"Avni Pllana" <avniu66@hotmail.com> wrote in message news:16882006.1173707869848.Java Mail.jakarta@nitrogen.mathforum.org... > > In the earlier question, we used a "mother" circle of > > D diameter and the 20 satellite circles progressively > > diminishing in diameter from d to (.9^19)*d, (and we > > called the largest d as #1, down to the smallest as > > #20). > > In what order do we place the satellite circles > > around the "mother" circle so that (a) the ratio D:r > > is largest, and (b) so that the ratio D:r is the > > smallest it can be ?, and (c) what are these ratios? > > > > A reminder that all the satellite circles must just > > touch the "mother" circle and also just touch the > > neighbouring satellite circle each side. > > Hi Bill, > > The maximum value for D:r we obtain for the order > > Na=[1 2 4 6 8 10 12 14 16 18 20 19 17 15 13 11 9 7 5 3], > > max(D:r) = 2.24194649536747 . > > > The minimum value for D:r we obtain for the order > > Nb=[1 17 2 18 3 16 4 19 5 20 6 15 7 14 8 13 9 12 10 11], > > min(D:r) = 1.94234849032733 . > > > Best regards, > Avni Hi, Bill and Avni. For the max(D:r), I agree also, neighbours must be as equal as possible For the minimum value, neighbours must be as unequal as possible, and it seems a good policy to put the small circles under the large ones, mindful of next-next neighbour overlap. I get Avnis min-number with his arrangement, and it seems hard to improve on it by machine. There are just too many possibilities with local ups and downs. Manual rearrangement can improve the packing, but only a little. Not being sure, how about min(D:r) = 1.935227285 for size-rank like this 1,20,8,11,10,12,9,13,7,14,5,15,4,16,3,17,2,18,6,19 to make a circle It is overlap-free, as I check it. Maybe a child could do it better :-) with a roundbottomed pot, a cup in it, and 20 marbles. Best regards, Hans.

"Avni Pllana" <avniu66@hotmail.com> wrote in message news:31912433.1174034096769.Java Mail.jakarta@nitrogen.mathforum.org... > > > > Hi, Bill and Avni. > > > > For the max(D:r), I agree also, neighbours must be as > > equal as possible > > > > For the minimum value, neighbours must be as unequal > > as possible, > > and it seems a good policy to put the small circles > > under the large ones, mindful of next-next neighbour > > overlap. > > > > I get Avnis min-number with his arrangement, and it > > seems hard to improve on it by machine. There are > > just too many possibilities > > with local ups and downs. > > > > Manual rearrangement can improve the packing, but > > only a little. > > Not being sure, how about > > min(D:r) = 1.935227285 > > for size-rank like this > > 1,20,8,11,10,12,9,13,7,14,5,15,4,16,3,17,2,18,6,19 to > > make a circle > > It is overlap-free, as I check it. > > > > Maybe a child could do it better :-) > > with a roundbottomed pot, a cup in it, and 20 > > marbles. > > > > Best regards, Hans. > > > > > > > Hi Hans, > > I found a slightly lower value for min(D:r), > > min(D:r) = 1.93437529786599 , > > N=[1,20,8,11,10,12,9,13,7,14,5,15,4,17,3,18,2,16,6,19]. > > I think the problem can be viewed as packing problem, and I am not sure whether children can do better. > > Best regards, > Avni Bill and Avni, what I meant by children was that some trial and error was involved. But Avni, I see the consistency in your sequence, and I get the same ratio you do, for your sequence. This looks like the minimum. Never say never:-) Hans.

"Avni Pllana" <avniu66@hotmail.com> wrote in message news:30380793.1174298945048.Java Mail.jakarta@nitrogen.mathforum.org... > > > Hi Hans, > > > > > > I found a slightly lower value for min(D:r), > > > > > > min(D:r) = 1.93437529786599 , > > > > > > > > > > > > N=[1,20,8,11,10,12,9,13,7,14,5,15,4,17,3,18,2,16,6,19 > > ]. > > > > > > I think the problem can be viewed as packing > > problem, and I am not sure whether children can do > > better. > > > > > > Best regards, > > > Avni > > > > Bill and Avni, what I meant by children > > was that some trial and error was involved. > > But Avni, I see the consistency in your sequence, > > and I get the same ratio you do, for your sequence. > > This looks like the minimum. > > Never say never:-) Hans. > > > > > Hi Hans, > > You are right ( Never say never:-) ). Using some logic I improved the minimum further > > min(D:r) = 1.93370059070196 , > > N=[1 17 2 18 4 20 6 14 8 12 10 11 9 13 7 15 5 19 3 16]. > > I think this should be the global minimum, but I can not provide a rigorous proof. > > Best regards, > Avni Avni !!! +++ The logic must have been something about as many and as close near-overlaps as possible. Respectfully, Hans.