Abstract
The finite dynamical system is a system consisting of some finite number of objects that take upon a value from some domain as a state, in which after initialization the states of the objects are updated based upon the states of the other objects and themselves according to a certain update schedule. This paper studies the subclass of finite dynamical systems the synchronous boolean finite dynamical system (synchronous BFDS, for short), where the states are boolean and the state update takes place in discrete time and at the same on all objects. The present paper is concerned with some problems regarding the behavior of synchronous BFDS in which the state update functions (or the local state transition functions) are chosen from a predetermined finite basis of boolean functions \documentclass[12pt]{minimal}
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\begin{document}$${\mathcal {B}}$$\end{document}. Specifically the following three behaviors are studied:Convergence. Does a system at hand converge on a given initial state configuration?Path Intersection. Will a system starting in given two state configurations produce a common configuration?Cycle Length. Since the state space is finite, every BFDS on a given initial state configuration either converges or enters a cycle having length greater than 1. If the latter is the case, what is the length of the loop? Or put more simply, for an integer \documentclass[12pt]{minimal}
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\begin{document}$$t$$\end{document}, is the length of loop greater than \documentclass[12pt]{minimal}
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\begin{document}$$t$$\end{document}?
The paper studies these questions in terms of computational complexity (in the case of Cycle Length using the decision version of the problem) and shows the following:The three problems are each \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {PSPACE}$$\end{document}-complete if the boolean function basis contains \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {NAND}$$\end{document}, \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {NOR}$$\end{document} or both \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {AND}$$\end{document} and \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {OR}$$\end{document}.The Convergence Problem is solvable in polynomial time if the set \documentclass[12pt]{minimal}
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\begin{document}$$B$$\end{document} is one of \documentclass[12pt]{minimal}
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\begin{document}$$\{\mathrm {AND}\}$$\end{document}, \documentclass[12pt]{minimal}
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\begin{document}$$\{\mathrm {OR}\}$$\end{document} and \documentclass[12pt]{minimal}
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\begin{document}$$\{\mathrm {XOR}, \mathrm {NXOR}\}$$\end{document}.If the set \documentclass[12pt]{minimal}
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\begin{document}$$B$$\end{document} is chosen from the three sets as in the case of the Convergence Problem, the Path Intersection Problem is in UP, and the Cycle Length Problem is in \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {UP}\cap \mathrm {coUP}$$\end{document}; thus, these are unlikely to be \documentclass[12pt]{minimal}
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\begin{document}$$\mathrm {NP}$$\end{document}-hard.
