How do I compute this non terminal positions evaluation, for tic-tac-toe?
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31-10-2019 - |
Pergunta
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.
Nenhuma solução correta