(related to Problem: The Six Frogs)
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 $\frac{n^2+n}2$ moves, of which $\frac{n^2-n}2$ will be leaps and $n$ simple moves. If $n$ be odd, we shall need $\frac{n^2+3n}2-4$ moves, of which $\frac{n^2-n}2$ 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 $\frac 12n$ 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 $\frac 12(n-1)$ 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.
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