On Feb 12, 4:24 pm, Samuel Karl Peterson
<[EMAIL PROTECTED]> wrote:
> C-like way to do it, (warning, untested in python):
>
> for i in range(8):
> for j in range(8):
> for offset_i in range(-1,2):
> for offset_j in range(-1, 2):
> row = i + offset_i
> col = j + offset_j
> if (row < 0 or row > 7) or (col < 0 or col > 8) \
> or ((row,col) == (i,j)):
> continue
> # else do something with board[row][col]
>
> I realize this is gross and un-Pythonic and does the same thing
It's also just a little bit inefficient. Never mind the algorithm,
we'll talk about that later. Let's just improve the code a little:
side_range = range(8)
delta_range = range(-1, 2)
for i in side_range:
for offset_i in delta_range:
row = i + offset_i
if not (0 <= row <= 7): continue
for j in side_range:
for offset_j in delta_range:
if not offset_i and not offset_j: continue
col = j + offset_j
if not(0 <= col <= 7): continue
# *now* it's safe to do something
# with (row, col)
Such hoisting of code out of inner loops is done for you by a C
compiler -- Python assumes you know what you are doing :-)
Now for the algorithm: all of that testing to see if you are about to
sail off the end of the world is a bit ugly and slow. You can use bit-
bashing, as Paul suggested, even though it's on Steven D'Aprano's list
of 6 deadly sins :-)
Another approach that has been used is to have a 9 x 9 storage area;
the squares off the edge contain a value that is neither "empty" nor
any value that represents a piece. So with careful coding, they
automatically fail tests like "can I capture the piece on this
square" [no, it's not a piece] and "can I move here" [no, it's not
empty]. If the game is chess, you need 10 x 10 to cater for those
pesky knights.
Cheers,
John
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