maccepts wR. (b) Show that A TM is Turing-recognizable. 2) Let INFINITE DFA = { | A is a DFA and L(A) contains an infinite number of strings}. For MSCS, Mega Collection of Solved and Unsolved Past and Current Mid and Final Term Exam Papers, Academic Research Term Papers, Assignments, Guidance, and all about MSCS. CS701 mid term paper shared by student. There is a single "line valid bit" and three bits, B0, B1,. Loading Autoplay When autoplay is enabled,. Problem 2 - Semi-Decidability (25 points) Show that the acceptance problem for Turing machines. Then, let M be a DFA that. Gaining skills and applying them well. 4 on input , where T decides E DFA iii. There is a single "line valid bit" and three bits, B0, B1,. Loading Unsubscribe from hhp3? Show more Show less. We show that EQ CFG is undecidable by showing that if it would be decidable, then so would ALL CFG, which is not true (Theorem 5. Problem 2 - Semi-Decidability (25 points) Show that the acceptance problem for Turing machines. 2 Consider the problem of determining whether a DFA and a regular expression are. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Show That SUB_DFA Is Decidable. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. (b) Show that L is decidable. Run M1 on w. Theory also is relevant to you because it shows you a new, simpler, and more elegant side of computers, which we normally consider to be complicated machines. Note: The next question is a programming question. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. Full text of "Introduction To Theory Of Computation" See other formats. Show that AεCFG is decidable. A Prunable State In A DFA Is Some State That Is Never Entered While Processing Any Input String. Say that a variable A in CFL G is usable if it appears in some derivation of some string w G. (a) Show that L is not regular. DFA = fhA;Bi j Aand Bare DFAs and L(A) = L(B)g is decidable. Submit to the decider for EQ DFA 3. A And B Are DFAs And L(A) C Problem 4 (5 Points). - 1796261. Show that ALLDFA is in P.

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