Show That Alldfa Is Decidable

10 Marks (PROBLEM 5. ) a) Give an example of a language that is NOT context free that can be accepted by a 2-PDA. Convert P0 into an equivalent CFG G 4. 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. Let SUBSETDFA-A, B L(B)). 24 on page 88 in the textbook. Given a CFG G and a variable A, consider the problem of testing whether A is usable. Assg 9 - Solution sketches with study points added. Run M on w. Show that A(CFG is decidable. 4 Let AεCFG = { G | G is a CFG that generates ε}. Let T = {(i, j, k)| i, j, k ∈ N}. 2 (Decidable Languages) Show that the following languages are decidable: (a) EQ. (b) Show that L is decidable. Please attach your answer to this in the file Spiral. CS701 ALL Current Mid Term Papers Fall 2016 And Past Mid Term Papers at One Place from 17 December 2016 to 29 December 2016. The Intel 80486 has an on-chip,unified cache. Show that the collection of decidable languages is closed under union. decidable, a decider for A NTM could be to decide A TM. A triangle in an undirected graph is a 3-clique. 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. Problem 3 (5 Points). 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. Run M on w. Loading Autoplay When autoplay is enabled,. Show that A is decidable. 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. Show that ALLDFA is decidable. Let T = “On input where A is a DFA i. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Show that, if P=NP. P is the class of all languages that are decidable by deterministic single-tape Turing machines running in polynomial time. Let build M0as follows: M0= \On input w: 1. Let A(CFG = { | G is a CFG that generates (}. Loading Unsubscribe from hhp3? Show more Show less. It contains 8 KBytes and has a four-way set-associative organization and a block length of four 32-bit words. Show that, if A is nonempty, A contains some string of length at most k. 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. Prove that ECFG is a decidable language. Then it accepts if L(D) is empty, otherwise it rejects. a Show that P is closed under complement and concatenation. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Show that A(CFG is decidable. 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. ThismeansthatforallDFAsA wewant †IfL(A)isaninflnitelanguagethenM(hAi)accepts †IfL(A)isaflnitelanguagethenM(hAi)rejects. Show that the following language is decidable (ie. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. 2) Let INFINITEDFA = { | A is a DFA and L(A) contains an infinite number of strings}. 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 AεCFG is decidable. Let ALLDFA-{(A): A Is A DFA And L(A) = Σ*). 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. 10 Marks (PROBLEM 5. The algorithm M 1 inputs hAi, where Ais a DFA. Homework 8Solutions. pdf), Text File (. Full text of "Introduction To Theory Of Computation" See other formats. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. 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. Gaining skills and applying them well. 1 Answer to Let ALL DFA = {hAi| A is a DFA and L(A) = Σ∗ }. Run M on w. (c) No, because the input is not in correct form: the second component of the input is missing. Loading Unsubscribe from hhp3? Show more Show less. Exercise 4. Show that ALLDFA is decidable. Give an example of a DFA Astallion that in in ALLDFA. DFA = fhA;Bi j Aand Bare DFAs and L(A) = L(B)g is decidable. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. Submit to the decider for EQ DFA 3. Notice that this claim follows from Exercise 1. - 1796261. The Intel 80486 has an on-chip,unified cache. Show that, if P=NP, then every language A ЄP, except A=ø and A=∑*,is NP-complete. 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. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. 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 Autoplay When autoplay is enabled,. Show that ALLDFA is in P. 3) Answer: Consider the following Turing machine M = “ on input , where A is a DFA 1. A and B are DFAs and L(A) C Problem 4 (5 points). Then it accepts if L(D) is empty, otherwise it rejects. , looping cannot happen. Show that A 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. 5 (EQUIVdfa is decidable), the condition L(A)=L(S) is decidable, and so determining if A belongs to ALLdfa is decidable too. Submit to the decider for EQ DFA 3. Show a Turing Machine accepts all DFAs. Convert P0 into an equivalent CFG G 4. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. 64 Let N be an NFA with k states that recognizes some language A. 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 Th(N,)is decidable. 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. Since EQ DFAis decidable, there is an algorithm. But we only show one here. Give an example of a DFA Astallion that in in ALLDFA. M 1 works as follows. If L(G) is empty, reject. tex and hw10. Express this problem as a language and show that it is decidable. A triangle in an undirected graph is a 3-clique. Then it accepts if L(D) is empty, otherwise it rejects. We show that Sis decidable. Think back to what you learned about NP-completeness. Show that for any two languages A and B a language J exists, where A≤T J and B≤T J. October 19, 2004. INTRODUCTION TO THE THEORY OF COMPUTATION. Show that, if P=NP. 