Come to think of it, maybe there's always a simplex of maximum volume whose vertices >are cube vertices. But there are other simplices of maximum volume. E.g., Conv({(0,0), (1,0), (1/2,1)}). --Dan
--true, the "other" ones arise if some "minor" is exactly 0. But there is always a maxvol simplex with only cube vertices.
So this is basically the Hadamard problem, and the maximum volume of an n-simplex is A003432 (n)/n!: 1 1 2 1/2 3 1/3 4 1/8 5 1/24 6 1/80 7 2/315 8 1/720 9 1/2520 10 1/11340 11 9/246400 12 3/394240 13 3/1971200 14 3/10250240 15 64/638512875 16 2/127702575 17 2/638512875 18 1/1876446000 19 78125/486580401635328 20 90625/3892643213082624 Charles Greathouse Analyst/Programmer Case Western Reserve University On Thu, Sep 26, 2013 at 12:06 PM, Warren D Smith <warren.wds@gmail.com>wrote:
Come to think of it, maybe there's always a simplex of maximum volume whose vertices >are cube vertices. But there are other simplices of maximum volume. E.g., Conv({(0,0), (1,0), (1/2,1)}). --Dan
--true, the "other" ones arise if some "minor" is exactly 0. But there is always a maxvol simplex with only cube vertices.
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Nice list! The nth item on the list below (starting from n=1) is (n-1)/n! (though I haven't checked all of them). This corresponds for example to the volume of the n-simplex in the n-cube [0,1]^n whose vertices are the origin and the n points whose coordinates are one 0 and the rest 1's. --Dan On 2013-09-26, at 11:47 AM, Charles Greathouse wrote:
So this is basically the Hadamard problem, and the maximum volume of an n-simplex is A003432 (n)/n!:
1 1 2 1/2 3 1/3 4 1/8 5 1/24 6 1/80 7 2/315 8 1/720 9 1/2520 10 1/11340 11 9/246400 12 3/394240 13 3/1971200 14 3/10250240 15 64/638512875 16 2/127702575 17 2/638512875 18 1/1876446000 19 78125/486580401635328 20 90625/3892643213082624
Charles Greathouse Analyst/Programmer Case Western Reserve University
It's (n-1)/n! only for n = 2, 3, 4. Charles Greathouse Analyst/Programmer Case Western Reserve University On Thu, Sep 26, 2013 at 6:57 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Nice list!
The nth item on the list below (starting from n=1) is (n-1)/n! (though I haven't checked all of them).
This corresponds for example to the volume of the n-simplex in the n-cube [0,1]^n whose vertices are the origin and the n points whose coordinates are one 0 and the rest 1's.
--Dan
On 2013-09-26, at 11:47 AM, Charles Greathouse wrote:
So this is basically the Hadamard problem, and the maximum volume of an n-simplex is A003432 (n)/n!:
1 1 2 1/2 3 1/3 4 1/8 5 1/24 6 1/80 7 2/315 8 1/720 9 1/2520 10 1/11340 11 9/246400 12 3/394240 13 3/1971200 14 3/10250240 15 64/638512875 16 2/127702575 17 2/638512875 18 1/1876446000 19 78125/486580401635328 20 90625/3892643213082624
Charles Greathouse Analyst/Programmer Case Western Reserve University
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Oh, well. Then I guess that simplex (whose volume is always (n-1)/n!) is a maximum configuration only for those values. --Dan Charles wrote: ----- It's (n-1)/n! only for n = 2, 3, 4. ----- On 2013-09-26, at 3:57 PM, Dan Asimov wrote:
The nth item on the list below (starting from n=1) is (n-1)/n! (though I haven't checked all of them).
This corresponds for example to the volume of the n-simplex in the n-cube [0,1]^n whose vertices are the origin and the n points whose coordinates are one 0 and the rest 1's.
--Dan
On 2013-09-26, at 11:47 AM, Charles Greathouse wrote:
So this is basically the Hadamard problem, and the maximum volume of an n-simplex is A003432 (n)/n!:
1 1 2 1/2 3 1/3 4 1/8 5 1/24 6 1/80 7 2/315 8 1/720 9 1/2520 10 1/11340
participants (3)
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Charles Greathouse -
Dan Asimov -
Warren D Smith