Give an example of a DFA Astallion that in in ALLDFA. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. (Proof idea) IWe construct a Turing machine S to decide the problem. 4 Let AεCFG = { G | G is a CFG that generates ε}. Show that Th(N,)is decidable. Show that EQ CFG is undecidable. Various application areas, such as modern cryptographic protocols, rely on theoretical principles that you will learn here. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. The following DFA (with an alphabet of 0 and 1) is an example of this. Prove that ECFG is a decidable language. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. Clearly, hMi2Sif and only if L(M) = L(N). Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Let CONNECTED={IG is a connected undirected graph}. 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. Show that the collection of decidable languages is closed under union. 2) Let INFINITEDFA = { | A is a DFA and L(A) contains an infinite number of strings}. give the pseudocode for function def deciderSUB DFA (G) and discussion of correctness. Show thatALLDFAis decidable. Exercise 4. b Let A be a decidable language and let D be a polytime decider for it. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. We show that Sis decidable. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. txt) or read online for free. Show that the collection of decidable languages is closed under union. Idea: If G is in CNF, then it takes at most 2n 1 steps to generate w. import Data. The Intel 80486 has an on-chip,unified cache. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. Step-by-Step Solution: Step 1 of 3. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. Show that INFINITEDFA is decidable. Slideshow 3925035 by eliza. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. Show that LEN_CFG is decidable. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. DFA is decidable. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. Use K to check if L(G) is empty. Then, let M be a DFA that. Since EQ DFAis decidable, there is an algorithm. E(dfa) is a decidable language. Prove that ECFG is a decidable language. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Show that ALLDFA is in P. Let build M0as follows: M0= \On input w: 1. Clearly, hMi2Sif and only if L(M) = L(N). Since EQ DFA is decidable, we assume that is an algorithm M (i. 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. Show that LEN_CFG is decidable. 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 SUB_DFA Is Decidable. equivalent. Give a CFG that (1) Show "all" DFA is decidable. Prove that ECFG is a decidable language. We now construct another algorithm M 1 to decide language ALL DFA. Berkshire Hathaway AGM is still the Warren Buffett show ; US election: Trump’s man in Vegas. CSCI 2670 Introduction to Theory of Computing October 13, 2005. Tariq Malik on December 17, 2016 at 11:59am in CS701 Theory of Computation; Back to CS701 Theory of Computation Discussions. 15 of Sipser book) Q16. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Give an example of a DFA Astallion that in in ALLDFA. (Hint: Look at the proof for EDFA to get an idea. One-to-one function f: T N. Step-by-Step Solution: Step 1 of 3. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Show that SUBSETDFA is decidable. Posted by + M. P is the class of all languages that are decidable by deterministic multi-tape Turing machines running in polynomial time. A Scottish start-up that is bidding to send satellites into orbit from the UK with a reusable rocket launcher is aiming to raise €5m through a listing on Malta’s new junior market. Show that INFINITEDFA is decidable. 1 (a) a Yes, because M on input 0100 ends in an accept state. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. Show that for any two languages A and B a language J exists, where A≤T J and B≤T J. (Proof idea) IWe construct a Turing machine S to decide the problem. 15 of Sipser book) Q16. (b) Show that L is decidable. E(dfa) is a decidable language. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. Show thatALLDFAis decidable. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. Show that A(CFG is decidable. If L(G) is empty, reject. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. Give a PDA that accepts Lpal2 b. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. Irish: ·sea Proverb: Bainfidh an fharraige a cuid féin amach; beidh a cuid féin ag an bhfarraige. Show that ALLDFA is decidable. 4 on input , where T decides E DFA iii. Step-by-Step Solution: Step 1 of 3. Videos recorded Spring 2014 for CSE355 at Arizona State University. Then it accepts if L(D) is empty, otherwise it rejects. Show that ALLDFA is decidable. Loading Autoplay When autoplay is enabled,. Announcements. Run M on w. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. Create a DFA B such that L(B) = Σ * 2. CSE 105 Sp04, Problem Set 3 Solutions 3 Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. Let T = {(i, j, k)| i, j, k ∈ N}. Show that SUB DFA is decidable, i. