The Halting Problem

1. The Halting Problem Sipser 4.2 (pages 173-182)

2. Taking stock All languages Turing-recognizable ? D Turing-decidable Context-free languages anbncn Regular languages 0n1n 0*1*

3. Are there problems a computer can’t solve?! • But they seem so powerful… • What about software verification? • Given a program and a specification of what it should do, can we check if it is correct?

4. What about deciding TMs? • ATM = {<M,w> | M is a TM and M accepts w} • Theorem 4.11: ATM is undecidable! • Is it even recognizable? • Let U = “On input <M,w>, where M is a TM: • Simulate M on input w. • If M ever enters its accept state, accept; if M ever enters its reject state, reject.”

5. Towards proving undecidability • Cantor 1873: How can we tell whether one infinite set is “larger” than another? • A function f from A to B is • One-to-one if f(x) ≠ f(y) if x≠y • (f never maps two elements to the same value) • Onto if every element of B is hit • A correspondence is a function that is both one-to-one and onto

6. Countable sets • Sets A and B have the same size if: • A and B are finite with the same number of elements • A and B are infinite with a correspondence between them • A set is countable if it is finite or has the same size as N • (natural numbers1, 2, 3,…)

7. For example… • E = { even natural numbers } is countable • Define f : N→ Eas f(n) = 2n

8. Diagonalization • Theorem 4.17: R is uncountable. • Proof: By contradiction. Assume there is a correspondence. We find a real number x ≠ f(n) for any natural number n.

9. Put your thinking caps on! • How do we show that: • Σ* is countable • assumeΣ = {0,1} • B = { all infinite binary sequences } is uncountable

10. Uh-oh… • Theorem 4.18: Some languages are not Turing-recognizable. • Proof: Let L be the set of all languages over Σ. Define a correspondence from L to B. Σ*= { ε, 0, 1, 00, 01, 10, 11, 000, 001,…} A = { 0, 00, 01, 000, 001,…} ΧA= 0 1 0 1 1 0 0 1 1 …

11. The Halting Problem • ATM = {<M,w> | M is a TM and M accepts w} • Theorem 4.11:ATMis undecidable. • Proof: By contradiction. Assume H is a decider for ATM. We construct a TM D = “On input <M>, where M is a TM: • Run H in input <M, <M>> • If H accepts, reject; if H rejects, accept.”

12. Huh? • What if we run D on <D>? • Then • Contradiction! Then H cannot exist and ATM is undecidable.

13. Where is the diagonalization? • Running a machine on its description

14. Where is the diagonalization? • Running H on a machine and its description

15. Where is the diagonalization? • Adding D to the picture

16. So have we found a language that is not Turing-recognizable? • Theorem 4.22: A language is decidable iff it and its complement are recognizable • Then ATMis not Turing-recognizable!

17. Updating the picture All languages Turing-recognizable ATM D Turing-decidable Context-free languages anbncn Regular languages 0n1n 0*1*