Timeline for 1 rectangle <= 4 squares
Current License: CC BY-SA 2.5
9 events
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| Mar 3, 2010 at 13:46 | history | edited | Yaakov Baruch | CC BY-SA 2.5 |
added 63 characters in body; added 2 characters in body
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| Mar 3, 2010 at 0:17 | comment | added | TonyK | I have just tried L=3 in one direction and L=5 in the other, and I still get 3.8. | |
| Mar 2, 2010 at 23:58 | comment | added | Yaakov Baruch | Correction: there are of course Ax5A and similar rectangles for L=3 (since m,n range from -3 to +3). Also, the limits 2 on Ax2A, 3 on Ax3A, 3.5 on Ax4A etc., while achievable by the program, nonetheless could make it run faster if imposed from the outset. | |
| Mar 2, 2010 at 23:14 | comment | added | F_G | I also bet on 3.75 (and have for the moment nothing better to report than 56/15 on the lower bound front). | |
| Mar 2, 2010 at 22:53 | history | edited | Yaakov Baruch | CC BY-SA 2.5 |
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| Mar 2, 2010 at 22:52 | comment | added | Yaakov Baruch | An idea: if L=4 there are rectangles of size Ax4A, 4AxA, Bx4B and 4BxB on which we can impose the known limit of 3.5, rather than 3.8 - well, that won't help any because the proof of a 3.5 on a 4x1 was all contained in a 4x4, and the L=4 grid contains the 4Ax4A and 4Bx4B needed for the proof! However, and THIS is the idea, if L=5 in one direction, the limit of 25/7 for Ax5A etc. is not contained in the grid and so imposing that could add the extra oomph to break the 3.8 barrier. I'd try L=2 in one direction and L=5 in the other, with those 25/7 limits... Even if that fails, I bet on 3.75. | |
| Mar 2, 2010 at 22:21 | comment | added | TonyK | I bet you it's 3.8. | |
| Mar 2, 2010 at 20:56 | history | edited | Yaakov Baruch | CC BY-SA 2.5 |
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| Mar 2, 2010 at 20:18 | history | answered | Yaakov Baruch | CC BY-SA 2.5 |