On 2011-03-17 15:04, Jason Grout wrote:
> def hash(self):
> h=0
> for i,j,entry in m.nonzero_entries(): # nonzero entries for sparse
> matrices
> h^=hash(entry)^i^j
> return h
Since you're only xorring, this will give a lot of collisions. I think
something like
def hash(self):
n = m.ncols()
h = (m.nrows() + m.ncols())**2 + n
for i,j,entry in m.nonzero_entries():
h ^= hash(entry)*(i*n + j + 1)
return h
has less chance of collisions. Of course everything should be computed
as a C long.
Jeroen.
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