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. P is the class of all languages that are decidable by deterministic multi-tape Turing machines running in polynomial time. Use K to check if L(G) is empty. 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. Show that the following language is decidable (ie. decidable, a decider for A NTM could be to decide A TM. Run M 1 on w. 4 Let AεCFG = { G | G is a CFG that generates ε}. Show that LEN_CFG is decidable. Submit to the decider for EQ DFA 3. There is a single “line valid bit” and three bits, B0, B1,. CSCI 2670 Introduction to Theory of Computing. 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 SUBSETDFA is decidable. Problem 3 (5 points). Let CONNECTED={IG is a connected undirected graph}. We construct a. Run M on w. The following DFA (with an alphabet of 0 and 1) is an example of this. Clearly, hMi2Sif and only if L(M) = L(N). Express this problem as a language and show that it is decidable. October 19, 2004. 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. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. 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. DFA is decidable. Let SUBSETDFA-A, B L(B)). (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. 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 LEN_CFG = { | G Is A CFG, K Elementof Z^noneg, And L(G) Intersection Sigma^k Notequalto (where Sigma Is The Alphabet Of G)}. Loading Autoplay When autoplay is enabled,. The sea will have its own, its share of tragedies. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. 24 on page 88 in the textbook. By Theorem 4. Gaining skills and applying them well. Show that ALLDFA is in P. (Proof idea) IWe construct a Turing machine S to decide the problem. Show that for any two languages A and B a language J exists, where A≤T J and B≤T J. " Note that U only recognizes A TM and does not decide A TM Because when we run M on w,. Question 2. 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. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. Give 5 languages that are in the class P. Show that ALLDFA is decidable. Note: The next question is a programming question. pdf using cvssubmit. Let build M0as follows: M0= \On input w: 1. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. 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 with an accepting state in the initial state. Homework Solution - Set 7 Due: Friday 10/24/08 1. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Being able to do the job. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. By Theorem 4. P is the class of all languages that are decidable by deterministic single-tape Turing machines running in polynomial time. Irish: ·sea Proverb: Bainfidh an fharraige a cuid féin amach; beidh a cuid féin ag an bhfarraige. Prove that ALLDFA is decidable. Loading Autoplay When autoplay is enabled,. A language L belongs to P iff L 2TIME(2n). , 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. Answer: Define the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. Give 5 languages that are in the class P. Else, accept. Course goals. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. 2) Let INFINITE DFA = { | A is a DFA and L(A) contains an infinite number of strings}. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. a Show that P is closed under complement and concatenation. Show That The Following. Since EQ DFA is decidable, we assume that is an algorithm M (i. Homework Solution - Set 7 Due: Friday 10/24/08 1. 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. DFA is decidable. Show that for any language A , a languages A and B that are Turing-incomparable-that is, where A≤T A. Tariq Malik on December 17, 2016 at 11:59am in CS701 Theory of Computation; Back to CS701 Theory of Computation Discussions. Show that NP is closed under union and concatenation. 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. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Let L 1 and L 2 be decidable languages, and M 1 and M 2 be the Turing machines that decide them. tex and hw10. DFA = fhA;Bi j Aand Bare DFAs and L(A) = L(B)g is decidable. Show that ALLDFA is decidable. The algorithm M 1 inputs hAi, where Ais a DFA. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. 3 Let ALLDFA = { | A is a DFA that recognizes (*}. Hint: In this case, all the states that are reachable from the start state are final states. 64 Let N be an NFA with k states that recognizes some language A. Let ALLDFA-{(A): A Is A DFA And L(A) = Σ*). 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. Let Mbe a Turing machine that decides this language. 