Machine that always halts
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In computability theory, a machine that always halts — also called a decider (Sipser, 1996) or a total Turing machine (Kozen, 1997) — is a Turing machine that halts for every input.
Because it always halts, the machine is able to decide whether a given string is a member of a formal language. The class of languages which can be decided by such machines is exactly the set of recursive languages. It is important to note, however, that, due to the Halting Problem, determining whether an arbitrary Turing machine halts on an arbitrary input is itself an undecidable decision problem.
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[edit] Functions computable by total Turing machines
In practice, many functions of interest are computable by machines that always halt. A machine can be forced to halt for every input by restricting its flow control capabilities so that no program will ever cause the machine to enter an infinite loop. As a trivial example, a machine implementing a finitary decision tree will always halt.
It is not required that the machine be entirely free of looping capabilities, however, to guarantee halting. If we restrict loops to be of a predictably finite size (not unlike the FOR loop in BASIC), we can express all of the primitive recursive functions (Meyer and Ritchie, 1967). An example of such a machine is provided by the toy programming language PL-{GOTO} of Brainerd and Landweber (1974).
We can further define a programming language in which we can ensure that even more sophisticated functions always halt. For example, the Ackermann function, which is not primitive recursive, nevertheless is a total computable function computable by a term rewriting system with a reduction ordering on its arguments (Ohlebusch, 2002, pp.67).
[edit] Relationship to partial Turing machines
A general Turing machine will compute a partial function. Two questions can be asked about the relationship between partial Turing machines and total Turing machines:
- Can every partial function computable by a partial Turing machine be extended (that is, have its domain enlarged) to become a total computable function?
- Is it possible to change the definition of a Turing machine so that a particular class of total Turing machines, computing all the total computable functions, can be found?
The answer to each of these questions is no.
The following theorem shows that the functions computable by machines that always halt do not include extensions of all partial computable functions, which implies the first question above has a negative answer. This fact is closely related to the algorithmic unsolvability of the Halting problem.
- Theorem. There are Turing computable partial functions that have no extension to a total Turing computable function. In particular, the partial function f defined so that f(n) = m if and only if the Turing machine with index n halts on input 0 with output m has no extension to a total computable function.
The proof proceeds by contradiction. If g were a total computable function extending f then g would be computable by some Turing machine; fix e as the index of such a machine. Build a Turing machine M, using Kleene's recursion theorem, which on input 0 simulates the machine with index e running on an index nM for M (thus the machine M can produce an index of itself; this is the role of the recursion theorem). By assumption, this simulation will eventually return an answer. Define M so that if g(nM) = m then the return value of M is m + 1. Thus f(nM), the true return value of M on input 0, will not equal g(nM). This contradicts the assumption that g extends f.
The second question asks, in essence, whether there is another reasonable model of computation which computes only total functions and computes all the total computable functions. Informally, if such a model existed then each of its computers could be simulated by a Turing machine. Thus if this new model of computation consisted of a sequence 
of machines, there would be a recursively enumerable sequence 
of Turing machines that compute total functions and so that every total computable function is computable by one of the machines Ti. This is impossible, because a machine T could be constructed such that on input i the machine T returns 
. Then T will be a total machine, because each Ti is total, and the function computed by T cannot appear in the list. This shows that the second question has a negative answer.
[edit] The set of indices of total Turing machines
The decision problem of whether the Turing machine with index e will halt on every input is not decidable. In fact, this problem is at level 
of the arithmetical hierarchy. Thus this problem is stricty more difficult than the Halting problem, which asks whether the machine with index e halts on input 0. Intuitively, this difference in unsolvability is because each instance of the "total machine" problem represents infinitely many instances of the Halting problem.
[edit] References
- Brainerd, W.S., Landweber, L.H. (1974), Theory of Computation, Wiley.
- Meyer, A.R., Ritchie, D.M. (1967), The complexity of loop programs, Proc. of the ACM National Meetings, 465.
- Sipser, M. (1996), Introduction to the Theory of Computation, PWS Publishing Co.
- Kozen, D.C. (1997), Automata and Computability, Springer.
- Ohlebusch, E. (2002), Advanced Topics in Term Rewriting, Springer.
| Automata theory: formal languages and formal grammars | |||
|---|---|---|---|
| Chomsky hierarchy |
Grammars | Languages | Minimal automaton |
| Type-0 | Unrestricted | Recursively enumerable | Turing machine |
| n/a | (no common name) | Recursive | Decider |
| Type-1 | Context-sensitive | Context-sensitive | Linear-bounded |
| n/a | Indexed | Indexed | Nested stack |
| Type-2 | Context-free | Context-free | Nondeterministic Pushdown |
| n/a | Deterministic Context-free | Deterministic Context-free | Deterministic Pushdown |
| Type-3 | Regular | Regular | Finite |
| Each category of languages or grammars is a proper subset of the category directly above it. | |||
