Timeline for How many matrices are possible for the given arrangement?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2012 at 13:56 | answer | added | jigsawmnc | timeline score: 0 | |
| Nov 12, 2012 at 10:32 | vote | accept | jigsawmnc | ||
| Nov 11, 2012 at 19:38 | comment | added | jigsawmnc | @Pietro Majer: Contiguous means that the rows and columns of the chosen sub-matrix should be adjascent. | |
| Nov 11, 2012 at 17:52 | comment | added | Pietro Majer | Could you clarify the term "contiguous sub-matrix"? | |
| Nov 11, 2012 at 17:48 | answer | added | Per Alexandersson | timeline score: 5 | |
| Nov 11, 2012 at 16:15 | history | edited | jigsawmnc | CC BY-SA 3.0 |
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| Nov 11, 2012 at 15:56 | comment | added | jigsawmnc | @Per Alexandersson: It is evident that we'll have (m - 1) * (n - 1) sub-matrices of order 2 * 2. My strategy is: start picking up 2 * 2 sub-matrices starting from top-left corner i.e. containing cells A(0,0), A(0,1), A(1,0), A(1,1) of main matrix; fill it with 0 and count the total number of matrices having these cells filled in this manner. Now, I proceed on to counting those matrices where the next 2 * 2 matrix i.e. A(0,1), A(0,2), A(1,1), A(1,2) are filled with 0. However, I am not able to count the repeating cases of 2nd matrix which already appeared in the first case. How to achieve this? | |
| Nov 11, 2012 at 15:49 | answer | added | verret | timeline score: 0 | |
| Nov 11, 2012 at 15:20 | comment | added | Per Alexandersson | It suffices that your matrix avoids $2 \times 2$-submatrices with all entries equal, since if you have a forbidden $3\times 3$-matrix or larger, it will automatically also contain a forbidden $2 \times 2$-matrix. Is it possible to count these using recursion maybe? | |
| Nov 11, 2012 at 14:15 | history | edited | jigsawmnc | CC BY-SA 3.0 |
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| Nov 11, 2012 at 14:04 | history | asked | jigsawmnc | CC BY-SA 3.0 |