Wikipedia:Reference desk/Archives/Mathematics/2013 February 28

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February 28

4 dimensional equivalents of the Johnson solids?

Is there a standard definition as to what would be the equivalent polychora to the Johnson solids? Would it be

1. Each of the polyhedra that make up the polychoron have to be regular such as a "hyper pyramid with a base of the octahedra" and a "hyper bi-pyramid with a base of a icosahedron" and

2. Each of the polyhedra has to be uniform?

3. Each of the polyhedra has to be a Johnson solid. I believe this is equivalent to requiring that all 2 dimensional faces have to be regular.Naraht (talk) 19:01, 28 February 2013 (UTC)

George Olshevsky uses #3. I'm not sure about other people. Double sharp (talk) 17:39, 2 March 2013 (UTC)

Well, if he does #3, then its at least in the argument.Naraht (talk) 11:45, 4 March 2013 (UTC)

My mind is fuzzy now, but I'd think a 4D Johnson polychora would be strictly convex, with all cells as convex uniform polyhedra or Johnson solids, but excluding all the convex uniform polychora. Tom Ruen (talk) 18:38, 2 March 2013 (UTC)

I was taking Convex as a given. I'm not sure what would be added to the Johnson solids if convex was not one of the characteristics, the Stellated octahedron is used in the article, but it really doesn't fit to me as part of the non-convex extension of the Johnson solids since all the faces are the same.

What's wrong with that? The irregular deltahedra also have all faces the same, but are still Johnson solids. Double sharp (talk) 12:05, 5 March 2013 (UTC)

Ah, like the Gyroelongated square bipyramid, still the Stellated ocahedron has all of its vertices identical, which by definition the Johnson Solids don't (J37 issues not withstanding)Naraht (talk) 18:01, 6 March 2013 (UTC)

It depends on what you mean by "Stellated octahedron". I think that while convex is not a characteristic of your "extended Johnson solids", the absence of self-intersections is. Then the stellated octahedron is a concave surface where some vertices have 3 triangles, others 8 (like a triakis octahedron with very tall pyramids). If you mean the actual stellated octahedron, that's a regular compound (and so wouldn't be an extended Johnson solid, even if you allow self-intersections). Double sharp (talk) 12:22, 7 March 2013 (UTC)

By Stellated Octohedron, I mean the "full one", not the Triakis. Which is why the use of Stellated Octahedron on the Johnson Solid page seems a little odd. If all that is to be avoided is convexity, perhaps another should be chosen there.Naraht (talk) 14:32, 7 March 2013 (UTC)

The broadest reasonable definition is "all 2-faces regular". That set includes (among others) the subsets "all cells Johnson" and "all cells uniform", and the latter includes "all cells regular". Seems to me we need at least four names, unless one or more of these sets is empty. —Tamfang (talk) 20:17, 17 November 2013 (UTC)

@Naraht You might want to take a look at http://eusebeia.dyndns.org/4d/crf. -- Toshio Yamaguchi 14:13, 6 March 2013 (UTC)

Thanx, I was partially wondering if his crf constructions represented the general opinion in the mathematical community.Naraht (talk)

I am not sure about that. I think a lot of the research in this area is carried out at a number of fora or mailing lists and people such as Wendy Krieger, George Olshevsky and Jonathan Bowers, who all seem to be at the forefront of the research on Polytopes seem to hardly publish their results in the traditional sense (for example in a peer-reviewed journal). If you define the mathematical community as those who publish their results in journals like Mathematics of Computation or similar journals, then I think the results of the people I mentioned are not settled inside the mathematical community. -- Toshio Yamaguchi 22:23, 6 March 2013 (UTC)

OK, thank you, I'll keep that in mind. Given that right now in the CRF they seem to be working on enumeration, I think we may be quite far from the equivalent to Zalgaller's work proving that the list of Enumerated Johnson Solids was complete. Polytopes strikes me as an area of mathematics where good work can be done without being in the realm of journals.Naraht (talk) 02:42, 7 March 2013 (UTC)

Percent chance for each individual

It seems to me the answer to the question is yes, but I've found that reality sometimes doesn't correspond to my intuition. If there are ten people in a group and six are going to be chosen at random all at the same moment, doesn't each individual have a 60% chance of being chosen? 67.163.109.173 (talk) 23:06, 28 February 2013 (UTC)

Yes. Looie496 (talk) 23:32, 28 February 2013 (UTC)

Should the six out of ten be chosen sequentially, initially all ten have a 60% chance of being chosen, then after the first choice the remaining nine each have a 5/9 chance of being chosen, and so on until finally the remaining four have zero chance of being chosen. But if all six are chosen at the same time, then obvious symmetry gives each the same chance of 60%.←86.186.142.172 (talk) 10:45, 1 March 2013 (UTC)

You might think the chance of being chosen first = 1/10, 2nd = 1/9, 3rd = 1/8, 4th = 1/7, 5th = 1/6, 6th = 1/5. However, for a given person, you only do the second draw if they weren't chosen the first time, so you multiply that 1/9 by 9/10 to get a 9/90, or 1/10th chance they will be chosen on the second draw, just like the first. This applies to all the draws, so we get 6×(1/10), or 6/10 chance they will be picked, in total, regardless of whether they are drawn at once or not. (If you don't make the adjustment I describe, then a person can be chosen twice or more, and the "chance of being chosen" reflects this, coming out around 0.845635. This is the chance of being chosen once, plus 2 times the chance of being chosen twice, plus three times the chance of being chosen thrice, etc. In fact, if you were to do one additional draw, this number comes out greater than 1.) StuRat (talk) 16:05, 1 March 2013 (UTC)

To clarify my post, I meant that "after the first choice the remaining nine each have a 5/9 chance of being chosen sometime in the subsequent process", and so on. ←86.186.142.172 (talk) 16:41, 1 March 2013 (UTC)

OK, have we answered this Q to your satisfaction ? StuRat (talk) 04:14, 2 March 2013 (UTC)

Thanks.

Resolved

67.163.109.173 (talk) 14:07, 2 March 2013 (UTC)