"Dr. Colombes" <[EMAIL PROTECTED]> writes:
> I'm looking for a good Python way to generate (enumerate) the 2**N
> tuples representing all vertices of the unit hypercube in N-dimensional
> hyperspace.
> For example, for N=4 the Python code should generate the following 2**N
> = 16 tuples:
Here's a recursive generator:
def hypercube(ndims):
if ndims == 0:
yield ()
return
for h in 1, -1:
for y in hypercube(ndims-1):
return (h,)+y
>>> print list(hypercube(4))
[(1, 1, 1, 1), (1, 1, 1, -1), (1, 1, -1, 1), (1, 1, -1, -1), (1, -1,
1, 1), (1, -1, 1, -1), (1, -1, -1, 1), (1, -1, -1, -1), (-1, 1, 1, 1),
(-1, 1, 1, -1), (-1, 1, -1, 1), (-1, 1, -1, -1), (-1, -1, 1, 1), (-1,
-1, 1, -1), (-1, -1, -1, 1), (-1, -1, -1, -1)]
>>>
> Maybe converting each integer in the range(2**N) to binary, then
> converting to bit string, then applying the "tuple" function to each
> bit string?
Yeah you could do that too.
--
http://mail.python.org/mailman/listinfo/python-list
