[math-fun] Mr. Perkins Quilt
This MPQ talk is similar to a problem I had considered - what is the smallest square needed to house all 1..n squares? e.g. with 1x1, we need a 1x1 square with 1x1 and 2x2, we need a 3x3 square with 1x1, 2x2 and 3x3, we need a 5x5 square. So a lower bound is 2n-1. However with all the squares from say 1 to 8, this packing fails. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
Although a different packing allows for 1..8 to be placed in a 15x15 sqtare. Is 2n-1 a upper bound then too? While I'm here, wrt Kepler's conjecture, is it true that the densest packing obeys the rule that every continuous subset of spheres must also have the densest packing possible? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Jon Perry Sent: 22 October 2003 12:33 To: Maths is Fun Subject: [math-fun] Mr. Perkins Quilt This MPQ talk is similar to a problem I had considered - what is the smallest square needed to house all 1..n squares? e.g. with 1x1, we need a 1x1 square with 1x1 and 2x2, we need a 3x3 square with 1x1, 2x2 and 3x3, we need a 5x5 square. So a lower bound is 2n-1. However with all the squares from say 1 to 8, this packing fails. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Wed, 22 Oct 2003 12:40:52 +0100 "Jon Perry" <perry@globalnet.co.uk> Although a different packing allows for 1..8 to be placed in a 15x15 sqtare. Is 2n-1 a upper bound then too? No. It can't be true for n > 9.
At 12:40 PM 10/22/03 +0100, Jon Perry wrote:
Although a different packing allows for 1..8 to be placed in a 15x15 sqtare. Is 2n-1 a upper bound then too?
No; you can show fairly easily that squares of sides 1..9 cannot fit in a square of side 17. The key lemma is that the three squares of sides 5, 6, and 7 cannot be packed into a 9x17 rectangle. The bound even fails numerically, when we get up to ten squares. The sum of the first ten squares is 385, which is bigger than 19x19. Note that the sum of the first n squares is a cubic, n^3/3 + n^2/2 + n/6. Therefore, the minimum side of a square in which they can be packed must increase faster than linear. The best you can hope for is a 3/2 power. -A
Wed, 22 Oct 2003 12:32:43 +0100 "Jon Perry" <perry@globalnet.co.uk> This MPQ talk is similar to a problem I had considered - what is the smallest square needed to house all 1..n squares? e.g. with 1x1, we need a 1x1 square with 1x1 and 2x2, we need a 3x3 square with 1x1, 2x2 and 3x3, we need a 5x5 square. So a lower bound is 2n-1. However with all the squares from say 1 to 8, this packing fails. It may be that I am off because I am on the road and just waking up, but this strikes me as a loose lower bound, since the _absolute_ minimum is sqrt(n*(n+1)*(2n+1)/6), and this is clearly not achievable for n>1. So any lower bound should be at least on the order of n^1.5.
participants (3)
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Allan C. Wechsler -
Jon Perry -
Michael B Greenwald