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February 17 Information

Distances on the 5-cell?

For a 5-cell with unit length edges, what are the following distances?

  • Center of the 5-cell to vertex?
  • Vertex to center of the opposite tetrahedron?
  • center of one edge to the center of the opposite triangle?

And for the lines for numbers 2 & 3, which one has its middle closer to the center of the 5-cell? My intuition says #3, but I'm not sure. Naraht ( talk) 14:48, 17 February 2016 (UTC) reply

Coordinatize the vertices in 5 dimensions as (00001), (00010), etc., (ok, rescale by sqrt(2) if you really want unit edges) and everything is easy. -- JBL ( talk) 15:14, 17 February 2016 (UTC) reply
For example, if you were working with the triangle in three dimensions, the three vertices are (001), (010) and (100) (all lying on the plane x+y+z=1). If you want to know the centroid of a face, you take the average of its vertices, so e.g. the midpoints of the edges are (1/2, 1/2, 0) and permutations. If you want to compute distances, you use the usual distance formula. This coordinatization is very easy to work with because it's symmetric and rational. -- JBL ( talk) 16:15, 17 February 2016 (UTC) reply
So the distance along the line from a vertex to the center of the opposite tetrahedron (vtet) would be the distance between (1,0,0,0,0) and (0,1/4,1/4,1/4,1/4) and the distance along a line from the center of the edge to the opposite triangle (etri) would be the distance from (1/2,1/2,0,0,0) to (0,0,1/3,1/3,1/3), right? So sqrt(5/4) vs. sqrt(5/6) for which obviously the first is longer as expected. The center of 5 cell (c5c) is at (1/5,1/5,1/5,1/5,1/5) the center of the vtet (cvtet) would be at (1/2,1/8,1/8,1/8,1/8) and the center of the etri (cetri) would be (1/4,1/4,1/6,1/6,1/6). Since every coordinate of cetri is closer to the c5c than those of cvtet, then it certainly is closer. (I started calculating the distances before I realized this). Fun to walk *up* a dimension to make everything easier. :) Naraht ( talk) 16:51, 17 February 2016 (UTC) reply
Yes, that looks right to me. It is perhaps also worth noting that the centroid of a regular n-simplex is much closer to the nearest facet than it is to the nearest vertex (and more so as n grows): this is clear in your example. And more generally it is closer to any given i-dimensional face than it is to any j-dimensional face if i > j. -- JBL ( talk) 21:54, 17 February 2016 (UTC) reply
Yup, and very generalizable...
For convenience, I turn to Coxeter's Regular Polytopes, Table I(ii). He uses edge length 2, so I divide each of his numbers by 2. From the centre to a vertex: √(2/5). From the centre to an edge: √(3/20). From the centre to a face: 1/√15 (at least I think that's what it says, the typesetting is a bit odd). From the centre to a cell: 1/√40. So your second number is √(2/5) + 1/√40 = √(5/8), and your third is √(3/20) + 1/√15 = √(5/12), if I haven't fumbled. Funny how each of Coxeter's numbers has a factor of √5 in the denominator and each of these sums has √5 in the numerator! — Tamfang ( talk) 04:16, 18 February 2016 (UTC) reply
Is it possible to talk about this topic without bringing in Coxeter's masterwork? :) Naraht ( talk) 15:26, 18 February 2016 (UTC) reply

R^(lsd) Tree

I know what an R tree is. I'm reading about Rlsd Trees with no definition of what lsd is an abbreviation for. Is that a common type of R tree? If so, how does it differ from a standard R tree or what defines the subset of R trees? 209.149.113.11 ( talk) 20:14, 17 February 2016 (UTC) reply

As near as I can tell the LSD stands for "local split decision". This is a rather specialized form of database indexing and this question might get a better response on the computing desk rather than here. Though if it was me I'd probably go straight to stack-overflow. -- RDBury ( talk) 23:02, 17 February 2016 (UTC) reply
Oh, gee. I always thought it stands for Lysergic acid diethylamide. (Sorry, someone had to make that joke, may as well be me...) -- Meni Rosenfeld ( talk) 14:08, 18 February 2016 (UTC) reply
I've always wondered whether certain parts of mathematics would be easier on various drugs, LSD seems appropriate for dealing with polytopes, most specifically the 4 dimensional equivalents of the Johnson solids Naraht ( talk) 18:39, 19 February 2016 (UTC) reply
Would those require regular faces, or regular cells? Double sharp ( talk) 14:06, 22 February 2016 (UTC) reply
From Wikipedia, the free encyclopedia
Mathematics desk
< February 16 << Jan | February | Mar >> February 18 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


February 17 Information

Distances on the 5-cell?