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Method II: Suppose on the contrary that L is regular. Assg 9 - Solution sketches with study points added. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. 15 of Sipser book) Q16. , looping cannot happen. Think back to what you learned about NP-completeness. 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. Problem 1 For this problem you may assume the containment diagram in Figure 4. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. Prove that ECFG is a decidable language. 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. Loading Autoplay When autoplay is enabled,. CS701 mid term paper shared by student. 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). Transcription. Lecture 32/65: Decidability and Decidable Problems hhp3. Show That SUB_DFA Is Decidable. 10 Marks (PROBLEM 5. Show that A is decidable. Create a DFA B such that L(B) = Σ * 2. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Gaining skills and applying them well. · billow, swell. Posted by + M. Exercise 4. (a) Show that L is not regular. 28) Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. Express this problem as a language and show that it is decidable. Create a DFA B such that L(B) = Σ * 2. Notice that this claim follows from Exercise 1. 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. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. The cache is organized into 128 sets. Submit to the decider for EQ DFA 3. Show That LEN_CFG Is Decidable. Show that Th(N,)is decidable. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. Show that the collection of decidable languages is closed under union. If L(G) is empty, reject. Show that ALLDFA is decidable. Run M on w. Since EQ DFAis decidable, there is an algorithm. CSCI 2670 Introduction to Theory of Computing October 19, 2004 Agenda Last week Variants of Turing machines Definition of algorithm This week Chapter 4 Decidable and - A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Full text of "Introduction To Theory Of Computation" See other formats. 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. 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 L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. Clearly, hMi2Sif and only if L(M) = L(N). Give an example of a DFA Astallion that in in ALLDFA. Run M 1 on w. 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. Sipser provides an algorithm for ALL NFA that runs in nondeterministic space O(n), so ALL. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. 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. pdf using cvssubmit. Homework Solution - Set 7 Due: Friday 10/24/08 1. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. Show a Turing Machine accepts all DFAs. Show that ALL DFA is decidable. · billow, swell. , looping cannot happen. It contains 8 KBytes and has a four-way set-associative organization and a block length of four 32-bit words. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. Construct DFA B that recognizes L(A) ii. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. 4 Let AεCFG = { G | G is a CFG that generates ε}. Since EQ DFA is decidable, we assume that is an algorithm M (i. Clearly, hMi2Sif and only if L(M) = L(N). Let SUBSETDFA-A, B L(B)). 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. Use K to check if L(G) is empty. FAFASD seeks to spread information, awareness, and hope for families impacted by fetal alcohol spectrum disorder. 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 ALL DFA = fhBijBgis DFA and L(B) = gis decidable. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Problem 3 (5 points). , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Create a DFA B such that L(B) = Σ * 2. tex and hw10. 4 on input , where T decides E DFA iii. Show that INFINITE DFA is decidable. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Use K to check if L(G) is empty. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. Show that the set of incompressible strings is un decidable. (c) No, because the input is not in correct form: the second component of the input is missing. Then, let M be a DFA that. 5 (EQUIVdfa is decidable), the condition L(A)=L(S) is decidable, and so determining if A belongs to ALLdfa is decidable too. com - id: 56456d-ZGMxZ. If L(G) is empty, reject. (b) Show that A TM is Turing-recognizable. Show that the collection of decidable languages is closed under union. Create a DFA B such that L(B) = Σ * 2. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Let CONNECTED={IG is a connected undirected graph}. 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. Marcela Reyna - Free download as PDF File (. Clearly, hMi2Sif and only if L(M) = L(N). 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)}. 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. A Prunable State In A DFA Is Some State That Is Never Entered While Processing Any Input String. Let T = “On input where A is a DFA i. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. Show that the following language is decidable (ie. Berkshire Hathaway AGM is still the Warren Buffett show ; US election: Trump’s man in Vegas. Assg 9 - Solution sketches with study points added. