Timeline for Number of turning points on a nondecreasing $n^2 \times n^2$ matrix
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2022 at 14:36 | comment | added | Wolfgang | Thanks, corrected. | |
| Jul 31, 2022 at 14:36 | history | edited | Wolfgang | CC BY-SA 4.0 |
corrected.
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| Jul 31, 2022 at 12:23 | comment | added | Gerry Myerson | Should be $\displaystyle n^3-n^2-{n(n-1)\over2}$. | |
| Jul 31, 2022 at 11:25 | history | edited | Wolfgang | CC BY-SA 4.0 |
corrected formula and final lines of matrix. But still O(n^3).
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| Jan 12, 2016 at 7:28 | comment | added | Wolfgang | For each $i=2,...,n-1$ there is exactly one antidiagonal $D_i$ with only $i$'s on it. For each $D_i$, and for the $D_1$ just above $D_2$, all its elements except the rightmost one are turning points. And this is best possible, as for any $M$ each column except the rightmost one can have at most $n-1$ turning points. | |
| Jan 12, 2016 at 5:25 | comment | added | Qin Jianbin | I don't think this answer works. @Wolfgang, would you explain it a bit more? | |
| Jan 11, 2016 at 13:26 | comment | added | Wolfgang | @EmanueleTron Does that work for you? For me, it doesn't. | |
| Jan 11, 2016 at 13:18 | comment | added | user41593 |
Do you mean \udots?
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| Jan 11, 2016 at 13:00 | history | answered | Wolfgang | CC BY-SA 3.0 |