Leaping frog Game
I wandered if anyone could give me as much as possible info on the task I need to investigate at:
I have worked out that frog 3 blue frogs and 3 green frogs it will take me 15 moves.
I have worked out that frog 3 blue frogs and 4 green frogs it will take me 19 moves.
I have worked out that frog 4 blue frogs and 4 green frogs it will take me 24 moves.
I have worked out that frog 4 blue frogs and 5 green frogs it will take me 29 moves.
I have worked out that frog 5 blue frogs and 5 green frogs it will take me 35 moves.
If n represents the number of frogs in a team then:
n = 1 we have 3 = n * 3 moves
n= 2 we have 8 = n * 4 moves
n = 3 we have 15 = n * 5 moves
n = 4 we have 24 = n * 6 moves
n = 5 we have 35 = n * 7 moves
n = 6 we have 48 = n * 8 moves
Or more generally if there are n frogs on each side, then the minimum number of moves will be n*(n+2)
This formula does not work if the number of frogs in each team is uneven
If n represents the number of frogs in a team and m the number of frogs in the other then:
n = 1 and m = 6 we have 13 = 1 * 6 + 1 + 6 moves
n = 4 and m = 3 we have 19 = 4 * 3 + 4 + 3 moves
n = 2 and m = 1 we have 5 = 2 * 1 + 2 + 1 moves
n = 5 and m = 6 we have 41 = 5 * 6 + 5 + 6 moves
n = 5 and m = 5 we have 35 = 5 * 5 + 5 + 5 moves
So if there are n frogs on one side and m on the other, then the minimum number of moves will be n*m+n+m
Other patterns notices but I can't explain why (please help!)
- For each frog extra that one team has, there will be a repetition of the jumps and slides. The string of jumps and slides is also symmetric.
- If there are the same number of frogs in each team, then each frog will have to move n+1 times.
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):
B B B _ G G G
1 2 3 4 5 6 7
Now move the pegs in the following positions:
3 → 5
2 ← 4 ← 6
1 → 3 → 5 → 7
2 ← 4 ← 6
3 → 5
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.
But surely the general form cannot be n² - 1 as this would give 8 (for n = 3) and the answer is 15 moves
It's actually 3^2 + 3 + 3
Originally Posted by Natasha1
Or generally MN + M + N where M = no. of frogs on one side, N = number on other side.
In the first case, M = 3 = N
So you get 15!
Which is what you have already...
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