1. Mapping Reducibility Sipser 5.3 (pages 206-210)

2. CS 311 Fall 2008 2 Computable functions Definition 5.17: A function f:Σ*→Σ* is a computable function if some Turing machine M, on every input w, halts with just f(w) on its tape. Example: The increment function inc++:{1}* → {1}* is Turing-computable

3. CS 311 Fall 2008 3 Incrementing inc++:{1}* → {1}* 1 1 1 1 1 1 Finite control Infinite tape 1

4. CS 311 Fall 2008 4 Transforming machines F = "On input : 1. Construct the machine M ∞ = "On input x : 1.Run M on x. 2.If M accepts, accept. 3.If M rejects, loop." 2. Output."

5. CS 311 Fall 2008 5 Mapping reducibility Definition 5.20: Language A is mapping reducible to language B, written A ≤ m B, if there is a computable function f:Σ* → Σ*, where for every w, w ∈ A ⇔ f(w) ∈ B A A B B f f

6. CS 311 Fall 2008 6 Problem reduction Theorem 5.22: If A ≤ m B and B is decidable, then A is decidable. A A B B f f

7. CS 311 Fall 2008 7 And… the contrapositive Theorem 5.22: If A ≤ m B and B is decidable, then A is decidable. Corollary 5.23: If A ≤ m B and A is undecidable, then B is undecidable.

8. CS 311 Fall 2008 8 A familiar mapping reduction… A TM = { | M is a TM and M accepts w } ≤ m HALT TM = { | M is a TM & M halts on input w } A TM HALT TM f f

9. CS 311 Fall 2008 9 A TM ≤ m HALT TM F = "On input : 1. Construct the machine M ∞ = "On input x : 1.Run M on x. 2.If M accepts, accept. 3.If M rejects, loop." 2. Output." A TM HALT TM f f

10. CS 311 Fall 2008 10 Similarly… Theorem 5.28: If A ≤ m B and B is Turing-recognizable, then A is Turing-recognizable. Theorem 5.29: If A ≤ m B and A is not Turing-recognizable, then B is not Turing-recognizable.

11. CS 311 Fall 2008 11 Solvable, half-solvable, hopeless Turing-decidable Turing- recognizable co-Turing- recognizable A DFA A TM E TM

12. CS 311 Fall 2008 12 EQ TM = { | L(M 1 ) = L(M 2 ) } is hopeless Theorem 5.30: EQ TM is neither Turing-recognizable nor co-Turing-recognizable Proof: –What if we show A TM ≤ m EQ TM ? A TM EQ TM f f A TM EQ TM

13. CS 311 Fall 2008 13 A TM ≤ m EQ TM G = "On input : 1.Construction the following two machines: M 1 = "On any input: 1.Accept.” M 2 = "On any input: 1.Run M on w. 2.If it accepts, accept." 2.Output.” A TM EQ TM f f A TM EQ TM

14. CS 311 Fall 2008 14 EQ TM is not Turing-recognizable Theorem 5.30: EQ TM is neither Turing-recognizable nor co-Turing-recognizable Proof: S how A TM ≤ m EQ TM A TM EQ TM f f A TM EQ TM

15. CS 311 Fall 2008 15 A TM ≤ m EQ TM G = "On input : 1.Construction the following two machines: M 1 = "On any input: 1.Reject.” M 2 = "On any input: 1.Run M on w. 2.If it accepts, accept." 2.Output.” A TM EQ TM f f A TM EQ TM

16. CS 311 Fall 2008 16 Solvable, half-solvable, hopeless Turing-decidable Turing- recognizable co-Turing- recognizable A DFA A TM E TM EQ TM