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Tournament Brackets (Double Elimination)?


summerpewp

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You do it whatever way you want to mate, my way is the fairest way which is why it is used internationally.

Just because it's used internationally in judo doesn't make it the de facto standard for all sports. Also you say it's the fairest, but what if the person who gains bronze won all their fights except the first round and in the first round they faced the eventual gold medalist? To make your way fair the runner up in the winner bracket should at least have to fight the winner of the losers bracket, since they have both only lost one fight each. With the winner of the lower bracket having actually won more fights (if I worked that out right  :-\). But that's just my opinion.

 

The method (generally) used in e-sports means the only way to win the tournament is to not lose twice. The winner of the lower bracket has to beat the winner of the upper bracket twice to gain the 'gold.

 

As for being easier that's purely down to the fact the governing bodies generally having the forethought to arrange tournaments with x^2 participants (4, 8, 16, 32), which makes it much simpler. Some numbers don't make it too difficult, for example 7, one team/player gets a BYE then it's a go from there. The difficult comes if you allow a stupid number of teams such as 5. You have the first round as 1 v 2, 3 v 4, 5 BYE. You still only have 3 fighters in the upper bracket for round 2.

 

It's possible to debate all day over the merits of each system. At the end of the day they are just different methods. Each of which has found it's niche market.

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yeah this debate isn't about which standard is used... my question stands present... :)

 

the link of the tournament system i posted for the software seems to work well, only tested up to 8 players.. since i'm not buying it.

 

so it can be done in a programming language, so it should be cable in php... just a matter of finding someone with the capabilities...

 

still open to some math for this matter

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There seem to have been very few people who have attempted a solution for a number of teams != 2^x. I did find this one example, but the balancing system may not be what your after. The simplest option would be to have a single elimination play-in round giving ((N=2^x) > num_teams) - num_teams BYEs, choosing the highest Seeded players if appropriate.

 

Edit: Obviously where x is the smallest possible value that makes the statement still true.

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  • 1 year later...

summerpewp did you find a solution in the end?

 

I have been working on a similar project but am unhappy with the results (excuse the punn) to date as everything is still a little too manual for my liking.

 

Qualifying example - http://www.thesnookerforum.com/Tournament-Brackets-Qualifying-Results.php?EventName=ChinaOpen2011

 

32 man Tournament example - http://www.thesnookerforum.com/Tournament-Brackets-Results.php?EventName=WelshOpen2011

 

16 man Tournament example - http://www.thesnookerforum.com/Tournament-Brackets-Results.php?EventName=WLBSA-southern-classic-2011

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