This reminds me of the beatpaths graphs I used to do for the NFL (http://beatpaths.com) - I'd then come up with power rankings by applying a topological sort. It was all sort of silly and tongue in cheek, but fun.
Hasse diagram of the 2008 Olympic medal table
21–30 of 37 posts
Re: Hasse diagram of the 2008 Olympic medal table
#22Neat. Has anyone created the diagram for 2012, 2016, or other historical Olympics?
There you go, quick and dirty.
Re: Hasse diagram of the 2008 Olympic medal table
#23> Different people have different conventions for how to rank the Olympic medal results. The newspaper in my country sometimes used another method, where they divide the medal count by the population of the country. In 2008 that put Jamaica at the top, followed by New Zealand. Both China and the U.S. were well down in the chart.
Adding a sheep-factor could help boost NZ up the list.
Re: Hasse diagram of the 2008 Olympic medal table
#24I 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, there's quite a bit of year-to-year variability. Teams go through cycles where they increase the budget to capitalize on a favorable situation, but then they get saddled with bad contracts and are disadvantaged in the draft, etc. Still, if you do this over the long term, there could some teams/GMs that are consistently good.
I remember that my first attempt generated really busy diagrams because the ordering is transitive, so if you simply draw an arrow whenever the ordering holds you end up with the transitive closure. A couple of hacks later (basically, add a test to check if this ordering can be obtained by triangulation, and if so, don't display it) I managed to end with the same kind of diagram as the OP.
Re: Hasse diagram of the 2008 Olympic medal table
#25Earlier quoted context omitted.
You're correct, and on this particular example you can see the problem immediately: is France in the same "rank" as Korea or Italy? Neither, the question is ill-posed.
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.
The point is that your idea of "ordinal rank with ties" isn't specified.
Your ordinal rank is essentially a question of "how many teams are ranked above this one". ie US and China are both 0, Russia is 1 because there is no 1 otherwise, GBR is 2, AUS/GER are tied for 3, Korea and France are 4, and Italy and Japan are 5. But there's no way to claim that Italy and Japan should be below France.
Re: Hasse diagram of the 2008 Olympic medal table
#26Earlier quoted context omitted.
You're correct, and on this particular example you can see the problem immediately: is France in the same "rank" as Korea or Italy? Neither, the question is ill-posed.
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 think you've missed the point that willun is making. It's in general impossible to assign a numeric "rank" to a partially ordered set such that the rank(A) > rank(B) iff A > B and rank(A) = rank(B) iff A is incomparable to B. I'm pointing out a concrete example where you cannot assign a rank to France that is consistent with the rank of Italy and the rank of Korea. Since France is incomparable to both Italy and Korea, it must have the same rank as both, but Korea is better ranked than Italy. Contradiction.
Re: Hasse diagram of the 2008 Olympic medal table
#27Earlier 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 think you've missed the point that willun is making. It's in general impossible to assign a numeric "rank" to a partially ordered set such that the rank(A) > rank(B) iff A > B and rank(A) = rank(B) iff A is incomparable to B. I'm pointing out a concrete example where you cannot assign a rank to France that is consistent with the rank of Italy and the rank of Korea. Since France is incomparable to both Italy and Kor…
I understand now. Thank you!
Re: Hasse diagram of the 2008 Olympic medal table
#28Earlier quoted context omitted.
You're correct, and on this particular example you can see the problem immediately: is France in the same "rank" as Korea or Italy? Neither, the question is ill-posed.
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.
Re: Hasse diagram of the 2008 Olympic medal table
#29why 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"?
Re: Hasse diagram of the 2008 Olympic medal table
#30why 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 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.