1. CSCI 2670 Introduction to Theory of Computing
November 3, 2005

2. Agenda Yesterday Today Reductions (Section 5.3) More on reductions
We will not be covering section 6.4
I will discuss some basic issues of this section when covering chapter 7
November 3, 2005

3. Announcement
Remember to let me know if you want to come to my pizza party next Wednesday
November 3, 2005

4. Mapping reducibility
Definition: Language A is mapping reducible to language B, written AmB, if there is a computable function f:* → *, where for every w,
w  A iff f(w)  B f
November 3, 2005

5. Why ≤m? In some sense, if A ≤m B, then A is “less powerful” than B
For example, we can map from CFG’s to decidable languages that aren’t CF, but not vice versa
Example
We can easily map any CFG C to ACFG
f(w) = <C,w>
w  C iff <C,w>  ACFG
You can use this example to help you remember how to use reductions
November 3, 2005

6. Mapping reductions & decidability
Theorem: If A m B and B is decidable, then A is decidable.
Proof: Let M be a decider for B and let f be a reduction from A to B.
Consider the following TM, N:
N = “On input w:
Compute f(w)
Run M on f(w) and report M’s output”
Then N decides A
November 3, 2005

7. Example
Let EV = {<A>| A is a DFA all strings in L(A) have an even number of 1’s}
How can we prove EV is decidable using a mapping reduction?
Consider the following DFA B 1
\{1}
L(B) = {w * | w has an even number of 1’s}
November 3, 2005

8. Mapping reduction of L
Use EQDFA = {<A,B> | A and B are DFA’s with L(A) = L(B)}
Mapping from EV to EQDFA
f(<A>) = <A,AB>
A has an even number of 1’s if and only if L(A) = L(AB)
I.e., A  EV iff f(A)  EQDFA
Since we know EQDFA is decidable and EV ≤m EQDFA, we now know EV is decidable
November 3, 2005

9. Mapping reductions & undecidability
Corollary: If A m B and A is undecidable, then B is undecidable.
We have been using this corollary implicitly already
November 3, 2005

10. Example We showed that HALTTM is undecidable by contradiction
Now let’s show it’s undecidable using mapping reduction
Need a function that takes input <M,w>
Halts if M accepts w, and loops if not
November 3, 2005

11. Mapping from ATM to HALTTM
F = “On input x
If x ≠ <M,w> for some TM M, output x
Otherwise, construct the following TM
M’ = “On input x:
1. Run M on x
2. If M accepts, accept
3. If M rejects, enter a loop”
2. Output <M’,w>”
If M accepts w M’ halts on w, otherwise M’ loops or generates a string not in HALTTM
I.e., <M,w>ATM iff <M’,w>HALTTM
November 3, 2005

12. HALTTM is undecidable We just showed that ATM ≤m HALTTM
Since we know ATM is undecidable, we can conclude that HALTTM is undecidable
November 3, 2005

13. Reductions & TM-recognizability
Theorem: If A m B and B is Turing-recognizable, then A is Turing-recognizable.
Proof: (same as decidable proof) Let M be a recognizer for B and let f be a reduction from A to B.
Consider the following TM, N:
N = “On input w:
Compute f(w)
Run M on f(w) and report M’s output”
Then N recognizes A
November 3, 2005

14. Reductions & non-TM-recognizability
Corollary: If A m B and A is not Turing-recognizable, then B is not Turing-recognizable.
Question: Which language have we seen that is not Turing-recognizable?
Answer: ATM
November 3, 2005

15. Proving non-Turing-recognizability
Question: If A m U is Ā m Ū?
Answer: Yes since x  A iff f(x)  U
How can we use this to prove non-Turing-recognizability?
Prove ATM m U or prove ATM m Ū
November 3, 2005

16. Is mapping reducibility enough?
Mapping reducibility does not completely capture our intuition about reductions
Example: ATM and ATM are not mapping reducible
ATM is Turing-recognizable and ATM isn’t
A solution to ATM would also provide a solution to ATM
November 3, 2005

17. Oracle
An oracle for a language B is an external device that is capable of reporting whether any string w is a member of B
We are not concerned how the oracle determines membership
An oracle Turing machine is a Turing machine that can query an oracle
The machine MB can query an oracle for the language B
November 3, 2005

18. Example
An oracle Turing machine with an oracle for EQTM can decide ETM
TEQ-TM = “On input <M>
Create TM M1 such that L(M1) = 
M1 has a transition from start state to reject state for every element of 
Call the EQTM oracle on input <M,M1>
If it accepts, accept; if it rejects, reject”
TEQ-TM decides ETM
ETM is decidable relative to EQTM
November 3, 2005

19. Turing reducibility
Definition: A language A is Turing reducible to a language B, written
A T B, if A is decidable relative to B
Previous slide shows ETM is Turing reducible to EQTM
Whenever A is mapping reducible to B, then A is Turing reducible to B
The function in the mapping reducibility could be replaced by an oracle
November 3, 2005

20. Applications If A T B and B is decidable, then A is decidable
If A T B and A is undecidable, then B is undecidable
If A T B and B is Turing-recognizable, then A is Turing-recognizable
If A T B and A is non-Turing-recognizable, then B is non-Turing-recognizable
November 3, 2005

21. Have a great weekend!
November 3, 2005