5.3 Mapping Reducibility (continued)


Turing Recognizable Reductions

If A mapping-reduces to B and B is Turing-Recognizable,
what do you know about A?

How can you use a mapping reduction
to show a problem is Turing-Recognizable?

	give a mapping reduction TO a Known-Recognizable problem

















If A mapping-reduces to B and A is not Turing-Recognizable,
what do you know about B?


How can you use a mapping reduction
to show a problem is not Turing-Recognizable?

	give a mapping reduction FROM a Known-Unrecognizable problem

















Reducing from NOT A[TM]


If A mapping-reduces to B,
what do you know about NOT-A and NOT-B?

If NOT-A mapping-reduces to NOT-B and A is not Turing-Recognizable,
what do you know about B?

How can you use a mapping reduction
to show a problem is not Turing-Recognizable?

	give a mapping reduction from
	the complement of a Known-Unrecognizable problem 
	to the complement of the problem

	NOT-A[TM] is a Known-Unrecognizable problem
	reduction from NOT-NOT-A[TM] (or A[TM]) to NOT-B
	is typically used to show B is Unrecognizable

















Equivalence is Unrecognizable


How do you prove that EQ[TM] is not Turing-Recognizable?

What reduction is equivalent to a reduction from NOT-A[TM] to EQ[TM]?

What's the intro for the proof?

	The TM F reduces A[TM] to NOT-EQ[TM].
	This shows that NOT-A[TM] reduces to EQ[TM].

Describe a mapping reduction from A[TM] to NOT-EQ[TM].

	F = "On input <M,w>, where M is a TM and w is a string:
	  1. Build two TMs M1 and M2.
	     M1 = "On any input:
		1. Reject."
	     M2 = "On any input:
		1. Run M on w and do what M does."
	  2. Output <M1,M2>."

If <M,w> is in A[TM], how do you show <M1,M2> is in NOT-EQ[TM]?

	M1 always accepts nothing.
	If <M,w> is in A[TM], M2 accepts everything.
	M1 and M2 are different, so <M1,M2> is in NOT-EQ[TM].

If <M,w> is not in A[TM], how do you show <M1,M2> is not in NOT-EQ[TM]?

	M1 always accepts nothing.
	If <M,w> is not in A[TM], M2 accepts nothing.
	M1 and M2 are the same, so <M1,M2> is not in NOT-EQ[TM].

What's the conclusion for the proof?

	A[TM] reduces to NOT-EQ[TM], so NOT-A[TM] reduces to EQ[TM].
	NOT-A[TM] is Unrecognizable, so EQ[TM] is Unrecognizable.

















Not Equivalent is Unrecognizable


How do you prove that NOT-EQ[TM] is not Turing-Recognizable?

What reduction is equivalent to a reduction from NOT-A[TM] to NOT-EQ[TM]?

What's the intro for the proof?

	The TM F reduces A[TM] to EQ[TM].
	This shows that NOT-A[TM] reduces to NOT-EQ[TM].

Describe a mapping reduction from A[TM] to EQ[TM].

	F = "On input <M,w>, where M is a TM and w is a string:
	  1. Build two TMs M1 and M2.
	     M1 = "On any input:
		1. Accept."
	     M2 = "On any input:
		1. Run M on w and do what M does."
	  2. Output <M1,M2>."

If <M,w> is in A[TM], how do you show <M1,M2> is in EQ[TM]?

	M1 always accepts everything.
	If <M,w> is in A[TM], M2 accepts everything.
	M1 and M2 are the same, so <M1,M2> is in EQ[TM].

If <M,w> is not in A[TM], how do you show <M1,M2> is not in EQ[TM]?

	M1 always accepts everything.
	If <M,w> is not in A[TM], M2 accepts nothing.
	M1 and M2 are different, so <M1,M2> is not in EQ[TM].

What's the conclusion for the proof?

	A[TM] reduces to EQ[TM], so NOT-A[TM] reduces to NOT-EQ[TM].
	NOT-A[TM] is Unrecognizable, so NOT-EQ[TM] is Unrecognizable.