I need help to understand this proof.
"First we show that if we have an enumerator E that enumerates a
language A, a TM M recognizes A. The TM M works in the following way.
PROOF M = "On input w:
1.Run E. Every time that E outputs a string, compare it with w.
If w ever appears in the output of E, accept."
Clearly, M accepts those strings that appear on E's list. "
If w doesn't appear in the output of E it doesn't appear in the E's list.
What is he triyng to say?
You have to prove both parts since it includes "if and only if" phrase.
First, you should show that if there exists an enumerator E which enumerates all strings in the language L, we can construct a recognizer for this language L.
This recognizer works with an input w ( a string ) and runs the E inside. E is an enumerator which generates all strings in L one by one. If the input string is equal to one of these generated strings then ACCEPT. If this language is infinite, then the recognizer may not halt, which is not a problem for a recognizer since it is not a decider.
Second part is, if L is Turing recognizable then there must be a Turing Machine M that recognizes L. An enumerator can be constructed as follows;
for k=1,2,3...
Run M on w1,w2,w3... in parallelized for k steps
if M accepts any of the wi then print wi on the printer.
The reason why we run them in parallelized with limited steps is the same reason why we prefer depth-limited search over depth-first search. It can go through an infinite dead path on the search graph.
Your theorem has two directions: if the "if" and the "only if". The proof is for the "if" direction.
Assuming you have an enumerator E for a language L, can you construct a Turing machine M that recognizes L? Yes, you can. Just define a Turing machine M that, on input string w, checks to see if w is ever in the output of E (which may be infinite). If it is, accept. If it isn't reject.
Since E is an enumerator for L, for any w in L, E eventually outputs w before halting (if it ever halts). Thus, M halts for every string in L. If w is not in L, either M never halts, or M rejects w.
Also, for M to be a decider, not just a recognizer for L, M must always halt.
Related
I have been asked to prove if the following set is decidible, semi-decidible or not semi-decidible:
In other words, it is the set of inputs such that exists a Turing Machine encoded with the natural y with input p that returns its input.
Consider the set K as the set of naturals such that the Turing machine encoded with x and input x stops. This is demonstrated to be a non-decidible set.
I think that what I need is to find a reduction of K to L, but I don't know how to prove that L is decidible, semi-decidible or not semi-decidible.
L may not look decidable at first glance, because there is this nasty unbounded quantifier included, which seems to make necessary a possibly infinite search when you look for a y satisfying the condition for a specific p.
However, the answer is much simpler: There is a turing machine M which always returns its input, i.e. M(p) = p holds for all p in the considered language. Let y be a code of M. Then you can use this same y for all p, showing that L contains all words of the language. Hence L is of course decidable.
In fact, this is an example to demonstrate the principle of extensionality (if two sets have the same elements and one is decidable, then the other is decidable too, even if it doesn't look so).
How is Turing Machine which accepts nothing is not Recursively Enumerable.
We will use an indirect argument to show that the language of encodings of Turing Machines that accept nothing cannot be recursively enumerable.
Lemma 1: if L and its complement are recursively enumerable, then L is recursive.
Proof: let M be a TM that enumerates L and M' be a TM that enumerates the complement of L. Given any string s, we can decide whether s is in L as follows. Begin running M and M', interleaving their executions so that each one eventually gets an arbitrary amount of runtime. If s is in L, M will eventually list it, at which point we know s is in L and we halt-accept. If s is not in L, M' will eventually list it, at which point we know s is not in L and we halt-reject. Thus, for any s, we can halt-accept if s is in L or halt-reject otherwise. Therefore, L and its complement are recursive.
Lemma 2: The language of encodings of Turing Machines that accept something is recursively enumerable.
Proof: The set of all Turing Machine encodings is countable, and so is the set of all possible tape inputs. Thus, the set (M, s) of pairs of machines and inputs is countable. We may therefore assume some ordering of these pairs p1, p2, ..., pk, ... For each pair p = (M, s), begin executing machine M on input s, interleaving the executions of pairs p1, p2, ..., pk, ... so each eventually gets an arbitrary amount of runtime. If pk enters the halt-accept state, we may immediately list M as a TM that accepts something (namely, the corresponding s), and we can even terminate all other running instances checking the same M (and forego starting any new ones). Any machine M that accepts some input will eventually be started and will eventually halt-accept on an input, so all machines are eventually enumerated.
Lemma 3: The language of encodings of Turing Machines that accept nothing is not recursive.
Proof: This is a direct result of Rice's Theorem. The property "accepts nothing" is a semantic property of the language itself and is true for some, but not all, languages; therefore, no TM can decide whether another TM accepts a language with the property or not.
Theorem: The language of encodings of Turing Machines that accept nothing is not recursively enumerable.
Proof: Assume this language is recursively enumerable. We have already proven in Lemma 2 that its complement is recursively enumerable. By Lemma 1, then, both languages are recursive. However, Lemma 3 proves that the language is not recursive. This is a contradiction. The only assumption was that the language is recursively enumerable, so that assumption must have been false: so the language is not recursively enumerable.
