Tournament Permutation Problem

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In summary, the conversation discusses the logistics of determining the best team out of ten teams in three nights. Suggestions for a fair scoring system include using the existing ranking, implementing a Swiss pairing system, or having a "last one standing" type of competition. However, each option has its own limitations and may not result in a completely fair outcome.
  • #1
DaveC426913
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Looking for a solution to play off ten darts teams on five boards in only 3 sessions.
I've got ten teams, five boards, and only three nights to figure out who's the best team.

We've had our regular season, so we have the teams ranked. (i.e. 1st place in regular season gets to play against last place, etc.)

Can we get a fair score out of 10 teams in three nights?

(Is that enough information? It's been so long. Curse you, Covid!)
 
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  • #2
What do you call a "fair score"?

Assuming each session only allows one competition per board: The largest possible connected set of teams (A played B, B played C, ...) is 23=8. Your 10 teams will be split into at least two groups who had no games against any team of the other group.
You can use the existing ranking as assistance, but then you'll have to figure out how much you want to rely on that and how much impact these three sessions should have. Any choice will favor some teams and disfavor others. "We use the existing ranking" is the least ambiguous option, but obviously that's not what you want.
 
  • #3
You could have the teams shoot darts at each other and the last one standing wins. :-)
 
  • #5
Single elimination appears to require 4 rounds, so unless someone plays twice in one night that won't fit.

You could use a Swiss pairing, which has the advantage of being non-elimination so everyone gets 3 matches.
 
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  • #6
mfb said:
Assuming each session only allows one competition per board: The largest possible connected set of teams (A played B, B played C, ...) is 23=8. Your 10 teams will be split into at least two groups who had no games against any team of the other group.

This doesn't sound right to me. If the first round is 1 plays 2, 3 plays 4 etc, then the next round is 2 plays 3, 3 plays 4,.., 10 plays 1, then 1 played against 2 played against 3 played against 4 played against 5 etc. You only need two rounds to get a connected set.
 
  • #7
pasmith said:
You could use a Swiss pairing, which has the advantage of being non-elimination so everyone gets 3 matches.
Yes a Swiss pairing system is designed for exactly this situation. The only disadvantage is that pairings cannot be decided in advance which can make it impractical where teams have to travel to each other.
 
  • #8
DaveC426913 said:
Can we get a fair score out of 10 teams in three nights?
Yes, but you might have a tie (## \lceil \log_2 n \rceil ## rounds are required to avoid this).
 
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  • #9
pbuk said:
Yes a Swiss pairing system is designed for exactly this situation. The only disadvantage is that pairings cannot be decided in advance which can make it impractical where teams have to travel to each other.
That should be all right. Most of us can climb the flight of stairs (to the lounge, with the fifth dart board) without too much forewarning. (We're all in the same club.) :wink:
 
  • #10
Office_Shredder said:
This doesn't sound right to me. If the first round is 1 plays 2, 3 plays 4 etc, then the next round is 2 plays 3, 3 plays 4,.., 10 plays 1, then 1 played against 2 played against 3 played against 4 played against 5 etc. You only need two rounds to get a connected set.
Oh right, I forgot the "with a unique winner" condition. In your example two teams winning both matches don't have a comparison. It's impossible to beat single-elimination or equivalent systems there.
 

Related to Tournament Permutation Problem

What is the Tournament Permutation Problem?

The Tournament Permutation Problem is a mathematical problem that involves arranging a set of objects or individuals in a specific order, such as in a tournament bracket. It is also known as the Round Robin Tournament Problem.

What is the objective of the Tournament Permutation Problem?

The objective of the Tournament Permutation Problem is to find all possible ways to arrange a set of objects or individuals in a round-robin tournament format, where each object or individual competes against every other object or individual exactly once.

How is the Tournament Permutation Problem different from other permutation problems?

The Tournament Permutation Problem is unique because it involves arranging objects or individuals in a specific order based on a round-robin tournament format, rather than just arranging them in a linear or circular order. This adds an additional layer of complexity to the problem.

What are some real-life applications of the Tournament Permutation Problem?

The Tournament Permutation Problem has various real-life applications, such as in sports tournaments, where teams or players compete against each other in a round-robin format. It is also used in scheduling and organizing events, such as conferences or workshops, where each participant needs to interact with every other participant.

What are some strategies for solving the Tournament Permutation Problem?

One strategy for solving the Tournament Permutation Problem is to use a recursive approach, where the problem is broken down into smaller subproblems until a solution is found. Another strategy is to use mathematical formulas and algorithms, such as the Berge-Tutte formula, to calculate the number of possible permutations. Additionally, computer programs and algorithms can be used to efficiently generate and evaluate possible permutations.

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