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Topic: 2010 World Championship (Read 16507 times) |
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Adanac
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Re: 2010 World Championship
« Reply #45 on: Jan 20th, 2010, 8:59am » |
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on Jan 20th, 2010, 7:56am, Fritzlein wrote:One way to make the parameter automatic based on the players who actually register rather than eyeballed by me after the fact would be to set it to
0.5 + ([highest rating] - [lowest rating])/400
a formula which works both for last year and this year. Last year it would have been 4.025, and this year 2.9675, i.e. basically 4 then and 3 now as I am recommending. |
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What if we have a strong tournament next year (2100 average rating) except we have 1 new player with a 1400 rating? Should we base the parameter upon (highest rating - lowest rating), or is it better to use standard deviation to avoid the outlier effect?
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| « Last Edit: Jan 20th, 2010, 8:59am by Adanac » |
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Janzert
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Re: 2010 World Championship
« Reply #46 on: Jan 20th, 2010, 10:10am » |
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I was quite confused reading this at first, and quite probably I still am, since it appears to me that the published rules never received this change in pairing and still list the simple "sum of opponent scores" as the first tiebreaker. Specifically in rule 6 of the swiss preliminary pairing algorithm.
Janzert
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Fritzlein
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Re: 2010 World Championship
« Reply #47 on: Jan 20th, 2010, 11:38am » |
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on Jan 20th, 2010, 8:59am, Adanac wrote:| What if we have a strong tournament next year (2100 average rating) except we have 1 new player with a 1400 rating? Should we base the parameter upon (highest rating - lowest rating), or is it better to use standard deviation to avoid the outlier effect? |
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You are right, the standard deviation is probably more robust than the range. For 2009 I calculate a standard deviation of 433 and for 2010 I get 297. So perhaps the exponent could be set to stdev/100.
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Fritzlein
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Re: 2010 World Championship
« Reply #48 on: Jan 20th, 2010, 11:42am » |
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on Jan 20th, 2010, 10:10am, Janzert wrote:| I was quite confused reading this at first, and quite probably I still am, since it appears to me that the published rules never received this change in pairing and still list the simple "sum of opponent scores" as the first tiebreaker. Specifically in rule 6 of the swiss preliminary pairing algorithm. |
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Uh oh. Tournament Director?!!?!!?
The good news is that the pairings of rounds 1, 2, and 3 will always be the same under the old SoS and new SoS, so it hasn't made any difference so far. But it could make a difference for pairing rounds 4 and 5, as well as influencing the final standings.
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| « Last Edit: Jan 20th, 2010, 11:42am by Fritzlein » |
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omar
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Re: 2010 World Championship
« Reply #49 on: Jan 20th, 2010, 12:25pm » |
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on Jan 20th, 2010, 10:10am, Janzert wrote:I was quite confused reading this at first, and quite probably I still am, since it appears to me that the published rules never received this change in pairing and still list the simple "sum of opponent scores" as the first tiebreaker. Specifically in rule 6 of the swiss preliminary pairing algorithm.
Janzert |
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Yes, I mentioned in the forum that the change Karl suggested for computing SoS looked simple enough that I could do it this year. I was going to update the rules page after implementing it, but forgot to. Maybe it turned out to be a good thing since my first implementation of the SoS formula was not quite right and a more general version of the SoS formula has been suggested now. So I'll update the rules page now.
But before I do I'll fix my implementation (forgot to divide the score by 2; I used win=2 points) and also use the more general formula with the factor in the exponent computed as: 0.5 + stdev(player ratings)/100.
Fortunately the rule change would not impact anything so far.
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| « Last Edit: Jan 20th, 2010, 12:27pm by omar » |
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Fritzlein
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Re: 2010 World Championship
« Reply #50 on: Jan 20th, 2010, 12:30pm » |
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on Jan 20th, 2010, 12:25pm, omar wrote:| and also use the more general formula with the factor in the exponent computed as: 0.5 + stdev(player ratings)/100. |
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There is no +0.5 any more, just stdev(player ratings)/100.
