# NC^{k}(NP)=AC^{k−1}(NP)

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## Abstract

It is shown for any *k*≥1, that the closure of NP under NC^{k} reducibility coincides with that of NP under AC^{k−1} reducibility, thereby giving an answer to a basic question that has been open for a long time. A similar result is shown for C_{=}P.

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© Springer-Verlag Berlin Heidelberg 1994