Timeline for Fill the board with zeroes, inverting the intersections of rows and columns
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2018 at 7:39 | comment | added | Brendan McKay | @IvanIzmestiev I meant to write $n^2-1$, but I'm sure you are correct about $(n-1)^2$. | |
| Feb 3, 2018 at 8:26 | comment | added | Ivan Izmestiev | I think, the rank is $(n-1)^2$ in the odd case. First, the corank is at least $2n-1$ because of those dependences (the sum of the matrix rows corresponding to any row or column of the board is a row consisting of ones). On the other hand, inside an $(n-1) \times (n-1)$ subboard we can recolor anything we want, so the rank is at least $(n-1)^2$. | |
| Feb 3, 2018 at 3:39 | comment | added | Brendan McKay | @IvanIzmestiev Is the rank $n-1$ in the odd case? | |
| Feb 2, 2018 at 13:06 | comment | added | Ivan Izmestiev | And here is a recoloring of the left upper cell in the case of an even $n$: $(n,1) + (n-1,1) + \cdots + (1,1) + (1,2) + \cdots + (1,n)$. | |
| Feb 2, 2018 at 13:02 | comment | added | Ivan Izmestiev | For $n$ odd the matrix is singular, because of the following identities between its rows: $(1,1) + (1,2) + \cdots + (1,n) = (2,1) + (2,2) + \cdots + (2,n)$. In the language of recolorings: if we recolor all "crosses" whose centers lie in the same row, then we recolor all cells (whatever row we choose). | |
| Feb 2, 2018 at 12:01 | history | answered | Brendan McKay | CC BY-SA 3.0 |