The exact "curvature" of the SoS calculation doesn't matter as much as the fact that it is curved. The whole point is to have differences near one's own strength count for more than differences far from one's own strength, and any exponent will do that. But as long as we are curving it at all, we might as well try to get the curvyness as close to correct as we can.
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| « Last Edit: Jan 20th, 2010, 12:34pm by Fritzlein » |
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omar
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Re: 2010 World Championship
« Reply #51 on: Jan 20th, 2010, 1:15pm » |
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I updated the implementation.
The stdev is 288.516
The exponent factor is: 3.385
The new SoS values can be seen here:
http://arimaa.com/arimaa/events/showGames.cgi?e=2010wc
chessandgo now has a SoS value of 0.075 in round 3.
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RonWeasley
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Re: 2010 World Championship
« Reply #52 on: Jan 21st, 2010, 6:35am » |
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on Jan 20th, 2010, 1:15pm, omar wrote:
From the previous discussion it looks like the 0.5 offset is still in the implementation and should not be. Exponent factor should be 2.885. I'm approving the stdev/100 exponent for the remainder of the tournament.
TD
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omar
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Re: 2010 World Championship
« Reply #53 on: Jan 21st, 2010, 1:59pm » |
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Thanks for the approval Ron and also noting the problem. Karl had also notified me about it. Should be fixed now.
Good thing we got this corrected and approved now. If we think of any other improvements after round 3, it will have to wait until next year.
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| « Last Edit: Jan 21st, 2010, 2:03pm by omar » |
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Adanac
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Re: 2010 World Championship
« Reply #54 on: Jan 24th, 2010, 8:40pm » |
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Standings after 7 games of round 3:
http://arimaa.com/arimaa/mwiki/index.php/2010_Open_Classic_Round_3#Round _3_Standings
Disclaimer: These pairings are not official and depend upon whether my interpretation of the asterisk pairings below are correct
If 99of9 wins tomorrow
Chessandgo vs. Adanac
Tuks vs. 99of9 (*)
Fritzlein vs. Nevermind
The_Jeh vs. Nombril
Simon vs. Omar
PMertens vs. Naveed (**)
Woh vs. Hippo
ChrisB vs. fritzlforpresident
If Omar wins tomorrow
Chessandgo vs. Adanac
Fritzlein vs. Nevermind
Tuks vs. The_Jeh
Omar vs. Nombril
PMertens vs. Naveed (**)
Simon vs. Hippo
Woh vs. 99of9 (***)
ChrisB vs. fritzlforpresident
(*) The pairing 3 vs. 6 would be Tuks vs. Nevermind but they have already played. And 3 vs. 7 would be Tuks vs. Nombril, but they’ve already played too. So Tuks vs. 99of9 represents a 3 vs. 8 pairing, which should force a 4 vs. 6 and 5 vs. 7 in the other two 2-1 games.
(**) PMertens would normally be scheduled against Hippo but because they’ve already played he is re-scheduled against Naveed in both scenarios.
(***) Uh oh. This pairing has already occurred, and the program will not be able to create a legal pairing. See the discussion below.
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| « Last Edit: Jan 25th, 2010, 7:13am by Adanac » |
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99of9
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Re: 2010 World Championship
« Reply #55 on: Jan 25th, 2010, 5:20am » |
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In your Omar wins scenario, you've got me playing Woh, but we've already played. Does that change it?
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Fritzlein
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Re: 2010 World Championship
« Reply #56 on: Jan 25th, 2010, 6:30am » |
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on Jan 25th, 2010, 5:20am, 99of9 wrote:| In your Omar wins scenario, you've got me playing Woh, but we've already played. Does that change it? |
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It appears that in the "omar wins" scenario, the pairing algorithm will break. For the preliminary tournament, we are not using the "global optimum floating elimination" pairing. Instead we are using relatively straightforward Swiss pairing which makes one pass through the players trying for a good pairing, and therefore may fail to produce any pairing at all.
If I understand the "omar wins" scenario, the algorithm will have successfully paired all players except 99of9, woh, ChrisB and fritzlforpresident. It will then try to pair 99of9. Since 99of9 vs. woh has already happened, it will skip that, and pair 99of9 vs. ChrisB. But then it is left with only woh and fritzlforpresident, who have already played each other. I believe then it exits with failure.
