The halting problem is Turing Recognizable. Consider a TM U that gets a pair as input and simulates the run of M on input w. If M accepts or rejects so does U. Consequently, is the halting problem recursive?
Summary. The language HALT corresponding to the Halting problem is recursively enumerable, but not recursive. In particular, the universal TM accepts HALT, but no TM can decide HALT.
Also Know, is the acceptance problem Turing recognizable? ACCEPTTM is not computable.
Beside this, how did Turing solve the halting problem?
In 1936, Alan Turing proved that the halting problem over Turing machines is undecidable using a Turing machine; that is, no Turing machine can decide correctly (terminate and produce the correct answer) for all possible program/input pairs.
How do you show that something is not Turing recognizable?
To prove that a given language is non-Turing-recognizable: Either do both of these: • Prove that its complement is Turing-recognizable. Prove that its complement is undecidable.
Related Question Answers
Can a human solve the halting problem?
Originally Answered: Can human brain solve turing halting problem? No, your brain can't solve the Halting Problem. What makes a problem unsolvable?
An unsolvable problem is one for which no algorithm can ever be written to find the solution. An undecidable problem is one for which no algorithm can ever be written that will always give a correct true/false decision for every input value. Is halting problem NP hard?
- If we had a polynomial time algorithm for the halting problem, then we could solve the satisfiability problem in polynomial time using A and X as input to the algorithm for the halting problem . - Hence the halting problem is an NP-hard problem which is not in NP. - So it is not NP-complete. What is Undecidability explain with halting problem?
The halting problem can be used to show that other problems are undecidable. Determining whether or not a function F is total is undecidable. Proof: Suppose that it is decidable. Assume there is a procedure TOTAL that takes as input a program P and outputs 'Yes' if P halts on all inputs and 'No' otherwise. Why is the halting problem so important?
The Halting problem lets us reason about the relative difficulty of algorithms. It lets us know that, there are some algorithms that don't exist, that sometimes, all we can do is guess at a problem, and never know if we've solved it. What is halting in Turing machine?
Halting means that the program on certain input will accept it and halt or reject it and halt and it would never go into an infinite loop. Basically halting means terminating. In terms of Turing machine, will it terminate when run on some machine with some particular given input string. Is Minecraft Turing complete?
If you're talking about infinite-sized redstone buildings, well, you can quite easily build conway's game of life in minecraft, which is turing complete. What does the Turing test prove?
What Is the Turing Test? The Turing Test is a deceptively simple method of determining whether a machine can demonstrate human intelligence: If a machine can engage in a conversation with a human without being detected as a machine, it has demonstrated human intelligence. What can a Turing machine not do?
Originally Answered: What are some functions that a turing machine cannot compute? Print all the real numbers between 0 and 1. This set is uncountable, but the tape in the turing machine is countable. Therefore the turing machine does not have the capacity to represent the output for this procedure. Who discovered the halting problem?
Alan Turing
What is the halting problem an example of?
The halting problem is an early example of a decision problem, and also a good example of the limits of determinism in computer science. Why Turing machine is more powerful than finite automata?
For example, a Turing machine is said to recognise a sequence of symbols written on the tape if it is started on the tape and halts in a special state called a final state. That is a Turing machine is more powerful than a finite state machine because it can count. What means Turing complete?
A Turing Complete system means a system in which a program can be written that will find an answer (although with no guarantees regarding runtime or memory). So, if somebody says "my new thing is Turing Complete" that means in principle (although often not in practice) it could be used to solve any computation problem. How does the universal Turing machine work?
In computer science, a universal Turing machine (UTM) is a Turing machine that simulates an arbitrary Turing machine on arbitrary input. The universal machine essentially achieves this by reading both the description of the machine to be simulated as well as the input to that machine from its own tape. Is it possible to create a computational machine that can solve the halting problem?
Hence there is no Turing machine that solves the halting problem. Note that for any computer program a Turing machine can be constructed that performs the task of the program. Thus the question of whether or not a program halts for a given input is nothing but the halting problem. Which language is accepted by Turing machine?
The turing machine accepts all the language even though they are recursively enumerable. Recursive means repeating the same set of rules for any number of times and enumerable means a list of elements. What are the reasons for a TM not accepting its input?
A TM rejects a string w if it enters its reject state. A TM loops on a string w if neither of these happens. A TM does not accept a string w if it either rejects w or loops infinitely on w. A TM does not reject a string w if it either accepts w or loops infinitely on w. What is Mpcp in TOC?
Modified Post Correspondence Problem (MPCP) Definition: first pair in the A and B lists must be the first pair in the solution, i.e., the problem is to determine if there is a sequence of zero or more integers i1, i2, …, im such that: w1wi1 wi2 … wim = x1xi1 xi2 … xim. Can the complement of a recognizable language that is not decidable be recognizable?
Theorem: A language is decidable iff both it and its complement are recognizable. Proof: Surely, a decidable language is recognizable. Moreover, if a language is decidable, then so is its complement, and hence that complement is recognizable. Why is ATM not decidable?
Because we know that ATM is recognizable, our theorem implies that ATM and ATM are both decidable. But we know that ATM is not decidable. This is a contradiction, hence ATM cannot be recognizable. The language ATM and its undecidability (including proof). What is Turing unrecognizable?
~ATM is the canonical example of a Turing-unrecognizable language. This means there does not exist a Turing Machine which will accept the set of all machine-string pairs <M,w> such that M does NOT halt when run on w. The proof of this is very short: Lemma: ATM is Turing-recognizable. Is the complement of ATM decidable?
Corollary 4.23: ATM is Turing-recognizable but not decidable, so its complement ATM is NOT Turing-recognizable. Are all languages Turing recognizable?
– There are uncountably many languages. – There are only countably many Turing-recognizable languages and only countably many co-Turing-recognizable languages. – Because there are only countably many Turing machines (up to renaming). Example: Every regular language L is decidable. What is a decidable language?
(definition) Definition: A language for which membership can be decided by an algorithm that halts on all inputs in a finite number of steps --- equivalently, can be recognized by a Turing machine that halts for all inputs. Is ATM Turing recognizable?
Recall that a language L is Turing recognizable if there is a Turing machine that accepts exactly the words in L, but can either reject or loop indefinitely on an input that's not in L. We will show that ATM , the complement of ATM , is not Turing-recognizable. What is a mapping reduction?
What is a mapping reduction? • A mapping reduction A ≤m B (or A ≤P B) is an algorithm (respectively, polytime algorithm) that can transform any instance of decision problem A into an instance of decision problem B, in such a way that the answer correspondence property holds. Can ATM reduce to its complement?
ATM can not reduce to its complement. From the lecture, we know the complement of L1 is decidable, so does L1. (b) L2 = { : N is a NFA and w ∈ L(N)} Decidable. 
