Re:Hasse diagram of the 2008 Olympic medal table
by grzm on Saturday September 16, 2017 @07:13PM
(#15267273)
I'm sure someone will pipe up with a more exact answer (likely using permutation and/or combination in their specific mathematical senses), but I believe it's because summing medals is algebraic, and with the given weights there's no way for the sum of any combination of non-identical medals that equals any other. The values of 𝛑 and 𝛑² (and the patient, unacknowledged workhorse 1) provide this. Looking at a simple case where s is the number of silvers and b the number of bronzes, there's no value of b that can equal any value of s·𝛑. You could use another triple of weights, such as (𝛑, e, 1) as well.
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Re:Hasse diagram of the 2008 Olympic medal table
by wtallis on Saturday September 16, 2017 @07:44PM
(#15267387)
Medals are awarded in integer quantities, and you don't want j·g+k·s+l·b=0 to be possible for integer j,k,l (except when all three medal counts are 0).
This is actually a slightly stronger condition than is needed, because no country can earn negative medals and there are a finite number of medals available, so you could actually have the weights for eg. gold and silver medals differ by a rational factor as long as the denominator was large enough.