57149995 >>57149874 I'm making a prop logic theorem prover. Knowing your stuff. Say that a variable A in CFL G is usable if it appears in some derivation of some string w G. 3) Answer: Consider the following Turing machine M = “ on input , where A is a DFA 1. CS701 ALL Current Mid Term Papers Fall 2016 And Past Mid Term Papers at One Place from 17 December 2016 to 29 December 2016. Q#4 Let LALL ={/M is a TM with i/p ∑ and L(M =∑*} Prove that LALL is not a co Turing recognizable. decidable, a decider for A NTM could be to decide A TM. Give 5 languages that are in the class P. Method II: Suppose on the contrary that L is regular. Berkshire Hathaway AGM is still the Warren Buffett show ; US election: Trump’s man in Vegas. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 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 that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. (Hint: Look at the proof for EDFA to get an idea. The Intel 80486 has an on-chip,unified cache. Show that ALLDFA is in P. Show that ALLDFA is in P. Given a CFG G and a variable A, consider the problem of testing whether A is usable. Run M1 on w. a Show that P is closed under complement and concatenation. txt) or read online for free. The algorithm M 1 inputs hAi, where Ais a DFA. 3) Let a 2-PDA be a pushdown automata with access to 2 stacks. Course goals. Show that, if A is nonempty, A contains some string of length at most k. Then, let M be a DFA that. Being able to do the job. 28) Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. Sipser provides an algorithm for ALL NFA that runs in nondeterministic space O(n), so ALL. Use K to check if L(G) is empty. (a) Show that L is not regular. M 1 works as follows. 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. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. 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. 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 =. Show that LEN_CFG is decidable. 17 December cs701 paper CS701 Q#1suppose there is a language L we know that L is Turing. We show that Sis decidable. Notice that this claim follows from Exercise 1. (b) No, because M on input 011 ends in a non-accept state. 4 on input , where T decides E DFA iii. (a) union: Construct TM M which decides the union of L1 and L2 : On input w: 1. Show that EQ CFG is undecidable. E(dfa) is a decidable language. Show that SUB_DFA is decidable. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Let build M0as follows: M0= \On input w: 1. Answer: Define the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. Show that the SPCP is decidable. P is the class of all languages that are decidable by deterministic single-tape Turing machines running in polynomial time. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. 3) Answer: Consider the following Turing machine M = " on input , where A is a DFA 1. Problem 3 (5 points). Yesterday Decidability and regular languages Today Putting things in perspective More on decidability and regular languages Decidability and context free grammars. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. Let T = “On input where A is a DFA i. Given a CFG G and a variable A, consider the problem of testing whether A is usable. Express this problem as a language and show that it is decidable. a Show that P is closed under complement and concatenation. Problem 1 For this problem you may assume the containment diagram in Figure 4. Formulate this problem as a language and show that it is decidable. Create a DFA B such that L(B) = Σ * 2. Exercise 4. Clearly, hMi2Sif and only if L(M) = L(N). There is a single “line valid bit” and three bits, B0, B1,. Show a Turing Machine accepts all DFAs. 15 of Sipser book) Q16. Able to give care, to show care, to just care. · billow, swell. It contains 8 KBytes and has a four-way set-associative organization and a block length of four 32-bit words. Let A = {hR,Si| R and S are regular expressions and L(R) L(S)}. Let SUB_DFA = { | A, B Are DFAs, And L(A) L(B)}. a Show that P is closed under complement and concatenation. CS701 mid term paper shared by student. 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 the SPCP 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)}. Think back to what you learned about NP-completeness. 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. We show that Sis decidable. Show that ALLDFA is decidable. Show that the following language is decidable (ie. decidable, a decider for A NTM could be to decide A TM. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. Show that A(CFG is decidable. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. Q#3 Let Alldfa={/A is a DFA and L(A)=∑*}Show that AllDFA is decidable. Note: The next question is a programming question. Use K to check if L(G) is empty. 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. cs701 fall 2016 mid term paper. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Give an example of a DFA Astallion that in in ALLDFA. Show that ALL DFA = fhBijBgis DFA and L(B) = gis decidable. 