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# MELJUN CORTES Automata Theory (Automata17)

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MELJUN CORTES Automata Theory (Automata17)

MELJUN CORTES Automata Theory (Automata17)

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• 1. MELJUN P. CORTES, MBA,MPA,BSCS,ACS CSC 3130: Automata theory and formal languages Undecidable Languages MELJUN CORTES
• 2. Are Turing Machines computers? input data instructions (program) computer output data A Turing Machine does not take the program as part of its input. ... but it can!
• 3. The Universal Turing Machine input x program 〈 M〉 U M on input x The universal TM U takes as inputs a program M and a string x and simulates M on x The program M itself is specified as a TM!
• 4. Turing Machines as strings ☐/☐R q0 0/0R 1/1R qacc M qrej • This Turing Machine can be described by the string 〈 M 〉 = ((q0, qacc, qrej), ((q0, q0, ☐/☐R), (q0, qacc, 0/0R), (q0, qrej, 1/1R)))
• 5. The universal Turing Machine 001 U input x ((q0, qacc, qrej), ((q0, q0, ☐/☐R), program 〈 M〉 q0 state U on input ( 〈 M 〉 , x) will simulate M on input x
• 6. Languages about automata • A TM can take another TM (or DFA, CFG) as input ADFA = {( 〈 D 〉 , w): decidDFA that accepts input w} D is a ab le APDA = {( 〈 P 〉 , w): Pdea PDA that accepts w} is cida b le ATM = {( 〈 M 〉 , w): M is a TM that accepts w} Which of these is decidable?
• 7. Languages about automata ATM = {( 〈 M 〉 , w): M is a TM that accepts w} • Idea: Simulate M on input w • Simulation is correct: – If M accepts w, we accept – If M rejects w, we reject – If M loops on w, we loop • This TM recognizes but does not decide ATM
• 8. Recognizing versus deciding The language recognized by a TM is the set of all inputs that make it reach qacc A TM decides language L if it recognizes L and does not loop on any input – This kind of TM is called a decider • The textbook uses different terminology: recognizable = recursively enumerable (r. e.) decidable = recursive
• 9. Undecidability • Turing’s Theorem: The language ATM is undecidable. • Before we prove this, let’s observe one thing: A Turing Machine M can be given its own description 〈 M 〉 as an input!
• 10. Proof of undecidability • Proof by contradiction: Suppose ATM is decidable. Then there is a TM H that decides ATM: accept if M accepts w 〈M〉 H reject if M rejects w 〈 M 〉 or M loops on w or M loops on 〈 M〉 What happens when w = 〈 M 〉 ? 〈M〉 ,w 〈M〉
• 11. Proof of undecidability 〈M〉, 〈M〉 accept if M accepts 〈M〉 reject if M rejects 〈 M 〉 or M loops on 〈M〉 H Let H’ be a TM that does the opposite of H acc H H’ acc rej rej
• 12. Proof of undecidability 〈M〉, 〈M〉 〈M〉, 〈M〉 H H’ accept if M accepts 〈M〉 reject if M rejects 〈 M 〉 or M loops on 〈M〉 if M rejects 〈 M 〉 or M loops on reject 〈M accepts if M 〉 accept 〈M〉
• 13. Proof of undecidability 〈M〉, 〈M〉 H’ if M rejects 〈 M 〉 or M loops on reject 〈M accepts 〈 M 〉 if M 〉 accept Let D be the following TM: 〈M〉 copy 〈M〉, 〈M〉 H’
• 14. Proof of undecidability 〈 M 〉 D if M rejects 〈M〉 reject or M accepts 〈 if M loops on 〈 M〉 M〉 accept What happens when M = D? If D accepts 〈 D 〉 D rejects 〈 D 〉 then so D does not exist! If D rejects 〈 D 〉 D accepts 〈 D 〉 then
• 15. Proof of undecidability: conclusion • Proof by contradiction – We assumed ATM was decidable – Then we had Turing Machines H, H’, D – But D does not exist! • Conclusion The language ATM is undecidable.
• 16. What happened? input w Turing Machine … ε 0 1 00 acc rej rej acc rej acc loop rej M3 rej loop rej rej … … M1 M2 … We can write an infinite table for every pair (M, w)
• 17. What happened? 〈 〈 M1 〉 M2 〉 M3 〉 M4 〉 … 〈 〈 acc loop rej acc rej acc rej acc rej loop rej … M3 loop … … M1 M2 Now let’s look only at those w that describe a TM M
