First, what the hell's a magic square?
A magic square is a 'matrix' (no not the movie) or a 2dimensional grid of numbers.
Take the simple case of 3x3 magic square. Here's one:
4

3

8

9

5

1

2

7

6

A Magic Square contains a certain bunch of numbers, in this case, 1..9, each of which has to be filled once into the grid. The 'magic' property of a Magic Square is that the sum of the numbers in the rows and columns and diagonals should all be same, in this case, 15.
Try making a 3x3 magic square yourself. It's not that easy. If it was easy, try a 4x4 grid with numbers 1..16. And what about 5x5, 6x6...? That's where computers come in! Okay, now, how do we go about programming something we hardly understand. The answer is : brute force. The computer may not be smart, but it can certainly do something you would take months to do, and that is, pure calculations  millions and millions of them. To have an analogy, suppose you are given two identical photos. It would take you just a glance of the eye to agree that they're same. The computer, on the other hand, would compare them dot by dot, pixel by pixel, and then agree that they're same. In the process, both you and the computer accomplish the same task, but in very different ways. So, the answer is brute force. Now, what's the question? (Hmm.. sounds familiar)
Okay, we've got a bunch of dumb numbers. We've got to arrange them in a matrix so that the sum is equal in all directions. Since we don't have a clue about any strategy, we'd just have to........ Try All Possibilities! Yes, and that's what a computer is meant to do  brute force. Okay, all this is fine, you say, but Now What?! Try all possibilities?!! Why, I don't even know how to shuffle a deck of cards... Did someone say Recursion?
Okay, try figuring it out yourself at this stage. You know everything there is to know  the concept of magic squares, the need to check all possibilities, and that the answer lies in recursion. You can easily make an extra simple function to check if a square is magical. So, basically, you've got to try all possibilities and permutations, and for each one, you have to test if the square is magical. I seriously suggest you stop at this point, and brainstorm yourself to extremes. Better yet, just finish the program (and notify me by email about how absolutely simple and silly a program it was).
Losers can read on: Having said all that I said, only the heart of the program is left to explain. The question boils down to, how do we accomplish permuting a set of numbers (or objects) using recursion? For simplicity sake, we work with a 1dimensional array. In the case of a 3x3 square, let's have a 9length singledimensional array. So we have numbers 1 to 9 and 9 locations to put them into. The permutations would be:
123456789 123456798 123456879 123456897 123456978 123456987 .... .... 987654321
Conversion from single to 2D array is fairly simple, and may be done within the magictesting function that we shall call 'TestMagic'.
As a preliminary exercise, could you program the following sequence...
111,112,113,121,122,123,131,132,133, 211,212,213,221,222,223,231,232,233, 311,312,313,321,322,323,331,332,333.
...using For loops? Try it.
Answer: it is as simple as:
for i = 1 to 3 for j = 1 to 3 for k = 1 to 3 print i,j,k
Now, try it using recursion.
Answer: Think of it this way: i loops from 1 to 3. For every value of i, j loops from 1 to 3. For every value of j, k loops from 1 to 3. For every value of k, the values are displayed. So, basically, these three loops perform very similar functions, which can therefore be reduced to a single recursive function with a single loop.
void Func(n) { for i = 1 to 3 { print i if (n<3) Func(n+1) } }
The function should be invoked with initial n value as 1, ie, Func(1).
Trace the program to figure out the output. For each pass through the loop,
the function's child's loop is completely executed, by means of the recursive
call. Just in order to confuse you even more, the same function can be written
in this way:
void Func(n) { if (n<4) { for i = 1 to 3 { print i Func(n+1) } } }
Anyway, coming back to our magic squares and our 9length array to be permuted....we can easily adopt the above functions, with one difference: the numbers should not repeat. If the numbers were allowed to repeat, we could easily write:
void Permute(n) { for i = 1 to 9 { A[n] = i if (n<9) Permute(n+1) else TestMagic } }
A[] is the 9length array.
The function should be called with n=1 initially. Function TestMagic checks
if the square is magical, and displays it if it is so.
However, we must not allow repetition, as per the rules and regulations of magic
square authorities and environmental agencies worldwide. There could be several
ways to perform this check, and is left open to you. What I did was:
I kept another 9length array, which contained information about which number
was used, and which was not.
Eg: Say B[] is the extra array. If B[1]=0, then, number 1 is still free. If
B[1]=1, then, number 1 is already used. Whenever we 'go to' a recursive call
and 'come back' from a recursive call, we should update this array. One possible
algorithm could be:
void Permute(n) { for i = 1 to 9 if B[i]=0 // if i is free { B[i]=1 A[n]=i if n<9 Permute(n+1) //recurse else TestMagic( ) //at the end B[i]=0 } }
The rest is left to you. Try the program, and taste the results yourself. There are totally 8 solutions for the 3x3 matrix with numbers 1..9. Did you get them all?
We have worked out this problem by realizing that we need to find all permutations of the numbers 1..9, and for each permutation, check if it satisfies the condition. There's another way of looking at it, which is a common viewpoint when talking about recursion. It is called 'BackTracking'. In effect, what our program does is, when it finds (at the 9th recursion) that a square is not magical (say), it 'goes back' to the previous instance of the function, and even more back if needed. This is called BackTracking and is used in all recursiveproblemsolving programs.
Even if you own a 800MHz m/c, you may have noticed that the calculation took atleast a few seconds, maybe longer. This is because the algorithm is not very efficient. It is possible to get all the results in under a second on any machine, if we tweak this algorithm here and there. That's what we'll do next (unless you're fed up of this, in which case, you could go on to tictactoe.)
Tweak
If you observe (and imagine) a bit... okay, more than just a bit.... you will realize that a LOT of the possibilities we test are actually quite useless. The first state is:
1

