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 2n - 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 ½n 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.
This complete general solution is published here for the first time.