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Leaping Frog Game

  1. Sep 23, 2006 #1
    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.
     
  2. jcsd
  3. Sep 23, 2006 #2
    No one? :-(
     
  4. Sep 23, 2006 #3

    0rthodontist

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    Well, from playing that there appears to be a simple strategy, namely:
    1. Let A = blue and B = green
    2. Make all possible jumps of color A over color B
    3. Make the one non-jumping move of color A.
    4. Exchange colors A and B and repeat from step 1
    You could probably prove that this works by induction though I suspect you might have a hard time showing that it's optimal.
     
  5. Sep 23, 2006 #4

    StatusX

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    Homework Helper

    How many total spaces have to be moved through? How many jumps will there be, and how will these reduce the number of moves required to reach the above total number of spaces?
     
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