Timeline for Finding integer points inside of a parallelogram
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2013 at 12:41 | answer | added | Alexey Ustinov | timeline score: 1 | |
| Aug 17, 2013 at 9:57 | comment | added | Ricardo Andrade | @Noam and Eric: If Noam Elkies' comment answers the question, may I suggest that Noam repost it as an answer? | |
| Aug 17, 2013 at 9:44 | history | edited | Eric Tressler | CC BY-SA 3.0 |
deleted 266 characters in body
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| Aug 17, 2013 at 6:08 | comment | added | Eric Tressler | @Noam Elkies: Thank you; I wasn't sure what terms to look for, but that seems like enough to solve my problem. | |
| Aug 17, 2013 at 6:03 | comment | added | Noam D. Elkies | This is a typical setting for lattice basis reduction (which in 2D is closely related to the Euclidean algorithm applied to $1$ and the slope of the parallelogram's longer edge, and to that slope's continued-fraction development). Apply a linear transformation $T$ taking the parallelogram to a unit square $S$, and ${\bf Z}^2$ to some lattice $L$. Find a reduced basis for $L$. It is then easy (assuming no precision issues with a point of $L$ coming very close to the boundary of $S$) to decide whether $S \cap L$ contains some point $p$, and if yes to find such $p$. Then recover $T^{-1}p$. | |
| Aug 17, 2013 at 6:02 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
removed deprecated tag 'geometry' and inapplicable tag 'diophantine-approximation'; added relevant tags; added link to original post on math.stackexchange
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| Aug 17, 2013 at 5:20 | history | asked | Eric Tressler | CC BY-SA 3.0 |