Answer
Move the frogs in
the following order: 2, 4, 6, 5,
3, 1 (repeat these moves in the same order twice more), 2, 4,
6.

This
is a solution in twenty-one moves—the fewest possible.

If *n*,
the number of frogs, be
even, we require ^{(n²+n)}/_{2}
moves, of which ^{(n²-n)}/_{2}
will be leaps and *n*
simple moves.

If *n*
be odd, we shall need (^{(n²+3n)}/_{2})-4
moves, of which ^{(n²-n)}/_{2}
will be leaps and 2*n*-4
simple moves.

In the even
cases write, for the moves, all the
even numbers in ascending order and the odd numbers in descending
order.

This series must be repeated ½*n*
times and followed by the even numbers in ascending order once
only.

Thus the solution for 14 frogs will be (2, 4, 6, 8, 10, 12, 14, 13, 11,
9, 7, 5, 3, 1) repeated 7 times and followed by 2, 4, 6, 8, 10, 12,
14 = 105 moves.

In the odd
cases, write the even numbers in
ascending order and the odd numbers in descending order, repeat this
series ½(*n*-1)
times, follow with the even
numbers in ascending order (omitting *n*-1),
the odd
numbers in descending order (omitting 1), and conclude with all the
numbers (odd and even) in their natural order (omitting 1 and *n*).

Thus for 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated 5
times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 = 73
moves.