I show a solution to that problem as follows:

Given a number x and a set S of n positive integers, MINIMUM is the problem of deciding whether x is the minimum of S. We can easily obtain an upper bound of n comparisons: find the minimum in the set and check whether the result is equal to x. Is this the best we can do? Yes, since we can obtain a lower bound of (n – 1) comparisons for the problem of determining the minimum and another obligatory comparison for checking whether that minimum is equal to x. A representation of a set S with n positive integers is a Boolean circuit C, such that C accepts the binary representation of a bit integer i if and only if i is in S. Given a positive integer x and a Boolean circuit C, we define SUCCINCT-MINIMUM as the problem of deciding whether x is the minimum bit integer which accepts C as input. For certain kind of SUCCINCT-MINIMUM instances, the input (x, C) is exponentially more succinct than the cardinality of the set S that represents C. Since we prove that SUCCINCT-MINIMUM is at least as hard as MINIMUM in order to the cardinality of S, then we could not decide every instance of SUCCINCT-MINIMUM in polynomial time. If some instance (x, C) is not in SUCCINCT-MINIMUM, then it would exist a positive integer y such that y < x and C accepts the bit integer y. Since we can evaluate whether C accepts the bit integer y in polynomial time and we have that y is polynomially bounded by x, then we can confirm SUCCINCT-MINIMUM is in coNP. If any single coNP problem cannot be solved in polynomial time, then P is not equal to coNP. Certainly, P = NP implies P = coNP because P is closed under complement, and therefore, we can conclude P is not equal to NP.

You could read the details in the link below…

]]>I’ve asked on Twitter about this — see thread and responses from Elsevier ]]>

https://web-beta.archive.org/web/20131022235944/http://www.elsevier.com/about/open-access/oa-and-elsevier/oa-license-policy#open-archive ]]>

Hilariously, the [Elsevier user license](https://www.elsevier.com/about/our-business/policies/open-access-licenses/elsevier-user-license) still says:

> Any translations, for which a prior translation agreement with Elsevier has not been established, must prominently display the statement: “This is an unofficial translation of an article that appeared in an Elsevier publication. Elsevier has not endorsed this translation.”

Given how much sense this makes without a right to redistribute, I guess they haven’t been very thorough :)

]]>These distributions form a convex region bounded by hyperplanes, so I just need to prove it for the vertices. For each such vertex, I express it as a linear combination of “triple distributions” (prob distributions on the region with mean (q-1)/3 and all probabilites equal to 0, 1/3, 2/3 or 1) and “reducible distributions” (ones where p_0=p_{q-2}=p_{q-1}=0 and p_1>=p_2>=…>=p_{q-3}).

Since triple distributions are clearly S3-symmetric marginals, and reducible distributions can be converted to things from the q-3 case (by subtracting 1 from everything) we are done by induction.

Devil’s in the details, of course. Hope to write it up this weekend.

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