## Can a decider be built with 2 “opposite” recognizers? - turing-machines

If I have 2 turing machines, A and B, let's say that A accepts every word in a language L that has a property P, while B does the opposite, it accepts every word that does not have the property P. I don't know if A and B would reject in their complementary cases, all I can say is that they are recognizers. Can I build a Turing Machine that decides if a word is in the language L?
I was thinking of making an abstract turing machine C composed of both A and B where, if a word "enters" C, then this word would "enter" both A and B at the same time and consequently accepting the moment A accepts and rejecting when B accepts, and thus, always halting? Does this follow? or I am missing something?
Sorry if I'm being too informal on my explanations.

## Related

### Does this proof really prove undecidability of halting?

I want to ask a couple questions about the following proof. The proof originally came from a textbook and then a question on stackoverflow below. How does this proof, that the halting problem is undecidable, work? Question 1: Does the proof below essentially make H a simulator for its input machine? In other words, is there an important difference between saying H = M and the following description from the proof? H([M,w]) = {accept if M accepts w} = {reject if M does not accept w.} Question 2: How is my following comments correct or incorrect? I thought the halting problem was the problem of deciding if a given machine will halt regardless of its output(accept/reject). If a solution exists for a halting problem, it has to be something that analyses source code like a compiler/decompiler/disassembler instead of actually running it. If it needed to run it, obviously it would never determine on a "no" answer. Noticing that apparent problem in the proof, the whole proof seems not to show undecidability of the halting problem. The proof instead seems to show this: The following algorithm will not halt: boolean D() { return not D(); } Following is the proof in question retyped from Intro to the Theory of Computation by Sipser. THE HALTING PROBLEM IS UNDECIDABLE Now we are ready to prove Theorem 4.11, the undecidability of the language ATM = {[M,w] | M is a TM and M accepts w}. PROOF: We assume that ATM is decidable and obtain a contradiction. Suppose that H is a decider for ATM. On input , where M is a TM and w is a string, H halts and accepts if M accepts w. Furthermore, H halts and rejects if M fails to accept w. In other words, we assume that H is a TM, where H([M,w]) = {accept if M accepts w} = {reject if M does not accept w.} Now we construct a new Turing machine D with H as a subroutine. This new TM calls H to determine what M does when the input to M is its own description . Once D has determined this information, it does the opposite. That is, it rejects if M accepts and accepts if M does not accept. The following is a description of D. D = "On input [M], where M is a TM: 1. Run H on input [M, [M]]. 2. Output the opposite of what H outputs; that is, if H accepts, reject and if H rejects, accept." Don't be confused by the idea of running a machine on its own description! That is similar to running a program with itself as input, something that does occasionally occer in practice. For example, a compiler is a program that translates other programs. A compiler for the language Pascal may itself be written in Pascal, so running that program on itself would make sense. In summary, D([M]) = { accept if M does not accept [M] = { reject if M accepts [M] What happens when we run D with its own description as input> In that case we get: D([D]) = {accept if D does not accept [D] = {reject if D accepts [D] No matter what D does, it is forces to do the opposite, which is obviously a contradiction. Thus neither TM D nor TM H can exist.

In other words, is there an important difference between saying H = M and the following description from the proof? The H machine is called Universal Turing Machine (UTM) and is able to simulate any other Turing Machine, including itself. If M is an Universal Turing Machine like H, it is ok to say H = M, otherwise this would be weird. I thought the halting problem was the problem of deciding if a given machine will halt regardless of its output(accept/reject). If a solution exists for a halting problem, it has to be something that analyses source code like a compiler/decompiler/disassembler instead of actually running it. If it needed to run it, obviously it would never determine on a "no" answer. That is why the proof works based on contradiction and it is kind hard to understand. Basically it assumes first that exists such a machine that answers "yes" or "no" to any given input. [Hypothesis] Let's call this machine Q. Assuming Q is valid and it is an UTM, it can simulate another machine S that works following the steps below: S reads an input (a program and its input) S duplicates the input it just read S calls Q passing the copied input S waits for Q to answer (and based on our hypothesis it always will) Let's imagine now the input Q(S, S). Q will receive the program S and the argument of S is S itself. This input will make S call Q indefinitely and will never stop. Since Q and S were legal programs but there is a kind of input that makes Q never stop, Q is a machine impossible to built and therefore it is impossible to decide if a program S stops or not. Therefore we have the proof that the halting problem is undecidable. Sipser explains it well. Read it again now and see if you catch the idea :) Now, on to your question again. The Turing Machine is our most powerful machine for representing problems. As a recognition machine, it has to go through the input and run the algorithm to determine if it is valid or not. It is impossible to know the output of an algorithm without running it. The compiler is just a translator of syntax and little semantics. It cannot foresee how one will use the program and what the output will be.

