Timeline for Tessellating $\mathbb{R}^n$ by bricks.
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2012 at 21:56 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
added 501 characters in body
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| Jul 12, 2012 at 21:31 | comment | added | Gerhard Paseman | I am starting to see that I can't just use bricks with each side of length 2. However, it may be possible to use a combination of cubes of side 2 and half-cubes in alternating layers. I will leave my answer up and edit it when I get a better one. Gerhard "Back To The Brick Pile" Paseman, 2012.07.12 | |
| Jul 12, 2012 at 21:19 | comment | added | Tapio Rajala | By looking at a point which is in $n+1$ bricks one sees that one of those neighboring bricks must have size at least $2n-1$. For $n = 2$ this agrees with the upper bound $2^n-1$. What happens for $n=3$? | |
| Jul 12, 2012 at 21:17 | comment | added | Will Sawin | In particular this bound is exactly $2^s-1$, which as far as we know is exact for all $s$. | |
| Jul 12, 2012 at 21:16 | comment | added | Gerhard Paseman | OK. Can I use 2x2 bricks and still get a linear upper bound eventually? Gerhard "Does Not Fit? Force It!" Paseman, 2012.07.12 | |
| Jul 12, 2012 at 21:01 | comment | added | user6976 | Every point belongs to exactly one brick or it is on the boundary of a brick. I do not think it is possible to use 1 by 1 bricks to tile the plane as required. Will is right: in order to get $2^n$: double the previous tessellation (rescale it by 2), then divide $\mathbb{R}^{n+1}$ by layers of thickness 1, each of the layers tessellate as $\mathbb{R}^{n}$ (the doubled tessellation multiplied by the unit interval), then shift the tessellation in odd layers by a half of the brick. The question is whether it is possible to get better bound on the sizes of bricks. | |
| Jul 12, 2012 at 21:01 | comment | added | Gerhard Paseman | I may have the wrong picture, but since he is summing lengths, I don't see why the bricks have to have a side longer than 2, so s(n) should still be of order O(n). Anyway, I have faith in Mark's not keeping quiet if I am wrong. Gerhard "Or If I Am Right" Paseman, 2012.07.12 | |
| Jul 12, 2012 at 20:49 | comment | added | Will Sawin | Isn't this just a proof of finiteness of $s(n)$? I am not sure but I think that thinking carefully about the scale factor here is how you get the $2^n$ upper bound. I think it's not too hard to show that the optimal laminated lattice-style construction has $s(n)=O(2^n)$, because every time you go up a dimension you cut the shortest path, at best, in half. | |
| Jul 12, 2012 at 20:45 | comment | added | Gerhard Paseman | Implicit in my understanding is that every point in R^n belongs to at least one brick. If this is not the case, I recommend "decorate regularly" instead of "tessellate". Gerhard "Ask Me About System Design" Paseman, 2012.07.12 | |
| Jul 12, 2012 at 20:36 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |