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The command Tetrahedron[] seems to work fine for the 3-d case. But how can we get hyper-tetrahedron for 4-d case in terms of inequality? Take for example the points:

P[0]={0,0,0,0}; P[1]={1,0,0,0}; P[2]={0,1,0,0}; P[3]={0,0,1,0}; P[4]={0,0,0,1};

The hyper-tetrahedron formed by these 5 points can be found out manually to be x1>0 && x2>0 && x3>0 && x4>0 && x1+x2+x3+x4<1. But how can we get it easily for any set of 5 points in 4-d using Mathematica?

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  • $\begingroup$ Wouldn’t this be the same question you asked before in terms of ConvexHull? Does the solution you received there not work here? $\endgroup$ Commented Jun 15, 2020 at 14:26
  • $\begingroup$ Note what you have here isn't a regular tetrahedron such as that returned by Tetrahedron[]. You're actually showing a Simplex[3]. If you want a regular one see here for the coordinates: en.wikipedia.org/wiki/5-cell $\endgroup$ Commented Jun 15, 2020 at 14:27
  • $\begingroup$ @MarcoB. No, it won't work for the 4-d cases $\endgroup$ Commented Jun 15, 2020 at 14:29
  • $\begingroup$ I was searching through the Wolfram library. Do you think Simplex[] will work for finding Hyper-tetrahedron? @MarcoB $\endgroup$ Commented Jun 15, 2020 at 14:31

1 Answer 1

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We can use Simplex instead:

Refine[RegionMember[Simplex[P /@ Range[0,4]], {x,y,z,w}], {x,y,z,w} ∈ Reals] 
x >= 0 && y >= 0 && z >= 0 && w >= 0 && w+x+y+z <= 1
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  • $\begingroup$ Thanks a lot. Is it possible to get rid of this (x|y|z|w) ∈ Reals part from the answer? $\endgroup$ Commented Jun 15, 2020 at 14:33
  • $\begingroup$ @SumitBanik See my edit. $\endgroup$ Commented Jun 15, 2020 at 14:34

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