A Comparison of the Work Done by Generalized Sequential Machines and Turing Machines

For a long time it has been accepted that Turing machines are more powerful than generalized sequential machines. Since a generalized sequential machine is also a Turing machine, the class of Turing machines is at least as powerful as the class of generalized sequential machines. The purpose of this paper is to examine the adverb "more" from the aspect of work accomplished. Three manifestations of work performed by machines are considered, namely (i) numerical functions, (ii) output tapes as a function of input tapes, and (iii) the sets of accepted tapes. For each of these expressions of work there is studied the question of the existence of codes so that the resulting set (i.e., the set of numerical functions, the set of input-to-output tape functions, and sets of accepted tapes respectively) for the class of Turing machines is transformed into a subset of the resulting set for the class of generalized sequential machines. In other words, the problem under investigation is to see if the behavior of Turing machines can be reinterpreted as behavior of generalized sequential machines. The main results achieved are as follows. (1) There exist effective mappings/ and g, f transforming a subset of the natural numbers into sequential input tapes and g transforming a set of sequential output tapes into the natural numbers, with the following property. To each partial recursive function h there corresponds a generalized sequential machine S so that for each natural number x, g(5(/(x))) = hix), where 5(/(x)) is the output from machine 5 upon application of the input tape/(x). Furthermore, all these machines S have a finite joint input alphabet and a finite joint output alphabet (Theorem 3). (2) Given any set of Turing machines in a finite joint alphabet there is a function / such that (a) / maps all Turing tapes one to one into sequential input tapes, and (b) for each Turing machine Z there is a generalized sequential machine S such that 5(/(A)) =Z(A), X being any Turing tape, 5(/(A)) being as above, and Z(A) being the analogous function to Z that 5(/(A)) is to 5 (Theorem 5). (3) Given any set of Turing machines there is a function / such that (a) / maps all Turing tapes one to one into sequential tapes; and (b) for each Turing machine Z there is a sequential machine 5 such that the set of tapes