Timeline for How many matrices are possible for the given arrangement?
Current License: CC BY-SA 3.0
6 events
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| Nov 11, 2012 at 16:48 | comment | added | Yuichiro Fujiwara | This is kind of irrelevant now that we know OP meant avoiding constant submatrices of adjacent rows and columns, but you can get a $3 \times 5$ matrix A avoiding constant $2 \times 2$ (general) submatrices by choosing $(1,0,0,0,1)$ as the third row; this choice is ok because your argument doesn't require the third row to have three $1$'s. | |
| Nov 11, 2012 at 16:13 | comment | added | verret | I assumed that you meant any sub-matrix, but now I see that you meant only submatrices consisting of adjacent rows and columns. Maybe you should clarify your question. | |
| Nov 11, 2012 at 16:11 | comment | added | jigsawmnc | What about the case when m = 5 & n = 5 and the matrix is: 0 1 0 1 0 | 0 1 0 1 0 | 0 1 0 1 0 | 0 1 0 1 0 | 0 0 1 1 0 The above is a valid case. | |
| Nov 11, 2012 at 16:11 | history | edited | verret | CC BY-SA 3.0 |
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| Nov 11, 2012 at 15:59 | history | edited | verret | CC BY-SA 3.0 |
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| Nov 11, 2012 at 15:49 | history | answered | verret | CC BY-SA 3.0 |