Combinatorial cube packings in cube and torus
| dc.creator | Sikirić, Mathieu Dutour | |
| dc.creator | Itoh, Yoshiaki | |
| dc.date | 2008-05-16 | |
| dc.date | 2008-09-24 | |
| dc.date.accessioned | 2026-07-07T10:04:31Z | |
| dc.date.available | 2026-07-07T10:04:31Z | |
| dc.description | We consider sequential random packing of cubes $z+[0,1]^n$ with $z\in \frac{1}{N}\ZZ^n$ into the cube $[0,2]^n$ and the torus $\QuotS{\RR^n}{2\ZZ^n}$ as $N\to\infty$. In the cube case $[0,2]^n$ as $N\to\infty$ the random cube packings thus obtained are reduced to a single cube with probability $1-O(\frac{1}{N})$. In the torus case the situation is different: for $n\leq 2$, sequential random cube packing yields cube tilings, but for $n\geq 3$ with strictly positive probability, one obtains non-extensible cube packings. So, we introduce the notion of combinatorial cube packing, which instead of depending on $N$ depend on some parameters. We use use them to derive an expansion of the packing density in powers of $\frac{1}{N}$. The explicit computation is done in the cube case. In the torus case, the situation is more complicate and we restrict ourselves to the case $N\to\infty$ of strictly positive probability. We prove the following results for torus combinatorial cube packings: We give a general Cartesian product construction. We prove that the number of parameters is at least $\frac{n(n+1)}{2}$ and we conjecture it to be at most $2^n-1$. We prove that cube packings with at least $2^n-3$ cubes are extensible. We find the minimal number of cubes in non-extensible cube packings for $n$ odd and $n\leq 6$. | |
| dc.description | 21 pages, 3 figures and 3 tables | |
| dc.identifier | https://arxiv.org/abs/0805.2493 | |
| dc.identifier | http://arxiv.org/abs/0805.2493 | |
| dc.identifier.uri | http://salesiana.dossiersoluciones.com/handle/123456789/169707 | |
| dc.subject | Combinatorics | |
| dc.subject | Metric Geometry | |
| dc.title | Combinatorial cube packings in cube and torus | |
| dc.type | text |