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February 17

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 R^lsd 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]