Timeline for Maximum number of distinct diagonals generated by permutations
Current License: CC BY-SA 3.0
13 events
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| Jan 18, 2017 at 13:02 | history | edited | Tony Huynh | CC BY-SA 3.0 |
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| Aug 10, 2010 at 12:17 | comment | added | Fedor Petrov | just a minor remark, the same without graphs: if two rows (say i-th and j-th) are equal to $r$, then each diagonal has at least two coincidences with $r$ (coincidence means coincidence of one of $n$ coordinate functionals). So, there are at least $n$ forbidden diagonals. If all rows are different, then all their complements are forbidden and we again have $n$ forbidden diagonals. | |
| Aug 10, 2010 at 8:30 | history | edited | Tony Huynh | CC BY-SA 2.5 |
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| Aug 10, 2010 at 7:37 | comment | added | Tony Huynh | Thanks Tsuyoshi! I edited the proof to handle that case separately. | |
| Aug 10, 2010 at 7:35 | history | edited | Tony Huynh | CC BY-SA 2.5 |
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| Aug 10, 2010 at 1:03 | comment | added | Tsuyoshi Ito | A clever argument! One minor comment: I do not think that the current proof of the lemma deals with the case where more than one vertex in [n]_c have exactly the same set of neighbors, but it can be fixed easily. | |
| Aug 10, 2010 at 0:30 | history | edited | Tony Huynh | CC BY-SA 2.5 |
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| Aug 10, 2010 at 0:24 | history | edited | Tony Huynh | CC BY-SA 2.5 |
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| Aug 9, 2010 at 12:37 | history | edited | Tony Huynh | CC BY-SA 2.5 |
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| Aug 9, 2010 at 12:24 | comment | added | Tony Huynh | I suppose that I cannot generate diagonals that have exactly $n-1$ ones in this way, since if I fix $n-1$ rows, then I automatically fix the last one. But it seems that I can still get $2^n-n$ diagonals. | |
| Aug 9, 2010 at 12:23 | comment | added | Robin Chapman | Not with $n=3$ you can't :-) | |
| Aug 9, 2010 at 12:20 | comment | added | damiano | This seems to have a slight problem with the case of the configurations with exactly one zero. | |
| Aug 9, 2010 at 12:18 | history | answered | Tony Huynh | CC BY-SA 2.5 |