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Announcements

Announcements. Homework HW8 due Tues 5/29 11am HW5 grades are out Monday is Memorial Day; no office hours, etc. Go see a parade, or some fireworks, or something TA Evaluations at the end of class ! There’s a lot of stuff in today’s lecture. CSE 105 Theory of Computability.

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Announcements

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  1. Announcements • Homework • HW8 due Tues 5/29 11am • HW5 grades are out • Monday is Memorial Day; no office hours, etc. • Go see a parade, or some fireworks, or something • TA Evaluations at the end of class! • There’s a lot of stuff in today’s lecture

  2. CSE 105Theory of Computability Dr. Alexander Tsiatas Spring 2012

  3. Preparing for DIAGONALIZATION proof!! Review

  4. Sizes of infinite sets: countably infinite • The natural numbers {0, 1, 2, 3, 4, …} are countably infinite. • Any set that’s the same size as the natural numbers is countably infinite. • 2 sets are the same sizeif there’s a correspondence between them. • Informal, but equivalent: a set is countably infiniteif you can list themsystematically like the natural numbers. • The list will eventually contain ALL elements

  5. Sizes of infinite sets: strings • For any alphabet (here, {0, 1}), the set of all strings is countably infinite: • ε, 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, 0000, ….

  6. Sizes of infinite sets: Turing machines • The set of all Turing machines is countably infinite: • You can just encode TM’s as strings.

  7. Sizes of infinite sets: Languages • Remember: a language is a set of strings. • Can we make a list of ALL languages? • Let’s try

  8. Sizes of infinite sets: Languages Claim: no matter the order you list languages, there is ALWAYS one that’s not on the list. BAD = {si | si is NOT in Li}

  9. So what?!?! • No matter how you try to list languages, there’s always a BAD language not on the list. • The set of languages is NOT countably infinite! The set of TM’s IS countably infinite! • |Languages| > |TM’s| • Pigeonhole principle: there are languages that are NOT accepted by any TM.

  10. ATM= {<M,w> | M is a TM, M accepts w} ATM is: • A language • An operation • A set of strings • A Turing Machine • None or more than one of the above

  11. The TM Acceptance ProblemATM

  12. ATM= {<M,w> | M is a TM, M accepts w} ATM is Recognizable • TRUE • FALSE • Other

  13. At last the proof you’ve all been waiting for! ATMIs UNDecidable

  14. The Game Plan: • Recall: ATM = {<M,w> | M is a TM, M accepts w} • Thm: ATM is undecidable • Proof by Contradiction: • Assume (towards contradiction) that ATM is decidable, so there exists some TM MATM decides ATM. • Want to show: A TM D that uses MATM as a subroutine, allowing D to do something impossible/contradictory. • D(w): //w is a string • What D is or what it does is totally up to us—but we will have to be very clever to come up with something that will cause an impossibility/contradiction. • Then, because we will have reached a contradiction, we will conclude that the assumption is false, and ATM is not decidable.

  15. Let’s look at MATM: • Assume (towards contradiction) that ATM is decidable, so there exists some TM MATM decides ATM. • Let TM G be a recognizer s.t. L(G) = {w | |w| is even} • What will happen if we run MATM(<G,111>)? • MATM accepts • MATM rejects • MATMloops • Not enough information • Other ATM = {<M,w> | M is a TM, M accepts w}

  16. Proof: • Thm: ATM = {<M,w> | M is a TM, M accepts w} is undecidable • Assume (towards contradiction) that ATM is decidable, so there exists some TM MATM that decides ATM. • Construct a TM D as follows: • D(<M>): //input is a string description of a TM • MATM(<M,<M>>) //MATM is a decider, so no infinite looping—it will tell us if <M> is in L(M) or not, even if M loops on <M> • If MATM accepts, we reject • If MATM rejects, we accept // we do the opposite of MATM • Now, it will take one more step to show how D can do something impossible