2

3

4

5

6

7

8

9

or, 1 2 3 4 5 6 7 8 9.
Now, the permuting begins by permuting 89, then 789, then 6789... which all takes time. All these are useless until we start permuting the top row, since 1+2+3=6, but we need 15. Get it? It's going to take a LOT of permutations before we finally backtrack to 3,2 and 1, which remain as sum 6 all this while. So, this 123 combination really sucks, and we should never allow it.
I wouldn't be describing this problem if I didn't know the answer, so here it is: While filling the numbers, at every row, we check if the sum is 15. So, like, before we go on to the second row (or the 4th element in our 1D array) we check the sum of the first row  if it isn't what we want, go back and permute... until the sum is 15. So, now the first row has a sum 15 (done very quickly since we permute only 3 locations) and all those useless permutations of the other two rows are not attempted. The same should happen for the second row, which saves some time, but not as much as that for the first row. Finally, after all 9 elements are inserted, check the columns and diagonals. Actually what happens is, if the sum of the first row elms is not 5, it backtracks until the sum is 15. If the sum is now 15, it 'goes forward' and checks the second row for sum 15. If second row sum is not 15, it backtracks to the first row again. If all permutations within row 2 are complete, it backtracks to row 1... The following function checks the sum at every row.
void Permute(n) { for i = 1 to 9 if B[i]=0 // if i is free { B[i]=1 A[n]=i if (n % 3 = 0) // if n is divisible by 3 { if RowSum() = 15 { if n<9 Permute(n+1) else // at the end TestMagic() } } else Permute(n+1) //recurse B[i]=0 } }
Tweak2
Even the above program is a bit wasteful. After all that work, you thought I'd give you a break, but no. Consider one of the cases in which the program successfully fills the entire 9 locations (The row sums are all 15):
1

5

9

2

6

7

3

4

8

This is by no means a magical square. Only the row sums are proper. What about the column sums? Look at column 1. Isn't it a little familiar? Right, we're back to the same old philosophies  this time, for the columns. A little thinking (for a guy like Einstein) or a lot of it (for you and me) will convince you that we also need to check and maintain column sums (as 15) along the way. So, we need to check both rows and columns. A little More thinking will convince you (poorly) that we need to fill the rows and columns alternatively to maximise efficiency (and confusion). At this point, keep in mind one of the greater truths that programmers (like you and me) live by: "The more complicated your program, the more you are respected." There are atleast 2 good ways to alternate rows and columns.

1

2

3

1

a

b

c

2

d

e

f

3

g

h

i

The first way: row1, column1, row2, column2, row3,column3. or, abcdgefhi. At c, check row1 sum. At g, check column1 sum. At f, check row 2 sum. At h, check column2 sum. At i check row3 and column3 sum and diagonals as well. So, with a very few 'fillings' we can extract the solutions.
The second way: In a spiral pattern: row1,column3, row3(reverse), column1(rev), row 2, column2 or, abcfihgde. At c, check row1. At i, check column3. At g, check row3. At d, check column1. At e, check row2, column2 and diagonals. And, there you have it. The solutions will pop out in a second.
And, Yes, you'll have to try the programs out yourself. I'm sick of this, so I'm moving on to TicTacToe... see you there!