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. 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 ALLDFA is decidable. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. Show that ALLDFA is in P. The algorithm M 1 inputs hAi, where Ais a DFA. 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. Use K to check if L(G) is empty. A And B Are DFAs And L(A) C Problem 4 (5 Points). Show that SUB DFA is decidable, i. Lecture 32/65: Decidability and Decidable Problems hhp3. Construct DFA B that recognizes L(A) ii. DFA is decidable. Show that the collection of decidable languages is closed under union. Let A(CFG = { | G is a CFG that generates (}. Answer: Deﬁne the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. Let Mbe a Turing machine that decides this language. · billow, swell. (Proof idea) IWe construct a Turing machine S to decide the problem. Use K to check if L(G) is empty. Give an example of a DFA Astallion that in in ALLDFA. 10 Marks (PROBLEM 5. Homework due next Tuesday (10/26) Slideshow 3839749 by aislin. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Clearly, hMi2Sif and only if L(M) = L(N). Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. If L(G) is empty, reject. tex and hw10. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Idea: If G is in CNF, then it takes at most 2n 1 steps to generate w. Show that ALLDFA is decidable. Then, let M be a DFA that. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. ) * (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. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. 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)}. The cache is organized into 128 sets. Q#4 Let LALL ={/M is a TM with i/p ∑ and L(M =∑*} Prove that LALL is not a co Turing recognizable. 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. One-to-one function f: T N. •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. Problem 1 For this problem you may assume the containment diagram in Figure 4. give the pseudocode for function def deciderSUB DFA (G) and discussion of correctness. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. Homework 8Solutions. CSCI 2670 Introduction to Theory of Computing. Show That SUBSETDFA Is Decidable. cs701 fall 2016 mid term paper. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. (b) Show that A TM is Turing-recognizable. Lecture 32/65: Decidability and Decidable Problems hhp3. Show that ALLDFA is in P. Submit to the decider for EQ DFA 3. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. Use K to check if L(G) is empty. Show that ALLDFA is decidable. Full text of "Introduction To Theory Of Computation" See other formats. There is a single "line valid bit" and three bits, B0, B1,. 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). decidable, a decider for A NTM could be to decide A TM. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Various application areas, such as modern cryptographic protocols, rely on theoretical principles that you will learn here. 17 December cs701 paper CS701 Q#1suppose there is a language L we know that L is Turing. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. 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. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. proof DFA defines same language as minimal DFA. Homework Solution - Set 7 Due: Friday 10/24/08 1. Keeping your cool. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. Give an example of a DFA Astallion that in in ALLDFA. Show that ALL DFA is decidable. Let Cbe the DFA obtained by interchanging accepting and rejecting states of B. 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. Create a DFA B such that L(B) = Σ * 2. We now construct another algorithm M 1 to decide language ALL DFA. 10 Marks (PROBLEM 5. com - id: 56456d-ZGMxZ. Express this problem as a language and show that it is decidable. 24 on page 88 in the textbook. Homework Solution - Set 7 Due: Friday 10/24/08 1. A triangle in an undirected graph is a 3-clique. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. A “good old boy” who can get along with others. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. Loading Unsubscribe from hhp3? Show more Show less. Formulate this problem as a language and show that it is decidable. Create a DFA B such that L(B) = Σ * 2. Let T = {(i, j, k)| i, j, k ∈ N}. Show that NP is closed under union and concatenation. Let CONNECTED={IG is a connected undirected graph}. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. Since EQ DFA is decidable, we assume that is an algorithm M (i. 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. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. 