General algorithm for a magic square


by Vineeth T
Tags: algorithm, magic, square
Vineeth T
Vineeth T is offline
#1
Dec28-12, 04:26 AM
P: 31
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 ?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
RamaWolf
RamaWolf is offline
#2
Dec28-12, 10:58 AM
P: 96
For a quick overview look at: http://en.wikipedia.org/wiki/Magic_square
Vineeth T
Vineeth T is offline
#3
Dec29-12, 02:50 AM
P: 31
Quote Quote by RamaWolf View Post
For a quick overview look at: http://en.wikipedia.org/wiki/Magic_square
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 ?

RamaWolf
RamaWolf is offline
#4
Dec29-12, 04:16 AM
P: 96

General algorithm for a magic square


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)

   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;
ramsey2879
ramsey2879 is offline
#5
Dec29-12, 08:11 AM
P: 891
Quote Quote by Vineeth T View Post
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 ?
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".
RamaWolf
RamaWolf is offline
#6
Dec29-12, 10:27 AM
P: 96
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)

   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;
RamaWolf
RamaWolf is offline
#7
Dec30-12, 08:09 AM
P: 96
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

   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;


Register to reply

Related Discussions
Magic Square- Java Engineering, Comp Sci, & Technology Homework 1
Help With Pseudo Coded Algorithm for The Diamond-Square Algorithm Engineering, Comp Sci, & Technology Homework 0
Apocalyptic Magic Square General Math 1
Magic Square question General Math 9
Magic square Brain Teasers 5