1. Theory of Computability
Giorgi Japaridze
Theory of Computability
Reducibility
Episode 3

2. The undecidability of the halting problem
Giorgi Japaridze Theory of Computability
Let HALTTM = {<M,w> | M is a TM and M halts on input w}
HALTTM is called the halting problem.
Theorem 3.1: HALTTM is undecidable.
Proof idea: Assume, for a contradiction, that HALTTM is decidable. I.e.
there is a TM R that decides HALTTM. Construct the following TM S:
S = “On input <M,w>, an encoding of a TM M and a string w:
1. Run R on input <M,w>.
2. If R rejects, reject.
3. If R accepts, simulate M on w until it halts.
4. If M has accepted, accept; if M has rejected, reject.”
If M works forever on w, what will S do on <M,w>?
Expicitly reject
If M accepts w, what will S do on input <M,w>?
Accept
If M explicitly rejects w, what will S do on <M,w>?
Expicitly reject
Thus, S decides the problem
ATM.
But this is impossible (Theorem 2.2)

3. Definition of mapping reducibility
Giorgi Japaridze Theory of Computability
We say that A is
mapping reducible to B, written AmB, if there is a computable
function f: ** such that, for every w*,
wA iff f(w)B.
The function f is called a mapping reduction (映射归约) of A to B.
Let A and B be sets of strings over an alphabet . * * A B f f

4. Exercise on mapping reducibility
Giorgi Japaridze Theory of Computability
In each case below, find (precisely define) a particular function that is
a mapping reduction of A to B. Both A and B are sets of strings over the
alphabet {0,1}.
A = {w | the length of w is even},    B = {w | the length of w is odd}
A = B = {w | the length of w is even}
A={0,1}*,   B={00,1,101}

5. Using mapping reducibility for proving decidability/undecidability
Giorgi Japaridze Theory of Computability
Theorem 3.2: If AmB and B is decidable, then A is decidable.
Proof: Let DB be a decider for B and f be a mapping reduction of
A to B. We describe a decider DA for A as follows.
DA= “On input w:
1. Compute f(w).
2. Run DB on input f(w) and do whatever DB does.”
Corollary 3.3: If AmB and A is undecidable, then B is undecidable.
Theorem 3.2 remains valid with “recognizable” instead of
“decidable”. So does Corollary 3.3.

6. A mapping reduction of ATM to HALTTM
Giorgi Japaridze Theory of Computability
For every TM M, let M* be the following TM:
M* = “On input x:
1. Run M on x.
2. If M accepts, accept.
3. If M rejects, enter an infinite loop.”
Thus,
If M accepts input x, then M*
If M explicitly rejects x, then M*
If M never halts on x, then M*
To summarize, M accepts x iff M*
accepts x
never halts on x
halts on x
Let then f be the function defined by f(<M,w>)=<M*,w>.
Is f computable?
Yes!
Obviously <M,w>ATM iff f(<M,w>)
i.e. f is a
HALTTM
mapping reduction of ATM to HALTTM
So, since ATM is undecidable, HALTTM is undecidable as well.

7. An oracle (谕示) for a set B is an external device that is capable
Turing reducibility
3.3.e
Giorgi Japaridze Theory of Computability
An oracle (谕示) for a set B is an external device that is capable
of reporting whether any string w is a member of B.
An oracle Turing machine (OTM) is a modified Turing machine that
has the additional capability of querying an oracle.
Example: Construct an OTM O such that, as long as the oracle of O
is for the acceptance problem ATM, O decides the non-acceptance
problem ATM.
O = “On input <M,w>, where M is a TM and w is a string:
1. Query the oracle to determine whether <M,w>ATM.
2. If the oracle answers NO, accept; if YES, reject.”
A set A is Turing reducible (图灵可归约) to set B, written ATB, iff
there is an OTM M such that, as long as the oracle of M is for B,
M decides A.

8. Using Turing reducibility
3.3.f
Giorgi Japaridze Theory of Computability
Theorem 3.4:
a) If ATB and B is decidable, then A is decidable.
Consequently,
b) If ATB and A is undecidable, then B is undecidable.
Proof (a): Assume M is an OTM that decides A as long as the oracle
of M is for B. If B is decidable, then we may replace the oracle for B
by an actual procedure that decides B. Thus we may replace M by an
ordinary TM that decides A.
Exercise: Show that the acceptance problem (ATM) is Turing
reducible to the halting problem (HALTTM).