Higher-Order Nonemptiness Step by Step [article]

Paweł Parys
2020 arXiv   pre-print
We show a new simple algorithm that checks whether a given higher-order grammar generates a nonempty language of trees. The algorithm amounts to a procedure that transforms a grammar of order n to a grammar of order n-1, preserving nonemptiness, and increasing the size only exponentially. After repeating the procedure n times, we obtain a grammar of order 0, whose nonemptiness can be easily checked. Since the size grows exponentially at each step, the overall complexity is n-EXPTIME, which is
more » ... own to be optimal. More precisely, the transformation (and hence the whole algorithm) is linear in the size of the grammar, assuming that the arity of employed nonterminals is bounded by a constant. The same algorithm allows to check whether an infinite tree generated by a higher-order recursion scheme is accepted by an alternating safety (or reachability) automaton, because this question can be reduced to the nonemptiness problem by taking a product of the recursion scheme with the automaton. A proof of correctness of the algorithm is formalised in the proof assistant Coq. Our transformation is motivated by a similar transformation of Asada and Kobayashi (2020) changing a word grammar of order n to a tree grammar of order n-1. The step-by-step approach can be opposed to previous algorithms solving the nonemptiness problem "in one step", being compulsorily more complicated.
arXiv:2009.08174v1 fatcat:6tul6byghnbkfo45swdkmitaue