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    Here is the "problem":
    Pick any square grid, and you have to find the amount of combinations for 2 adjacent dots that can go on the grid(horizontal, vetical and diagonal).



    So for example, for a 3x3 square as above, the total amount would come to 20 (I would advise just doing this one or the 2x2 one to check the amount).

    And by hand, for a 2x2 square the total would be 6, for a 4x4 square it would be 42 and for a 5x5 it is 72.

    Now, we were given this problem for one lesson in GCSE 2 years ago in year 10 to give us a little practice on what our 4th module coursework could be like. Somehow, I came up with a formula that worked for 2x2, 3x3, 4x4, 5x5 and I calculated 6x6, 7x7 and 8x8 with this formula (I dont want to do these ones by hand). I, nor my teacher, knew why this formula worked (I might have an idea now though).

    I want to see if anyone else can come up with this formula, I'm not going to give it to you because I want to see if you can derive it. If you cant then ill just post it and you can see what you think.

    Note: This isnt me trying to get you to do my coursework for me, as this was from 2 years ago and im in year 13 now

    Say if you dont actually get the problem.










    Note: I'm not actually doing this coursework, it was a practice GCSE coursework from 2 years ago
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    Do you want to restrict it to people doing GCSE? It should be straightforward for many of the posters here.
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    EDIT: See above

    Hint: Consider different 'types' of squares.

    Extension:
    - What above m by n rectangles?
    - What above sets of 3 (or k) adjacent squares?
    - What if we're on a torus (each edge connects to the other side)?
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    search it on the internet
    • Wiki Support Team
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    (Original post by Skye333)
    search it on the internet
    Spoils the fun a bit, no?
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    (Original post by generalebriety)
    Spoils the fun a bit, no?
    ahh, the joy comes from missing the problem/cheating the system :awesome:
    • TSR Group Staff
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    It's a fairly straightforward problem, the answer comes out as a neat quadratic: 2(2n-1)(n-1), where n is the side length of the grid. You could spend a lot of time looking at different variations of the problem, but at the end of the day all you're really doing is counting permutations. Not all that interesting from a mathematical point-of-view.
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    All very well, though 2 years ago I did not look at the problem like this but rather borrowed something else

    \displaystyle\frac{(2x - 1)!}{(2x - 3)!}}

    where x = length of the grid

    from permutations.

    I cant really see why this works, other than translating the 2d grid into a 1d grid (a grid with only 1 row and length which is 2 times, minus 1 of the original length of the box) and working from there. But I still cant see it.

    (Original post by DFranklin)
    Do you want to restrict it to people doing GCSE? It should be straightforward for many of the posters here.
    no, its for everyone
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    (Original post by Dez)
    It's a fairly straightforward problem, the answer comes out as a neat quadratic: 2(2n-1)(n-1), where n is the side length of the grid. You could spend a lot of time looking at different variations of the problem, but at the end of the day all you're really doing is counting permutations. Not all that interesting from a mathematical point-of-view.
    true!!!its more like a "riddle" or a brain teaser not a mathematical problem
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    (Original post by Dez)
    It's a fairly straightforward problem, the answer comes out as a neat quadratic: 2(2n-1)(n-1), where n is the side length of the grid. You could spend a lot of time looking at different variations of the problem, but at the end of the day all you're really doing is counting permutations. Not all that interesting from a mathematical point-of-view.
    (Original post by ny2ko)
    true!!!its more like a "riddle" or a brain teaser not a mathematical problem
    There's no need for this kind of speak; this is actually a fairly difficult and fairly interesting genuine mathematical problem for your average year 10 student, with a lot of scope for extension.
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    (Original post by ny2ko)
    true!!!its more like a "riddle" or a brain teaser not a mathematical problem
    see my post above yours and then explain please.

    *edit* we didnt actually DO the problem in year 10, it was just a little teaser to what our coursework could be like.
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    Wouldn´t it be combinations? The order would matter wouldn´t it, otherwise there would be 12 different positions for a 2x2 square if we were counting permutations.
    • TSR Group Staff
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    It's just permutations. If you work through each square systematically then you'll count each pair exactly twice, so at the end you just have to divide everything by half.
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    (Original post by Dez)
    ...divide everything by half...
    :p:
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    (Original post by Dez)
    It's just permutations. If you work through each square systematically then you'll count each pair exactly twice, so at the end you just have to divide everything by half.
    The thing is this is a 2d grid and we are counting diagonals too, so I cant really see how you can use permutations (other than the fact that I came up with the formula by chance using the permutation formula).
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    (Original post by ShortRef)
    The thing is this is a 2d grid and we are counting diagonals too, so I cant really see how you can use permutations (other than the fact that I came up with the formula by chance using the permutation formula).
    Every square along a corner is connected to three other squares. Every non-corner square along an edge is connected to five other squares. Every square not on a corner or the edge is connected to eight other squares. Count all these up, then note that you've double-counted everything, so divide by 2.
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    It's about selecting 2 points. The first point can be selected in O(n^2) ways. The second in only O(1) ways. Hence the total number of ways must be O(n^2). Also note that it is polynomial.

    After that we can use the first 3 terms: 0, 6, 20 to find a degree O(n^2) polynomial which factors neatly.
 
 
 
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