Re: A square packing problem
- From: klewis@xxxxxxxxxxxxxxxx (Keith A. Lewis)
- Date: Mon, 17 Oct 2005 18:40:28 +0000 (UTC)
jmd <jdroz@xxxxxxx> writes in article <26214232.1129444364601.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx> dated Sun, 16 Oct 2005 02:32:14 EDT:
>What is the minimum of the sum of the areas of the members of a set of squares that cannot be packed in a unit square?
>
>The answer seems to be 1/2, but I don't see how to prove the difficult
>direction. (That a set of squares with total area <=1/2 can always be packed in
>a unit square.)
Suppose you have a set of squares S with total area = 1/2.
Consider the largest square in the set, of area (s1)^2. Put it in one
corner of your unit square. You now have 1/2-(s1)^2 in S\s1 still to pack.
If s1 > 1/2, you have at least 3 remaining empty squares of side (1-s1)
into which to pack it.
If s1 <= 1/2, you have 1 empty square of side (1-s1) and at least 2 empty
squares of side s1.
Just a start.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
.
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