Dave Angel d at davea.name
Fri Nov 2 20:56:40 CET 2012

```On 11/02/2012 03:19 PM, foster63 at 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.

--

DaveA

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