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What would be the consequence and meaning of existence of polynomial reduction of #P-complete problem into NP problem (not NP-complete problem)?

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    You have to be substantially more specific about what you mean by 'polynomial reduction' because #P problems aren't decision problems, so the usual notions of 'reduction' don't directly apply. Given that $P^{#P}=PH$, though (for a suitable definition of $P^{#P}$), it seems like one natural consequence of such a reduction would be the collapse of the polynomial hierarchy to at most $\Sigma_2$ ($=P^{NP}$). – Steven Stadnicki Jul 11 '12 at 04:44
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    @Steven: $P^{# P} \supseteq PH$ but equality is not known. – sdcvvc Oct 16 '12 at 17:30
  • @sdcvvc D'oh, good catch - I'm not sure where I got the equality from. Though I think my conclusion still holds, regardless. – Steven Stadnicki Oct 16 '12 at 18:26

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