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. Creating DFA to prove closure properties. give the pseudocode for function def deciderSUB DFA (G) and discussion of correctness. Please attach your answer to this in the file Spiral. Let SUB_DFA = { | A, B Are DFAs, And L(A) L(B)}. Keeping a good spirit, one of optimism and openness. Show that ALLDFA is decidable. Exercise 4. Show that, if P=NP. CSCI 2670 Introduction to Theory of Computing. txt) or read online for free. Prove that ECFG is a decidable language. CS701 mid term paper shared by student. Videos recorded Spring 2014 for CSE355 at Arizona State University. Yesterday Decidability and regular languages Today Putting things in perspective More on decidability and regular languages Decidability and context free grammars. Show that the set of incompressible strings is un decidable. Lecture 32/65: Decidability and Decidable Problems hhp3. Idea: If G is in CNF, then it takes at most 2n 1 steps to generate w. Use the closure properties of regular languages and context-free languages to show that Lne is not regular 9. 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. Loading Autoplay When autoplay is enabled,. Transcription. 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. CSCI 2670 Introduction to Theory of Computing. , Turing machine that halts on all inputs) that decides if two given DFAs accepts the same language. A “good old boy” who can get along with others. CS701 - Theory of Computation Mid Term Paper Fall 2016 From 17 December 2016 to 03 January 2017. 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. Show That LEN_CFG Is Decidable. Submit to the decider for EQ DFA 3. com - id: 56456d-ZGMxZ. Run M on w. 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 Mbe a Turing machine that decides this language. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. Show that ALLDFA is in P. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. Let A = {hR,Si| R and S are regular expressions and L(R) L(S)}. Showing up. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. By Theorem 4. 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. Give a PDA that accepts Lpal2 b. Submit to the decider for EQ DFA 3. CSCI670CSCI670IntroductiontoTheoryofIntroductiontoTheoryofComputingComputingComputingComputingOctober13005AgendaAgenda•Yesterday-Decidabilityandregularlanguages. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. A triangle in an undirected graph is a 3-clique. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. 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. 1 (a) a Yes, because M on input 0100 ends in an accept state. Loading Unsubscribe from hhp3? Show more Show less. We construct a. (Proof idea) IWe construct a Turing machine S to decide the problem. Last week Variants of Turing machines Definition of algorithm This week Chapter 4 Decidable and undecidable languages The halting problem. Irish: ·sea Proverb: Bainfidh an fharraige a cuid féin amach; beidh a cuid féin ag an bhfarraige. a Show that P is closed under complement and concatenation. Announcements. 3) Answer: Consider the following Turing machine M = " on input , where A is a DFA 1. Formulate this problem as a language and show that it is decidable. Show thatALLDFAis decidable. Show That SUBSETDFA Is Decidable. Homework 8Solutions. Problem 1 For this problem you may assume the containment diagram in Figure 4. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. (b) Show that L is decidable. Homework Solution - Set 7 Due: Friday 10/24/08 1. 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. 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. Convert P0 into an equivalent CFG G 4. 2) Let INFINITEDFA = { | A is a DFA and L(A) contains an infinite number of strings}. The following DFA (with an alphabet of 0 and 1) is an example of this. Show that the collection of decidable languages is closed under union. 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. , 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. Keeping your cool. Homework due next Tuesday (10/26) Slideshow 3839749 by aislin. Let Mbe a Turing machine that decides this language. 57149995 >>57149874 I'm making a prop logic theorem prover. com - id: 56456d-ZGMxZ. Let Lpal2 be the language wwRvvR:w, v E E' a. Show that the SPCP is decidable. Express this problem as a language and show that it is decidable. Show that SUBSETDFA is decidable. Submit to the decider for EQ DFA 3. To test this condition, we can design a >TM T that uses a marking algorithm similar to that used in Example 3. Show That ALLDFA Is Decidable. Assg 9 - Solution sketches with study points added. Show that AεCFG is decidable. 3 Let ALLDFA = { | A is a DFA that recognizes (*}. Show that ALLDFA is decidable. Assg 9 - Solution sketches with study points added. 24 on page 88 in the textbook. Method II: Suppose on the contrary that L is regular. 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. Clearly, hMi2Sif and only if L(M) = L(N). Show That SUBSETDFA Is Decidable. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. Use the closure properties of regular languages and context-free languages to show that Lne is not regular 9. Since EQ DFA is decidable, we assume that is an algorithm M (i. E(dfa) is a decidable language. Showing up. import Data. Notice that this claim follows from Exercise 1. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. DFA = fhA;Bi j Aand Bare DFAs and L(A) = L(B)g is decidable. