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Math Help - Leaping frog Game

  1. #1
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    Leaping frog Game

    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.
    Last edited by Natasha1; September 23rd 2006 at 03:06 AM.
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  2. #2
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    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.

    Last edited by Soroban; September 23rd 2006 at 09:56 AM. Reason: error
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  3. #3
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    Thanks Soroban,

    But surely the general form cannot be n² - 1 as this would give 8 (for n = 3) and the answer is 15 moves
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  4. #4
    Member classicstrings's Avatar
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    Quote Originally Posted by Natasha1 View Post
    Thanks Soroban,

    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

    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...

    The Frog Puzzle: A Solution

    Toads And Frogs puzzle
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