problem X reduces to Problem Y solving X would be easy, if we knew how to solve Y

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1 CPS220 Reducibility A reduction is a procedure to convert one problem to another problem, in such a way that a solution to the second problem can be used to solve the first problem. The conversion itself must be doable by a Turing Machine that halts. That is, there must be a concrete algorithm for a conversion. REDUCTION: problem X reduces to Problem Y means solving X would be easy, if we knew how to solve Y Example 1. The problem of having a Ferrari reduces to the problem of obtaining $200K. Proof: Say that we could obtain $200K. The following algorithm leads us to drive a Ferrari: 1. Go to a Ferrari dealer. 2. uy a Ferrari (trivial, as several good models cost < $200K) Corollary. $200K is hard to obtain. Proof. ecause if it was easy, then having a Ferrari would be easy. Since that is not the case, the corollary follows. Example 2. The problem of traveling from oston to anywhere warm reduces to the problem of deciding on a warm location and buying a plane ticket to that location. This problem of buying a plane ticket reduces to the problem of earning the money for the ticket. This problem of earning the money reduces to the problem of finding a job. This idea of reducibility plays a part in classifying problems: if a A is reducible to and is decidable, A also is decidable. <A> Reduction algorithm <> Algorithm to decide yes no Algorithm to decide A 1

2 If A is undecidable and reducible to, is undecidable. Why? ecause if is actually decidable then we could build a machine and decide A therefore must not be decidable. HALT TM = { (M, w) M is a TM and halts on input w } Note: related to the A TM problem - A TM = { <M, w> M is a TM that s w } which determines whether a machine s a particular string (ance problem) Example. HALT TM = { (M, w) M is a TM and halts on input w } Theorem. HALT TM is undecidable. Proof. We will show that solving A TM would be easy, if we could solve HALT TM. Then, since we know that solving A TM is hard, we conclude that solving HALT TM is hard. More precisely, we will reduce A TM to HALT TM, which means that if HALT TM is decidable, then A TM is decidable which it isn t. < M,w > < M,w > Machine H to decide HALT TM No halt halt Simulate M on w Let a machine H decide HALT TM. Here is a machine S that decides A TM : A := On input (M, w), 1. Run H on input (M, w) 2. If H s, 3. If H s, simulate M on input w until it halts 4. If M s, ; if M s, Proof by contradiction. How? Classifying the Computability of Problems A language can be in one of the following categories: Decidable These are the problems that we call solvable by an algorithm. To prove a language decidable, find an algorithm run by a Turing Machine that always halts. 2

3 Recognizable To prove a language recognizable, find a Turing Machine that s all strings of the language and does not any strings out of the language. The Turing Machine is allowed to loop forever instead of ing. To prove the language not decidable, use a reduction from a recognizable problem that is not decidable. Non-recognizable Those are the hopeless problems! To prove a language is not recognizable, show that it is the complement of a language that is not decidable. Or, use a reduction from a non-recognizable language. E TM = { M M is a TM and L(M) = } Theorem. E TM is undecidable. Proof. We will reduce A TM to E TM. Say that there is a machine R deciding E TM. So R can decide, given a machine description (M), whether L(M) =. For a given string w, consider the following machine: M (w) := On input x, 1. If x w,. 2. Otherwise, simulate M on w. If M s,. If M s,. Notice that M (w) has w hardwired in it. What is L(M (w))? It is simply { w }, if M s w. ut it is if M does not w. < M,w > < M,w > Construct machine M (w) M E TM Given the description of M as an input, R s if M s w, and s if M s w. So here is a machine that decides A TM : S := On input (M, w), 3

4 1. Construct the machine M (w) as described above 2. Run R (E TM ) on input <M > 3. If R s,. If R s,. means its empty or w x means its not empty and x=w Since R always halts and decides whether L(M (w)) =, S always halts and decides whether M s w. EQ TM = { M, N M and N are TMs, and L(M) = L(N) } Theorem. EQ TM is undecidable. Proof. We will reduce E TM to EQ TM. Say that a machine R decides EQ TM. Here is a machine S that decides E TM : S := On input M, where M is a TM, 1. Run R on input (M, M 1 ) where M 1 is a trivial TM that s all inputs. 2. If R s,. If R s,. < M > < M > <N> Machine R to decide EQ TM Machine S to decide E TM Proof by contradiction: if EQ TM is possible then it would be possible (by simple reduction) to create E TM - however we know that E TM is undecidable therefore it must be impossible to create EQ TM Theorem. The complement of EQ TM, L(N) } is undecidable. EQ TM = { M, N M and N are TMs and L(M) Proof. Say it is decidable, and R is a machine that on input (M, N), it s if L(M) = L(N). We show how to construct S deciding A TM : S := On input (M, w), Construct the descriptions of the following machines: M 1 := On any input, 4

5 M 2 := On any input, Simulate M on w, and if M s,. Pass (M 1, M 2 ) to R. If R s,. If it s,. If R s, then L(M 2 ) = {w}, so M s w. Corollary. Therefore, both EQ TM and EQ TM are non-recognizable. <A> Mapping Reducibility eing able to reduce problem A to problem by using a mapping reducibility means that a computable function (reduction) exists that converts instances of problem A to instances of problem. Reduction algorithm Algorithm to decide A <> Algorithm to decide yes no Computable functions * A function f : " # "* is a computable function if some Turing machine M, on every input w, halts with just f(w) on its tape. - All math operations are computable functions. It is possible to make a machine that takes <M,N> and halts with M+N on its tape. - Computable functions may be transformations of machine descriptions. Formal Definition of Mapping Reducibility Language A is mapping reducible to language ( A " m ) if there is a computable function f : "* # " *, where for every w, w " A # f (w) " The function f is called the reduction of A to. 5

6 A f f Theorem 5.22 and is decidable, then A is decidable. Corollary 5.23 and A is undecidable, then is undecidable. <A> Reduction algorithm f(w) f(<a>)=<> Algorithm to decide yes no Algorithm to decide A Theorem 5.28 and is Turing-recognizable, then A is Turing-recognizable. Corollary 5.29 and A is not Turing-recognizable, then is not Turing-recognizable. Example 5.24 Reduction from A TM to HALT TM to prove HALT TM is undecidable Previously used: < M,w > < M,w > Machine H to decide HALT TM No halt halt Simulate M on w 6

7 Mapping reducibility from A TM to HALT TM must present a computable function f that takes <M,w> and returns output <M,w >, where M,w " A TM if and only if M #, w # " HALT TM f = On input <M,w>: 1. Construct the following machine M M = On input x: 1. Run M on x. 2. If M s,. 3. If M s, enter a loop. 2. Output <M,w>. Mapping function < M,w > < M,w > <M,w> Machine to decide HALT TM 7