### Turing machine decidability ambiguous cases

1) Is a Turing machine M that accepts the language L = {ε}, accepting no entry? In one hand, I think it can be false because the empty word could be an entry, but in another i think this could possibly be an indecidable problem. 2) Is every Turing machine whose language is decidable stops on any input ? Same idea, intuitively I would have say yes, due to the definition of decidable, but I don't know, something trouble me. 3) Is the language of the palindromes decidable whatever the aphabet ? For this one, I have almost no doubt that it's False, because with Rice's Theorem we can prove that, this probleme is indecidable.

1) I am not sure how to parse this but if a TM accepts the language consisting only of the empty set, it will eventually halt-accept on a blank tape. Whether that counts as an entry or not depends on your definition of "entry". I would count it as an entry, so I would answer "no". 2) The language consisting of only the empty string is decidable. However, we can write a TM that halt-accepts the empty string only and goes into an infinite loop for all other inputs. What is meant by "whose language" is debatable but for TMs that encode partial functions I would call the language of that TM the set of strings which it halt-accepts on, so I would answer "no". 3) It seems to me that, given an alphabet with n symbols, you can always construct a single-tape deterministic TM with O(n) states which halt-accepts on palindromes over that alphabet and halt-rejects other strings, thus deciding the language of palindromes over the alphabet. I would answer "yes", as long as the terms have their usual meanings. Note that Rice's theorem does not apply; it would apply to the problem of deciding whether a TM accepts the language of palindromes over an alphabet, but actually deciding whether something is a palindrome is of course possible (PDAs do it).

### PushDown Automaton (PDA) for L={a^(n)b^(n)c^(n)|n>=1}

I am on a fool's errand trying to construct a Pushdown automaton for the non-context-free language L={a^(n)b^(n)c^(n)|n>=1} and thought of two approaches. First approach:- I thought that for every 'a' in string I will push 3 'a' into the stack and for every 'b' in the string, I will pop 2 'a' from the stack now for every 'c' in the string I will still have 1 'a' in the stack. Problem with the First approach:- the language generated becomes something like this L={a^(p)b^(m)c^(n)| p>=1 and could not determine how m and n can be defined} Second approach:- We know that L={ a^(n)b^(m)c^(m)d^(n) | n>=0 } is a context-free language and L={ wxw | w∈(a,b)* } is also context-free language. So, I thought L={ a^(n)b^(m)b^(m)c^(n) | n>=1 and m=floor((n+1)/2) } Problem with the Second approach:- don't know if we can calculate floor(n+1/2) in the PDA without disturbing the elements of the stack. Please help in determining how m and n can be defined in the first approach and how can I find floor((n+1)/2) in the PDA. JFLAP files available for both if needed.

As Ami Tavory points out, there is no PDA for this language because this language is not context-free. It is easy to recognize this language if you use a queue instead of a stack, use two stacks, or use a Turing machine (all equivalent). Queue machine: Enqueue as as long as you see as, until you see a b. Dequeue as and enqueue bs as long as you see bs, until you see a c Dequeue bs as long as you see cs. Accept if you end this process with no additional input and an empty queue. Two-stack PDA: Use the first stack to make sure a^n b^n by pushing a when you see an a and popping a when you see a b; Use the second stack to make sure b^n c^n by pushing b when you see a b and popping b when you see a c; Accept if both stacks are empty at the end of this process. Turing machine: Ensure a^n ... c^n by replacing each a with A and erasing a matching c; Ensure A^n b^n by erasing matching pairs of A and b; Accept if at the end of this process you have no more A and no more b, i.e., the tape has been completely cleared.

One reason you've not managed to construct a pushdown automaton for this language, is because there isn't any. The Bar Hillel pumping lemma shows this. To outline the proof, suppose it can be done. Then, for some p, each string larger than p can be partitioned to uvwxy, s.t., |vwx| < p |vx| > 1 uvnwxny is also accepted by the automaton, for any n. The first rule implies that vwx can't span the three regions, only at most two (for large enough strings). The second and third rules now imply that you can pump so that the un-spanned region is smaller than the at least one of the other regions.

### How to know if a Turing Machine is a decider?

Is there an easy way to approach this? Can I find if it is a decider by looking at the diagram of the Turing Machine?

A decider is a machine that halts on all inputs. There is no general way to prove whether a given machine halts on all inputs. If you have a specific machine, you can try to formally prove that all paths of execution halt. For example, if your machine's read head always moves right on each transition (never left), and halts when there is no more input, then for all finite inputs the machine will halt. A simpler example would be a machine that has only one state: halt.

