# Participation graphs



## clement (Oct 8, 2010)

Hello guys

Following the work of Stefan (http://www.speedsolving.com/forum/showthread.php?20492-Blonk-numbers-and-Center-of-the-Cubieverse), I wanted to derive some visuals of this.
My graph is defined as:
- the vertices are all the competitors who did at least 20 competitions (data from ~one month ago, sorry)
- an edge between two competitors is weighted with the number of competitions they made together

To represent the graph, I choose Himmeli (http://www.finndiane.fi/software/himmeli).
The graph is built as the atoms of a molecule who tried to minimize its energy: the stronger is the link between two atoms, the more likely they will be close.
The result is here: http://s89pq2.1fichier.com/Speedcubers_all.pdf

I also displayed only the minimum spanning tree of this graph (http://en.wikipedia.org/wiki/Minimum_spanning_tree)(well, with maximizing the sum of the weights).
Result here: http://8kceui.1fichier.com/Speedcubers_mst.pdf

For the vertices:
circle = Europe
square = North America
triangle = Asia

Enjoy !


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## Olivér Perge (Oct 8, 2010)

Great job, Clément!

Good to see these connections visually.

Haha, Bence is connected to us via me.  You are welcome!


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## Stefan (Oct 8, 2010)

Very nice . Looks like I'm pretty much the center (see first image) and the main connection between Europe and America (see second image), thanks to my good connections to Ron and Bob. Or did you somehow put me there on purpose?


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## Tim Reynolds (Oct 8, 2010)

Looks cool! One question: I don't see Lars Vandenbergh on there, is that a mistake?

It looks from the weighting of the edges that Ton and Ron have the most competitions in common. I'd be curious to see what the ranking for "most competitions in common" looks like.


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## keemy (Oct 8, 2010)

on the 2nd graph you can see Bob has the most connections, (meaning he is the most popular person to have your greatest number of comps in common with) pretty cool but what else would you expect from Bob.


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## Stefan (Oct 8, 2010)

keemy said:


> on the 2nd graph you can see Bob has the most connections, (meaning he is the most popular person to have your greatest number of comps in common with)


 
Not necessarily. I'm connected with Bob there, but my most common partner is Ron. So I don't count for Bob. Neither do Tyson, Dave and perhaps Kian and Timothy (who seem to be undecided). Shelley on the other hand doesn't lose anyone that way.


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## keemy (Oct 9, 2010)

Stefan said:


> Not necessarily. I'm connected with Bob there, but my most common partner is Ron. So I don't count for Bob. Neither do Tyson, Dave and perhaps Kian and Timothy (who seem to be undecided). Shelley on the other hand doesn't lose anyone that way.


 
oh your right I feel silly but counting for Tim, Bob and Haiyan are tied at 10 competitions in common each, which makes Bob and Shelley both tied at 11 strongest connections


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