Timeline for Higher dimensional Rubik's cube group
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2022 at 18:41 | comment | added | Stefan Kohl♦ | @peawormsworth Well -- the colored "squares" here are cubes. So, there are 4 of them around each corner. | |
| Sep 5, 2022 at 9:14 | comment | added | peawormsworth | I think it should be "Each corner stone has 6 visible colored squares". Where you said "4" Because there are 4 cubes around each corner. Cubes corners each have 3 colored tiles. So 4 cube corners attached will have 4*3 = 12 colored tiles. But the cubes are flush against each other, so each tile will be paired with another. So there are really just 12/2 = 6 tiles around each vertex of the tesseract rubik's cube. | |
| Nov 9, 2013 at 12:58 | comment | added | Jason Pioneer | Oh thats great! Where can i find more information on the number of $k$-faces of a $n$-dimensional cube? The number of corners and edges is obvious. | |
| Nov 9, 2013 at 0:34 | comment | added | Logan M | @JasonPioneer That seems correct to me. The full expression should be (assuming I haven't made a typo) $$\displaystyle (A_n \wr S_{2^n}) \times \left(\prod_{i=2}^{n-1} S_i \wr S_{{n \choose i} 2^{i}} \right).$$ | |
| Nov 9, 2013 at 0:07 | comment | added | Jason Pioneer | So i think the first factor of this group in the n-dimensional case will be $A_n \wr S_{2^n}$. | |
| Nov 8, 2013 at 23:58 | comment | added | Stefan Kohl♦ | @LoganMaingi: Sure. -- Thanks for spotting this! | |
| Nov 8, 2013 at 23:57 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Meant A_4 and S_3, not C_4 and C_3
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| Nov 8, 2013 at 19:33 | comment | added | Logan M | I agree with your argument, but if by $C_n$ in your final formula you mean the cyclic group of order $n$, then I don't see exactly where that is coming from. Naively I would think that for the corners the group should be $A_4$, and for the cubies with 3 stickers it should be $S_3$. More generally I'd expect for a $d$-dimensional cube that the corners correspond to $A_d$ and all other types of pieces correspond to $S_k$ for $k=2,\ldots,d-1$. Could you explain how you get that (or if I'm misunderstanding your claim)? | |
| Nov 7, 2013 at 19:17 | vote | accept | Jason Pioneer | ||
| Nov 7, 2013 at 19:12 | comment | added | Jason Pioneer | Hm im pretty happy with that answer so far,because you have the calculation of the stuff i had a problem imagining. | |
| Nov 7, 2013 at 10:37 | history | answered | Stefan Kohl♦ | CC BY-SA 3.0 |