John,
I am not aware of any standard solution to this problem. With 28 possible matches, of course there can be no nice soution where the whole tournament can be played out with 5 games per round. One possibility is the following, where the last column gives the players who have a bye.
(3 8) (4 1) (4 7) (7 3) (1 8) - (2 5 6)
(8 2) (6 4) (1 2) (1 6) (8 4) - (3 5 7)
(4 5) (2 7) (2 3) (7 5) (3 4) - (1 6 8)
(8 7) (1 5) (5 6) (7 1) (8 6) - (2 3 4)
(5 3) (6 2) (5 2) (3 6) - (1 4 7 8)
(2 4) (6 7) (5 8) (3 1) - ( )
So the first four rounds are as you specify above, then the penultimate round has only 4 players, while the last round has everyone. If you were willing to play 6 players and 6 games per round then this second schedule may be slightly better.
(3 8) (1 2) (4 7) (1 8) (3 7) (2 4) - (5 6)
(2 8) (4 6) (1 5) (4 8) (2 5) (1 6) - (7 3)
(4 5) (2 7) (3 6) (5 7) (3 4) (2 6) - (1 8)
(7 8) (1 3) (5 6) (1 7) (3 5) (6 8) - (2 4)
(1 4) (6 7) (5 8) (2 3) - ( )
If it is possible for all players to turn up to all rounds, then it's easy as you play 2 normal rounds on the same day. If you want 3 players per round then there will be a number of good solutions. For example with 7 players, the plan below determines which three players turn up to each round.
1 2 4
2 3 5
3 4 6
4 5 7
5 6 1
6 7 2
7 1 3
Each time you play the three matches that are possible, so round 1 would be (1 2) (1 4) & (2 4).
Hope that helps,
Ian.
PS I have moved your request here since it's not possible to reply to postings in the "comments" area of the message board. Please consider registering if you want to reply, as it's perfectly safe.