TM decides language L iff 1- the strings L put M into the Accept state 2- the strings NOT IN L put M into the Reject state TM M recognizes language L iff 1- the strings L put M into the Accept state 2- the strings NOT IN L - EITHER put M into the Reject state - OR cause M to loop #Wanhui Qiao you are saying "TM changes between two states and both states are neither accepting nor rejecting states, is this Turing decidable?" then it is definitely going infinitely i.e it is going into loop which is Turing recognizable.

You can prove in general that DECIDER_tm = { <M> : TM M is a decider } is undecidable. Prove by contradiction. Assume it is decidable and let R be a decider for DECIDER_tm. Construct S decider for HALT_tm using R as a subroutine . S = on input <M,w> 1. construct M_w = " on input x" run M on w if M accepts accept. if M rejects reject. 2. Run R on M_w 3. If R accept => accept, if R rejects => reject. Note that if M accepts or rejects w, M_w halts on all input, R accepts since M_w is a decider. If M loops on w, M_w loops on all input, R rejects M_w. We have built a decider for HALT_tm since we know HALT_tm is undecidable our assumption was wrong => DECIDER_tm is undecidable.

### proving if a Turnig machine accept a string is undecidable

I have seen the proof of ATM = {〈M,w〉 | M is a TM and M accepts w} is undecidable. initially we build another Turing machine H with input <M,w> it accept if M accept w otherwise reject then we make another Turing machine D which on input <M> 1.run H on <M,<M>> 2.output the opposite of what H outputs. That is if H accepts reject and if H reject accept I dont understand how is it possible that a Turing machine as input gets its own description and then reject it! Can you please explain it?

I'm not sure where exacltly you got stuck, so I'll more or less try to reiterate your question, hopefully filling any gaps. The idea is, that we want to show that ATM is undecidable, i.e. that there is no TM H, that when given any another TM M and any other input w to this TM M as input can decide if the TM M will accept input w. To show that such a TM H doesn't exist we make the ASSUMPTION such a machine H exists. Which at least with enough phantasy might not seem to be a much too unreasonable guess. The next Idea is to just by deducing the consequences the existence of such a TM H would have lead our argumentation the to a contradiction. Thus proving that our initial assumption such a TM H might exist must be wrong. Meaning there doesn't exist any TM H following our specification given above. So now on to the tricky part: As we assumed there exists a TM H that when given ANY TM M and ANY input w to this TM M as input can decide wether M will accept w nobody can forbid us to just feed a description of the machine M itself als input w into M itself. We even made the assumption we might use ANY input. As you already stated above the idea of feeding any machine as input into another doesn't seem to bee too easy to accomplish, we will rather try to encode the description of the second TM on our first TM's tape. At a first glance it might not seem to clear if it is really possible to be able to encode any TM on the tape, but in fact it is possible. One of the possible schemes to accomplish this goes by the name of Gödelisierung named after the mathematician Kurt Gödel. The encoding itself is also often refered to as Gödelnumber. Given our TM H that when given the description of ANY TM M and ANY input w to M as input decides wether M will accept w, using our freedom already mentioned above we will simply use TM M's description as input to itself. Thus constructing a TM H' that when given any description of a TM M as input will decide if TM M working on its own description as input will accept or not. As you hopefully see starting from our assumption a TM H as specified above we haven't done any "forbidden" move and we surely won't do, if we use our newly constructed TM H' to construct a TM D that when given ANY TM M as input wil just return the opposite output than our machine H'. This new TM D, when given the description of ANY TM M as input will accept > if the M doesn't accept its own description as input and reject > if the TM M accepts its own description as input. As you can see we still have the freedom to feed any (description of a) TM M as input into D, so happily using our freedom once again nobody can ever forbid to use the description of the TM D itself as input. According to our construction D works on ANY TM M as input, so it surely has to be able to work on itself. But exactly here comes the twist. As a matter of fact to determine the outcome ouf our run of D on its own description as input we just have to possibilities: D will either accept or reject. Now let's examine both cases: In case D accepts it's own description as input by our construction this would mean D would have to reject and vice versa. Remember it was the TM H' that acceptet ANY TM that accepted it's own description as input and we just inverted it in our construction. Thus leading us to the odd fact that D would have to reject its own description as input if and only if it accepted it. As this obviously can't be posssible you have to find the root of the error we made in our construction. Following the whole path we made up again this leads us our initial deduction that there was a TM H able to decide for ANY TM M and ANY input w if M will accept w, which was implied by our assumption that ATM was decidable.

The input to a machine has no special status, and doesn't impose any requirements. A machine is free to reject itself. :) Cause i don't feel like building a Turing machine, here's an oversimplified example in Java (which is close enough to Turing completeness that it serves to illustrate): class SelfReject { public boolean equals(Object other) { ... } public boolean accepts(Object input) { return (!this.equals(input)); } } An object of this type would accept any input but an object equal to itself.