Hello, Natasha!

Edit: my formula was wrong . . . I've corrected it.

This is a classic peg-jumping puzzle. .I worked on this many years ago.

Traditionally, there is an __equal__ number of pegs (frogs) on each side.

If there are *n* pegs on each of the two sides,

. . the number of moves is: .(n+1)² - 1

. . and there is a lovely pattern to the moves.

Example: n = 3. .There are 3 blue pegs and 3 green pegs.

Number the seven **positions** (not the pegs): Code:

B B B _ G G G
1 2 3 4 5 6 7

Now move the pegs in the following __positions__: Code:

4
/
3 → 5
\
2 ← 4 ← 6
/
1 → 3 → 5 → 7
/
2 ← 4 ← 6
\
3 → 5
/
4

The first "4" means "move the peg in position #4".

. . Of course, there is no peg to move in position #4.

I included that initial "4" only to complete the diamond pattern.

Then it says: move 3, move 5, move 6, . . .

. . For each move, there is exactly one possible move.

Try it . . . With a little practice, you can solve it at dazzling speed.