Timeline for Independence number of a grid like graph
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2023 at 12:26 | comment | added | Fedor Petrov | For $d=2$ and $n=(k+1)q+r$ you can easily partition the grid without $r\times r$ square by $(n^2-r^2)/(k+1)$ $1 \times (k+1)$ rectangles, and cover the remaining $r\times r$ square by $r$ such rectangles, this gives the upper bound $r+(n^2-r^2)/(k+1)$. It is achieved if you consider the points with prescribed sum of coordinates modulo $k+1$, the value of this remainder is choose so that the whole diagonal of the remaining $r\times r$ square is chosen | |
| Jan 26, 2023 at 11:36 | history | edited | Pritam Majumder | CC BY-SA 4.0 |
added 1 character in body
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| Jan 26, 2023 at 10:49 | history | asked | Pritam Majumder | CC BY-SA 4.0 |