Any human TD confronted with the problem of 99of9 vs. woh being a repeat would not cross score boundaries to create two 1-2 vs. 0-3 pairings. A human TD would backtrack one step to undo Simon vs. Hippo, and then complete the pairing with Simon vs. woh, 99of9 vs. Hippo, and ChrisB vs. fritzlforpresident.
I think we may have an appeal to RonWeasly brewing unless omar can pull out the loss today.
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Adanac
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Re: 2010 World Championship
« Reply #57 on: Jan 25th, 2010, 7:09am » |
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on Jan 25th, 2010, 6:30am, Fritzlein wrote:
It appears that in the "omar wins" scenario, the pairing algorithm will break. For the preliminary tournament, we are not using the "global optimum floating elimination" pairing. Instead we are using relatively straightforward Swiss pairing which makes one pass through the players trying for a good pairing, and therefore may fail to produce any pairing at all.
If I understand the "omar wins" scenario, the algorithm will have successfully paired all players except 99of9, woh, ChrisB and fritzlforpresident. It will then try to pair 99of9. Since 99of9 vs. woh has already happened, it will skip that, and pair 99of9 vs. ChrisB. But then it is left with only woh and fritzlforpresident, who have already played each other. I believe then it exits with failure.
Any human TD confronted with the problem of 99of9 vs. woh being a repeat would not cross score boundaries to create two 1-2 vs. 0-3 pairings. A human TD would backtrack one step to undo Simon vs. Hippo, and then complete the pairing with Simon vs. woh, 99of9 vs. Hippo, and ChrisB vs. fritzlforpresident.
I think we may have an appeal to RonWeasly brewing unless omar can pull out the loss today. |
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Yes, I double-checked the results and 99of9 would be in 11th place with an SoS of 1.8753 and Woh would be in 14th with 1.6048. I thought matching the 1-2 and 0-3 players superceded repeat pairings but I was wrong. Fritzlein's interpretation above is the correct one.
This is hypothetical because, *statistically*, this is the less likely of the two scenarios but...We're fortunate that woh vs. fritzlforpresident has already occurred because that breaks the algorithm and would allow RonWeasley to manually create a more logical pairing. If woh vs. ffp had not previously been paired, then there would be a debate about whether to invoke this disclaimer from the bottom of the tournament rules.
"The director will also make the final decision on matters of pairing and color assignment. However, the pairings and color assignments are done as much as possible using a program and the need for intervention should be rare."
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| « Last Edit: Jan 25th, 2010, 7:25am by Adanac » |
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Fritzlein
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Re: 2010 World Championship
« Reply #58 on: Jan 25th, 2010, 10:12am » |
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on Jan 25th, 2010, 7:09am, Adanac wrote:| We're fortunate that woh vs. fritzlforpresident has already occurred because that breaks the algorithm and would allow RonWeasley to manually create a more logical pairing. If woh vs. ffp had not previously been paired, then there would be a debate about whether to invoke this disclaimer from the bottom of the tournament rules. |
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It may be that Omar has extended the pairing algorithm to backtrack until it finds at least one legal pairing, albeit not the global optimum. I think he mentioned working on such code. In that case the algorithm might undo 99of9 vs ChrisB, pair 99of9 vs. fritzlforpresident and ChrisB vs. woh, and exit with success. If that happens, I will appeal the pairing, since there is a simple change that does not cross score boundaries. (and incidentally would be the second legal pairing discovered by a backtracker)
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RonWeasley
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Re: 2010 World Championship
« Reply #59 on: Jan 25th, 2010, 10:31am » |
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on Jan 25th, 2010, 6:30am, Fritzlein wrote:
Any human TD confronted with the problem of 99of9 vs. woh being a repeat would not cross score boundaries to create two 1-2 vs. 0-3 pairings. A human TD would backtrack one step to undo Simon vs. Hippo, and then complete the pairing with Simon vs. woh, 99of9 vs. Hippo, and ChrisB vs. fritzlforpresident.
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Thank you for seeing this in advance. If this happens, based on the current consensus, I approve the suggested pairing for Round 4.
TD
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