As there is only one skeleton possible for 3 tables, it is clear there cannot be a Howell of the desirable form. There are of course other solutions to this problem.
The first solution is based on the use of the only skeleton available and using it as a round robin draw with all tables playing the same boards on each round.
Round Boards * Table 1 Table 2 Table 3
1 1-6 6-1 5-2 4-3
2 7-12 6-2 1-3 5-4
3 13-18 6-3 2-4 1-5
4 19-24 6-4 3-5 2-1
5 25-30 6-5 4-1 3-2
* Five board sets may be used.
This movement, as such, is not balanced, but by playing half the boards in each direction at table 2 or 3 gives perfect comparisons. Table 3 is the better choice as one of the pairs goes into the next round at that table. The advantage of this method is the boards can be scored after each round (barometer scoring).
This second movement is better in that there is less board sharing. However the board movement is not orderly. On the other hand it is not as unordered as those in the third movement. The major objection to this second movement is that the comparison of pairs is not equal, but this can be improved by switching the boards on table 1.
Round Table 1 Table 2 Table 3
1 6-1 5-2 4-3
1 4 2
2 6-2 1-3 5-4
2 4 3
3 6-3 2-4 1-5
3 1 2
4 6-4 3-5 2-1
4 1 3
5 6-5 4-1 3-2
5 5 5
The board sets are shown under the pairings, these sets usually consisting of 4 to 6 boards.
The third movement has perfect comparison of players, and has no board sharing, but the disadvantage is the board moving.
Round 1 2 3 4 5
Table 1 6-1 6-2 6-3 6-4 6-5
Boards 1&2 3&4 5&6 7&8 9&10
Table 2 4-3 5-4 1-5 2-1 3-2
Boards 3&9 1&5 3&7 5&9 1&7
Table 3 5-2 1-3 2-4 3-5 4-1
Boards 6&8 8&10 10&2 2&4 4&6
The boards shown above refer to sets of 2 or 3 boards.
