Help me finish this Python 3.x self-challenge.

Question

This is not homework.

I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion.

After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please.

from itertools import permutations
from operator import mul
from functools import reduce
glob_lst = []
def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0)
oneToNine = list(range(1, 10))
twoToNine = oneToNine[1:]
for perm in permutations(oneToNine, 9):
    for n in twoToNine:
        glob_lst = perm[1:n]
        #print(glob_lst)
        if not divisible(n):
            continue
    else:
        # Is invoked if the loop succeeds
        # So, we found the number
        print(perm)

Thanks!

Answer 1

Here's a short solution, using itertools.permutations:

from itertools import permutations

def is_solution(seq):
    return all(int(seq[:i]) % i == 0 for i in range(2, 9))

for p in permutations('123456789'):
    seq = ''.join(p)
    if is_solution(seq):
        print(seq)

I've deliberately omitted the divisibility checks by 1 and by 9, since they'll always be satisfied.

Answer 2

Here's my solution (not as elegant as Mark's, but it still works):

from itertools import permutations

for perm in permutations('123456789'):
    isgood = 1
    for i in xrange(9):
        if(int(''.join(perm[:9-i])) % (9-i)):
            isgood = 0
            break
    if isgood:
        print ''.join(perm)