  17. Pause to Examine TM D D(<M>): //input is a string description of a TM • MATM(<M,<M>>) //MATM is a decider, so no infinite looping—it will tell us if <M> is in L(M) or not, even if M loops on <M> • If MATM accepts, we reject • If MATM rejects, we accept // we do the opposite of MATM • Let Mx be a TM, where L(Mx) = {}. What happens when we input Mx to D like this: D(<Mx>)? • D accepts • D rejects • D loops • Not enough information • Other A ATM = {<M,w> | M is a TM, M accepts w}

  18. Pause to Examine TM D D(<M>): //input is a string description of a TM • MATM(<M,<M>>) //MATM is a decider, so no infinite looping—it will tell us if <M> is in L(M) or not, even if M loops on <M> • If MATM accepts, we reject • If MATM rejects, we accept // we do the opposite of MATM ML(w): • Go into an infinite loop • What happens when we input ML to D like this: D(<ML>)? • D accepts • D rejects • D loops • Not enough information • Other A A ATM = {<M,w> | M is a TM, M accepts w}

  19. Pause to Examine TM D D(<M>): //input is a string description of a TM • MATM(<M,<M>>) //MATMis a decider, so no infinite looping—it will tell us if <M> is in L(M) or not, even if M loops on <M> • If MATM accepts, we reject • If MATM rejects, we accept // we do the opposite of MATM • Let Mvbe a TM where L(Mv) = Σ*.What happens when we input Mv to D like this: D(<Mv>)? • D accepts • D rejects • D loops • Not enough information • Other B ATM = {<M,w> | M is a TM, M accepts w}

  20. Pause to Examine TM D D(<M>): //input is a string description of a TM • MATM(<M,<M>>) //MATM is a decider, so no infinite looping—it will tell us if <M> is in L(M) or not, even if M loops on <M> • If MATM accepts, we reject • If MATM rejects, we accept // we do the opposite of MATM • What happens when we input D to D like this: D(<D>)? • D accepts • D rejects • D loops • Not enough information • Other E ATM = {<M,w> | M is a TM, M accepts w}

  21. Proof: • Thm: ATM = {<M,w> | M is a TM, M accepts w} is undecidable • Assume (towards contradiction) that ATM is decidable, so some TM MATM decides ATM. • Construct a TM D as follows: • D(<M>): //input is a string description of a TM • MATM(<M,<M>>)– MATM is a decider, so it will either accept or reject (no infinite looping) • If MATM accepts, we reject • If MATM rejects, we accept // we do the opposite of MATM • Run D(<D>). Observe that D(<D>) should accept when D(<D>) rejects, and D(<D>) should reject when D(<D>) accepts, a contradiction. Therefore the assumption is false, and ATM is undecidable.

  22. Proving it is undecidable The Halting ProblemHALTTM

  23. The Halting Problem • HALTTM = {<M,w> | M is a TM and M halts on input w} • Doesn’t say if M accepts or rejects w, just that it halts • Imagine we have a hypothetical TM Mhalt that decides HALTTM. • Could we use Mhaltto build a decider for ATM? • (recall: ATM = {<M,w> | M is a TM and M accepts w}) Showing HALT is undecidable by reduction from ATM

  24. Thm. HALTTM is undecidable. • Proof by contradiction. • Assume that HALTTM is decidable, and some TM Mhaltdecides it. • Construct a TM MATM that decides ATM: • But is ATM is undecidable, a contradiction. So the assumption is false and HALTTM is undecidable.

  25. We just did a reduction! • We will do a LOT more of this next week!

  26. TA Evaluations!!!!!111 • Class number: CSE 105 • TA’s name: SrdjanKrsticOR Wenbo Zhao • Evaluate whoever you had the MOST contact with • Only evaluate ONE of the TA’s • If you didn’t have much TA contact: leave it blank • When finished: bring to front and you can go

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