Deriving Howell movements

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Having already observed that the Howell movement has a horizontal skeleton and a vertical skeleton, it may be intuitively reasoned that if all the skeletons can be found, then all the Howell movements can be found. As the horizontal skeleton defines the starting positions of the pairs, this skeleton will be called the super-skeleton (SS). The vertical skeleton must be balanced and is the foundation of the movement, and so will be called the base-skeleton (BS).

Any skeleton can act as a SS or as a BS. That is, the super-imposing of any skeleton upon another produces a Howell movement. The result will not necessarily be a good Howell (a Howell where no tables share boards) and only occasionally will it be a perfect Howell (a Howell with all the relay boards on one relay when the number of tables is odd; and on two relays where the number of tables is even - see the 4 and 5 table examples given above).

It should be noted that any balanced BS is in fact a Barometer or Butler-scoring-method movement. In these movements, all the tables play all the boards for each round at the same time. This is the trivial case of superimposing a BS on itself (or BS = SS). The next problem is to derive all the skeletons. This is a relatively simple process. The first thing to do is to set up a table of all combinations of the different "differences" of the pairs (ignoring the highest numbered pair and 1).

First write down the trivial situation numbering from left to right and back again.

e.g.      7-2      6-3      5-4

Then add 1 to both the N-S and E-W pairs, reduce modulus 2T-l (ie. divide by 2T-1 and use the remainder), ignore the pairings where pair 1 occurs, and then enter it as the next entry in that column. The number of rows in each column will always be equal to 2T-3 (T = number of tables).

e.g.   for 4 tables the differences are:

2   3   1

7-2   6-3   5-4

2-4   7-4   6-5

3-5   2-6   7-6

4-6   3-7   3-2

5-7   5-2   4-3

To extract the skeletons from this table, one pair of numbers is taken from each column such that no number is mentioned twice.

Therefore, all the skeletons for 4 tables are

a)   8-1   7-2   6-3   5-4

b)   8-1   2-4   3-7   6-5

c)   8-1   5-7   2-6   4-3

By eliminating the first row in the difference table, and placing an imaginary mirror between the middle two rows like this:

2-4   7-4   6-5

3-5   2-6   7-6

4-6   3-7   3-2

5-7   5-2   4-3

it can be seen that the skeleton (c) is in fact a mirror image of skeleton (b). This situation always exists irrespective of the number of tables