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. Show That The Following. Lecture 32/65: Decidability and Decidable Problems hhp3. Method II: Suppose on the contrary that L is regular. Prove that ALLDFA is decidable. a Show that P is closed under complement and concatenation. Show that INFINITE DFA is decidable. Irish: ·sea Proverb: Bainfidh an fharraige a cuid féin amach; beidh a cuid féin ag an bhfarraige. Show that SUBSETDFA is decidable. Use K to check if L(G) is empty. If L(G) is empty, reject. A language L belongs to P iff L 2TIME(2n). Various application areas, such as modern cryptographic protocols, rely on theoretical principles that you will learn here. Let SUBSETDFA-A, B L(B)). 3) Consider counting the ordered pairs of integers in the Cartesian plane. 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. 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 that ALLDFA is decidable. One-to-one function f: T N. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Full text of "Introduction To Theory Of Computation" See other formats. 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 SUB_DFA Is Decidable. A triangle in an undirected graph is a 3-clique. Show that SUB_DFA is decidable. October 13, 2005. Loading Autoplay When autoplay is enabled,. Give an example of a DFA Astallion that in in ALLDFA. pdf - Free download as PDF File (. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. Let A(CFG = { | G is a CFG that generates (}. A Prunable State In A DFA Is Some State That Is Never Entered While Processing Any Input String. cs701 fall 2016 mid term paper. Run M on w. •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. Hint: First mark each variable that can yield ( by one derivation, then mark those variables that can yield ( by more derivations. The Intel 80486 has an on-chip,unified cache. But we only show one here. CSCI670CSCI670IntroductiontoTheoryofIntroductiontoTheoryofComputingComputingComputingComputingOctober13005AgendaAgenda•Yesterday-Decidabilityandregularlanguages. Homework due next Tuesday (10/26) Slideshow 3839749 by aislin. Videos recorded Spring 2014 for CSE355 at Arizona State University. A and B are DFAs and L(A) C Problem 4 (5 points). (b) Show that L is decidable. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. decidable, a decider for A NTM could be to decide A TM. Videos recorded Spring 2014 for CSE355 at Arizona State University. 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 that ALLDFA is in P. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. Let SUBSETDFA-A, B L(B)). 5 (EQUIVdfa is decidable), the condition L(A)=L(S) is decidable, and so determining if A belongs to ALLdfa is decidable too. (Hint: Look at the proof for EDFA to get an idea. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Let CONNECTED={IG is a connected undirected graph}. 3 Let ALLDFA = { A | A is a DFA and L(A) = Σ∗}. Let ALLDFA-{(A): A Is A DFA And L(A) = Σ*). But we only show one here. Yesterday Decidability and regular languages Today Putting things in perspective More on decidability and regular languages Decidability and context free grammars. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Show that the set of incompressible strings is un decidable. 24 on page 88 in the textbook. proof DFA defines same language as minimal DFA. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. Run M on w. One-to-one function f: T N. } Using the procedure given in class, convert the regular expression aa∪bb into an NFA, and then a DFA, M 1. Show that the SPCP is decidable. Idea: If G is in CNF, then it takes at most 2n 1 steps to generate w. The sea will have its own, its share of tragedies. Else, accept. 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. E(dfa) is a decidable language. tex and hw10. Loading Autoplay When autoplay is enabled,. Show that EQ CFG is undecidable. A triangle in an undirected graph is a 3-clique. (Proof idea) IWe construct a Turing machine S to decide the problem. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. · billow, swell. 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. By Theorem 4. 24 on page 88 in the textbook. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. A Prunable State In A DFA Is Some State That Is Never Entered While Processing Any Input String. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. 28) Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Thelanguageweareconcernedwithis INFINITEDFA =fhAi: A isaDFAandL(A)isaninﬂnitelanguageg: Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. 4 Let AεCFG = { G | G is a CFG that generates ε}. Show that, if A is nonempty, A contains some string of length at most 2k d. 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. Hint: In this case, all the states that are reachable from the start state are final states. 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 ALL DFA is decidable. Lecture 32/65: Decidability and Decidable Problems hhp3. 57149995 >>57149874 I'm making a prop logic theorem prover. · billow, swell. Question 2. M 1 works as follows. Note: The next question is a programming question. (a) Show that L is not regular. It contains 8 KBytes and has a four-way set-associative organization and a block length of four 32-bit words. Get best Help for Others questions and answers in computer-architecture Page-2877, step-by-step Solutions, 100% Plagiarism free Question Answers. pdf - Free download as PDF File (. Show that ALLDFA is decidable. Show that ETM, the complement of. Sipser Problem 3. