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General algorithm for a magic square

  1. Dec 28, 2012 #1
    Is there any algorithm to form a magic square of any size with a desired magic sum ?
    Also can we make a magic square not only with the numbers from 1 to n2
    but using any random numbers ?
     
  2. jcsd
  3. Dec 28, 2012 #2
  4. Dec 29, 2012 #3
    But in this webpage,no general methods of construction of a magic square of any numbers are given.
    Can you give me a better one ?
     
  5. Dec 29, 2012 #4
    There is not one algorithm for all n (a natural number); I think, there are at least 3:
    for odd numbers, for double even ( ie for numbers with 4 as factor) and for even numbers.

    Below you find my little algorithm (written in ARIBAS) to generate an odd magic square;
    example for n = 11; for simplicity of the algorithm, a 'vector' is used to store the 'square' .


    ==> MagicSquareOdd(11).
    -: (68, 81, 94, 107, 120, 1, 14, 27, 40, 53, 66, 80, 93, 106, 119, 11, 13, 26,
    39, 52, 65, 67, 92, 105, 118, 10, 12, 25, 38, 51, 64, 77, 79, 104, 117, 9, 22,
    24, 37, 50, 63, 76, 78, 91, 116, 8, 21, 23, 36, 49, 62, 75, 88, 90, 103, 7, 20,
    33, 35, 48, 61, 74, 87, 89, 102, 115, 19, 32, 34, 47, 60, 73, 86, 99, 101, 114,
    6, 31, 44, 46, 59, 72, 85, 98, 100, 113, 5, 18, 43, 45, 58, 71, 84, 97, 110,
    112, 4, 17, 30, 55, 57, 70, 83, 96, 109, 111, 3, 16, 29, 42, 56, 69, 82, 95,
    108, 121, 2, 15, 28, 41, 54)

    Code (Text):

       n1:=n - 1; n2:=n*n; n3:=n1 + n1 + 1;
       realloc(a, n2);
       i:=1;
       j:= n1 div 2;
       a[j]:=i;
       while i < n2 do
          i:=i + 1;
          if j mod n = n1 then
             k:=j - n3;  
          else
             k:=j - n1;
          end;
          if k < 0 then
             k:=k + n2;
          end;
          if a[k] = 0 then
             j:=k;
          else
             j:=j + n;      
          end;
          a[j]:=i;
       end;
     
     
  6. Dec 29, 2012 #5
    What do you mean by "general methods"? I consider "A method of constructing majic squares of even number of rows" and "Method for constructing a majic square of odd order" to be "general methods".
     
  7. Dec 29, 2012 #6
    Below you find my little algorithm (written in ARIBAS) to generate an double-even magic square of side length n (ie 4 divides n);

    example for n = 4; for simplicity of the algorithm, a 'vector' is used to store the 'square'


    MagicSquareDoubleEven(4).
    -: (16, 2, 3, 13, 5, 11, 10, 8, 9, 7, 6, 12, 4, 14, 15, 1)

    Code (Text):

       d:= n div 4; d2:=d + d; n2:=n * n;
       realloc(a,n2);
       k:=1 + n2; k2:=0;
       for i:=1 to d do
          for j:=1 to d do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k;
          end;
          for j:=1 to d2 do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k2;
          end;
          for j:=1 to d do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k;
          end;      
       end;
       for i:=1 to d2 do
          for j:=1 to d do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k2;
          end;
          for j:=1 to d2 do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k;
          end;
          for j:=1 to d do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k2;
          end;      
       end;  
       for i:=1 to d do
          for j:=1 to d do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k;
          end;
          for j:=1 to d2 do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k2;
          end;
          for j:=1 to d do
             k:=k - 1; k2:=k2 + 1; a[k2 - 1]:=k;
          end;      
       end;    
     
     
  8. Dec 30, 2012 #7
    Below you find my little algorithm (written in ARIBAS) to generate an simple-even magic square of side length n (ie n div 2 must be odd);

    example for n = 6; the algorithm uses my 'MagicSquareOdd' (see above)
    to form a matrix a and an auxiliary matrix 'aux' for the needed inter change of
    some elements of matix a

    MagicSquareSimpleEven(6)

    ((35, 1, 6, 26, 19, 24),
    (3, 32, 7, 21, 23, 25),
    (31, 9, 2, 22, 27, 20),
    (8, 28, 33, 17, 10, 15),
    (30, 5, 34, 12, 14, 16),
    (4, 36, 29, 13, 18, 11))

    A good companion for this algorithm was: http://www.hp-gramatke.de/magic_sq/index.htm

    Code (Text):

       k:= n div 2; k2:= k * k; d:= (k - 1) div 2;
       a:=alloc(array,n,alloc(array,n));
       aux:=alloc(array,k,alloc(array,n));
       for i:=0 to d - 1 do
          for j:=0 to d - 1 do
             aux[i][j]:=1;
          end;
       end;
       for j:=0 to d - 1 do
          aux[d][1+j]:=2;
       end;
       for i:=d + 1 to k - 1 do
          for j:=0 to d - 1 do
             aux[i][j]:=3;      
          end;
       end;
       for i:=0 to k - 1 do
          for j:=0 to d - 2 do
             aux[i][n-j-1]:=4;
          end;
       end;
       realloc(b, k2);
       b:=MagicSquareOdd(k);
       for i:=0 to k - 1 do
          for j:=0 to k - 1 do  
             a[i][j]:=b[i*k+j]; a[i][j+k]:=2*k2+b[i*k+j];
             a[i+k][j]:=3*k2+b[i*k+j]; a[i+k][j+k]:=k2+b[i*k+j];
          end;
       end;
       for i:=0 to k - 1 do
          for j:=0 to n - 1 do
             if aux[i][j] > 0 then
                d:=a[i][j]; a[i][j]:=a[i+k][j]; a[i+k][j]:=d;
             end;
          end;
       end;
     
     
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