Comparison of pairs

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By expanding the above movement in a different way, it is possible to examine the comparison of pairs with each other throughout the event.

Board sets:   1   2   3   4   5   6   7   
Rounds                        
1      8-1         6-3      7-2   4-5   
2      5-6   8-2         7-4      1-3   
3      2-4   6-7   8-3         1-5      
4         3-5   7-1   8-4         2-6   
5      3-7      4-6   1-2   8-5         
6         4-1      5-7   2-3   8-6      
7            5-2      6-1   3-4   8-7   

In the case of pair 1 it can be seen that on set one of the boards, they are compared with pairs 6, 4and 7. On set 2 with 2, 7and 5 etc.

Note that each pair is compared with all 7 other pairs and there are 7 sets of boards. Each pair is compared with 3 other pairs on each set of boards. Therefore, for the comparisons to be equal, each pair should be compared with every other pair 3 times.

A movement where each pair is compared with every other pair the same number of times is called a balanced Howell. This is the case with the above movement and can be checked by the reader quite easily. For example, pair 1 is compared with pair 6 on sets 1, 3 and 4.

For a shorthand method of testing the comparisons, it is necessary to examine the shorthand representation of the Howell.
Eliminating any reference to the placement of the boards, the resultant diagram is called a skeleton.

8-1   6-3   7-2   4-5

or by re-arranging the result is:

N-S   8   7   6   5

E-W   1   2   3   4

The perfect sequence of numbers should be noted (pairs 6 and 3 are repositioned and pairs 4 and 5 are switched) but it also should be realised that this perfect ordering is not necessary, and in general, is not there. Note the switching of 4 and 5 makes the movement unbalanced.

The quality that is essential for this to be a valid skeleton is that the algebraic differences between each pair of numbers are themselves different. The highest numbered pair and pair 1 are ignored for these purposes (i.e. 8 and 1 in this case).

5 minus 4 equals 1

6 minus 3 equals 3

7 minus 2 equals 5

Note that the difference between 7 and 2 is actually 2 (the numbers run in sequence 6, 7, 1, 2, 3...), so the three differences are 1, 2 and 3. These are all different.

Examining the expansion of the movement given for the comparisons, it is possible to find a second skeleton under board set 1.

N-S   5   2   3

E-W   6   4   7

Diff's:   1   2   3

It is this vertical skeleton that has to be examined to determine if in fact the Howell is balanced, not the horizontal skeleton. Notice that the table with pairs 4 and 5 was switched in the diagram of the first (horizontal) skeleton. This means that the 5-6 couple must be also switched (the 5-6 couple is also difference 1) in the vertical skeleton.

The resultant skeleton (eliminating the highest numbered pair as it is not used here) would be:

N-S      6   2   3

E-W   1   5   4   7

Taking all the algebraic differences of the N-S pairs, and all the algebraic differences of the E-W pairs, it is possible to determine if the movement is balanced or not.

N-S differences:   6-2=4;   6-3=3;   3-2=1

E-W differences:   5-1=4;   4-1=3;   7-1=6;   5-4=1;   7-5=2;   7-4=3

As a difference of 1 is equivalent to a difference of 6 etc., the total number of occurrences of each difference would be:

Differences:   1&6   2&5   3&4

Totals:   3   1   5

These are not equal; therefore the movement is not balanced. Switching the 6 and 5 back the result is:

N-S      5   2   3   
E-W   1   6   4   7
   

N-S differences:   5-2=3;   5-3=2;   3-2=1.

E-W differences:   6-1=5;   4-1=3;   7-6=1;   6-4=2;   7-1=6;   7-4=3.

This gives:
Differences:   1&6   2&5   3&4   
Totals:       3    3    3   

All the totals are equal therefore the movement is balanced. Unfortunately this is only possible for an even number of tables, where these totals will always be equal to T-1 where T = the number of tables.

For an odd number of tables to be balanced, these totals must be equal to T or T-2. Even then, the balance is not perfect, for if it were, it would be necessary for the totals to equal T-1 and this is not possible.

It is possible on the other hand, to find a Howell for an odd number of tables where the switching of the pairs at one table will reverse the comparisons. This means that if half the boards at that table are played N-S on each round, and the other half are played E-W, then the resultant movement will be balanced. For example, switching table 4 in the following 5-table movement will achieve this effect.

10-1   8-9   6-4   4R   7x3   2-5