"""
Cube Digit Pairs
Project Euler Problem #90
by Muaz Siddiqui

Each of the six faces on a cube has a different digit (0 to 9) written on it;
the same is done to a second cube. By placing the two cubes side-by-side in
different positions we can form a variety of 2-digit numbers.

For example, the square number 64 could be formed:

In fact, by carefully choosing the digits on both cubes it is possible to
display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36,
49, 64, and 81.

For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on
one cube and {1, 2, 3, 4, 8, 9} on the other cube.

However, for this problem we shall allow the 6 or 9 to be turned upside-down
so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows
for all nine square numbers to be displayed; otherwise it would be impossible
to obtain 09.

In determining a distinct arrangement we are interested in the digits on each
cube, not the order.

{1, 2, 3, 4, 5, 6} is equivalent to 3, 6, 4, 1, 2, 5} 1, 2, 3, 4, 5, 6} is
{distinct from 1, 2, 3, 4, 5, 9}

But because we are allowing 6 and 9 to be reversed, the two distinct sets in
the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the
purpose of forming 2-digit numbers.

How many distinct arrangements of the two cubes allow for all of the square
numbers to be displayed? """
from euler_helpers import timeit
from itertools import combinations
@timeit
def answer():
    squares = [(0,1), (0,4), (0,6), (1,6), (2,5), (3,6), (4,6), (8,1)]
    cube = list(combinations([0,1,2,3,4,5,6,7,8,6], 6))

    return sum(1 for i,c1 in enumerate(cube) for c2 in cube[:i] if all(x in c1 and y in c2 or x in c2 and y in c1 for x, y in squares))