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CSCI 2670 Introduction to Theory of Computing

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Theorem: If A m B and B is Turing-recognizable, then A is Turing-recognizable. ... If A T B and B is decidable, then A is decidable ... – PowerPoint PPT presentation

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Title: CSCI 2670 Introduction to Theory of Computing

1
CSCI 2670 Introduction to Theory of Computing
November 3, 2005
2
Agenda
• Yesterday
• Reductions (Section 5.3)
• Today
• More on reductions
• Section 6.3
• We will not be covering section 6.4
• I will discuss some basic issues of this section
when covering chapter 7

3
Announcement
• Remember to let me know if you want to come to my
pizza party next Wednesday

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

5
Why m?
• In some sense, if A m B, then A is less
powerful than B
• For example, we can map from CFGs to decidable
languages that arent CF, but not vice versa
• Example
• We can easily map any CFG C to ACFG
• f(w) ltC,wgt
• w ? C iff ltC,wgt ? ACFG
to use reductions

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 Ms output
• Then N decides A

7
Example
• Let EV ltAgt A is a DFA all strings in L(A)
have an even number of 1s
• How can we prove EV is decidable using a mapping
reduction?
• Consider the following DFA B

L(B) w ?? w has an even number of 1s
8
Mapping reduction of L
• Use EQDFA ltA,Bgt A and B are DFAs with L(A)
L(B)
• Mapping from EV to EQDFA
• f(ltAgt) ltA,A?Bgt
• A has an even number of 1s 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

9
Mapping reductions undecidability
• Corollary If A ?m B and A is undecidable, then
B is undecidable.
• We have been using this corollary implicitly

10
Example
• We showed that HALTTM is undecidable by
• Now lets show its undecidable using mapping
reduction
• Need a function that takes input ltM,wgt
• Halts if M accepts w, and loops if not

11
Mapping from ATM to HALTTM
• F On input x
• If x ? ltM,wgt 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 ltM,wgt
• If M accepts w M halts on w, otherwise M loops
or generates a string not in HALTTM
• I.e., ltM,wgt?ATM iff ltM,wgt?HALTTM

12
HALTTM is undecidable
• We just showed that ATM m HALTTM
• Since we know ATM is undecidable, we can conclude
that HALTTM is undecidable

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 Ms output
• Then N recognizes A

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?

15
Proving non-Turing-recognizability
• Question If A ?m U is A ?m U?
• Answer Yes since x ? A iff f(x) ? U
• How can we use this to prove non-Turing-recognizab
ility?
• Prove ATM ?m U or prove ATM ?m U

16
Is mapping reducibility enough?
• Mapping reducibility does not completely capture
• Example ATM and ATM are not mapping reducible
• ATM is Turing-recognizable and ATM isnt
• A solution to ATM would also provide a solution
to ATM

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

18
Example
• An oracle Turing machine with an oracle for EQTM
can decide ETM
• TEQ-TM On input ltMgt
• 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 ltM,M1gt
• If it accepts, accept if it rejects, reject
• TEQ-TM decides ETM
• ETM is decidable relative to EQTM

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

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

21
Have a great weekend!