- #1
Natasha1
- 493
- 9
I wandered if anyone could give me as much as possible info on the task I need to investigate at:
http://maths.fallibroome.cheshire.sch.uk/leapfrog.swf
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:
For:
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:
For:
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.
http://maths.fallibroome.cheshire.sch.uk/leapfrog.swf
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:
For:
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:
For:
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.
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