Timeline for Overlaying two domino-like constructions such that all individual pairs of domino-like cells in the overlay have matching symbols
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2020 at 14:35 | comment | added | Ben Barber | @MartinSleziak, I'm not sure if it's more remarkable that I had it or that I could find it. | |
| May 4, 2020 at 14:34 | history | edited | Ben Barber | CC BY-SA 4.0 |
restored the image
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| May 2, 2020 at 22:54 | comment | added | Martin Sleziak | The link in the post no longer works. And I do not see the picture in the Wayback Machine, either. Perhaps you still have the picture available somewhere? | |
| Oct 13, 2012 at 4:30 | vote | accept | Perpetuum | ||
| Oct 12, 2012 at 18:36 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Fixed image linking.
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| Oct 12, 2012 at 12:12 | comment | added | Ben Barber | I can't get the image to display in the answer. If anyone with the power to do so can sort it out for me (and possibly explain how to do it properly) it would be much appreciated. | |
| Oct 12, 2012 at 12:05 | history | edited | Ben Barber | CC BY-SA 3.0 |
giving up
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| Oct 12, 2012 at 11:15 | comment | added | Ben Barber | I was ignoring translations of the smaller square, assuming that when we came to overlap the larger rectangles then the small rectangle would have to be in the corner of each of the larger rectangles. I now see that this is incorrect, as, for example, two $1 \times n$ rectangles can be overlapped to form a $k=1$ rectangle in many ways. So the condition I've given is necessary, but not sufficient, and a complete answer needs to consider all of the (finitely many) different ways that two rectangles can overlap. | |
| Oct 12, 2012 at 10:22 | comment | added | Perpetuum | @Ben Barber I'm not sure I understand your simplification? If we take two 3x3 squares squares, and we set k = 2, there should only be one way to overlay the two 3x3 squares (not accounting for rotational and reflection symmetries). However, a k = 2 square should be able to fit into a 3x3 square in many different ways. | |
| Oct 12, 2012 at 9:34 | history | answered | Ben Barber | CC BY-SA 3.0 |