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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

526.0. "Magic squares" by VOGON::CATTERMOUL (Richard REO F/M8 830-4564) Fri Jun 27 1986 14:30

Using a few simple rules, it is possible to build up magic squares 
8x8, 16x16 etc using the basic 4x4 magic square.

The smallest even number magic square is the 4x4:

	               9 12  6  7
                       8  5 11 10
                      15 14  4  1
                       2  3 13 16

As you can see, it uses the numbers 1 to 16, The sums across, down
and diagonally = 34 = (1 + 16)*2.

To construct an 8x8 using the numbers 1 to 64, our magic sum = 
			(1+64)*2 = 130.

Construct four 4x4 squares by adding 16 successively to each number in
the 1st square (the above one), and then 16 to the numbers in the 2nd
square and so on. Thus you get:

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

          41 44 38 39    57 60 54 55
          40 37 43 42    56 53 59 58
          47 46 36 33    63 62 52 49
          34 35 45 48    50 51 61 64

Divide each square into 4 quadrants. Take each of the 16 numbers from the
top left hand quadrants of the four squares and position them in a 4x4
square according to the order of the numbers in the basic 4x4. i.e. take
the numbers 
          5, 8, 9, 12, 21, 24, 25, 28, 37, 40, 41, 44, 53, 56, 57, 60. 
position: 1  2  3   4   5   6   7   8   9  10  11  12  13  14  15  16

Therefore, 5 goes in position 1, 8 goes in position 2, 9 goes in position 3
etc. 60 goes in position 16. The result is:

	37 44 24 25
        28 21 41 40
        57 56 12  5
         8  9 53 60

Do the same for the 16 numbers in the top right corner of the four squares.
Similarly for the lower left and right quadrants. The result is:

	37 44 24 25  38 43 23 26  
        28 21 41 40  27 22 42 39
        57 56 12  5  58 55 11  6
         8  9 53 60   7 10 54 59

        34 47 19 30  33 48 20 29
        31 18 46 35  32 17 45 36
        62 51 15  2  61 52 16  1
         3 14 50 63   4 13 49 64

Thus, using the numbers 1 to 64, we have produced 4 magic squares each of
which produce the same sum of 130 across, down and diagonally.

The same rules apply to building a 16 x 16 square using the numbers 1 to 256.
The result is sixteen 4x4 squares whose sum across, down and diagonally for
each of the 4x4s is (1 + 256)*2 = 514.

To the 64 numbers positioned as above [square A], add 64 to each. 
The resulting 16x16 square [square B] uses 65 to 128. 
To square B, add 64 to each number to generate square C; it will use
the numbers 129 to 192. 
Finally, add 64 to each number in square C to make square D which
uses 193 to 256.

Divide each 16x16 square into sixteen little 4x4s and label them thus:
	
Square A:
        a      b      c      d

1	37 44  24 25  38 43  23 26  
        28 21  41 40  27 22  42 39

2       57 56  12  5  58 55  11  6
         8  9  53 60   7 10  54 59

3       34 47  19 30  33 48  20 29
        31 18  46 35  32 17  45 36

4       62 51  15  2  61 52  16  1
         3 14  50 63   4 13  49 64

                     Similarly for squares B, C and D.

Take the 4x4 squares a1 in big squares A, B, C and D and position them 
according to the order given in the basic 4x4 to make a new 4x4 square.
It will add across, down and diagonally to 514.

        37  44  101 108                         9 12  6  7
        28  21   92  85     position according  8  5 11 10
       165 172  229 236     to order of        15 14  4  1
       156 149  220 213                         2  3 13 16

i.e. smallest number (21) in position 1, next (28) in position 2. 256
goes in position 16 to give:

                   149 172  92 101
                   108  85 165 156
                   229 220  44  21
                    28  37 213 236

I leave it as an exercise for the reader to construct the remaining 15
4x4 squares.
Richard Cattermoul
    
T.RTitleUserPersonal
Name
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526.1Help on Latin and Orthogonal Latin Squares in note # 421THEBUS::KOSTASKostas G. Gavrielidis <o.o> Wed Jul 09 1986 13:2912
    re. .0
    
    Richard, given your exceptional explanation on how to create Magic
    squares, practically of any size, could I ask you to take a look
    at Latin and Orthogonal Latin Squares on note # 421, and may be
    you can make the same magic as you did for the Magic squares.
    
    Thanks,
    
    Kostas G.
    
526.2Here's an exotic oneMODEL::YARBROUGHThu Jan 15 1987 17:3417
The latest issue of ABACUS has an article on magic squares that is both 
historically and mathematically interesting. The central idea is the 
formation of squares that are magic both arithmetically and lexically. For 
example:

	five		twenty-two	eighteen
	twenty-eight	fifteen		two
	twelve		eight		twenty-five

Numerically, the rows, columns, and diagonals add to 45. At the same time, 
the number of letters (ignoring hyphens) in each row, column, and diagonal
is twenty-one! 

The author of the article gives numerous examples in several languages. In 
order to comprehend the historical aspects of the subject, you will have to 
read the article. The square above is from a rare book, lost for nearly 100 
years, on the worship of trees.
526.3Solution for odd-sided sequare of any sizeAISG::KCHENThu Mar 22 1990 16:4124
I am surprised that nobody has mentioned a solution for sqaures with
odd number of cells on each side (e.g., 3 by 3, 7 by 7, etc.).
I learned the solution as a kid, and have always thought that
it must have been documented somewhere. FWIW, following is
a pseudo algorithm that will fill an odd-sided magic sqaure 
of any size.

Assumption: the size of the square is N by N, where N is any
            odd number. For the purpose of this solution, treat 
	    the square as cyclic. I.e., row N is considered to be
            next to row 1, and column N is considered to be next
            to column 1. 

Algorithm:

Let current_cell := the cell at top row, center column;
For I := 1 to N**2
  Begin
   Put I in current_cell;
   Case 
    The north-east neighbor of current_cell NE is empty: current_cell := NE;
    The south neighbor of current_cell S is empty: current_cell := S;
    Otherwise: exit;
  End;	    
526.4AISG::KCHENThu Mar 22 1990 18:486
    By the way, I do not have a formal proof that the algorithm
    always works. It was just that the several squares of various 
    sizes that I have tried all came out correct. Come to think of 
    it, I shouldn't have claimed that it works for odd-sided squares 
    of ANY SIZE. That's a bit too strong without a proof. Maybe I'll 
    just leave the proof as a simple excercise for those curious readers. ;-)