Timeline for Can an arbitrary collection of circles of total area 1/2 fit into a circle of area 1?
Current License: CC BY-SA 2.5
11 events
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| Feb 17, 2012 at 22:55 | comment | added | Victor Protsak | Hi Kevin, I am sorry I don't follow MO as ardently as before, so I just briefly saw your questions yesterday and you got it before I could respond. "Width" is in the x-direction, "height" is in the y-direction and the greedy algorithm works. The SE comment is right on target. Best, Victor | |
| Feb 17, 2012 at 15:34 | comment | added | Kevin Buzzard | Oh, I see now: someone at stackexchange pointed out to me that the point is that if there aren't enough squares left to fill half the rectangle, just shrink the rectangle until there are and then you're done because you've put all the squares in. | |
| Feb 10, 2012 at 20:49 | comment | added | Kevin Buzzard | ...slightly confused about what "the strip" is, and very confused about why you know that, once you have removed the square of side length $c$, the remaining squares can be assumed to cover at least half of the area of "the strip". Can you clarify? I'm keen to understand your argument. My impression is that Wadim understands it, so it's probably fine, but I don't understand it :-( Cheers, Kevin | |
| Feb 10, 2012 at 20:47 | comment | added | Kevin Buzzard | Victor -- sorry to bother you -- but I can't follow the proof of the Packing Lemma. This came up in some undergraduate maths puzzle-solving group here in London. The trivial comments are that you talk about the "top" and the "width" and "horizontal strips" etc etc without ever explaining what the orientation of everything is. But I think I can guess my way around these things. The thing that really bothers me though is the application of the inductive hypothesis. You say "By the inductive assumption, at least half of the strip, area-wise, can be packed with squares from F". But I am... | |
| Jun 8, 2010 at 7:05 | comment | added | Wadim Zudilin | Following the link in Apery's question, I found an interesting "circle" of packing problems, Jung's and Borsuk's problem. The results are nice by themselves (and maybe not well related to the OP, my findings include dx.doi.org/10.2307/2031795 and dx.doi.org/10.1017/S0305004100032849). But is it possible to prove a similar lemma for right $n$-gons instead of squares, under the same assumption of equal orientation (so, they are congruent)? If yes... (I don't need to continue). | |
| May 16, 2010 at 2:53 | comment | added | Wadim Zudilin | Victor, I just feel sad about the whole answering story for this question. | |
| May 16, 2010 at 2:21 | comment | added | Victor Protsak | Hi Wadim, I am sorry that you felt that way about people's reaction to your answer - I wish you'd left it there, it certainly looked very interesting to me! FWIW, someone downrated my post, too! Having spent last several hours chasing references, I've found out that much is known about these kinds of problems. Apparently, there is an online algorithm for squares with the same constant 2, which implies constant 4 for circles. ("Online" means that squares are given in a sequence, with the total area fixed ahead of time, and must be packed sequentially with no later changes allowed). | |
| May 16, 2010 at 1:20 | comment | added | Wadim Zudilin | @Victor: I am convinced. But since I voted on your answer already when you posted it, I can't put you higher. In any case, I find this "democratic" voting very inadequate. | |
| May 15, 2010 at 18:31 | history | edited | Victor Protsak | CC BY-SA 2.5 |
rewrote, with the new proof of a more general result
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| May 12, 2010 at 21:20 | history | edited | Victor Protsak | CC BY-SA 2.5 |
accomodated the OP's change in formulation; added an explanation of a problem with the proof
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| May 12, 2010 at 10:27 | history | answered | Victor Protsak | CC BY-SA 2.5 |