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Topic: Queen Alice 1st Anniversary Tournament
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| How about: 30 days each player per game + 1 additional day per move made? |
This was proposed before, I like the idea, but I won't have time to implement it in time, maybe I'll do it for next year's tournament.
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Hi Miguel,
Regarding “random pairing�, I've been trying to figure out if this will produce the right winner in an n^3 Swiss system. (n= no of players). Apologies if this is rather late, but better late than never.
Imagine a simple tournament where there are n=16 unique players. Strongest (called #1) = 2500 rating, weakest (called #16) = 1000 rating, and all 16 players ratings are = 2500, 2400, 2300, etc. down to 1000. According to the n^3 method, there will be 3 rounds (since 3^3 = 27 > 16).
If there are only 3 rounds, it is not clear to me if #1 (the strongest player) will necessarily emerge the winner, even if all 16 players performance follows rating. For example, suppose if due to random pairing, # 1 met #10, #12 and #15 (and so, score maximum 6 points). But a weaker player, say #3 also met #4, #8, #9 (and also score maximum 6 points). Since #1 and #3 ties with 6 points each, how will the system determine the winner? Note #3 met with stronger opponents (#4, 8, 9 is stronger than #10, 12, 15).
If the tie-break is based on the strength of opposition, I think #1 will surely feel he is unfairly treated, because he was not given the opportunity to meet stronger opposition, due to random pairing.
There are 2 ways to overcome this problem – either increase the number of rounds by 1 more (which can extend the tournament), or change the pairing system, so that #1 (and everybody else) is given the opportunity to meet stronger opponents. The latter can use what I call the “(n/2+1)� pairing system, which involve the following simple steps.
Step 1 – rank all 16 players by rating #1 to #16.
Step 2 – split into 2 equal size list (here, 8 players each), using (n/2+1) rule, so that Top 8 goes to List 1, Bottom 8 goes to List 2. Note n/2+1 = 16/2+1 = 9, so, Player #9 is the cut-off, to become the first highest ranked player in the 2nd list.
Step 3 – With the 2 lists, match the top players from each list. So, #1 will play #9, #2 vs #10, #3 vs #11, and so on, whilst ensuring that no one plays each other twice.
Step 4 – let them play, and for the next Round pairing, repeat the steps 1-3.
Using these steps, you will find that #1 will be paired with #9 in Round 1, and win. Round 2, he will be paired with #5, and win. Round 3, he will be paired with #3, and win. So, he got maximum 6 points against #9, #5, #3.
What about player #2? He will be first paired with #10 and win, Round 2, against #6 and win. Round 3, against #4 and win. So, he also got maximum 6 points against #10, #6 and #4.
Then, the tie-break can reflect the strength of the opposition, and we will declare #1 winner as he met stronger opposition. In my mind, this seems fairer.
Miguel, although this is wordy, I hope you understand and find the pairing / sorting concepts straightforward and practical. I personally feel this is fairer than random pairing, and I hope to see the best player winning the first Queen Alice tournament!
Cheers.
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SengTeoh, I think you are right. The n^3 estimation for the number of rounds and players is not correct, it works better as n gets bigger, but for small number of rounds is a really bad estimation.
I did a bit more research online and found another estimation that at least in the simulations I'm running here gives a single winner every time. The formula is 2^n, n being again the number of rounds. So for a 6 round tournament we should not have more than 64 players. If we add one more round we can handle 128 players and we are past that number already. Besides, I prefer to have an even number of rounds, so that most players get to play the same number of games with both colors, so I need to add 2 more rounds.
To summarize, I will change the number of rounds to 8, and that will be ok for up to 256 players.
Now on to the pairing algorithm. I am aware of the pairing method you suggest, but I decided to go with random pairings because they are much faster to compute. With all the conditions that need to be checked it may take the web server several minutes to compute the pairings for the next round. If random pairings turn out to be a complete disaster I will change things so that next time the pairings are done offline so that I have the liberty of doing it on my own computer and not the server.
Miguel
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changed to 3days/move? I am out
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It does seem unfair to set the tournament up with one time control and then change it because a few people complained.
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