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The halting problem is undecidable, meaning that no general algorithm exists that solves the halting problem for all possible program–input pairs. The problem comes up often in discussions of computability since it demonstrates that some functions are mathematically definable but not computable .
This means that this gives us an algorithm to decide the halting problem. Since we know that there cannot be such an algorithm, it follows that the assumption that there is a consistent and complete axiomatization of all true first-order logic statements about natural numbers must be false.
The halting problem (determining whether a Turing machine halts on a given input) and the mortality problem (determining whether it halts for every starting configuration). Determining whether a Turing machine is a busy beaver champion (i.e., is the longest-running among halting Turing machines with the same number of states and symbols).
The partial function computed by the algorithm represented by a string a is denoted F a. This proof proceeds by reductio ad absurdum: we assume that there is a non-trivial property that is decided by an algorithm, and then show that it follows that we can decide the halting problem, which is not possible, and therefore a contradiction. Let us ...
No halting probability is computable. The proof of this fact relies on an algorithm which, given the first n digits of Ω, solves Turing's halting problem for programs of length up to n. Since the halting problem is undecidable, Ω cannot be computed. The algorithm proceeds as follows.
The problem does not have to be computable; the oracle is not assumed to be a Turing machine or computer program. The oracle is simply a "black box" that is able to produce a solution for any instance of a given computational problem: A decision problem is represented as a set A of natural numbers (or strings). An instance of the problem is an ...
There are decision problems that are NP-hard but not NP-complete such as the halting problem. That is the problem which asks "given a program and its input, will it run forever?" That is a yes/no question and so is a decision problem. It is easy to prove that the halting problem is NP-hard but not NP-complete.
Thus the halting problem is an example of a computably enumerable (c.e.) set, which is a set that can be enumerated by a Turing machine (other terms for computably enumerable include recursively enumerable and semidecidable). Equivalently, a set is c.e. if and only if it is the range of some computable function.