I have to prove that the language L = {< M >: |L(M)| <= 2016} is NOT semi-decidable. Now I thought of doing it like this:
Take a random alfabet E. Now there are an infinite number of words in E. We can only conclude that |L(M)| <= 2016 by passing every word from E* as an input to M. But because there are an infinite number of words, this means that we would have to pass an input to M an infinite number of times. But this implies that Turing Machine that performs these checks ends up in an infinite loop, and thus never returns accepts nor rejects it's input. The language L is thus not semi-decidable.
But I think that this might not be formal enough? Mainly because I just assume that the Turing Machine that checks this language will let M run on every word from E*. Is this assumption valid, or should I be more formal in this?
Consider the complement of L: the language of Turing-machine encodings whose languages contain more than 2016 strings. We can easily prove this language is semi-decidable. Given this fact, assume that L is also semi-decidable. Because L and its complement are both semi-decidable, we have an effective procedure for deciding each: run the TMs for L and its complement alternately, and one will eventually list the input.
To see that the complement of L is semi-decidable, imagine a TM that executes one step of M on every possible input, alternating moves so that each string is eventually processed by arbitrarily many moves of the TM. The order in which it would visit the inputs would be like:
e,
e, 0,
e, 0, 1,
e, 0, 1, 00,
e, 0, 1, 00, 01, ...
Clearly, any string of input will eventually be processed every possible number of times. Our machine, which is simulating M running on all possible input strings, one step at a time, will eventually find 2017 strings have cause M to halt-accept, or it will run forever. If it finds 2017 strings that works, it halt-accepts.
It is easy to say that {} and {a,b}* are not P complete because other problems in P can't be reduced to these because {} can't accept anything and {a,b}* cannot reject anything. So, proper mapping can't be done with a reduction function.
But I'm stuck with proving that every other problem in P is P-complete.
You have to be careful when talking about P-completeness because this means different things to different people based on what type of reductions you're allowing. I'm going to assume that you're talking about using polynomial-time reductions. In that case, choose any language L ∈ P other than ∅ or {a, b}*. Now pick any language M in P that you like. Here's a silly reduction from M to L:
Given an input string w, decide whether w in M in polynomial time (this is possible because M ∈ P.)
If w ∈ M, output any string w ∈ L that you'd like (at least one exists because L is nonempty.)
Otherwise, w ∉ M, so output any string w ∉ L that you'd like (at least one exists, because L isn't {a, b}*.
This reduction takes polynomial time because each step takes polynomial time, so it's a polynomial-time reduction from an arbitrary P language to L. Therefore, L is P-complete with respect to polynomial-time reductions.
Generally speaking, when you talk about notions of completeness, you have to make sure that your reductions are given fewer computational resources than the class of solvers that you're using, or you can do weird things like what's described here that make reductions essentially useless.
So I am reviewing my notes for this problem, and I cant seem to understand how this problem works. Say we have M, and M accepts an input that makes it visit every non-halting state.
I convinced myself that this problem is decidable, but I am having trouble proving so. A rough outline of my answer would be : Assume we have a TM T that has only one halting state, and if it wants to go through all the states it needs to pass through this halt state and we somehow need to show how they cycle through all the states as such.
Any help would be beneficial, thanks!
I think you'll find the answer is actually that it's undecidable. Why? Well this would let you solve the halting problem.
You are given a TM M and an input x and an oracle Q for the problem you describe. Can we solve the halting problem for M with input x using oracle Q?
First, we connect a new TM N to the front of M. Here's what N does:
- deletes the tape contents
- writes x onto the tape
M halts on x iff NM halts on all inputs. This should be easy to see since N leaves the tape exactly as M would have seen if the input had been x. We can design N in such a way that all states of N are visited.
Now, modify M into M' by adding a second tape. The second tape will be used to keep track of the highest-numbered state of M' we have visited. Add transitions to M' which read the nth state from the secondary tape and cause M' to transition to the (n+1)st state. The transition from the highest-numbered state of M' should return to the initial state of M' and write something like "finished" on the secondary tape. When M' sees "finished" on the secondary tape, it behaves just like M did and only considers the primary tape.
So M' does exactly what M' does, except that it first visits all the states of M and then resets after which it behaves just like M.
M' halts on x iff M halts on x. Also NM' halts iff M halts on x.
Finally we're ready for the proof. The oracle Q accepts NM' iff M halts on x. Q accepts NM' if NM' accepts an input y that causes NM' to visit all states. But:
- all inputs y cause NM' to visit all states (since all states of N are visited by construction, and all states of M' are visited by modification of M); so Q is really just answering the question "does NM' accept any strings?"
- NM' erases y from the input tape, writes x and hands it off to M'. So the input y doesn't matter; if NM' accepts any input, it accepts all of them. And it accepts all of them if M' accepts x.
- M' accepts the same language as M.
So Q applied to NM' does indeed tell us if M halts on x. Then Q solves the halting problem.