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. Let T = {(i, j, k)| i, j, k ∈ N}. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. (b) Show that L is decidable. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. 3) Answer: Consider the following Turing machine M = “ on input , where A is a DFA 1. 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. (Proof idea) IWe construct a Turing machine S to decide the problem. Create DFA that accept language where number of 0's is even and after every 1 goes 0. Homework Solution - Set 7 Due: Friday 10/24/08 1. 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. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. Show That LEN_CFG Is Decidable. From input hB;Cia Turing machine constructs a DFA Drecognizing L(B)\L(C). Show that ALLDFA is decidable. 4 Let AεCFG = { G | G is a CFG that generates ε}. Let SUB_DFA = { | A, B Are DFAs, And L(A) L(B)}. 17 December cs701 paper CS701 Q#1suppose there is a language L we know that L is Turing. Notice that this claim follows from Exercise 1. E(dfa) is a decidable language. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. (b) No, because M on input 011 ends in a non-accept state. (a) Show that L is not regular. (b) Show that L is decidable. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. CSCI 2670 Introduction to Theory of Computing. cs701 fall 2016 mid term paper. We show that Sis decidable. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. Show That ALLDFA Is Decidable. 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. 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. Show that ALLDFA is in P. Then it accepts if L(D) is empty, otherwise it rejects. Show that Th(N,)is decidable. txt) or read online for free. 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 that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Hint: First mark each variable that can yield ( by one derivation, then mark those variables that can yield ( by more derivations. Slideshow 3925035 by eliza. Notice that this claim follows from Exercise 1. Use the closure properties of regular languages and context-free languages to show that Lne is not regular 9. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. Using the procedure given in class, convert the regular expression. Loading Autoplay When autoplay is enabled,. Being there. 3 Let ALLDFA = { | A is a DFA that recognizes (*}. Create a DFA B such that L(B) = Σ * 2. Problem 1 For this problem you may assume the containment diagram in Figure 4. proof DFA defines same language as minimal DFA. 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. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. Last week Variants of Turing machines Definition of algorithm This week Chapter 4 Decidable and undecidable languages The halting problem. But we only show one here. Notice that this claim follows from Exercise 1. ) * (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. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Convert P0 into an equivalent CFG G 4. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Problem 3 (5 points). Express this problem as a language and show that it is decidable. Show that SUB DFA is decidable, i. Show that the SPCP is decidable. 5 (EQUIVdfa is decidable), the condition L(A)=L(S) is decidable, and so determining if A belongs to ALLdfa is decidable too. import Data. M 1 works as follows. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. ) * (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. By Theorem 4. Show that NP is closed under union and concatenation. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. Show that the collection of decidable languages is closed under union. 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. CS701 mid term paper shared by student. (a) Show that L is not regular. Say that a variable A in CFL G is usable if it appears in some derivation of some string w G. If T rejects, “reject” 2. Showing up. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Show that ALLDFA is in P. Being able to do the job. 173 and the fact that each region is no-empty. Announcements. Question 2. Thelanguageweareconcernedwithis INFINITEDFA =fhAi: A isaDFAandL(A)isaninﬂnitelanguageg: Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. 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. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. Videos recorded Spring 2014 for CSE355 at Arizona State University. What to turn in Save your work in the hw10 folder and turn in the files hw10. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Show that ALLDFA is decidable. Show that ALLDFA is decidable. 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. A triangle in an undirected graph is a 3-clique. 1 (a) a Yes, because M on input 0100 ends in an accept state. 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). Loading Autoplay When autoplay is enabled,. