Example: The halting problem is
partially computable
. To determine HALTS(P,D), simply call P(D). Then, HALTS(P,D) halts and outputs Yes if P(D) halts, and loops otherwise. … If a problem is not even partially computable, there is no way of checking even a YES answer.
What problems are not computable?
A non-computable is a problem for which there is no algorithm that can be used to solve it. An example of a non-computable is
the halting problem
. Hyper computation is more powerful than a Turing Machine and has the capability of solving problems that the Turing Machine can’t.
Is the halting problem solvable?
Because the
halting problem is not solvable on a Turing machine
, it is not solvable on any computer, or by any algorithm, given the Church-Turing thesis. Many other unsolvability results are derived starting from the ones given here.
Is this problem computable?
A mathematical problem is computable if it can be solved in principle by a computing device
. Some common synonyms for “computable” are “solvable”, “decidable”, and “recursive”. Hilbert believed that all mathematical problems were solvable, but in the 1930’s Gödel, Turing, and Church showed that this is not the case.
Is the halting problem non-computable?
There are many problems for which there is no algorithm. In fact, the number of things which can be computed is infinitesimal compared with the number of things one might like to compute but which cannot be computed. To prove a problem P is
non-computable
, … But we know the Halting Problem is non-computable.
What kind of problem is the halting problem?
In computability theory, the halting problem is
the problem of determining
, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever.
What type of problem is the halting problem?
unsolvable algorithmic problem
is the halting problem, which states that no program can be written that can predict whether or not any other program halts after a finite number of steps. The unsolvability of the halting problem has immediate practical bearing on software development.
What is an undecidable problem example?
Examples – These are few important Undecidable Problems:
Whether a CFG generates all the strings or not
? As a CFG generates infinite strings, we can’t ever reach up to the last string and hence it is Undecidable. … Since we cannot determine all the strings of any CFG, we can predict that two CFG are equal or not.
Which problems are decidable?
Definition:
A decision problem that can be solved by an algorithm that halts on all inputs in a finite number of steps
. The associated language is called a decidable language. Also known as totally decidable problem, algorithmically solvable, recursively solvable.
Are all decision problems computable?
If this decision problem were decidable then the function that yields the answer of the function problem is computable. Every decision problem can
be converted
into the function problem of computing the characteristic function of the set associated to the decision problem.
Why is the halting problem undecidable?
Example: the halting problem in computability theory
Alan Turing proved in 1936 that
a general algorithm running on a Turing machine that solves the halting problem for all possible program-input pairs necessarily cannot exist
. Hence, the halting problem is undecidable for Turing machines.
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.
Is halting problem recursively enumerable?
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. There are languages which are not recursively enumerable, in particular the language NOTRE in the proof.
Are all functions are computable?
Every such function is computable
. It is not known whether there are arbitrarily long runs of fives in the decimal expansion of π, so we don’t know which of those functions is f. Nevertheless, we know that the function f must be computable.)
How do you prove halting problems?
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. The given this program for the Halting Problem, we could construct the following string/code Z: Program (String x) If Halt(x, x) then Loop Forever Else Halt.
How do you know if something is computable?
Now, consider g(p). As φp(x)↓ for all x≥1,
g(p)=1 if and only if φp(p)↓ by
the definition of φp, which is actually the function g. Hence, if g would be computable, the halting problem would be computable as well. Therefore, we reach a contradiction.
