🚀 go-pugleaf

Dave Hamilton wrote:

> >First, notice that you can label a 3x3 board with numbers 1..9 in the "magic
> >square" fashion (ie. where every row, column and diagonal yields 15 when summed).
> 
> What layout would meet this requirement?

It's pretty easy to figure out.  You know that some
combinations of two numbers can't go on the same row,
column or diagonal, because their sum would be >= 15.
9 can't go with 8, 7 or 6; and 8 can't go with 7.
Since the center square is on the same row, column or
diagonal with every other square, that means the
largest number that can go in the center is 5.

 | | 
-+-+-
 |5| 
-+-+-
 | | 

9 can't go with 8, 7 or 6.  But if 9 goes in a corner,
there are only two squares that aren't on the same row,
column or diagonal.  So 9 has to go on a side.

 |9| 
-+-+-
 |5| 
-+-+-
 | | 

The numbers that can go on the same row as 9 are 2 and 4.
(1's taken, 3 would require another 3, 5's taken, and
anything larger would be too big).  Since the board is
still symmetrical, we can place them on the same row in
either order.

2|9|4
-+-+-
 |5| 
-+-+-
 | | 

Now sum the diagonals.

2|9|4
-+-+-
 |5| 
-+-+-
6| |8

Finally, fill in the rest.

2|9|4
-+-+-
7|5|3
-+-+-
6|1|8

Larry Coon
University of California
larry@assist.org
and lmcoon@home.com