Timeline for Is there always a maximum anti-rectangle with a corner square?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2013 at 1:01 | comment | added | Joseph O'Rourke | @GerryMyerson: $\{a,b,c,1,5,12\}$ is not an antirectangle, correct. And $\{a,b,c,1,12\}$ is an antirectangle, but not a maximal one, because there is another set (now also illustrated) of six mutually invisible squares: $\{a,b,c,1,6,12\}$. | |
| Nov 12, 2013 at 0:59 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Second image showing a maximal antirectangle.
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| Nov 11, 2013 at 22:37 | comment | added | Gerry Myerson | I don't get it. $\{\,a,b,c,1,5,12\,\}$ isn't an anti-rectangle, since there's a rectangle containing $b$ and 5, no? And $\{\,a,b,c,1,12\,\}$ includes a corner, indeed two corners, 1 and 12, so if it's maximal then it is the largest antirectangle that includes a corner. | |
| Nov 11, 2013 at 14:14 | comment | added | Erel Segal-Halevi | Yes, this is correct. | |
| Nov 11, 2013 at 14:04 | comment | added | Joseph O'Rourke | @ErelSegal-haLevi: Thanks for clarifying. And if the top $(5,6,7,8)$ were just a straight segment, then the max size is $5$, but some include a corner. | |
| Nov 11, 2013 at 13:52 | comment | added | Erel Segal-Halevi | If I see correctly, $\{a,b,c,1,12\}$ is an antirectangle of size 5, but it is not maximum size, because you can add a square at the corner to the left of $5$ (maybe slide $a$ slightly to the left so that they don't intersect). | |
| Nov 11, 2013 at 12:57 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |