All of glauberdebona's Comments + Replies

Correcting myself: "doubly" just added above.

For the time-complexity of the verifier to grow only linearly with , we need also to assume the complexity of the reduction  is  for a fixed , regardless of , which I admit is quite optimistic. If the time-complexity of the reduction  is doubly exponential in , the (upper bound of the) time-complexity of the verifier will be exponential in , even with the stronger version of Reduction-Regularity. 

1glauberdebona
Correcting myself: "doubly" just added above.

By "unconditionally", do you mean without an additional assumption, such as Reduction-Regularity?

I actually have a result in the other direction (an older version of this post). That is, with a slightly stronger assumption, we could derive a better upper bound (on ) for the time-complexity of the verifier.  The stronger assumption is just a stronger version of Reduction-Regularity, requiring  -- that is,  replaces  in the inequality. It could still lead to an exponential ... (read more)

4Eric Neyman
Echoing Jacob, yeah, thanks for writing this! Since there are only exponentially many circuits, having the time-complexity of the verifier grow only linearly with O(log1/ε) would mean that you could get a verifier that never makes mistakes. So (if I'm not mistaken) if you're right about that, then the stronger version of reduction-regularity would imply that our conjecture is equivalent to NP = coNP. I haven't thought enough about the reduction-regularity assumption to have a take on its plausibility, but based on my intuition about our no-coincidence principle, I think it's pretty unlikely to be equivalent to NP = coNP in a way that's easy-ish to show.
2Jacob_Hilton
Yes, by "unconditionally" I meant "without an additional assumption". I don't currently see why the Reduction-Regularity assumption ought to be true (I may need to think about it more).

Your comment seems to imply that the conjecture's "99%" is about circuits  for which  is false. Otherwise, it would be impossible for a  to miss 100% of random reversible circuits without missing the circuits  for which  is true. In the conjecture, should "random reversible circuits " be read as "random reversible circuits  for which  is false"? It does not change much indeed, but I might have to correct my presentation of the conjecture here.

2Jacob_Hilton
The statements are equivalent if only a tiny fraction (tending to 0) of random reversible circuits satisfy P(C). We think this is very likely to be true, since it is a very weak consequence of the conjecture that random (depth-~O(n)) reversible circuits are pseudorandom permutations. If it turned out to not be true, it would no longer make sense to think of P(C) as an "outrageous coincidence" and so I think we would have to abandon the conjecture. So in short we are happy to consider either version (though I agree that "for which P(C) is false" is a bit more natural).

I think I have an idea to make that 99% closer to 1. But its development became too big for a comment here, so I made a post about it.

Thanks for the feedback! The summary here and the abstract in the draft paper have been updated; I hope it is clearer now.

Hi everyone,

I’m Glauber De Bona, a computer scientist and engineer. I left an assistant professor position to focus on independent research related to AGI and AI alignment (topics that led me to this forum). My current interests include self-location beliefs, particularly the Sleeping Beauty problem.

I’m also interested in philosophy, especially formal reasoning -- logic, Bayesianism, and decision theory. Outside of research, I’m an amateur chess player.