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Topic: Sweeping a cube (Read 802 times) |
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JocK
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Sweeping a cube
« on: Dec 7th, 2004, 10:44am » |
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What is the largest volume that fits inside a unit cube in any orientation, and that can be moved around inside this cube so as to sweep it completely?
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| « Last Edit: Dec 9th, 2004, 12:57pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
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Re: Sweeping a cube
« Reply #1 on: Dec 9th, 2004, 1:55am » |
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Oh, that seems tough... 
To avoid mis-interpretations: "can freely rotate" means "no 2 points of the body are more than a unit length apart"?
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Barukh
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Re: Sweeping a cube
« Reply #2 on: Dec 9th, 2004, 5:54am » |
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Here is one idea (assuming my interpretation of “freely rotate” is correct).
Take any figure F that satisfies Sweeping a Square. Let the area of F be S. If we dilate F by a factor of k < 1, we get a similar figure F’. In it, every 2 points are at the distance no more than k. The area of F’ is Sk2.
Now, on F’ as a base, build a right generalized cylinder of height h. If h2 + k2 [le] 1, no two points in this solid will be more than a unit apart. So, we may achieve the volume of Sk2[sqrt](1-k2). This reaches the maximum 2S/[sqrt]27.
Taking S = 0.6866… (the best known so far) gives 0.264…
I will try to visualize this.
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JocK
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Re: Sweeping a cube
« Reply #3 on: Dec 9th, 2004, 12:53pm » |
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on Dec 9th, 2004, 1:55am, Barukh wrote:| Oh, that seems tough... |
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C'mon Barukh, a piece of cake for u! (you have solved the 2D case already and I won't bother you with the 4D case...  )
on Dec 9th, 2004, 1:55am, Barukh wrote:| To avoid mis-interpretations: "can freely rotate" means "no 2 points of the body are more than a unit length apart"? |
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Correct. I will change the wording in the riddle from 'freely rotate inside a unit cube' into: 'can be placed inside a unit cube in any orientation'.
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| « Last Edit: Dec 9th, 2004, 12:55pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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JocK
Uberpuzzler
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Posts: 877
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Re: Sweeping a cube
« Reply #4 on: Dec 9th, 2004, 1:55pm » |
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on Dec 9th, 2004, 5:54am, Barukh wrote:Here is one idea (assuming my interpretation of “freely rotate” is correct).
Take any figure F that satisfies Sweeping a Square. Let the area of F be S. If we dilate F by a factor of k < 1, we get a similar figure F’. In it, every 2 points are at the distance no more than k. The area of F’ is Sk2.
Now, on F’ as a base, build a right generalized cylinder of height h. If h2 + k2 [le] 1, no two points in this solid will be more than a unit apart. So, we may achieve the volume of Sk2[sqrt](1-k2). This reaches the maximum 2S/[sqrt]27.
Taking S = 0.6866… (the best known so far) gives 0.264…
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That's a start! 
Would you have applied the same methodology to find a solution to 2D problem Sweeping a Square given that the 1D problem is trivially solved by a line segment of unit lenth, you would have ended up with a square with unit diagonal and area 1/2. Not bad for a first guess, but not optimal...
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
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Re: Sweeping a cube
« Reply #5 on: Dec 10th, 2004, 11:04am » |
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on Dec 9th, 2004, 12:53pm, JocK wrote:| C'mon Barukh, a piece of cake for u! (you have solved the 2D case already... |
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You give me too much credit - after all, it was SWF's idea.
Quote:| ...and I won't bother you with the 4D case... ) |
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Do you promise?
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JocK
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Re: Sweeping a cube
« Reply #6 on: Dec 11th, 2004, 7:33pm » |
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on Dec 10th, 2004, 11:04am, Barukh wrote:
Do you promise?
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Considering various spatial dimensions N = 1, 2, 3, ... might actually help. Reasonable target values for the volume fraction in various dimensions can be defined as follows:
A (hyper)cube with unit diagonal (volume N-N/2) consitutes a lower bound, and a (hyper)sphere with unit diameter (volume ([sqrt][pi] / 2)N/Gamma(1 + N/2) ) an upper bound to the solution of the (hyper)cube sweeping problem.
N low.bound upp.bound
1 - - - 1.000 - - - 1.000
2 - - - 0.500 - - - 0.785
3 - - - 0.192 - - - 0.524
4 - - - 0.063 - - - 0.308
5 - - - 0.018 - - - 0.164
The midpoint between this upper bound and this lower bound represents a volume fraction that you should be able to reach or exceed... (at least for all dimensions up to N=5).
Good luck!
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
Uberpuzzler
Gender:
Posts: 2276
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Re: Sweeping a cube
« Reply #7 on: Dec 12th, 2004, 1:02am » |
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on Dec 11th, 2004, 7:33pm, JocK wrote:A (hyper)cube with unit diagonal (volume N-N/2) consitutes a lower bound, and a (hyper)sphere with unit diameter (volume ([sqrt][pi] / 2)N/Gamma(1 + N/2) ) an upper bound to the solution of the (hyper)cube sweeping problem.
N low.bound upp.bound
1 - - - 1.000 - - - 1.000
2 - - - 0.500 - - - 0.785
3 - - - 0.192 - - - 0.524
4 - - - 0.063 - - - 0.308
5 - - - 0.018 - - - 0.164
The midpoint between this upper bound and this lower bound represents a volume fraction that you should be able to reach or exceed... (at least for all dimensions up to N=5). |
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Hmm, that's interesting... For n=2, we get the SWF's proposed solution.
But I don't see why this should be necessarily true for higher dimensions? Probably, my imagination deceives me... 
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Barukh
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Re: Sweeping a cube
« Reply #8 on: Feb 19th, 2005, 5:21am » |
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Is this thread dead? I am still not able to find something better...
More than 2 months have passed since the last post. Maybe, a hint is relevant?
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JocK
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Re: Sweeping a cube
« Reply #9 on: Feb 19th, 2005, 1:35pm » |
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on Feb 19th, 2005, 5:21am, Barukh wrote:Maybe, a hint is relevant?
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OK -- at the danger of giving too much away -- here is a 2nd hint (the first hint being that one should strive for volumes of at least 0.358..):
Think about an octant of a sphere with unit radius. Such a body fits snuggly in the corner of cube, and has a volume as large as 0.524. However, some points on this octant are further than a unit distance apart. So you have to cut off some part(s)...
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| « Last Edit: Feb 19th, 2005, 1:37pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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