• 18. What happened? 〈 〈 M1 〉 M2 〉 M3 〉 M4 〉 … 〈 〈 acc loop rej acc rej acc rej acc rej rej acc rej … … D … … … M1 M2 If ATM is decidable, then TM D is in the table
• 19. What happened? 〈 〈 M1 〉 M2 〉 M3 〉 M4 〉 … 〈 〈 acc loop rej acc rej acc rej acc rej rej acc rej D … … … M1 M2 D does the opposite of the diagonal entries 〈 M 〉 D accept if M rejects or loops on 〈 M 〉 reject if M accepts 〈M〉
• 20. What happened? … 〈 〈 M1 〉 M2 〉 M3 〉 M4 〉 〈 D 〉 〈 〈 acc loop rej acc rej acc rej acc … acc rej rej acc rej … ? D loop … … M1 M2 We run into trouble when we look at (D, 〈 D 〉 )!
• 21. Unrecognizable languages The language ATM is recognizable but not decidable. • How about languages that are not recognizable? ATM = {( 〈 M 〉 , w): M is a TM that does not accept w} = {( 〈 M 〉 , w): M rejects or loops on input w} The language ATM is not recognizable.
• 22. Unrecognizable languages • General Theorem If L and L are both recognizable, then L is decidable. • We know ATM is recognizable, so if ATM were also, then ATM would be decidable • Impossible by Turing’s Theorem
• 23. Unrecognizable languages If L and L are both recognizable, then L is decidable. • Proof idea w M accept if w ∈ L rej/loop if w ∉ L accept M’ accept if w ∉ L rej/loop if w ∈ L reject w w
• 24. Unrecognizable languages If L and L are both recognizable, then L is decidable. • Proof attempt Problem: If M loops on w, we Let M = TM for L, M’ = TM for L will never get to step 2! On input w, ① Simulate M on input w. If it accepts, accept. ② Simulate M’ on input w. If it accepts, reject.
• 25. Bounded simulation If L and L are both recognizable, then L is decidable. • Proof Let M = TM for L, M’ = TM for L On input w, For k = 0 to infinity Simulate M on input w for k steps. If it accepts, accept. Simulate M’ on input w for k steps. If it accepts, reject.
• 26. Another undecidable language HALTTM = {( 〈 M 〉 , w): M is a TM that halts on input w} HALTTM is an undecidable language • To prove this, we will show that If HALTTM can be decided, so can ATM. … so by Turing’s Theorem it is undecidable.
• 27. Undecidability of halting If HALTTM can be decided, so can ATM. • Suppose H decides HALTTM 〈M〉,w H accept if M halts on w reject if M loops on w • We want to use H to design D for ATM 〈M〉,w ? accept if M accepts w reject if M rejects or loops on w
• 28. Undecidability of halting 〈M〉,w 〈M〉,w acc H rej H accept if M halts on w reject if M loops on w acc simulate M on w rej D accept if M accepts w reject if M rejects or loops on w
• 29. More undecidable problems L1 = { 〈 M 〉 : M is a TM that accepts input ε} L2 = { 〈 M 〉 : M is a TM that accepts some input} L3 = { 〈 M 〉 : M is a TM that accepts all inputs} L4 = { 〈 M, M’ 〉 : M and M’ accept the same inputs} Which of these are recognizable?
• 30. Gödel’s incompleteness theorem Some mathematical statements are true but not provable. • Proof due to Turing: If every true statement is provable, then ATM can be decided.
• 31. Proof sketch of Gödel’s theorem • Suppose every true statement has a proof • For every TM M and input w, one of the statements M accepts w or M does not accept w is true, so it must have a proof
• 32. Proof sketch of Gödel’s theorem • Consider this algorithm for ATM: On input 〈 M 〉 , w: For k = 0 to infinity For all possible proofs p of length k If p is a proof of “M accepts w”, accept. If p is a proof of “M does not accept w”, reject. One of them is true, so it has a proof p Eventually, the algorithm will find this proof
• 33. Proof sketch of Godel’s theorem • Consider this algorithm for ATM: On input 〈 M 〉 , w: For k = 0 to infinity For all possible proofs p of length k If p is a proof of “M accepts w”, accept. If p is a proof of “M does not accept w”, reject. Recall that TMs can check if a proof p is correct