CSE 105 Theory of Computation

1. CSE 105Theory of Computation Alexander Tsiatas Spring 2012 Theory of Computation Lecture Slides by Alexander Tsiatas is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Based on a work at http://peerinstruction4cs.org. Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org.

2. What we did last time ATMis UNDECIDABLE

3. ATM is undecidable • Proof by contradiction • Assume ATM is decidable • There’s a TM MATM that decides it • Use MATM to create another TM D in a clever way: • Running D(<D>) creates a paradox • D accepts <D> if and only if D rejects <D>

4. 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.

5. Proving it is undecidable The Halting ProblemHALTTM

6. 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

7. 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.

8. We just did a reduction! • We showed that *if* we have a solution to HALTTM, then we have a solution to ATM. • What did we show exactly? • ATM reduces to HALTTM. • HALTTM reduces to ATM. • HALTTM and ATM reduce to each other. • None of the above or more than one of the above.

9. We showed some languages are not decidable… …Are there Languages That Are NotTuring-recognizable?

10. Are there languages that areNOT recognizable? • YES, based on diagonalization and comparing infinite sets • |Languages| > |TM’s|, so there must be a language not recognized by a TM

11. Could ATMbe not recognizable?{<M,w> | M is a TM, M accepts w} • No. We proved it was recognizable on Thurs. • B(<M,w>): • Run M(w) and accept if it accepts, reject if it rejects • (If it loops that’s ok) • Thought experiment: • What about the complement of ATM? • Lets just guess that it is also recognizable, and has a TM C that recognizes it • Could we use B and C to build a decider(!) for ATM?

12. Thm.: ATM is not recognizable Proof by contradiction • Assume ATM is recognizable, so some TM C recognizes it. • We know that ATM is recognizable, so let Bbe a TM that recognizes it. • MATM-Decider (<M,w>): • But ATM is undecidable, a contradiction. So the assumption is false and ATM is not recognizable

13. Recognizable and co-recognizable • Same method for arbitrary L: • L is recognizable but not decidable => L is not recognizable • Class of recognizable languages is not closed under complement • Same construction for arbitrary L: • L is recognizable AND co-recognizable => L is decidable

14. EQTM, SubsetTM So, what does the Venn Diagram look like? Co-recognizable A B Co-recognizable Decidable Recognizable Decidable Recognizable C D Recognizable Recognizable Co-recognizable Co-recognizable Decidable Decidable

15. Changing one problem into another Reductions More examples 15

16. ETM = { <M> | M is a TM, and L(M) = {} } • Undecidable! Proven in textbook

17. EQTM= { <M1,M2> | M1, M2 are TMs and L(M1) = L(M2) } • Thm. EQTMis undecidable. • Proof by contradiction. • Assume that EQTM is decidable, and some TM MEQ decides it. • Construct a TM METM that decides ETM: • METM(<M,w>): • Run MEQ(<M,Mx>) • If MEQ accepts, then accept. If it rejects, then reject. • But is ETM is undecidable, a contradiction. So the assumption is false and EQTM is undecidable. What Mx makes this proof work? • (a) L(Mx) = { w | w is in Σ* } (b) L(Mx) = EQTM • (c) L(Mx) = {} (d) L(MX) = ETM

18. ETM reduces to EQTM (a) “If A reduces to B, we can use a solution to B to solve A.” We just did a reduction! • We showed that *if* we have a solution to EQTM, then we have a solution to ETM. • What did we show exactly? • ETM reduces to EQTM. • EQTM reduces to ETM. • EQTM and ETM reduce to each other. • None of the above or more than one of the above.

19. Changing one problem into another Reductions Using ATMMore examples 19

20. Thm.: TWO = {<M> |  for all w in L(M), w has exactly two 1s} is undecidable • Proof by contradiction: • Assume TWO is decidable by TM DTWO. Use DTWOto construct TM DATM that decides ATM. • DATM (<M, w>): • Construct M'(x): • Count the number of 1s in x.  If it is 2, accept. • Simulate M(w). If M(w) accepts, accept. If it rejects, reject. (might loop) • Call DTWO(<M'>)  //DTWOis a decider for TWO • If DTWO(<M'>) accepts, reject, else accept. • But ATM is undecidable, a contradiction. So the assumption is false and T is undecidable. QED. What is L(M’)? (a) Σ* (b) {x | x has exactly two 1s}(c) Σ * if M(w) accepts, and {x|x has exactly two 1s} if M(w) does not accept(d) Σ * if M(w) accepts, and the empty set if M(w) rejects. 20

21. We just did a reduction! • We showed that if we have a solution to TWO, then we have a solution to ATM. • What did we show exactly? • ATM reduces to TWO. • TWO reduces to ATM. • TWO and ATM reduce to each other. • None of the above or more than one of the above.

22. Thm. T = {<M> | M is a TM and both “101” and “111” are in L(M)} is undecidable • Proof by contradiction: (Reduce from ATM.) • Assume T is decidable by TM MT. Use MT to construct TM X that decides ATM. • X(<M,w>): • Construct Z(m): • If m != “111” and m != “101” then reject. • Else: Run M(w), if it accepts then accept. If it rejects then reject (might loop in which case obviously Z loops). • Run MT(<Z>). If it accepts then accept, otherwise reject. • But ATM is undecidable, a contradiction. So the assumption is false and T is undecidable. QED. What is L(Z)? (a) Σ * (b) {“101”, “111”}(c) empty set if M(w) rejects, and {“101”,”111”} if M(w) accepts.(d) Σ * if M(w) accepts, and {“101”,”111”} if M(w) does not accept 22

23. We just did a reduction! • We showed that if we have a solution to T, then we have a solution to ATM. • What did we show exactly? • ATM reduces to T. • T reduces to ATM. • T and ATM reduce to each other. • None of the above or more than one of the above.

24. Thm. T = {<M> | M is a TM that accepts wR whenever it accepts w} is undecidable • Proof by contradiction: (Reduce from ATM.) • Assume T is decidable by TM MT. Use MT to construct TM X that decides ATM. • X(<M,w>): • Construct Z(m): • If m != “01” and m!= “10” then reject • If m == “01” accept • ??? • Run MT(<Z>). If it accepts then accept, otherwise reject. • But ATM is undecidable, a contradiction. So the assumption is false and T is undecidable. QED. How do we finish Z? • Run M(w), if it accepts then accept. If it rejects then reject (might loop in which case obviously Z loops). • Run M(w), if it accepts then reject. If it rejects then accept (might loop in which case obviously Z loops). 24

25. We just did a reduction! • We showed that if we have a solution to T, then we have a solution to ATM. • What did we show exactly? • ATM reduces to T. • T reduces to ATM. • T and ATM reduce to each other. • None of the above or more than one of the above.