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Hasse diagram of the 2008 Olympic medal table

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Re: Hasse diagram of the 2008 Olympic medal table

#31

Post by wtallis »

grzm wrote: post #30
obel1x wrote: post #29

why does allocating "pi^2 for a gold, pi for a silver and 1 for a bronze" mean "there could never be any tie in the total scores except when two countries had exactly the same medal counts in all three categories"?

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…

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.

Re: Hasse diagram of the 2008 Olympic medal table

#32

Post by grzm »

wtallis wrote: post #31
grzm wrote: post #30

Earlier quoted context omitted.

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…

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…

Cheers!

Re: Hasse diagram of the 2008 Olympic medal table

#33

Post by Magnap »

esfandia wrote: post #24

I did this exact thing (even the part where I displayed the results using Graphviz) to come up with a partial ordering of General Managers of Major League Baseball teams. The partial order was simply: GM_of_Team A > GM_of_Team B iff (endofseason_ranking (A) > endofseason_ranking(B)) and (budget(A) So this would capture how much "bang for the buck" a GM is getting. I did it for a couple of years, and not surprisingly,…

The theoretical property those hacks were aiming at showing is called the transitive reduction. And you don't need any hacks to calculate it, graphviz comes with the `tred` tool to do so.

Re: Hasse diagram of the 2008 Olympic medal table

#35

Post by iainmerrick »

kzrdude wrote: post #28
dxbydt wrote: post #19

Earlier quoted context omitted.

since france isn't connected by arrow to korea, or vice-versa, the article rightly says you can't determine who'se ranked higher. you have (13>7, 23=23,31<40) in this case. you clearly want the first inequality to prevail.

I see that France dominates 2 fewer nations than South Korea. Does this number have any meaning? In a way, South Korea did better in this way.

That sounds familiar to the various voting systems that satisfy the Condorcet criterion (basically, ignoring all votes for other parties, do people prefer A to B?). There are a lot of different ways of picking a winner when two or more candidates are tied under Condorcet -- ranked pairs, beat paths, etc, including rules like the one you suggest.

But in voting systems, Arrow's Theorem says none of those methods can be truly satisfactory, at least if you accept Arrow's list of requirements.

Similarly for Olympics ranking, I don't think any method for breaking ties will be able to satisfy all the requirements of a good ranking (e.g., removing a country from competition shouldn't change any other rank orderings).

So personally I don't think any of those tie-breaking methods really have any meaning. You might as well just toss a coin.

Re: Hasse diagram of the 2008 Olympic medal table

#37

Post by nicky0 »

te wrote: post #5

Where I'm from, it's "inarguable" that [1,0,1] > [0,2,0]; that is, that a gold and a bronze is better than two silvers. This would break some of the ties shown in the diagram.

Your statement is unclear; are you asserting the truth or false of the inequality? (The unclear word is "inarguable" - do you mean there is no argument in favour or no argument against)
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