In article <c0ggdc$rm2$1@xxxxxxxxxxxxxxx>,
perl-golf@xxxxxxxxxxxxx (Ton Hospel) writes:
> which corresponds to the bottom case in my drawing, so I have that
> case covered. It's not enough however. Consider the top case,
> and try to handle the 7 divisor:
>
> 14 = 7+7 => (7-1)+(7+1) => missing 7, oops
> => (7-0.5)+(7+0.5) => not integer, oops
> The other split:
> 14 = 2+2...2 (7 times) => (2-3)+(2-2)+(2-1)+2+(2+1)+(2+2)+(2+3) =>
> 2-3 is smaller than one, even smaller than zero, oops
Ah, I reacted to quick, and missed your trick of canceling
out zero and negatives. So actually you end up with 14 = 2+3+4+5,
which corresponds to the my top case. So your argument can be made
precise (you must prove you get all splitups and that the mapping is
one to one) and the two ways will turn out to be equivalent.
Making it precise is easy: extend an even trapezium whose bottom
width is n+1 with: -n, -n+1,...0,...n-1, n. That will make the
height odd, and then you can work with the bottom transform only
(the "middle" will remain above zero).
The resulting proofs are about equally complex though.
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