For a 5-cell with unit length edges, what are the following distances?

  • Center of the 5-cell to vertex?
  • Vertex to center of the opposite tetrahedron?
  • center of one edge to the center of the opposite triangle?

And for the lines for numbers 2 & 3, which one has its middle closer to the center of the 5-cell? My intuition says #3, but I'm not sure. Naraht ( talk) 14:48, 17 February 2016 (UTC) reply

Coordinatize the vertices in 5 dimensions as (00001), (00010), etc., (ok, rescale by sqrt(2) if you really want unit edges) and everything is easy. -- JBL ( talk) 15:14, 17 February 2016 (UTC) reply
For example, if you were working with the triangle in three dimensions, the three vertices are (001), (010) and (100) (all lying on the plane x+y+z=1). If you want to know the centroid of a face, you take the average of its vertices, so e.g. the midpoints of the edges are (1/2, 1/2, 0) and permutations. If you want to compute distances, you use the usual distance formula. This coordinatization is very easy to work with because it's symmetric and rational. -- JBL ( talk) 16:15, 17 February 2016 (UTC) reply
So the distance along the line from a vertex to the center of the opposite tetrahedron (vtet) would be the distance between (1,0,0,0,0) and (0,1/4,1/4,1/4,1/4) and the distance along a line from the center of the edge to the opposite triangle (etri) would be the distance from (1/2,1/2,0,0,0) to (0,0,1/3,1/3,1/3), right? So sqrt(5/4) vs. sqrt(5/6) for which obviously the first is longer as expected. The center of 5 cell (c5c) is at (1/5,1/5,1/5,1/5,1/5) the center of the vtet (cvtet) would be at (1/2,1/8,1/8,1/8,1/8) and the center of the etri (cetri) would be (1/4,1/4,1/6,1/6,1/6). Since every coordinate of cetri is closer to the c5c than those of cvtet, then it certainly is closer. (I started calculating the distances before I realized this). Fun to walk *up* a dimension to make everything easier. :) Naraht ( talk) 16:51, 17 February 2016 (UTC) reply
Yes, that looks right to me. It is perhaps also worth noting that the centroid of a regular n-simplex is much closer to the nearest facet than it is to the nearest vertex (and more so as n grows): this is clear in your example. And more generally it is closer to any given i-dimensional face than it is to any j-dimensional face if i > j. -- JBL ( talk) 21:54, 17 February 2016 (UTC) reply
Yup, and very generalizable...
For convenience, I turn to Coxeter's Regular Polytopes, Table I(ii). He uses edge length 2, so I divide each of his numbers by 2. From the centre to a vertex: √(2/5). From the centre to an edge: √(3/20). From the centre to a face: 1/√15 (at least I think that's what it says, the typesetting is a bit odd). From the centre to a cell: 1/√40. So your second number is √(2/5) + 1/√40 = √(5/8), and your third is √(3/20) + 1/√15 = √(5/12), if I haven't fumbled. Funny how each of Coxeter's numbers has a factor of √5 in the denominator and each of these sums has √5 in the numerator! — Tamfang ( talk) 04:16, 18 February 2016 (UTC) reply
Is it possible to talk about this topic without bringing in Coxeter's masterwork? :) Naraht ( talk) 15:26, 18 February 2016 (UTC) reply

R^(lsd) Tree

I know what an R tree is. I'm reading about Rlsd Trees with no definition of what lsd is an abbreviation for. Is that a common type of R tree? If so, how does it differ from a standard R tree or what defines the subset of R trees? 209.149.113.11 ( talk) 20:14, 17 February 2016 (UTC) reply

As near as I can tell the LSD stands for "local split decision". This is a rather specialized form of database indexing and this question might get a better response on the computing desk rather than here. Though if it was me I'd probably go straight to stack-overflow. -- RDBury ( talk) 23:02, 17 February 2016 (UTC) reply
Oh, gee. I always thought it stands for Lysergic acid diethylamide. (Sorry, someone had to make that joke, may as well be me...) -- Meni Rosenfeld ( talk) 14:08, 18 February 2016 (UTC) reply
I've always wondered whether certain parts of mathematics would be easier on various drugs, LSD seems appropriate for dealing with polytopes, most specifically the 4 dimensional equivalents of the Johnson solids Naraht ( talk) 18:39, 19 February 2016 (UTC) reply
Would those require regular faces, or regular cells? Double sharp ( talk) 14:06, 22 February 2016 (UTC) reply

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