Mathematicians Please !

[sirlearie]

golf trip problem.
8 players .
4 rounds , each fourball format ( ie. 2 teams of four ).
ideally to get every player to play with all other 7 players
at least once ( easy ) but .....
if possible without playing more than twice with each ?
not sure it can be done but help with a grid scheme would be appreciated !!

thanks .
sirlearie.

[Buenchico]

ABCD v EFGH

AFCH v EBGD

ABGH v CDEF

ADGF v CBEH

[jim360]

Where's the AE pairing?

[Buenchico]

Damn! I was concentrating on ensuring that no two players came together more than twice! I'll have to rethink it now!

[Buenchico]

Er, I now think that it's impossible!

[Calibax]

Trickier than it looks isn't it?
I think each player needs to be paired with five of the other seven twice and with the remaining two, once only. I can't think of a reason that this should be impossible but nether can I come up with a grid that satisfies the condition!

[Buenchico]

According to my calculations, you either have to accept that each player will be in the same team as someone else 3 times or, as I've got it above, that each player is always on the opposite side to one other player. In my example A & E always oppose each other, as do B & F, C & G and D & H. In some circumstances that might actually be quite a fun way to play, e.g. if A & E are a husband and wife pairing who normally play together, B & F are brothers with a strong rivalry between them, etc,

[dr b]

A would play with:

ABCD
AEFG
ABCH
AEFD

plays twice with B, C, D, E, F, once with G and H.

Am I missing anything? probably.

[dr b]

Oops, yes, missing a lot. B plays with C 4 times, for example.

[jim360]

By my reckoning it's impossible to keep the condition "no more than twice" for each player. I'd love to come up with a proof of this, though.

[factor-fiction]

I have just seen this thread after work and I thought it seemed familiar- and I found this
http://www.theanswerbank.co.uk/Quizzes-and-Puzzles/Question1162765.html

[mibn2cweus]

For four rounds you need 16 players to fill the opposing team spots. With only seven other players to pool from, one or two of them are going to have to play on the opposing team three or four times.

[mibn2cweus]

On the other hand, if you only play three rounds . . .

ABCD EFGH
ACEG BDFH
AFGH BCDE

A 3 0

B 1 2

C 2 1

D 1 2

E 1 2

F 1 2

G 2 1

H 1 2

[sirlearie]

thanks all for your interest .
yes , factor fiction , well remembered . indeed I did ask this before but still couldn't work out formula which ensured all played each other equally ( or as near as ) .
will look at above answers for some inspiration .
thanks again.
sirlearie.