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Movement for Rotating-Team-of-Four Tournament

#1 User is offline   plum_tree 

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Posted 2015-December-01, 11:22

Can someone provide a link to the movement for a Rotating-Team-of-Four tournament for 4 tables? This is for a fun tournament at our local club. Instead of playing an Individual Tournament, the idea is to play with your regular partner the whole time, but during the course of the tournament, every other partnership will be your teammates once and your opponents once. With 4 tables it is 8 pairs, and say, 6 boards against every pair, would equate to 42 boards to complete the tournament.

Any help will be appreciated.
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#2 User is offline   gordontd 

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Posted 2015-December-01, 12:32

I haven't come across such a thing but I imagine it would work to play a barometer Howell movement (two sets of boards would be best, but they needn't be duplicated if you don't have the facilities for that) and nominate tables 1 & 3 to be playing a match at any one time and tables 2 & 4 correspondingly.
Gordon Rainsford
London UK
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#3 User is offline   RMB1 

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Posted 2015-December-01, 14:55

View Postgordontd, on 2015-December-01, 12:32, said:

I haven't come across such a thing but I imagine it would work to play a barometer Howell movement (two sets of boards would be best, but they needn't be duplicated if you don't have the facilities for that) and nominate tables 1 & 3 to be playing a match at any one time and tables 2 & 4 correspondingly.

I thought that would work, but the 4-table "Flower" Howell movement (below) does not do what we want. There is no round where 1 and 2 are NS/EW at tables 1/3 or 2/4.

8v1 2v7 3v6 5v4
8v2 3v1 4v7 6v5
8v3 4v2 5v1 7v6
8v4 5v3 6v2 1v7
8v5 6v4 7v3 2v1
8v6 7v5 1v4 3v2
8v7 1v6 2v5 4v3
Robin

"Robin Barker is a mathematician. ... All highly skilled in their respective fields and clearly accomplished bridge players."
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#4 User is offline   gordontd 

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Posted 2015-December-02, 04:52

I've found a different 4-table Howell movement that looks to me as though it would work, whether you have tables 1 & 2 playing the same match or 1 & 3:

8 v 1 ... 5 v 6 ... 2 v 4 ... 3 v 7
8 v 2 ... 6 v 7 ... 3 v 5 ... 4 v 1
8 v 3 ... 7 v 1 ... 4 v 6 ... 5 v 2
8 v 4 ... 1 v 2 ... 5 v 7 ... 6 v 3
8 v 5 ... 2 v 3 ... 6 v 1 ... 7 v 4
8 v 6 ... 3 v 4 ... 7 v 2 ... 1 v 5
8 v 7 ... 4 v 5 ... 1 v 3 ... 2 v 6

It looks like the important thing is that the switch table should be at some point other than between pairs T/2 & T/2 + 1
Gordon Rainsford
London UK
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