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

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.

Incorrect Orthogonal Projections

  It is my belief that the orthogonal projections of the section titled Generalized Hypercubes is incorrect for all shapes where p>2. My reasoning is that an equidistant shape should have vertices consisting of exactly two connected line segments for γ₂, each vertex of γ₃ should have exactly 3, ..., γ_x should have exactly x. Instead, all shapes where p>2 have either double those when p=2, or some other inconsistent/increasing amount.

The pattern that should be followed is the expansion of pascal's triangle from binomial (for simplex at p=1), to the total number of elements being equal to powers of 3 when p=2 (as seen in the graph for the section titled Faces), and following the same pattern of (p+1)^x for all higher dimensional shapes where γ=x. (It should be noted that the simplex is currently NOT recognized as a non-dimensional point whose elements would complete the binomial coefficients on pascal's triangle.)

Following this pattern, the maximum number of outer points of the orthogonal projection seem to be equal to n-1 faces, and the total number of vertices (having p connected line segments each) should be equal to pⁿ, or the value when m=0 for any given n-cube. Also, the number of nth elements is equal to the nth column of the table for exponential coefficients, so the number of edges for the shape γ₂ when p=3 should be 6 and not 18.

After being corrected, the shape for γ₂ when p=3 should project a shape similar to a half-hexagon folded into a hollow tetrahedron without a bottom face. This would give the shape the required 9 vertices (3 per plane with each consisting of 2 line segments) while maintaining the 6 required edges for its 1 flat face existing in (p=3)-dimensional space.

If someone is able to better explain this, or knows a proof, it would be greatly appreciated. Webdnd (talk) 23:23, 20 September 2024 (UTC)