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. 15 of Sipser book) Q16. 4 Let AεCFG = { G | G is a CFG that generates ε}. Problem 2 - Semi-Decidability (25 points) Show that the acceptance problem for Turing machines. Transcription. Q#4 Let LALL ={/M is a TM with i/p ∑ and L(M =∑*} Prove that LALL is not a co Turing recognizable. CS701 ALL Current Mid Term Papers Fall 2016 And Past Mid Term Papers at One Place from 17 December 2016 to 29 December 2016. Give an example of a DFA Astallion that in in ALLDFA. , looping cannot happen. , 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 INFINITEDFA is decidable. Homework 8Solutions. Show that the SPCP is decidable. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. We now construct another algorithm M 1 to decide language ALL DFA. show that TRIANGLE Є powered TRIANGLE={IG contains a triangle}. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. 4 on input , where T decides E DFA iii. 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. 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 the set of incompressible strings is un decidable. Run M 1 on w. Show that ALLDFA is in P. Give a CFG that (1) Show "all" DFA is decidable. Give a CFG that (1) Show "all" DFA is decidable. Let Mbe a Turing machine that decides this language. Loading Autoplay When autoplay is enabled,. Therefore, A NTM cannot be in PSPACE, so A NTM is not PSPACE-complete. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Show that SUB DFA is decidable, i. cs701 fall 2016 mid term paper. 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. 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 EQ CFG is undecidable. Various application areas, such as modern cryptographic protocols, rely on theoretical principles that you will learn here. txt) or read online for free. Show that A(CFG is decidable. } Using the procedure given in class, convert the regular expression aa∪bb into an NFA, and then a DFA, M 1. DFA is decidable. Construct a PDA P0 such that L(P0) = L(P)\L(M) 3. Let CONNECTED={IG is a connected undirected graph}. Show that the set of incompressible strings is un decidable. IScan the input string hG;wi, determining whether the input constitutes a valid CFG. Show that the following language is decidable: SIMPDFA-[A) : A is a DFA with no. The following DFA (with an alphabet of 0 and 1) is an example of this. Showing up. Express this problem as a language and show that it is decidable. A language L belongs to P iff L 2TIME(2n). 28) Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. DFA is decidable. P is the class of all languages that are decidable by deterministic multi-tape Turing machines running in polynomial time. Show that the collection of decidable languages is closed under union. 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 ETM, the complement of. 15 of Sipser book) Q16. October 13, 2005. Homework 8Solutions. Let Lpal2 be the language wwRvvR:w, v E E' a. There is a single “line valid bit” and three bits, B0, B1,. Give 5 languages that are in the class P. proof DFA defines same language as minimal DFA. equivalent. M 1 works as follows. If T rejects, “reject” 2. Show that ALLDFA is decidable. pdf - Free download as PDF File (. Show that ALLDFA is reducible to EDFA by constructing a reduction f from ALLDFA to EDFA (in a form of a Turing machine) such that 〈A〉 ∈ ALLDFA ↔ f(〈A〉) ∈ EDFA Expert Answer (Solved) : Write a function that has one input, an integer number n, and one output, an array factors. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 is decidable. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. The following DFA (with an alphabet of 0 and 1) is an example of this. 24 on page 88 in the textbook. Show that A(CFG is decidable. Let Lpal2 be the language wwRvvR:w, v E E' a. Show that ALL NFA = fhMi: L(M) for the NFA Mwhose input alphabet is gis in PSPACE. A triangle in an undirected graph is a 3-clique. (b) No, because M on input 011 ends in a non-accept state. Let SUB_DFA = { | A, B Are DFAs, And L(A) L(B)}. Using the procedure given in class, convert the regular expression. If M accepts w, accept; if M rejects w, reject. Show that T is countable. 3) Answer: Consider the following Turing machine M = “ on input , where A is a DFA 1. (Hint: Look at the proof for EDFA to get an idea. 3 Let ALLDFA ={ (A) I A is a DFA and L(A) = Sigma* }. •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. Show that NP is closed under union and concatenation. Show that INFINITEDFA is decidable. We show that Sis decidable. Exercise 4. CSCI 2670 Introduction to Theory of Computing. (Families Affected by Fetal Alcohol Spectrum Disorder). Since EQ DFA is decidable, we assume that is an algorithm M (i. Solution Outline: (a) Method I: Let D i,i ≥ 1 be the DFA that recognizes the language {1i}. b Let A be a decidable language and let D be a polytime decider for it. Convert P0 into an equivalent CFG G 4. 