math is an attitude
 
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Sunday, March 26, 2006 - 06:46 AM

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Magic SquaresThis is one of the best methods used to create a magic square of any order. I will start with an example of 3*3 grid to explain the method.


We will create the following magic square step by step by putting 1 digit (between 1 to 9) at a time:
| 4 | 9 | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 |
The sum is 15. For any magic square, the sum is n*(n2 + 1)/2. and hence in our case (of 3*3 matrix) it is 3*5 = 15.

Step 1: Start process from the middle square in the lowest row and write 1 there (always!)
It will look like as below
 C1   C2   C3
|   |    |   | R1
|   |    |   | R2
|   |  1 |   | R3
Step 2: Now move 2 steps above )straight) and 1 step right and write the next number 2(in the sequence)
|   |    | 2 |
|   |    |   |
|   |  1 |   |
Now repeat Step 2 to write next number 3. What is this, i am out of grid!!
Step 3: We have to move 2 steps above but the grid is closed and it will take us out of the grid. Now 1 step above will take us to the lowermost row (R1, assuming R1 and R3 are attached virutally in a cylinderical shape) and 2nd step above will take us to the lowest row +1 (that is row R2). Now we have move towards right by 1 step and again it takes us out of the Grid. So we move to the left most column of the same row (R2,C1 again by assuming a virutal cylinderical shape).
And yes, this cell MUST be empty so we can place our number here.
YES, it is empty so place number 3 here.
|   |    | 2 |
| 3 |    |   |
|   |  1 |   |
Now in similar fashion (step 2) we will try to place 4 but we see that the position (by this method) is already occupied by another number (1 in this case) so what to do?
If such a condition occurs (which happens frequently in larger squares) then place the number just above the previous number, i mean in this case just above number 3, as shown below:
| 4 |    | 2 |
| 3 |    |   |
|   |  1 |   |
By using this methodology, we can complete the magic square as below:
Placing number 5:
| 4 |    | 2 |
| 3 | 5  |   |
|   |  1 |   |
Placing number 6:
| 4 |    | 2 |
| 3 | 5  |   |
|   |  1 | 6 |
Placing number 7:
| 4 |    | 2 |
| 3 | 5  | 7 |
|   |  1 | 6 |
Placing number 8:
| 4 |    | 2 |
| 3 | 5  | 7 |
| 8 |  1 | 6 |
Placing number 9:
| 4 | 9  | 2 |
| 3 | 5  | 7 |
| 8 | 1 | 6 |
Simple Huh!


Comments

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Re: Unique method to find Magic Square of any order
by puneet478
on Mar 28, 2006

Please write to puneet478@yahoo.com in case you need further clarifications.


Re: Unique method to find Magic Square of any order
by ThosRTanner
on Apr 04, 2006

One method I learnt was to start with '1' in the centre of the bottom line and go 1 step diagonally down and right, unless that space was filled, in which case go up one. This produces an identical result for 3x3, but gives a somewhat different result for 5x5 (still magic, but not a reflection or rotation)


Re: Unique method to find Magic Square of any order
by puneet478
on Aug 21, 2006
<u>How to create Magic Square of an even order using the same method</u>



1. Start by filling the magic square with the numbers in order, starting in the upper left corner with one, going across to the right, and beginning a new row when you reach the end.

2. Change the order of the numbers in the two diagonals. Thats it !!

For eg:

For a 4*4 grid, Switch the numbers on the diagonals--1 and 16, 11 and 6 on one diagonal; 7 and 10, 4 and 13 on the other. That's it!

Enjoy!

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