|
Main Menu
Who's Online
There are 120 unlogged users and 0 registered users online.
You can log-in or register for a user account here.
Sponsored Links
|
Sunday, March 26, 2006 - 06:46 AM
This 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 | | R3Step 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! |
User's Login
Sponsored Links
Other Stories
|
Comments
Please write to puneet478@yahoo.com in case you need further clarifications.
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)
Only logged in users are allowed to comment. register/log in