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In 2 dimensions, it is easy to show that the answer is no. I use
the pigeon-hole principle for the proof. Consider the two opposed corners
where there is no vacancy. Each 4x4 corner is filled by 8 dominos, and
the inner corners touch, so the remaining vacancies are in two bounded 4x4
sections, with 15 dominos remaining. The two vacant sections must be
filled symmetrically, since no domino can intersect both sections. This
is impossible with an odd number of dominos.
Going to 3 dimensions and the possibility of inserting a domino
edgewise, or some such trick, the answer would be different.
Regards,
Jack
On Tue, 20 Aug 2002, John S. Denker wrote:
> You are given a supply of 31 ordinary dominos. You are also
given a checkerboard with the ordinary arrangement of colored> akin to a physics principle hiding in there. Hint, hint.
squares, of a size-scale such that each domino just covers
two squares. Can you arrange the dominos so that all squares
are covered except for the two squares at the opposite ends
of the main NW/SE diagonal?
This seems like a pure math/logic puzzle, but there's something