This time around its something a bit tricky generating an odd magic square. Details are given everywhere. The wiki entry is sufficient, check it out http://en.wikipedia.org/wiki/Magic_square.

I base on the same logic which is commonly followed the one up one right stepping strategy. Here is the pseudocode,

//n is the dimension of magic square for a 3*3 square n=3;

int magic[n][n]={0};

curRow=0, curCol=1;

oldRow=0, oldCol=0;

count =0;

for(i=1;i<=n;i++)

{

for(j=1;j<=n;j++)

{

//See if we are into an empty cell

if( magic[curRow][curCol]==0 )

{

magic[curRow][curCol] = ++count;

}

else

{

//Move down from the previous row but keep the same column

magic[(oldRow+1)%n][oldCol] = ++count;

curRow = (oldRow+1)%n;

curCol = oldCol;

}

//Store the last row and column index

oldRow=curRow;

oldCol=curCol;

//Move one up and one right

curRow=(curRow-1)%n;

curCol=(curCol+1)%n;

//Wrap around the size

if(curRow<0)

curRow =n-1;

}

}

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