Timeline for Combinatorial Problem
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 2, 2011 at 0:54 | vote | accept | sks | ||
| Nov 1, 2011 at 23:50 | comment | added | fedja | Going over the perimeter from $(0,0)$, so that each vertex will require just one multiplication, 2 additions and one comparison, of course, not recomputing the full products, which would be $n^2$. | |
| Nov 1, 2011 at 23:29 | comment | added | fedja | Just try the $n$ vertices one by one! | |
| Nov 1, 2011 at 22:37 | history | answered | fedja | CC BY-SA 3.0 |