Timeline for Area of a lattice polygon in terms of its width
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 2, 2013 at 0:32 | comment | added | Nikita Kalinin | @Ilya: it is not a simple problem, my solution is not true. | |
| Apr 1, 2013 at 23:50 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
added 60 characters in body
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| Mar 30, 2013 at 12:44 | comment | added | Nikita Kalinin | @Ilya: I had started this bounty before I realized that it is a simple problem, and I was very nervous =)) Now I can not cancel it. Concerning the first comment: an affine transformation preserves the lattice width, it is enough for me. | |
| Mar 30, 2013 at 8:35 | comment | added | Ilya Bogdanov | In fact, this bound is tight. Look at the triangle wth vertices $(0,0)$, $(2,1)$ and $(1,2)$. Its area is $3/2$, while its "integral width" is 2. To see that, first notice that all its altitude lengths are greater than 1, hence it is enough to check only the vectors of length less than 2. This check is straightforward. So, what is this bounty for? | |
| Mar 30, 2013 at 8:32 | comment | added | Ilya Bogdanov | You need to be a bit more careful, since the affine transforms do not preserve scalar product. So in fact you need to put the vector orthogonal to $v$ into $(-1,1)$. | |
| Mar 29, 2013 at 17:26 | history | answered | Nikita Kalinin | CC BY-SA 3.0 |