Josh wrote:
> [EMAIL PROTECTED] (Tanguy) wrote in message
>news:<[EMAIL PROTECTED]>...
>
>>I would like to thank you for sharing your thoughts and I would like
>>also to answer some issues
>>you've raised.
>>Firstly, even if the lead is harshly disputed and even if the boats
>>are nearly similar, the boats
>>are sufficiently away from each other at the finish line that it is
>>easy to say who's the winner.
>>Thus a tie is impossible.
>>Secondly, the order in the distribution of points is not taken into
>>account. As you intuited,
>>"210", "201", "120", "102", "021", "012" means merely that "of three
>>boats, one of them one
>>twice, another one won once, and the third lost all three races".
>>I am still looking for a way to guess the number of patterns, rather
>>than the indivual outcomes
>>from a given number of races or trials.
>>
>>Tanguy Arzel
>
>
> OK, three trivial corrections I HAVE to make :)
>
> (1) Louis Vuitton Cup (not Vitton)
> (2) Hauraki Gulf (not bay)
> (3) Scenarios (not scenarii)
>
> Pedantic, I know...
>
> If I understand your problem correctly, you are not interested in the
> boats that actually win the races, but in the different classes of
> results. So the "210", "201", "120", "102", "021", "012" are all to be
> considered the same thing. Is this right?
>
> If this is the case then we can apply a little graph theory to the
> problem. Each boat is a vertex/node/point of the graph and each race
> is represented by an edge/line/connection between each of the points.
> Each edge is directed, i.e. it points from, say, vertex A to B and not
> from B to A. Suppose that an edge directed from A to B means that boat
> A beat boat B in a race. Since each boat races every other boat in the
> group, there must be an edge connecting each pair of vertices. This is
> called a tournament graph (not a surprising name, eh?).
>
> This graph is explained in Eric's world of mathematics
>
> http://mathworld.wolfram.com/Tournament.html
>
> and the sequence of distinct graphs is given by
>
> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000568
>
> Since there are only 9 syndicates in the Louis Vuitton Cup, the
> sequence given in the previous web page should answer your question.
> There also some references given if you want to find an extended
> sequence
>
> I hope this helps!
>
> Regards,
>
>
> Josh
While your description of the graph is correct, the results given in
Harary and Palmer, and on the att webpage you cite, are not solutions to
this particular problem. In other words, there is a theorem that says
for n = 5 there are exactly 12 "tournaments" of a certain type, but this
is not the number of possible standings asked for.
My enumeration for n = 5 yields 9 possible standings (below). If I have
omitted one (or three!) please let me know.
Neil Henry, Virginia Commonwealth University
Standings for n = 5 round-robin competition:
4 3 2 1 0
4 3 1 1 1
4 2 2 2 0
4 2 2 2 1
3 3 3 1 0
3 3 2 2 0
3 3 2 1 1
3 2 2 2 1
2 2 2 2 2
.
.
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