
Hasse diagram of the 2008 Olympic medal table - ColinWright
https://tartarus.org/~simon/2008-olympics-hasse/
======
dxbydt
After unambiguously specifying the partial ordering for the triple, its very
gratifying to note how each node gets an ordinal rank in the Hasse. So you can
talk about USA > Russia ( rank 0 vs rank 1) but can't say anything about USA
vs China ( both rank 0) - you can do this simply in terms of the ordinal rank
and forget all about the medal tally.

A few years back, there was a data science problem to predict which startup
had the best chances of exit, based on similar set of presumably
unquantifiable predictors. There were some 10000 startups, and for each
startup you knew who the VC's were, number of rounds of funding, valuation
etc. Though it seemed quite an amorphous problem, we hit upon a very similar
partial order, that made all the startup nodes fall into a Hasse and you could
practically read off from top to bottom which startup would succeed and which
ones would fail.

At that time, I remember thinking how you could do the same exact thing for
humans too. Quantitative reductionism at its finest, sure, but when it works
it's pretty amazing.

~~~
wilun
> After unambiguously specifying the partial ordering for the triple, its very
> gratifying to note how each node gets an ordinal rank in the Hasse. So you
> can talk about USA > Russia ( rank 0 vs rank 1) but can't say anything about
> USA vs China ( both rank 0) - you can do this simply in terms of the ordinal
> rank and forget all about the medal tally.

I think you can't do that on a partial order.

~~~
tgb
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.

~~~
dxbydt
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.

~~~
tgb
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.

~~~
dxbydt
I understand now. Thank you!

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vorg
> 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.

~~~
pishpash
A more helpful metric would be medals divided by delegation size.

~~~
Someone
That partly measures a policy choice "how good must athletes be for us to send
them to the Olympics?"

Different countries make different choices there. Small poorer countries, for
example, aren't likely to have any athlete who can get to the final in the 100
meters, but yet may want to send an athlete to that event, simply because it
is one of the cheapest events to send someone for, and they don't want to send
a zero sized team (for example because they want to earn 'legitimacy points'
as a country or because their Olympic committee wants to make a trip to the
Olympics)

The USA, on the other hand, isn't likely to send anybody who can't make the
final in that event. Part of the reason for that is that they try hard good
candidates, but part also simply is that they can make way more throws of the
genetic dice, and thus are more likely to hit on an outlier who performs
exceptionally well.

~~~
pishpash
If you have a notion of "country" that you're measuring things against, then
the genetic darts they have is just an intrinsic property. Normalizing by
total population is almost certainly not the right metric, but to your point,
there could exist more sophisticated distribution-based metrics that better
capture the distribution mean which is what you might be after.

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esfandia
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) < budget(B))

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.

~~~
Magnap
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.

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te
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.

~~~
nicky0
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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tunesmith
This reminds me of the beatpaths graphs I used to do for the NFL
([http://beatpaths.com](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.

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obel1x
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"?

~~~
grzm
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.

~~~
wtallis
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.

~~~
grzm
Cheers!

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yeldarb
Neat. Has anyone created the diagram for 2012, 2016, or other historical
Olympics?

~~~
shloub
There you go, quick and dirty.

[http://paste.awesom.eu/MtSW](http://paste.awesom.eu/MtSW)

[https://imgur.com/a/YpPqy](https://imgur.com/a/YpPqy)

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autokad
i always favored the 5 gold / 3 silver / 1 bronze scoring, the way track and
fields meets were scored in high school.

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mlamat
[http://www.medalspercapita.com/](http://www.medalspercapita.com/)

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pishpash
Somebody discovered partial ordering.

