Talk:Hypercube

Does anybody know how to transpose this table for better display when page is enlarged?

Hypercube elements $E_{m,n}\!$ (sequence A038207 in the OEIS)

			m	0	1	2	3	4	5	6	7	8	9	10
n	n-cube	Names	Schläfli Coxeter	Vertex 0-face	Edge 1-face	Face 2-face	Cell 3-face	4-face	5-face	6-face	7-face	8-face	9-face	10-face
0	0-cube	Point Monon	( )	1
1	1-cube	Line segment Ditel	{}	2	1
2	2-cube	Square Tetragon	{4}	4	4	1
3	3-cube	Cube Hexahedron	{4,3}	8	12	6	1
4	4-cube	Tesseract Octachoron	{4,3,3}	16	32	24	8	1
5	5-cube	Penteract Deca-5-tope	{4,3,3,3}	32	80	80	40	10	1
6	6-cube	Hexeract Dodeca-6-tope	{4,3,3,3,3}	64	192	240	160	60	12	1
7	7-cube	Hepteract Tetradeca-7-tope	{4,3,3,3,3,3}	128	448	672	560	280	84	14	1
8	8-cube	Octeract Hexadeca-8-tope	{4,3,3,3,3,3,3}	256	1024	1792	1792	1120	448	112	16	1
9	9-cube	Enneract Octadeca-9-tope	{4,3,3,3,3,3,3,3}	512	2304	4608	5376	4032	2016	672	144	18	1
10	10-cube	Dekeract Icosa-10-tope	{4,3,3,3,3,3,3,3,3}	1024	5120	11520	15360	13440	8064	3360	960	180	20	1

Tesseract

Why is this article different from that of the Tesseract? It feels like roughly the same thing, at least close enough to be merged. Language Boi (talk) 20:33, 22 December 2023 (UTC)

That one is only about four-dimensional things. This one is about the generalization to any number of dimensions. Four dimensions is still low enough that the regular polytopes there are special — in five or more dimensions there are always exactly three regular polytopes (the hypercube being one of them) but just as three dimensions has the five Platonic solids, four dimensions has six regular polytopes. So I think that as part of this set it has enough independent notability as a mathematical topic to justify a separate article from the general hypercube article. And certainly the same is true for the section on cultural uses. —David Eppstein (talk) 20:37, 22 December 2023 (UTC)

Ah, I see. That should probably be clarified at the top of the article though--because I would probably use the two words interchangeably. Whether that's correct or not I don't know, but I know it's common. Language Boi (talk) 21:21, 23 December 2023 (UTC)

You mean like the clarification that already exists in the hatnote at the top of the article? Or the one already in the caption of the figure at the top of the article? —David Eppstein (talk) 22:39, 23 December 2023 (UTC)

Text

@David Eppstein: "Increasing axis vectors" means exactly what it means, increasing the amount of axes in any dimension. Examples include x axis, y axis, z axis, w axis, v axis, etc... Adding a coordinate axis does increases the vertices of a hypercube by a multiplication of two. Moreover the text is useful because it links to 4-cube, 5-cube and so on and makes clear the relationship between the number of vertices and the hypercube. Lebesgue measure is not irrelevant, as the text is in specific relationship to a hyper-volume, it's not WP:SUBMARINE because it's directly linked to the text. 21:04, 17 May 2024 (UTC) Des Vallee (talk) 21:04, 17 May 2024 (UTC)

You do know that squares and cubes do not need to be aligned to the axes of any particular coordinate system, right? Perhaps you should also know that providing formulas for the number of faces of different dimensions, including vertices, is already done in the "faces" section, and that your attempt to add a very partial piece of that information counting only the vertices is misplaced in a section about coordinates. —David Eppstein (talk) 21:14, 17 May 2024 (UTC)

More natural viewpoint

The section Faces contains this fragment:

"The number of the $m$-dimensional hypercubes (just referred to as $m$-cubes from here on) contained in the boundary of an $n$-cube is

$E_{m,n}=2^{n-m}{n \choose m}$, where ${n \choose m}={\frac {n!}{m!\,(n-m)!}}$ and $n!$ denotes the factorial of $n$."

But there is no good reason to limit the counted faces to the boundary.

The n-cube is a perfectly fine polytope, and it has exactly one additional face beyond those on the boundary: its single n-dimensional face.

What's more, this corresponds to the case above where m = n, and it is easy to see that the very same formula $2^{n-m}{n \choose m}$ is then equal to 1, the correct count.