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A comment on Markets are efficient if and only if P = NP (2010). Parent comment

Re:Markets are efficient if and only if P = NP (2010) by Ar-Curunir on Friday June 01, 2018 @03:20AM (#17203277)
> I'm not sure I understand the author's implication that the energy used in the past to calculate the current price is equivalent to the energy to verify the current price.

That's just what P = NP means: the cost to verify a solution is the same as the cost of finding one.

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Re:Markets are efficient if and only if P = NP (2010) by lend000 on Friday June 01, 2018 @03:53AM (#17203418)
Yes, but how the author relates that to a market's continuous price calculations is beyond me.
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Re:Markets are efficient if and only if P = NP (2010) by babypistol on Friday June 01, 2018 @05:27AM (#17203817)
In my opinion P = NP does not mean that. There still might be a big polynomial difference between verifying a solution and finding one.
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Re:Markets are efficient if and only if P = NP (2010) by Ar-Curunir on Friday June 01, 2018 @11:46AM (#17206437)
Of course, the concrete costs can be different; I was talking asymptotically :)
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Re:Markets are efficient if and only if P = NP (2010) by babypistol on Friday June 01, 2018 @01:14PM (#17207585)
Well that's the problem, man.

And asymptotically is not the right word either, because n^3 is asymptotically different from n^2. You probably meant poly-time reducible, whichs is quite different from "the same".

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