4 ms·
I don't know the exact answer, but the lower bound I computed had over a quarter million digits. See [1] 1: http://algassert.com/puzzle/2014/07/08/Boxing-Megas
by Strilanc 8y ago
I don't know the exact answer, but the lower bound I computed had over a quarter million digits. See [1]
1: http://algassert.com/puzzle/2014/07/08/Boxing-Megaspheres.html http://algassert.com/puzzle/2014/07/08/Boxing-Megaspheres.ht...
- OscarCunningham 8y agoI can improve the lower bound a bit. Let V be the volume (content) of a hypersphere of radius 0.99 (twice the radius of the spheres in the question). The total volume in which the centres of the spheres can lie is 0.01^1000000. So if the densest packing uses n spheres, then we must have nV >= 0.01^1000000 or else there would be a point at least 0.99 away from the centre of any sphere, and we could add an extra one there. Using Stirling's approximation to bound V above gives that n is greater than 10^388118.6...
- deleted 8y ago[deleted]
- Strilanc 8y agoI don't follow the logic here. How are you getting the equation nV >= 0.01^1000000? Shouldn't there be a packing efficiency constant in there? It almost looks like you're trying to pour the spheres like a liquid.
- OscarCunningham 8y agoImagine just putting the spheres in one at a time until you can't put in any more. Why can't you put in any more? It must be that every point of [0.495,0.505]^1000000 is within 0.99 of the centre of one of the spheres. Otherwise you could put in a new sphere with that point as its centre. This means that if we imagine spheres of radius 0.99 around each of our original spheres (which only have radius 0.495) those big spheres must cover all of [0.495,0.505]^1000000. Hence their combined volume must be as large as the volume of that box. Note that I defined V to be the volume of a sphere with radius 0.99, not diameter 0.99 as in the statement of the problem.
- Strilanc 8y agoThanks, that's much clearer and I understand now. That's a good simple lower bound.