ph0t3k - 7 months ago 50

Python Question

My goal is to find any permutation of a diagonal in a nxn matrix (2 <= n <= 15). The matrix consists of zeros and ones.

Currently I do it like this:

`indices = [[j for j, x in enumerate(row) if x == 1]`

for row in self.matrix]

cart = list(itertools.product(*indices))

cart = [list(tup) for tup in cart]

cart = filter(lambda dia: len(list(set(dia))) == len(dia), cart)

return cart

This works fine if the matrix is not too large, but otherwise it fails with:

MemoryError

So is there a way to avoid the whole computation of cart? so that it for example one permutation is found, the computation stops?

Answer

Simply make all the evaluations lazy by not calling `list`

on the result of `itertools.product`

and using `itertools.ifilter`

in place of `filter`

:

```
from itertools import ifilter, product
indices = [[j for j, x in enumerate(row) if x == 1] for row in self.matrix]
cart = product(*indices)
found_cart = next(ifilter(lambda dia: len(set(dia)) == len(dia), cart), None)
```

`next`

returns the first case where the predicate in `ifilter`

is `True`

or returns `None`

in case there is no matching item.

The *computation* stops once a matching item is found.

Source (Stackoverflow)