🚀 go-pugleaf

I wrote:

>borie@ATHOS.CS.UA.EDU (Richard Borie) writes:

>I'd think not, assuming you want a finite number of squares.
>Here's a proof, but beware - I don't do this for a living...

Oops!  I'm glad about that...

>subsquares placed along the top edge of the hollow.  But placing two
>(or more) unequally-sized squares in a hollow means that you can never
>decrease the number (N) of distinct colored depths in S - in the best
>case, you'll keep N constant (if the left square matches the level of
>the left neighbor, and the right square matches the level of the right
>neighbor).  If either of the new squares are next to a wall, the best
>you can do is (still) keep N constant.  Obviously *both* squares being
>placed in the hollow can't be next to the left and right walls.

This is wrong, sorry.  If you place 2 squares in a single hollow,
it is possible to decrease N by 1.

Should have been obvious, really - my "proof" also showed that
you couldn't tile a rectangle.

Cheers,

    Mike.
--
Mike McGaughey			AARNET:	mmcg@bruce.cs.monash.oz.au

"His state is kingly; thousands at his bidding speed,
 And post o'er land and ocean without rest."		- Milton.