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.
Functions computable by total Turing machinesIn practice, many functions of interest are computable by machines that always halt. A machine that uses only finite memory on any particular input can be forced to halt for every input by restricting its flow control capabilities so that no input 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). Relationship to partial Turing machinesA general Turing machine will compute a partial function. Two questions can be asked about the relationship between partial Turing machines and total Turing machines:
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.
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 Failed to parse (Missing texvc executable; please see math/README to configure.): M_1,M_2,\ldots of machines, there would be a recursively enumerable sequence Failed to parse (Missing texvc executable; please see math/README to configure.): T_1,\ldots T_2,\ldots 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 Failed to parse (Missing texvc executable; please see math/README to configure.): T could be constructed such that on input i the machine T returns Failed to parse (Missing texvc executable; please see math/README to configure.): T_i(i)+1\, . This machine cannot be equivalent to any machine T on the list: suppose it were on the list at index j. Then Failed to parse (Missing texvc executable; please see math/README to configure.): T_j(j)=T_j(j)+1\, , which does not return an integer result. Therefore it can't be total, but the function by construction must be total (if total functions are recursively enumerable, then this function can be constructed), so we have a contradiction. This shows that the second question has a negative answer. The set of indices of total Turing machinesThe 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 Failed to parse (Missing texvc executable; please see math/README to configure.): \Pi^0_2 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. References
it:Macchina che termina sempre ru:Машины, которые всегда останавливаются | ||||||||||||||||||||||||||||||||||||||||||||||
