1. Mapping Reducibility
Sipser 5.3 (pages )

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. Incrementing inc++:{1}* → {1}* Infinite tape 1 1 1 1 ⨆ ⨆ ⨆ ⨆
Finite control

4. Transforming machines
F = "On input <M>:
Construct the machine
M∞ = "On input x:
Run M on x.
If M accepts, accept.
If M rejects, loop."
Output <M∞>."

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 B f f

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

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. A familiar mapping reduction…
ATM = {<M,w> | M is a TM and M accepts w} ≤m HALTTM = {<M,w> | M is a TM & M halts on input w}
ATM
HALTTM f
<M,w>
<M∞,w> f
<M,w>
<M∞,w>

9. ATM ≤m HALTTM F = "On input <M>: M∞ = "On input x:
Construct the machine
M∞ = "On input x:
Run M on x.
If M accepts, accept.
If M rejects, loop."
Output <M∞>."
ATM
HALTTM f
<M,w>
<M∞,w> f
<M,w>
<M∞,w>

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. Solvable, half-solvable, hopeless
ATM
ADFA
ETM
co-Turing-
recognizable
Turing-
recognizable
Turing-decidable

12. EQTM = {<M1,M2>| L(M1)=L(M2)} is hopeless
Theorem 5.30:
EQTM is neither Turing-recognizable nor
co-Turing-recognizable
Proof:
What if we show ATM ≤m EQTM?
ATM
EQTM f
<M,w>
<M1,M2> f
EQTM
ATM
<M,w>
<M1,M2>

13. ATM ≤m EQTM G = "On input <M,w>:
Construction the following two machines:
M1 = "On any input:
Accept.”
M2 = "On any input:
Run M on w.
If it accepts, accept."
Output <M1,M2>.”
ATM
EQTM f
<M,w>
<M1,M2>

14. EQTM is not Turing-recognizable
Theorem 5.30:
EQTM is neither Turing-recognizable nor
co-Turing-recognizable
Proof:
Show ATM ≤m EQTM
ATM
EQTM f
<M,w>
<M1,M2> f
EQTM
ATM
<M,w>
<M1,M2>

15. ATM ≤m EQTM G = "On input <M,w>:
Construction the following two machines:
M1 = "On any input:
Reject.”
M2 = "On any input:
Run M on w.
If it accepts, accept."
Output <M1,M2>.”
ATM
EQTM f
<M,w>
<M1,M2> f
EQTM
ATM
<M,w>
<M1,M2>

16. Solvable, half-solvable, hopeless
EQTM
ATM
ADFA
ETM
co-Turing-
recognizable
Turing-
recognizable
Turing-decidable