173 and the fact that each region is no-empty. FAFASD seeks to spread information, awareness, and hope for families impacted by fetal alcohol spectrum disorder. Then, {(hD ii, )} i≥1 constitutes an infinite collection of distinguishable strings. (b) No, because M on input 011 ends in a non-accept state. Show that for any two languages A and B a language J exists, where A≤T J and B≤T J. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. Create DFA that accept language where number of 0's is even and after every 1 goes 0. Show that ALLDFA is in P. Turing-decidable language Answer: A language A that is decided by a Turing machine; i. INTRODUCTION TO THE THEORY OF COMPUTATION. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. 3) Consider counting the ordered pairs of integers in the Cartesian plane. 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. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. decidable, a decider for A NTM could be to decide A TM. Problem 3 (5 points). 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. All DFA = {| A is DFA and L(A)= ∑ *} show that is AllDFA decidable (Exercise 4. By Theorem 4. Yesterday Decidability and regular languages Today Putting things in perspective More on decidability and regular languages Decidability and context free grammars. Think back to what you learned about NP-completeness. Able to give care, to show care, to just care. Thelanguageweareconcernedwithis INFINITEDFA =fhAi: A isaDFAandL(A)isaninflnitelanguageg: Want: To show that INFINITEDFA is decidable, meaning to construct a TM M that decides INFINITEDFA. com - id: 56456d-ZGMxZ. Show a Turing Machine accepts all DFAs. CS701 mid term paper shared by student. If L(G) is empty, reject. Clearly, hMi2Sif and only if L(M) = L(N). Show that T is countable. 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. 15 (a) We want to show that if L 1 and L 2 are decidable, then L 1 [L 2 = L 3 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. If L(G) is empty, reject. 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. Q#4 Let LALL ={/M is a TM with i/p ∑ and L(M =∑*} Prove that LALL is not a co Turing recognizable. Give a PDA that accepts Lpal2 b. Being able to do the job. Express this problem as a language and show that it is decidable. Show That The Following. Able to give care, to show care, to just care. 5 Let ETM = { M | M is a TM and L(M ) = ∅}. ) * (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. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. 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. We show that Sis decidable. Question 2. Give an example of a DFA Astallion that in in ALLDFA. From input hB;Cia Turing machine constructs a DFA Drecognizing L(B)\L(C). Clearly, hMi2Sif and only if L(M) = L(N). , looping cannot happen. Note: The next question is a programming question. Show that ALLDFA is in P. pdf), Text File (. Show that the set of incompressible strings is un 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. 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. decidable, a decider for A NTM could be to decide A TM. Show that ALLDFA is decidable. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. M 1 works as follows. Show that SUBSETDFA is decidable. Show that EQ CFG is undecidable. proof DFA defines same language as minimal DFA. Show that INFINITEDFA is decidable. Keeping your cool. Let CONNECTED={IG is a connected undirected graph}. Step-by-Step Solution: Step 1 of 3. Decidable Problems for Context Free Languages Theorem A CFG is decidable, where A CFG = fhG;wijG is a CFG generating wg. The algorithm M 1 inputs hAi, where Ais a DFA. One-to-one function f: T N. , exhibit a decision procedure for this language): L = {ha,b,ci : a, b and c are regular expressions and a2 ∪b2 = c2. Since EQ DFA is decidable, we assume that is an algorithm M (i. Convert P0 into an equivalent CFG G 4. Notice that this claim follows from Exercise 1. Show that LEN_CFG is decidable. A “good old boy” who can get along with others. We show that Sis decidable. b Let A be a decidable language and let D be a polytime decider for it. (In each transition, we can read the top of both stacks and push something on top of both stacks, if we choose. Keeping a good spirit, one of optimism and openness. Construct a PDA P such that L(P) = fw j w is a palindromeg 2. a Show that P is closed under complement and concatenation. Let ALLDFA = { | A is a DFA and L(A) = Σ* }. Sipser Problem 3. Show that ALLDFA is in P. 3 Let ALLDFA = { A | A is a DFA and L(A) = Σ∗}. Problem 1 For this problem you may assume the containment diagram in Figure 4. ThismeansthatforallDFAsA wewant †IfL(A)isaninflnitelanguagethenM(hAi)accepts †IfL(A)isaflnitelanguagethenM(hAi)rejects. Answer: Define the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. The following DFA (with an alphabet of 0 and 1) is an example of this. ) * (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. We now construct another algorithm M 1 to decide language ALL DFA. The sea will have its own, its share of tragedies. Irish: ·sea Proverb: Bainfidh an fharraige a cuid féin amach; beidh a cuid féin ag an bhfarraige. Show that A is decidable. Use the closure properties of regular languages and context-free languages to show that Lne is not regular 9.