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In computer science, computability theory is the branch of the theory of computation that studies which problems are computationally solvable using different models of computation.

Computability theory differs from the related discipline of computational complexity theory, which deals with the question of how efficiently a problem can be solved, rather than whether it is solvable at all.

To explore this area, computer scientists invented automata theory which addresses problems such as the following: Given a formal language, and a string, is the string a member of that language? This is a somewhat esoteric way of asking this question, so an example is illuminating. We might define our language as the set of all strings of digits which represent a prime number. To ask whether an input string is a member of this language is equivalent to asking whether the number represented by that input string is prime. Similarly, we define a language as the set of all palindromes, or the set of all strings consisting only of the letter 'a'. In these examples, it is easy to see that constructing a computer to solve one problem is easier in some cases than in others.

But in what real sense is this observation true? Can we define a formal sense in which we can understand how hard a particular problem is to solve on a computer? It is the goal of computability theory of automata to answer just this question.

An example of such a language is the set of all strings consisting of the letters 'a' and 'b' which contain an equal number of the letter 'a' and 'b'. To see why this language cannot be correctly recognized by a finite state machine, assume first that such a machine $M$ exists. $M$ must have some number of states $n$. Now consider the string $x$ consisting of $(n+1)$ 'a's followed by $(n+1)$ 'b's.

As $M$ reads in $x$, there must be some state in the machine that is repeated as it reads in the first series of 'a's, since there are $(n+1)$ 'a's and only $n$ states by the pigeonhole principle. Call this state $S$, and further let $d$ be the number of 'a's that our machine read in order to get from the first occurrence of $S$ to some subsequent occurrence during the 'a' sequence. We know, then, that at that second occurrence of $S$, we can add in an additional $d$ (where $d\; >\; 0$) 'a's and we will be again at state $S$. This means that we know that a string of $(n+d+1)$ 'a's must end up in the same state as the string of $(n+1)$ 'a's. This implies that if our machine accepts $x$, it must also accept the string of $(n+d+1)$ 'a's followed by $(n+1)$ 'b's, which is not in the language of strings containing an equal number of 'a's and 'b's. In other words, $M$ cannot correctly distinguish between a string of equal number of 'a's and 'b's and a string with $(n+d+1)$ 'a's and $n+1$ 'b's.

We know, therefore, that this language cannot be accepted correctly by any finite state machine, and is thus not a regular language. A more general form of this result is called the Pumping lemma for regular languages, which can be used to show that broad classes of languages cannot be recognized by a finite state machine.

However, it turns out there are languages that cannot be decided by push-down automaton either. The result is similar to that for regular expressions, and won't be detailed here. There exists a Pumping lemma for context-free languages. An example of such a language is the set of prime numbers.

Because Turing machines have the ability to "back up" in their input tape, it is possible for a Turing machine to run for a long time in a way that is not possible with the other computation models previously described. It is possible to construct a Turing machine that will never finish running (halt) on some inputs. We say that a Turing machine can decide a language if it eventually will halt on all inputs and give an answer. A language that can be so decided is called a recursive language. We can further describe Turing machines that will eventually halt and give an answer for any input in a language, but which may run forever for input strings which are not in the language. Such Turing machines could tell us that a given string is in the language, but we may never be sure based on its behavior that a given string is not in a language, since it may run forever in such a case. A language which is accepted by such a Turing machine is called a recursively enumerable language.

The Turing machine, it turns out, is an exceedingly powerful model of automata. Attempts to amend the definition of a Turing machine to produce a more powerful machine are surprisingly met with failure. For example, adding an extra tape to the Turing machine, giving it a 2-dimensional (or 3 or any-dimensional) infinite surface to work with can all be simulated by a Turing machine with the basic 1-dimensional tape. These models are thus not more powerful. In fact, a consequence of the Church-Turing thesis is that there is no reasonable model of computation which can decide languages that cannot be decided by a Turing machine.

The question to ask then is: do there exist languages which are recursively enumerable, but not recursive? And, furthermore, are there languages which are not even recursively enumerable?

The halting problem is one of the most famous problems in computer science, because it has profound implications on the theory of computability and on how we use computers in everyday practice. The problem can be phrased:

- Given a description of a Turing machine and its initial input, determine whether the program, when executed on this input, ever halts (completes). The alternative is that it runs forever without halting.

Here we are asking not a simple question about a prime number or a palindrome, but we are instead turning the tables and asking a Turing machine to answer a question about another Turing machine. It can be shown (See main article: Halting problem) that it is not possible to construct a Turing machine that can answer this question in all cases.

That is, the only general way to know for sure if a given program will halt on a particular input in all cases is simply to run it and see if it halts. If it does halt, then you know it halts. If it doesn't halt, however, you may never know if it will eventually halt. The language consisting of all Turing machine descriptions paired with all possible input streams on which those Turing machines will eventually halt, is not recursive. The halting problem is therefore called non-computable or undecidable.

An extension of the halting problem is called Rice's Theorem, which states that it is undecidable (in general) whether a given language possesses any specific nontrivial property.

A simple example of such a language is the complement of the halting language; that is the language consisting of all Turing machines paired with input strings where the Turing machines do not halt on their input. To see that this language is not recursively enumerable, imagine that we construct a Turing machine $M$ which is able to give a definite answer for all such Turing machines, but that it may run forever on any Turing machine that does eventually halt. We can then construct another Turing machine $M\text{'}$ that simulates the operation of this machine, along with simulating directly the execution of the machine given in the input as well, by interleaving the execution of the two programs. Since the direct simulation will eventually halt if the program it is simulating halts, and since by assumption the simulation of $M$ will eventually halt if the input program would never halt, we know that $M\text{'}$ will eventually have one of its parallel versions halt. $M\text{'}$ is thus a decider for the halting problem. We have previously shown, however, that the halting problem is undecidable. We have a contradiction, and we have thus shown that our assumption that $M$ exists is incorrect. The complement of the halting language is therefore not recursively enumerable.

- $1\; +\; \{1\; over\; 2\}\; +\; \{1\; over\; 4\}\; +\; cdots$

time to run. This infinite series converges to 2 time units, which means that this Turing machine can run an infinite execution in 2 time units. This machine is capable of deciding the halting problem by directly simulating the execution of the machine in question. By extension, any convergent series would work. Assuming that the series converges to a value $n$, the Turing machine would complete an infinite execution in $n$ time units.

- Automata theory
- Abstract machine
- List of undecidable problems
- Computational complexity theory
- Computability logic
- Important publications in computability

- Michael Sipser (1997).
*Introduction to the Theory of Computation*. PWS Publishing. ISBN 0-534-94728-X. Part Two: Computability Theory, chapters 3–6, pp.123–222. - Christos Papadimitriou (1993).
*Computational Complexity*. 1st edition, Addison Wesley. ISBN 0-201-53082-1. Chapter 3: Computability, pp.57–70.

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Last updated on Sunday July 06, 2008 at 18:39:09 PDT (GMT -0700)

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Last updated on Sunday July 06, 2008 at 18:39:09 PDT (GMT -0700)

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