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What are the implications of the halting problem?
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.
Is halting problem unsolvable?
But the halting problem is unsolvable, which means that it is impossible to test if an arbitrary Turing machine T halts on an arbitrary input x.
Is halting problem proof by contradiction?
Theorem (Turing circa 1940): There is no program to solve the Halting Problem. Proof: Assume to reach a contradiction that there exists a program Halt(P, I) that solves the halting problem, Halt(P, I) returns True if and only P halts on I. Contradiction.
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.
Can an undecidable problem ever be solved?
Definition: A decision problem is a problem that requires a yes or no answer. Definition: A decision problem that admits no algorithmic solution is said to be undecidable. No undecidable problem can ever be solved by a computer or computer program of any kind. It means we can never find an algorithm for the problem.
What is halting problem give an example?
There are many equivalent formulations of the halting problem; any set whose Turing degree equals that of the halting problem is such a formulation. Examples of such sets include: {i | program i eventually halts when run with input 0} {i | there is an input x such that program i eventually halts when run with input x}.
Why is the halting problem a difficult problem?
The difficulty in the halting problem lies in the requirement that the decision procedure must work for all programs and inputs. A particular program either halts on a given input or does not halt. Consider one algorithm that always answers “halts” and another that always answers “does not halt”.
Is there an algorithm to determine when a program halts?
But if the program does not halt, it is unknown whether the program will eventually halt or run forever. Turing proved no algorithm exists that always correctly decides whether, for a given arbitrary program and input, the program halts when run with that input.
Which is an equivalent formulation of the halting problem?
There are many equivalent formulations of the halting problem; any set whose Turing degree equals that of the halting problem is such a formulation. Examples of such sets include: {i | program i eventually halts when run with input 0}. {i | there is an input x such that program i eventually halts when run with input x}.
Is it possible to run a program that does not halt?
While deciding whether these programs halt is simple, more complex programs prove problematic. One approach to the problem might be to run the program for some number of steps and check if it halts. But if the program does not halt, it is unknown whether the program will eventually halt or run forever.