Answer
:
The fewest possible
moves in which this puzzle can
be solved
are 118.
I
will give the complete solution.
The black figures on
white discs move
in the directions of the hands of a clock, and the white figures on
black
discs the other way. The following are the numbers in the order in
which
they move.
Whether you have to make a simple move or a leaping move
will
be clear from the position, as you never can have an alternative.
The
moves enclosed in brackets are to be played five times over: 6, 7, 8,
6,
5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1),
6,
5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3,
2,
12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11,
2.
We
thus have made 118 moves within the conditions, the black frogs have
changed places with the white ones, and 1 and 12 are side by side in
the
positions stipulated.
The general
solution in the case of this puzzle is
3n^{2}
+ 2n
 2
moves, where the number of frogs of each colour is n.
The law governing
the sequence of moves is easily discovered by an examination of the
simpler cases, where n
= 2, 3, and 4.
If, instead of 11
and 12 changing places, the 6
and 7 must
interchange,
the expression is n^{2}
+ 4n
+ 2 moves. If we give n
the value 6,
as in the example of the Frogs' Ring, the number of moves would be 62.
