Heikki Levanto wrote:
> I am sure there is a mathematically sound way to measure
> how symmetric the evaluation is, but my math is a bit rusty,
> so I am asking if someone can come up with a good way. After
> that, I'm asking if various programmers would be willing to
> run this test, and publish the results?
A simple idea:
1. Even if you have 3 different types of symmetry ..
a. X symmetry (x,y) v.s. (bs+1-x, y)
b. Y symmetry (x,y) v.s. (x, bs+1-y)
c. Center symmetry (x,y) v.s. (bs+1-x, bs+1-y)
(bs+1 = board size + 1 assuming 1-based indices for clarity.)
.. it is clear from their expression that the third is implied
by the first 2.
2. You could use a simple measure of skewness:
http://www.itl.nist.gov/div898/handbook/eda/section3/eda35b.htm
Note that skewness measures the *lack* of symmetry.
Two measures: One for X and one for Y
3. Possible objections: Since these measures use the third
moment of a distribution, they are very sensible to the deviation
for the mean. In other words: skewness between the 2nd and 18th row
of a 19x19 board weight much more than between the 9th and 11th.
To compensate this, you can compute another estimator with the
rows (same for the columns) inverted in each half board.
Toggle columns:
1 <-> 9 and 11 <-> 19
2 <-> 8 and 12 <-> 18
...
so you would have 4 estimators: X-direct, X-inverted,
Y-direct, Y-inverted. Use the highest, i.e., the worst.
Jacques.
Java is a religion. Asm hackers don't spend valuable picoseconds
arguing with Believers.
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