1 Introduction to Computability Theory Lecture12: 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. Introduction 2

A 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

In 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

Using 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 5

1.We know that A is undecidable. 2.We want to prove B is undecidable. 3.We assume that B is decidable and use this assumption to prove that A is decidable. 4.We conclude that B is undecidable. Note: The reduction is from A to B. Schematic of a Reduction 6

1.We know that A is undecidable. The only undecidable language we know, so far, is whose undecidability was proven directly. (In the discussion you also proved directly that is undecidable). So we pick to play the role of A. 2.We want to prove B is undecidable. Demonstration 7

2.We want to prove B is undecidable. We pick to play the role of B that is: We want to prove that is undecidable. 3.We assume that B is decidable and use this assumption to prove that A is decidable. Demonstration 8

3.We 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. 4.We conclude that B is undecidable. Demonstration 9

Consider Theorem is undecidable. Proof By reducing to. The “Real” Halting Problem 10

Assume 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 11

Recall 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 12

Assume 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 13

S= “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.) 14

In 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 15

Note: 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 16

1.We know that A is undecidable. Now we pick to play the role of A. 2.We want to prove B is undecidable. We pick to play the role of B, that is: We want to prove that is undecidable. 3.We assume that B is decidable and use this assumption to prove that A is decidable. Demonstration 17

3.We 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. 4.We conclude that B is undecidable. Demonstration 18

Let 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 19

If 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 20

How 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 21

Faithful 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 22

We 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 23

Discussion 24

The 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 25

It 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 26

We continue to demonstrate reductions by showing that the language, defined by is undecidable. Theorem is undecidable. The TM Emptiness Problem 27

The proof is by reduction from : 1.We know that is undecidable. 2.We want to prove is undecidable. 3.We assume toward a contradiction that is decidable and devise a decider for. 4.We conclude that is undecidable. Proof Outline 28

Assume 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 29

Given 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 30

What 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 31

We start with a TM satisfying. Description of___ 32

Now we add a filter to divert all inputs but w. Description of___ 33 filter no yes

TM has a filter that rejects all inputs excepts w, so the only input reaching M, is w. Therefore, satisfies: Proof 34

Here 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 35

This 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 36

S= “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 37

Recall 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 38

We continue to demonstrate reductions by showing that the language : is undecidable. Theorem is undecidable. _______ is undecidable 39

The proof is by reduction from : 1.We know that is undecidable. 2.We want to prove is undecidable. 3.We assume that is decidable and devise a decider for. 4.We conclude that is undecidable. Proof Outline 40

Consider an instance of A TM. Once again, the idea is to transform TM M to another TM,, such that is regular if and only if. Once we have such a machine we can run a decider for with as input, and accept, if the decider accepts. Proof 41

Here the idea is to devise a TM that satisfy:. Description of___ 42

TM M 2 is obtained as follows: 1.Start with M. 2.Add a filter in front of M such that for every input x : 2.1 If x is of the form 0 n 1 n - accept. 2.2 Send w to M – If M accepts - accept. Description of___ 43

We start with a TM satisfying. Description of___ 44

Add a filter to accept all inputs of form. Description of___ 45 filter no yes

“On input x : 1. If x has the form, accept. 2. If x does not have this form, run M on w and accept if M accepts w. ” Note: If M accept w then. If M does not accepts w then. Proof 46

Now consider S : “On input : 1. Construct as described using. 2. Run R on. 3. If R accepts (Meaning ) - accept, otherwise ( ) - reject.” Proof 47

Consider Theorem is undecidable. TM Equivalence is Undecidable 48

The proof is by reduction from. We have to show that if is decidable, so is. The idea is very intuitive: In order to check that a language of TM M is empty, as required by, we will check whether M is equivalent to a TM that rejects all its inputs. Proof 49

S= “On input, where M is a TM: 1. Run R on input where is a TM that rejects all its inputs. 2. If R accepts – accept, otherwise – reject. If R decides, S decides. Since is undecidable, so is. QED Proof 50

Mapping Reductions are the most common reductions in Computer Science. In this lecture we define mapping reductions and demonstrate the way in which they are used. Mapping Reductions 51

The idea of a mapping reduction is very simple: If the instances (candidate elements) of one language, say A, are mapped to the instances of another language, say B, by a computable mapping M in a way that iff, then a decider for B can be used to devise a decider for A. Mapping Reductions 52

Let A and B be two languages over. A function is a computable function if there exists a TM M such that for every, if M computes with input w it halts with on its tape. Computable Functions 53

1. Let m and n be natural numbers, let be a string encoding both m and n, and let be the string representing their sum. The function,, is computable. 2. The function is a computable function. Can you devise TM-s to compute f and g ? Examples of Computable Functions 54

3. Let be an encoding of TM M and let be an encoding of another TM M’ satisfying: TM M’ never makes two steps in the same head direction. The function t defined below is computable: Examples of Computable Functions 55

Let A and B be two languages over. A computable function is a mapping reduction from A to B if for every it holds that follows: For each it holds that iff. The function f is called reduction of A to B. The arguments of the reduction are often called instances. Mapping Reductions 56

If there exists a mapping reduction from A to B, We say that A is mapping reducible to B and denote it by. If language A is mapping reducible to language B, then a solution for B, can be used to derive a solution for A. This fact is made formal in the following theorem: Mapping Reductions 57

If and B is decidable, then A is decidable. Theorem 58

Let M be a decider for B and let f be the reduction from A to B. Consider TM N : N= “On input w : 1. Compute. 2. Run M on and output whatever M outputs. “ Clearly N accepts iff. QED Proof 59

If and A is undecidable, then B is undecidable. Corollary 60

The previous corollary is our main tool for proving undecidability. Notice that in order to prove B undecidable we reduce from A which is known to be undecidable to B. The reduction direction is often a source of errors. A similar tool is used in Complexity theory. Discussion 61

At the beginning of this lecture we looked at and proved that it is undecidable. Now, we prove this theorem once more by demonstrating a mapping reduction from to. The Halting Problem Revisited 62

The mapping reduction is presented by TM F : F= “On input : 1. Construct TM M’. M’= “On input x : 1. Run M on x. 2. if M accepts accept. 3. If M rejects, enter a loop. “ 2. Output. “ The Halting Problem Revisited 63

What happens if the input does not contain a valid description of a TM? By the specification of we know that in this case. Therefore in this case TM F should output any string s satisfying. The Halting Problem Revisited 64