Fred,

I would suggest something like this:

Day 1: (1,2,3,4), (5,6,7,8), (9,10,11,12)
Day 2: (1,2,7,11), (3,5,6,12), (4,8,9,10)
Day 3: (1,6,7,10), (2,3,8,9), (4,5,11,12)

Then the pairs of people who play together twice are:
(1,2), (5,6), (9,10), (Day 1 and Day 2)
(2,3), (6,7),(11,12) (Day 1 and Day 3)
(1,7), (5,12), (8,9) (Day 2 and Day 3)

On any pair of days you will have to have at least 3 pairs of people who play together both days. This is simply because if you split 4 people into 3 teams at least one team will have 2 players (this idea is important enough to have a name, it is called the "pigeon hole principle") This schedule has the minimal number of pairs of people who play together twice.

Karen

Twelve is a hard number for golf schedules. It isn't possible to have no duplicates, even if only on two days. Six duplicate pairs is a theoretical minimum. The reason is that on the second day, each group of 4 must contain two players who were together on day 1. After selecting one from each of the foursomes from the first day -- the only way to get no duplicates -- the fourth player in the group came from the same foursome as one of the first three. The same reasoning holds for the third day.

Let's suppose first there are 9 players in three threesomes per day. Later a fourth player will be added to each threesome. A schedule for 9 players with no duplicates is

Day 1: 123, 456, 789
Day 2: 147, 258, 369
Day 3: 159, 267, 348

Now add three more players A, B and C, one to each group on each day, however you think works best. It looks like six duplicate pairs is easy to achieve in this way.

Best of luck, and have fun.
Victoria