On 02/11/12 19:56, Dave Angel wrote:
On 11/02/2012 03:19 PM, foste...@gmail.com wrote:
Hi All,
As part of a Nim solver I'm playing around with I'm trying to code this Haskell
snippet:
options [x] = zero : [ [y] | y <- [1..x - 1] ]
options (x:xs) = map (++ xs) (options [x]) ++ map (x:) (options xs)
in Python. So far I have this, which works OK, but somehow doesn't feel right:
def options( heaps ):
if heaps == []: return []
head, tail = heaps[:1], heaps[1:]
# Calculate all possible moves which is the sum of
# prepending all possible head "moves" to the tail
# and appending all possible tail "moves" to the head
return [ [h] + tail for h in range( head[0] ) ] \
+ [ head + t for t in options( tail ) ]
Is there anything anyone could recommend to make it more "Pythonic" or more
functional. It looks clumsy next to the Haskell.
Regards
etc.
You'd save people a lot of time if you'd specify that the parameter
heaps is a list of ints, perhaps initially [1,3,5,7] or [3, 4, 5]
depending on which variation of Nim you're trying to. There are many.
One variant is that some versions of Nim say the winner is the player
who does NOT take the last piece. I'll assume that the goal is to end
up with [0,0,0,0]
My main problem with studying your code is that brute force is totally
unnecessary; there's a fairly simple strategy for winning at Nim.
Certainly it's simple enough to have perfect strategy without any computer.
A "good" move is any one where the xor of all the items in the list ends
up as zero. There is always at least one move for an "ungood" position
that results in a "good" one. Thus the strategy is to go from good to
good, with the opponent always stuck on an ungood one.
Hi Dave,
Thanks for the comments. Yes, I should have specified that the input is
a list of ints giving the size of each heap, and the return value should
be a list of all game positions reachable from the input position.
At the moment I'm not concentrating on any particular Nim flavour, just
trying to enumerate all possible moves from a given position.
I know that there's an easier way to determine winning-losing positions,
but my question was more about programming style than Nim strategy.
My code to calculate the "nim-value" looks like this:
def nim_val( heaps ):
return functools.reduce( operator.xor, heaps, 0 )
Assuming that we're playing "non-misere" Nim then a zero nim-value is a
lose for the player *about* to play.
Regards
Simon
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