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(Moving Counter Problem)
If you mark off ten divisions on a sheet of paper to represent the
chairs, and use eight numbered counters for the children, you will have
a fascinating pastime. Let the odd numbers represent boys and even
numbers girls, or you can use counters of two colours, or coins.
The puzzle is to remove two children who are occupying adjoining chairs
and place them in two empty chairs, _making them first change sides_;
then remove a second pair of children from adjoining chairs and place
them in the two now vacant, making them change sides; and so on, until
all the boys are together and all the girls together, with the two
vacant chairs at one end as at present. To solve the puzzle you must do
this in five moves. The two children must always be taken from chairs
that are next to one another; and remember the important point of making
the two children change sides, as this latter is the distinctive feature
of the puzzle. By "change sides" I simply mean that if, for example, you
first move 1 and 2 to the vacant chairs, then the first (the outside)
chair will be occupied by 2 and the second one by 1.


There are a good many different solutions to this puzzle. Any contiguous
pair, except 7-8, may be moved first, and after the first move there are
variations. The following solution shows the position from the start
right through each successive move to the end:--
. . 1 2 3 4 5 6 7 8
4 3 1 2 . . 5 6 7 8
4 3 1 2 7 6 5 . . 8
4 3 1 2 7 . . 5 6 8
4 . . 2 7 1 3 5 6 8
4 8 6 2 7 1 3 5 . .

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