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Posers and Puzzles

Posers and Puzzles

  1. 10 Dec '04 19:49
    As everyone knows, a 2x2 magic square is not possible in 2 dimensions. Similarly in 3 dimensions a 3x3x3 strict magic cube ,in which all row sums, lateral row sums, column sums, and diagonal sums , in each of the 3x3 squares in each of the 3 layers of each plane( x-y plane, y-z plane and z-x plane), aswell as the 4 body- diagonal sums are each equal(i.e. equal to 42), - is NOT possible. I have found a proof for the impossibility of the magic cube of order 3 i.e. 3x3x3 magic cube..
    AS in 2 dimensions the smallest order of possible magic square is 2+1 i.e. 3 only, , my conjecture is that , in 3- dimensions , the smallest strict magic cube can be of order 3+1 i.e.4 or above only. There are some websites giving magic cubes of any order but they are not strict magic cubes in the sense that sums along all rows, lateral rows, columns , diagonals, body -diagonals ,are not equal.
  2. Standard member TheMaster37
    Kupikupopo!
    11 Dec '04 13:33
    You could try to prove it by making a base for the space of magic cubes (or the 64 dimensional real space it's equivalent to). Just to make sure i haven't lost you here;

    A base is a set of 4x4x4 cubes so that every magic cube is a combination of sums of these cubes (with the possibility to multiply every cube with a constant.

    A base for the three-dimensional real space would be <(1,0,0), (0,1,0), (0,0,1)>. Every triple (x,y,z) of coordinates in space can be made with those three elements.

    Now you can represent a cube by simply writing the numbers in it in a row. You would get a 4*4*4=64 long vector. You can make a base for those cubes by making a base for the 64-dimensional real space. Magic cubes have a similar base, only one with restrictions on the elements of the base (the sums of diagonals, vertical...whatever must be equal).

    Let's see...a magic cube has 49 sums that have to be equal (i can be mistaken about this, i'm not sure if i haven't counted any twice).

    In math that means you have 49 restrictions on your set of base-elements. That leaves 15 dimensions of freedom, wich means you'll have 15 base-elements for your magic cubes.

    I humbly apologise if anything i said is not true.
  3. 13 Dec '04 04:28 / 2 edits
    Originally posted by sarathian
    I have found a proof for the impossibility of the magic cube of order 3 i.e. 3x3x3 magic cube..

    My conjecture is that , in 3- dimensions , the smallest strict magic cube can be of order 3+1 i.e.4 or above only.
    These are called perfect magic cubes.

    If you have indeed proved the impossibility of the 3x3x3 case, what is the probability that your 'conjecture' is true?

    Anyway, check out http://mathworld.wolfram.com/news/2003-11-18/magiccube/

  4. Standard member TheMaster37
    Kupikupopo!
    13 Dec '04 14:27
    Ok, i've done some calculating:

    3x3 magic squares have a base of three elements (eg, you can choose 3 numbers in the sqaure, and then the rest is fixed).

    4x4 magic squares have a base of 8 elements.

    If i haven't made a mistake, 4x4x4 magic cubes have a 18-dimensional base. I said 15 before, but i forgot that several equations are abundant.

    I can show you how to make a base for whatever size magic cube/square you want. It involves heavy calculation though

    I'll use magic squares for example (sides of length three). I'll use the following natation: (1, 2, 3, 4, 5, 6,7,8,9) will be a square with a 1 in the top left, 3 in the top right, and a 7 in the bottom left.

    Say you want the sums of the sides to be X. You can choose the first two spaces freely, say a and b and choose the 4th space to be c (Not the third, because that one is already set because of your choice of a and b).

    The third coordinate is set, namely X-a-b. You see that now the 7th coordinate is set as well: X-a-c. Now the 5th coordinate is set (because the diagonal also has to have sum X) : -X +2a+b+c. You can now complete the rest of the square (using horizontals and verticals) wich gives you the following square (draw it for a clear idea):

    (a, b, X-a-b, c, -X+2a+b+c, 2X-2a-b-2c, x-a-c, 2X-2a-2b-c, -2X+3a+2b+2c)

    The only sum we haven't checked yet is the first diagonal. If we add up the numbers on the diagonal (1st, 5th and 9th element) we get -3X+6a+3b+3c. To be a magic square this needs to equal X.

    X = -3X+6a+3b+3c => X = 1/4 * (6a +3b +3c)

    Now we're ready to make the base-elements. We're going to find squares A,B and C so that a*A + b*B + c*C is the square we have now (basically, we seperate our square in a square with b=c=0, a square with a=c=0 and a square with a=b=0).

    You'll see quickly that they are the following squares (Note that X is determined by a,b and c!)

    A = (1, 0, 1/2, 0, 1/2, 1, 1/2, 1, 0)

    B= (0, 1, -1/4, 0, 1/4, 1/2, 3/4, -1/2, 1/2)

    C= (0, 0, 3/4, 1, 1/4, -1/2, -1/4, 1/2, 1/2)

    (You can check it yourself: a*A + b*B + c*C gives our square with sums X = 1/4*(6a+3b+3c) )

    ANY magic 3x3 square is of the form a*A + b*B + c*C with a, b and c anything you want.

    This method can be used for 4x4x4 magic cubes as well, or even higher dimensional things with longer sides. The problem is that you'll lose oversight on what is set by your choices. I think you'll have 18 free choices in your magic cube, then the rest is set.

    PS There ARE 2x2 magic squares and 3x3x3 magic cubes! They simply aren't interesting because all the numbers are the same)