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From: borie@ATHOS.CS.UA.EDU (Richard Borie)
Newsgroups: comp.theory
Subject: Partitioning squares into unequal squares
Message-ID: <9105242046.AA15308@athos.cs.ua.edu>
Date: 28 May 91 14:01:37 GMT
Sender: daemon@ucbvax.BERKELEY.EDU
Reply-To: Richard Borie <borie%athos.cs.ua.edu@VM1.NoDak.EDU>
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Is it possible to partition the interior of a square into
smaller squares, no two of which have the same size?
(Notice that it is trivial to partition a square with side 2
into 4 squares each with side 1, hence the requirement of distinct sizes.)
The closest I have been able to get is to partition a 32-by-33
rectangle into squares with sides 1, 4, 7, 8, 9, 10, 14, 15, and 18.

This problem is related to that of finding two (dual?) partial orders
over the same set of elements,
such that a chain in one order corresponds to an antichain in the other,
and such that the sum of the elements along any maximal chain or antichain
is the same.
(Intuitively, the squares are the elements of the partial orders,
one partial order is the relation "is to the east of"
and the other is the relation "is to the south of".)