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Lecture15: . Reductions. Prof. Amos Israeli. The rest of the course deals with an important tool in Computability and Complexity theories, namely: . Reductions. .. . The reduction technique enables us to use the undecidability of to prove many other languages undecidable.. ID: 257219

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## Presentations text content in 1 Introduction to Computability Theory

1

Introduction to Computability Theory

Lecture15:

Reductions

Prof. Amos Israeli

Slide2The rest of the course deals with an important tool in Computability and Complexity theories, namely: Reductions. The reduction technique enables us to use the undecidability of to prove many other languages undecidable.

Introduction

2

Slide3A reduction always involves two computational problems. Generally speaking, the idea is to show that a solution for some problem A induces a solution for problem B. If we know that B does not have a solution, we may deduce that A is also insolvable. In this case we say that B is reducible to A.

Introduction

3

Slide4In the context of undecidability: If we want to prove that a certain language L is undecidable. We assume by way of contradiction that L is decidable, and show that a decider for L, can be used to devise a decider for . Since is undecidable, so is the language L.

Introduction

4

Slide5Using a decider for L to construct a decider for , is called reducing L to . Note: Once we prove that a certain language L is undecidable, we can prove that some other language, say L’ , is undecidable, by reducing L’ to L.

Introduction

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Slide6We know that A is undecidable. We want to prove B is undecidable. We assume that B is decidable and use this assumption to prove that A is decidable.We conclude that B is undecidable.Note: The reduction is from A to B.

Schematic of a Reduction

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Slide7We know that A is undecidable. The only undecidable language we know, so far, is whose undecidability was proven directly. So we pick to play the role of A.We want to prove B is undecidable.

Demonstration

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Slide8We want to prove B is undecidable. We pick to play the role of B that is: We want to prove that is undecidable. We assume that B is decidable and use this assumption to prove that A is decidable.

Demonstration

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Slide9We assume that B is decidable and use this assumption to prove that A is decidable.In the following slides we assume (towards a contradiction) that is decidable and use this assumption to prove that is decidable.We conclude that B is undecidable.

Demonstration

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Slide10ConsiderTheorem is undecidable.ProofBy reducing to .

The “Real” Halting Problem

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Slide11Assume by way of contradiction that is decidable.Recall that a decidable set has a decider R: A TM that halts on every input and either accepts or rejects, but never loops!.We will use the assumed decider of to devise a decider for .

Discussion

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Slide12Recall the definition of :Why is it impossible to decide ?Because as long as M runs, we cannot determine whether it will eventually halt.Well, now we can, using the decider R for .

Discussion

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Slide13Assume by way of contradiction that is decidable and let R be a TM deciding it. In the next slide we present TM S that uses R as a subroutine and decides . Since is undecidable this constitutes a contradiction, so R does not exist.

Proof

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Slide14S=“On input where M is a TM: 1. Run R on input until it halts. 2. If R rejects, (i.e. M loops on w ) - reject.(At this stage we know that R accepts, and we conclude that M halts on input w.) 3. Simulate M on w until it halts. 4. If M accepts - accept, otherwise - reject. “

Proof (cont.)

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Slide15In the discussion, you saw how Diagonalization can be used to prove that is not decidable. We can use this result to prove by reduction that is not decidable.

Another Example

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Slide16Note: Since we already know that both and are undecidable, this new proof does not add any new information. We bring it here only for the the sake of demonstration.

Another Example

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Slide17We know that A is undecidable. Now we pick to play the role of A.We want to prove B is undecidable. We pick to play the role of B, that is: We want to prove that is undecidable. We assume that B is decidable and use this assumption to prove that A is decidable.

Demonstration

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Slide18We assume that B is decidable and use this assumption to prove that A is decidable.In the following slides we assume that is decidable and use this assumption to prove that is decidable.We conclude that B is undecidable.

Demonstration

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Slide19Let R be a decider for . Given an input for , R can be run with this input :If R accepts, it means that .This means that M accepts on input w. In particular, M stops on input w. Therefore, a decider for must accept too.

Discussion

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Slide20If however R rejects on input , a decider for cannot safely reject: M may be halting on w to reject it. So if M rejects w, a decider for must accept .

Discussion

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Slide21How can we use our decider for ?The answer here is more difficult. The new decider should first modify the input TM, M, so the modified TM, , accepts, whenever TM M halts. Since M is a part of the input, the modification must be a part of the computation.

Discussion

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Slide22Faithful to our principal “ If it ain’t broken don’t fix it”, the modified TM keeps M as a subroutine, and the idea is quite simple:Let and be the accepting and rejecting states of TM M, respectively. In the modified TM, , and are kept as ordinary states.

Discussion

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Slide23We continue the modification of M by adding a new accepting sate . Then we add two new transitions: A transition from to , and another transition from to . This completes the description of . It is not hard to verify that accepts iff M halts.

Discussion

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Slide24Discussion

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Slide25The final description of a decider S for is:S=“On input where M is a TM: 1. Modify M as described to get . 2. Run R, the decider of with input . 3. If R accepts - accept, otherwise - reject. ”

Discussion

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Slide26It should be noted that modifying TM M to get , is part of TM S, the new decider for , and can be carried out by it.It is not hard to see that S decides . Since is undecidable, we conclude that is undecidable too.

Discussion

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Slide27We continue to demonstrate reductions by showing that the language , defined by is undecidable.Theorem is undecidable.

The TM Emptiness Problem

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Slide28The proof is by reduction from :We know that is undecidable. We want to prove is undecidable. We assume toward a contradiction that is decidable and devise a decider for .We conclude that is undecidable.

Proof Outline

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Slide29Assume by way of contradiction that is decidable and let R be a TM deciding it. In the next slides we devise TM S that uses R as a subroutine and decides .

Proof

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Slide30Given an instance for , , we may try to run R on this instance. If R accepts, we know that . In particular, M does not accept w so a decider for must reject .

Proof

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Slide31What happens if R rejects? The only conclusion we can draw is that . What we need to know though is whether .In order to use our decider R for , we once again modify the input machine M to obtain TM :

Proof

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Slide32We start with a TM satisfying .

Description of___

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Slide33Now we add a

filter to divert all inputs but w.

Description of___

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filter

no

yes

Slide34TM has a filter that rejects all inputs excepts w, so the only input reaching M, is w.Therefore, satisfies:

Proof

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Slide35Here is a formal description of : “On input x : 1. If - reject . 2. If - run M on w and accept if M accepts. ” Note: M accepts w if and only if .

Proof

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Slide36This way, if R accepts, S “can be sure” that and accept. Note that S gets the pair as input, thus before S runs R, it should compute an encoding of .This encoding is not too hard to compute using S’s input .

Proof

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Slide37S=“On input where M is a TM: 1. Compute an encoding of TM . 2. Run R on input . 3. If R rejects - accept, otherwise - reject.

Proof

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Slide38Recall that R is a decider for . If R rejects the modified machine , , hence by the specification of , , and a decider for must accept .If however R accepts, it means that , hence , and S must reject . QED

Proof

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