Timeline for Occupy a field with your choice of tetromino
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 15, 2015 at 3:49 | vote | accept | Ben Frankel | ||
| Sep 15, 2015 at 3:49 | comment | added | Ben Frankel | @Marconius, Your new T solution is also mine. I read through your J/L proof and I feel like I agree. Great job, and by the way, here's a rotationally symmetric solution that I found for S/Z: i.imgur.com/rdqlnNY.png. If you're interested in another puzzle like this, I found that if you consider S=Z, then S/Z is still possible. | |
| Sep 15, 2015 at 0:21 | history | edited | Marconius | CC BY-SA 3.0 |
improved solution for T tetromino
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| Sep 14, 2015 at 22:28 | comment | added | Marconius | @BenFrankel - I think I can see why the T solution is not optimal - the T's are paired so that the corners meet to enclose a single empty cell - this is sub-optimal. I may have over-relied on symmetry. | |
| Sep 14, 2015 at 21:47 | comment | added | Ben Frankel | I'm going to read through your JaiL proof and see if I agree. In the meantime, since it may seem as if the question has been answered in its entirety, I'll let you know that it hasn't. Your T solution is not optimal. | |
| Sep 14, 2015 at 21:28 | comment | added | Marconius | @IanMacDonald - Thanks. Drawing were done in Excel, using primarily cell background colours and borders, with adjusted row height (24) and column width (4). I am curious whether the S/Z and T tilings are minimal or not - may need to write a computer program for this ;-) | |
| Sep 14, 2015 at 20:47 | comment | added | Ian MacDonald | +1 Yup. That JL is exactly what I found out, I'm just unable to make pretty pictures. :) | |
| Sep 14, 2015 at 20:34 | history | edited | Marconius | CC BY-SA 3.0 |
add proof that J/L tetrominos cannot tile the board
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| Sep 13, 2015 at 17:49 | comment | added | Marconius | Good observation. I am still working on the J/L tilings. | |
| Sep 13, 2015 at 16:38 | comment | added | Ben Frankel | Very cool! I have a different solution for S/Z and T. I noticed that your S/Z tiling could be expanded upwards +3 tiles at a time, and rightwards +2/+3 tiles at a time thus solving S/Z in the case of boards sized $3n+2$ by $5m$ or $5m-2$. | |
| Sep 13, 2015 at 16:27 | history | edited | Ben Frankel | CC BY-SA 3.0 |
added 4 characters in body
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| Sep 13, 2015 at 16:03 | history | edited | Marconius | CC BY-SA 3.0 |
minor wording change
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| Sep 13, 2015 at 15:21 | history | answered | Marconius | CC BY-SA 3.0 |
