Timeline for Traversing the infinite square grid
Current License: CC BY-SA 3.0
20 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Feb 27, 2012 at 12:51 | comment | added | François Brunault | @mmm. Thanks for the edit. Your proof seems correct and I now understand your idea of making the stepsize arbitrarily large in order to make the induction work. This is a quite nice and surprising result ! | |
| Feb 27, 2012 at 12:17 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 11:57 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 11:53 | comment | added | mmm | Ofcourse all squares count as visited. See my description of full path. NEXT TIME YOU INCREASE m to contain all cells, ofcourse. | |
| Feb 27, 2012 at 3:53 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 3:43 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 3:16 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 3:14 | vote | accept | mmm | ||
| Feb 27, 2012 at 3:08 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 3:03 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 2:57 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 27, 2012 at 2:15 | comment | added | Brendan McKay | I agree with François. You are using squares outside the $m\times m$ part in order to visit one square inside it, but you don't seem to be counting those outside squares as being visited. But the way you stated your problem that doesn't seem to be allowed. How do you know that when you visit the next square inside the $m\times m$ part you won't need to use one of the squares outside it that you used before? | |
| Feb 27, 2012 at 1:43 | vote | accept | mmm | ||
| Feb 27, 2012 at 3:03 | |||||
| Feb 27, 2012 at 1:37 | vote | accept | mmm | ||
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| Feb 27, 2012 at 1:36 | comment | added | mmm | @FrançoisBrunault All the original visited squares lie within the mxm square. We visit only one cell, (x,y) within the original mxm square. By alternating n times, we move n steps closer to the mxm-square, but the possible next step, moves 3*n, so if we moved far enough away initially then 3n-n>m, and we'd only need to step once inside the mxm square. – mmm | |
| Feb 27, 2012 at 1:28 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 26, 2012 at 21:54 | comment | added | François Brunault | I'm not convinced by your argument. It seems to work for visiting one given cell, but when you repeat the process, how do you make sure the squares you need are free ? For example, you may need to use the same axis many more times. | |
| Feb 26, 2012 at 18:35 | history | edited | mmm | CC BY-SA 3.0 |
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| Feb 26, 2012 at 18:23 | history | answered | mmm | CC BY-SA 3.0 |