PBTD for 8 teams with more venues

[Innesw20]

In the game of lawn bowls, if there are only 8 teams in a competition there would frequently be more rinks available (the standard green has eight) than are required to play each round (4).

The mathematicians have shown that there does not exist a PBTD for 8 teams and 4 venues.

I was wondering whether a PBTB-equivalent for 8 teams could be constructed where the number of venues is 5 instead of 4? If not, would 6 venues allow it? Or is it just not mathematically possible? This is essentially easing the restriction on the number of venues that applies in the standard definition of PBTDs.

[Ian Wakeling]

The PBTD might not be necessary.  Could you use the regular balanced schedule instead.  Go here and enter 8 items and 'balanced'.

[Innesw20]

Afraid not - it does need to be partitioned: a team at a venue at most once in rounds 1-4 and at most once in rounds 4-7.

I imagine it would look something like this, if it can work for 5 venues (rounds in rows, venues in cols):
(   ) (1 x) (x x) (x x) (x x)
(x x) (x x) (   ) (x x) (1 x)
(x x) (   ) (1 x) (x x) (x x)
(1 x) (x x) (x x) (x x) (   )
(x x) (x x) (1 x) (   ) (x x)
(x x) (   ) (x x) (x x) (1 x)
(   ) (x x) (x x) (1 x) (x x)
Ideally each team should play on each venue at least once in the competition, but that may be asking too much.

[Ian Wakeling]

The more venues there are, the easier things get.  With 6 venues you can have:

(H A) (---) (F G) (B C) (---) (D E)
(---) (H B) (---) (F E) (D G) (A C)
(E G) (D F) (H C) (---) (A B) (---)
(---) (A E) (---) (H D) (F C) (B G)
(F B) (C G) (D A) (---) (H E) (---)
(D C) (---) (B E) (A G) (---) (H F)

which leaves the following 4 pairings to be played in the final round.

(A F) (B D) (C E) (G H)


If you have 7 venues, then there is the Room Square.