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 3_n_^{2} + 2_n_ - 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} + 4_n_ + 2 moves. If we give _n_ the value 6,
as in the example of the Frogs' Ring, the number of moves would be 62.
For a general solution of the case where frogs of one colour reverse
their order, leaving the blank space in the same position, and each frog
is allowed to be moved in either direction (leaping, of course, over his
own colour), see "The Grasshopper Puzzle" in _A. in M._, p. 193.