Timeline for Integer decomposition of dilated integral polytopes
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2019 at 0:59 | answer | added | David Handelman | timeline score: 3 | |
| May 17, 2015 at 15:27 | vote | accept | hypercube | ||
| May 16, 2015 at 19:08 | comment | added | Francisco Santos |
@JonMarkPerry: that the RHS is contained in the LHS is obvious, but it is not obvious that every point in the LHS decomposes as required in the RHS. Consider the following example. Let $P$ be the regular simplex inscribed in the unit cube, with vertices (0,0,0), (1,1,0), (1,0,1) and (0,1,1). The point $(1,1,1)$ is in $2P$ but it cannot be decomposed as the sum of two integer points in $P$. This is usually expressed as $P$ is not integrally closed or $P$ does not have the IDP property.
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| May 16, 2015 at 19:07 | answer | added | Francisco Santos | timeline score: 7 | |
| Apr 29, 2015 at 17:03 | comment | added | Per Alexandersson | I have worked a bit with IDP and in my experience, intuition doesn't really work. In dimension 2, all integer polytopes have IDP; counterexamples live in higher dimensions, as mentioned. | |
| Apr 29, 2015 at 16:59 | comment | added | JMP | by sort of 'linear extension' it looks pretty true. e.g. (n+2)(x,y,z)+(d-(n+2))(x,y,z) is going to equal d(x,y,z), which needs a little bit of proving. 1-on-1 integer mapping seems to fail however, and I am more concerned with cases such as (a,b,c)+(d,e,f)=(g,h,i) | |
| Apr 29, 2015 at 16:56 | comment | added | hypercube | @JonMarkPerry Is it trivially obvious that it holds or doesn't? Can you elaborate a little more please? | |
| Apr 29, 2015 at 16:54 | comment | added | hypercube | @PerAlexandersson I completely agree. If $P$ has IDP or is integrally closed then the decomposition is possible. Are there conditions when this happens? Is there a construction for the decomposition? What if it doesn't have IDP? I've looked here and they prove that $kP$ will have IDP for all $k \geq n - 1$, but I don't think that's enough to prove this. They also provide an example where $2P\cap\mathbb{Z}^n \neq P\cap\mathbb{Z}^n + P\cap\mathbb{Z}^n$, but for $P$ with $\dim(P) \geq 7$. Is there an easy counterexample in my case? | |
| Apr 29, 2015 at 4:43 | comment | added | JMP | is this not trivially obvious? | |
| Apr 28, 2015 at 20:32 | comment | added | Per Alexandersson | This is closely related to the integer decomposition property (IDP), or being "integrally closed" or being "a normal polytope". | |
| Apr 28, 2015 at 20:31 | review | First posts | |||
| Apr 28, 2015 at 20:32 | |||||
| Apr 28, 2015 at 20:27 | history | asked | hypercube | CC BY-SA 3.0 |