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CSE 105 theory of computation

CSE 105 theory of computation. Fall 2019 https://cseweb.ucsd.edu/classes/fa19/cse105-a/. Today's learning goals Sipser Ch 4.1, 4.2. Trace high-level descriptions of algorithms for computational problems.

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CSE 105 theory of computation

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  1. CSE 105 theory of computation Fall 2019 https://cseweb.ucsd.edu/classes/fa19/cse105-a/

  2. Today's learning goals Sipser Ch 4.1, 4.2 • Trace high-level descriptions of algorithms for computational problems. • Use counting arguments to prove the existence of unrecognizable (undecidable) languages. • Use diagonalization in a proof of undecidability.

  3. Undecidable? • There are many ways to prove that a problem is decidable. • How do we find (and prove) that a problem is notdecidable?

  4. Counting arguments Before we proved the Pumping Lemma … We proved there was a set that was not regular because Uncountable All sets of strings All Regular Sets Countable

  5. Counting arguments Uncountable All sets of strings Why is the set of Turing-recognizable languages countable? • It's equal to the set of all TMs, which we showed is countable. • It's a subset of the set of all TMs, which we showed is countable. • Each Turing-recognizable language is associated with a TM, so there can be no more Turing-recognizable languages than TMs. • More than one of the above. • I don't know. All Turing-recognizable sets Countable

  6. Satisfied? • Maybe not … • What's a specific example of a language that is not Turing-recognizable? or not Turing-decidable? • Idea: consider a set that, were it to be Turing-decidable, would have to "talk" about itself, and contradict itself!

  7. ATM Recall ADFA = {<B,w> | B is a DFA and w is in L(B) } Decider for this set simulates arbitrary DFA ATM = {<M,w> | M is a TM and w is in L(M) } Decider for this set simulates arbitrary TMs ???

  8. ATM ATM = {<M,w> | M is a TM and w is in L(M) } Define the TM N = "On input <M,w>: • Simulate M on w. • If M accepts, accept. If M rejects, reject."

  9. ATM ATM = {<M,w> | M is a TM and w is in L(M) } Define the TM N = "On input <M,w>: • Simulate M on w. • If M accepts, accept. If M rejects, reject." What is L(N)?

  10. ATM ATM = {<M,w> | M is a TM and w is in L(M) } Define the TM N = "On input <M,w>: • Simulate M on w. • If M accepts, accept. If M rejects, reject." Does N decide ATM?

  11. ATM ATM = {<M,w> | M is a TM and w is in L(M) } Define the TM N = "On input <M,w>: • Simulate M on w. • If M accepts, accept. If M rejects, reject." Conclude: ATM is Turing-recognizable. Is it decidable?

  12. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that it is. Call MATM the decider for ATM: For every TM M and every string w, • Computation of MATM on <M,w> halts and accepts if w is in L(M). • Computation of MATM on <M,w> halts and rejects if w is not in L(M). MATM ≠ N

  13. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM Define the TM D = "On input <M>: • Run MATM on <M, <M>>. • If MATM accepts, reject; if MATM rejects, accept."

  14. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM Define the TM D = "On input <M>: • Run MATM on <M, <M>>. • If MATM accepts, reject; if MATM rejects, accept." Which of the following computations halt? Computation of D on <X> Computation of D on <Y> where Y is TM with L(Y) =Σ* Computation of D on <D> All of the above.

  15. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM Define the TM D = "On input <M>: • Run MATM on <M, <M>>. • If MATM accepts, reject; if MATM rejects, accept." Consider running D on input <D>. Because D is a decider: • either computation halts and accepts … • or computation halts and rejects …

  16. Diagonalization proof: ATM not decidable Sipser 4.11 Diagonalization??? Self-reference "Is <D> an element of L(D)?" Assume, towards a contradiction, that MATM decides ATM Define the TM D = "On input <M>: • Run MATM on <M, <M>>. • If MATM accepts, reject; if MATM rejects, accept." Consider running D on input <D>. Because D is a decider: • either computation halts and accepts … • or computation halts and rejects …

  17. ATM • Recognizable • Not decidable Fact (from discussion section): A language is decidable iff it and its complement are both recgonizable. Corollary: The complement of ATM is unrecognizable.

  18. Decidable vs. undecidable Which of the following languages is undecidable? • INFINITEDFA = { <A> | A is a DFA and L(A) is an infinite language} • STM = { <M> | M is a TM and M has exactly 7 states} • RevDFA = { <B> | B is a DFA and for all strings w, B accepts w iff it accepts wR} • RecTM = { <X> | X is a TM and L(X) is recognizable} • DecTM = { <Y> | Y is a TM and L(Y) is decidable}

  19. So far

  20. Do we have to diagonalize? • Next time: comparing difficulty of problems.

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