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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