Timeline for Dissecting a square
Current License: CC BY-SA 3.0
21 events
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| S Nov 9, 2017 at 20:46 | history | suggested | jeq | CC BY-SA 3.0 |
Copied image to imgur.com, as it was not being displayed because of the new https rule. Added link to original image source.
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| Nov 9, 2017 at 20:16 | review | Suggested edits | |||
| S Nov 9, 2017 at 20:46 | |||||
| Jun 13, 2014 at 1:14 | comment | added | JRN | Regarding Stewart and Wormstein's paper, more info about the second author can be found here. | |
| Jun 12, 2014 at 14:06 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Aug 9, 2013 at 6:20 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| May 30, 2011 at 8:10 | vote | accept | Colin D Wright | ||
| May 30, 2011 at 8:08 | comment | added | Colin D Wright | @Aaron Re your comment and Beni's situation: noted & understood. Re the original question, also understood. That's the culture of the place, and that's what I didn't know. My original cases didn't include a figure of 16 cells in 4 left justified rows of 4,3,5,4 - and I can say that without actually understanding your description. My original three infinite families were very naive, and only ever had 2, 4 or 8 pieces. I've tried your 128 solution but can't make it work, but that's no indication either way. | |
| May 29, 2011 at 22:26 | comment | added | Aaron Meyerowitz | I was brief and unartful because I wanted to get to the math part. I was more reporting on the attitudes I've found then my own feelings. I will say that I would have been annoyed if I was Beni and worked on an answer twice to get "You're getting closer". I think you would have gotten the same feedback (and not risked getting the problem closed) had you said: I see these cases... Incidentally, did your three cases (under strict rules) include a figure of 16 cells in 4 left justified rows of 4,3,5,4 ? Also, do you think 128 is possible as I suggested? Not having drawn it I wasn't sure. | |
| May 29, 2011 at 11:30 | vote | accept | Colin D Wright | ||
| May 29, 2011 at 11:31 | |||||
| May 29, 2011 at 11:30 | comment | added | Colin D Wright | @Aaron - this is why I'm pleased (in retrospect) that I asked the question in the vague way I did - the answers that I've had are incredibly illuminating. Thank you for your time - I hope you've found it interesting. It's taught me a lot. @Daniel - I was offended by the way it was phrased, but I understand that offence often arises from a mismatch of culture. I need to become enculturated, so I swallowed hard and reacted in what I hope is a positive and appropriate manner. | |
| May 29, 2011 at 7:44 | comment | added | Daniel Litt | @Aaron: This is a cool answer. I urge you to rephrase the beginning of comment (1), however, even though the OP seems not to have been offended. | |
| May 29, 2011 at 6:03 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| May 29, 2011 at 5:22 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| May 26, 2011 at 20:56 | comment | added | Colin D Wright | My infinite families are as follows. Take any non self-intersecting path from the center to the border such that a 180 degree rotation does not intersect. That divides the square into two. Similarly for a 90 degree rotation, which divides the square into 4. Now divide into four squares, and for each small square, divide diagonally with a curve that has 180 degree symmetry. That divides the square into 8. The sporadic is the trivial case - one piece. But fedja has just demonstrated a 16 piece dissection, so now I know I know nothing. | |
| May 26, 2011 at 20:23 | comment | added | Aaron Meyerowitz | @Colin I forgive you and I don't disagree! I'm just telling you what I've seen. I'd still like to know what your 3 families and sporadic case are. | |
| May 26, 2011 at 19:31 | comment | added | Douglas Zare | @Colin D Wright: This is not a discussion site. This is a question and answer site. Say what you know if you want me to spend time thinking about it. | |
| May 26, 2011 at 19:29 | comment | added | Colin D Wright | So thank you for your comprehensive reply, and for the references. They are most valuable. I'll go away and reconsider my approach. Next time I have a question I'll consider my phrasing more carefully. I will certainly question whether this is the right forum. | |
| May 26, 2011 at 19:26 | comment | added | Colin D Wright | "If you want respect on this site then stop hinting ..." - forgive me, but I'm accustomed to sites where one creates discussion so people can contribute to finding a solution. This site feels different, and I'm certainly learning quickly that it has a different ethos. My apologies. I've been intrigued to see the different results where the partitioning of the square has not been into sets homeomorphic to a disk up to a set of Lebesgue measure zero. Seeing the speculation has been of use to me, and would have been missed had I been more specific. | |
| May 26, 2011 at 19:13 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| May 26, 2011 at 17:19 | comment | added | Mark Bennet | To fill out: Take a path from a point on the boundary to the centre. Rotate it by a quarter turn, half turn, three quarters about the centre of the square. A condition is required so that these paths do not cross each other, but if they don't they divide the square into four congruent pieces. For eight piece solutions, divide into four by horizontal and vertical straight lines through the centre. Join the centre to the edge with a path having twofold rotational symmetry around the midpoint of the diagonal from centre to vertex and lying within the quarter. Replicate in the other quarters. | |
| May 26, 2011 at 14:20 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |