To get for example all bit vectors of size 3 one can say
CartesianProduct(range(2), range(2), range(2)).list()
To get this done for given n one can say at least
a=CartesianProduct(range(2)).list()
for n in range(1,n):
a=[flatten(x) for x in CartesianProduct(a, range(2)).list()]
Is there any direct (and quite possibly faster) way to accomplish this?
--
Jori Mäntysalo
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