Timeline for Picking collections to get over half the number of each type of object
Current License: CC BY-SA 2.5
13 events
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Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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Aug 18, 2010 at 0:48 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
Aug 17, 2010 at 7:38 | history | edited | Casebash | CC BY-SA 2.5 |
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Aug 17, 2010 at 7:15 | answer | added | Artem Kaznatcheev | timeline score: 0 | |
Aug 17, 2010 at 5:41 | answer | added | Gerry Myerson | timeline score: 1 | |
Aug 17, 2010 at 4:29 | comment | added | Gerry Myerson | I think it's an interesting question if you fix $k$ and imagine $n$ large compared to $k$. For $k=1$ it's trivial; you may need $\lceil n/2\rceil$ boxes, you never need more. But for $k=2$ it's already not clear (to me) what the answer might be. I can see where you may need $(n/2)+1$, I can't see if you may need more. | |
Aug 16, 2010 at 20:46 | comment | added | Casebash | @Steve, @Dan: Updated the question | |
Aug 16, 2010 at 20:45 | history | edited | Casebash | CC BY-SA 2.5 |
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Aug 16, 2010 at 14:51 | comment | added | Steve Kass | @Casebash: Not sure what your comment is saying. Dan is correct. Your question isn't clear, so Dan may not have answered it. Do you mean "Given a known distribution, what is the number B (which depends on the distribution) so that every subset of B boxes contains at least half of each type, and B is minimal (some B-1 boxes fail to contain half of each type)?" If this is the question, there are some distributions for which B=n (everything in one box and n-1 empty boxes), and some distributions for which B = n/2 (identical boxes). What is your question? | |
Aug 16, 2010 at 10:43 | comment | added | Casebash | @Dan: The boxes can contain mixed types | |
Aug 16, 2010 at 10:43 | history | edited | Casebash | CC BY-SA 2.5 |
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Aug 16, 2010 at 10:41 | comment | added | Dan Brumleve | If k = n and each box contains a different kind of object then we need all boxes. | |
Aug 16, 2010 at 10:22 | history | asked | Casebash | CC BY-SA 2.5 |