How do I compute this non terminal positions evaluation, for tic-tac-toe? Looking at this solution on page 3 but having trouble understanding the evaluation function specifically.   


  For non terminal positions, we use a linear evaluation function
  defined as
  
  Eval (s) = 3[X2(s)]+X1(s)−(3[O2(s)]+O1(s))
  
  We define X n ( s ) as the number of rows, columns, and diagonals in
  state s with exactly n X’s and no O’s, and similarly define O n ( s )


Looking at this state 's' for example (these problems accounts for symmetry):

|X| | |
| |O| |
| | | |


It shows an evaluation of -1  

However, got 0:  

X2(s) = 0 because there are no rows or columns with 2 X's.
X1(s) = 2 because the top row and left column both have 1 X (but the diagonal has an O)

O2(s) = 0 because there are no rows or columns with 2 O's.
O1(s) = 2 because the middle row and middle column both have 1 O (but the diagonal has an X)

Plugging in:  

0+2 - (0+2) = 0


So, I thought I am missing symmetry but I just get a bigger scale that still = 0's.

I must be overlooking something simple but not sure.

Any help is appreciated.
O1(s) = 3. You missed another diagonal.