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Convert P0 into an equivalent CFG G 4. A and B are DFAs and L(A) C Problem 4 (5 points). Prove that ECFG is a decidable language. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. ) a) Give an example of a language that is NOT context free that can be accepted by a 2-PDA. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. Submit to the decider for EQ DFA 3. If L(G) is empty, reject. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. CSCI 2670 Introduction to Theory of Computing October 13, 2005. Show That SUB_DFA Is Decidable. Problem 3 (5 Points). Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Show that SUBSETDFA is decidable. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. By Theorem 4. A language L belongs to P iff L 2TIME(2n). The following DFA (with an alphabet of 0 and 1) is an example of this. A “good old boy” who can get along with others. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. Convert P0 into an equivalent CFG G 4. Yesterday Decidability and regular languages Today Putting things in perspective More on decidability and regular languages Decidability and context free grammars. Sipser Problem 3. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. Discrete Homework 4 - Travis Montey COT 4210 Homework#4 1 Let ALLDFA = cfw_| A is a DFA and L(A = Show that ALLDFA is decidable a Let T = On input. decidable, a decider for A NTM could be to decide A TM. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. Businessman with a ‘golden touch’ who gave the presidential candidate his place in Sin City. Show that A(CFG is decidable. Tariq Malik on December 17, 2016 at 11:59am in CS701 Theory of Computation; Back to CS701 Theory of Computation Discussions. (Hint: Look at the proof for EDFA to get an idea. Give a PDA that accepts Lpal2 b. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. 64 Let N be an NFA with k states that recognizes some language A. Say that a variable A in CFL G is usable if it appears in some derivation of some string w G. Show that ALLDFA is in P. (b) Show that A TM is Turing-recognizable. Homework 8Solutions. (a) Show that L is not regular. Answer: Deﬁne the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. , 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. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Show That SUBSETDFA Is Decidable. Videos recorded Spring 2014 for CSE355 at Arizona State University. Create DFA that accept language where number of 0's is even and after every 1 goes 0. 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. There is a single “line valid bit” and three bits, B0, B1,. Show that ALLDFA is decidable. Businessman with a ‘golden touch’ who gave the presidential candidate his place in Sin City. 2) Let INFINITEDFA = { | A is a DFA and L(A) contains an infinite number of strings}. 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. (Proof idea) IWe construct a Turing machine S to decide the problem. Does A TM belong to P. Show that EQ CFG is undecidable. b Let A be a decidable language and let D be a polytime decider for it. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Use the closure properties of regular languages and context-free languages to show that Lne is not regular 9. We show that Sis decidable. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. [10 marks] Solution: The following Turing machine decides ALL DFA: M= \on input hBiwhere Bis a DFA: 1. 17 December cs701 paper CS701 Q#1suppose there is a language L we know that L is Turing. A triangle in an undirected graph is a 3-clique. Show that A(CFG is decidable. We construct a. Notice that this claim follows from Exercise 1. Show that NP is closed under union and concatenation. Berkshire Hathaway AGM is still the Warren Buffett show ; US election: Trump’s man in Vegas. Create a DFA B such that L(B) = Σ * 2. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Transcription. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. 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. Submit to the decider for EQ DFA 3. If L(G) is empty, reject. Loading Unsubscribe from hhp3? Show more Show less. Hint: In this case, all the states that are reachable from the start state are final states. Say that a variable A in CFL G is usable if it appears in some derivation of some string w G. Slideshow 3925035 by eliza. Show that Th(N,)is decidable. Then, {(hD ii, )} i≥1 constitutes an inﬁnite collection of distinguishable strings. From input hB;Cia Turing machine constructs a DFA Drecognizing L(B)\L(C). 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. 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. There is a single "line valid bit" and three bits, B0, B1,. Run M 1 on w. (c) No, because the input is not in correct form: the second component of the input is missing. 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. ThismeansthatforallDFAsA wewant †IfL(A)isaninﬂnitelanguagethenM(hAi)accepts †IfL(A)isaﬂnitelanguagethenM(hAi)rejects. · billow, swell. Problem 1 For this problem you may assume the containment diagram in Figure 4.