Round Robin Tournament Scheduling
Schedules - You must register to Post and Download => Requests => Topic started by: sezuh on April 22, 2012, 12:17:19 PM
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Hi Ian,Many many thanks for previous help,very much appreciated. :)
May I request something similar to round robin schedule but much more bigger than any thing I read in this forum.
Here it goes;would it be even possible to have "combin(51,3)=20825" scheduled in 17 group of three with 1225 rows? :-[
Much obliged for any comment or help on this problem.
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Hi Sezuh,
Welcome back. I don't know of a solution to your question, but it does seem possible.
As 51 is of the form 6i+3, then there will be a solution in 25 rows where all possible pairs occur once. This is the generalization of Kirkman's schoolgirl problem (http://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem).
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Hi Ian,
Thanks for your time,The 17 threesome with 25 rows you already gave me in my 1 tread last year.
This one is massive 17 threesome with 1225 rows ,if you can suggest a website or anything would be very very much appreciated.
Apart from Round Robin Tournament Scheduling ,or Kirkman's school problem, in what subject I can search it ?
Thanks again for your invaluable help.
Regards
Sezuh
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Sezuh,
You need to start investigating the academic literature as I don't think you will find the answer in text books or on web sites. Look for papers on things such as resolvable 3-designs, triple systems, or sets of all triples. 'resolvable' is the key word here, as this corresponds to your requirement that each row contains each of the 51 players once.
Ian.
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Thanks Ian,
I'll see what can I find,inthe mean time thanks ever so much for all your help and advice,much obliged.
Sezuh