Finding unique solution puzzle python

Question

x = np.array([[0,1,11],[0,2,11],[0,3,10],[0,4,10],[0,5,9],[0,6,9],[1,7,9],
              [1,5,11],[1,6,11],[2,7,11],[2,8,10]])

I'm pretty new to this so i'm gonna call things like this [element1,element2,element3]

i have an array as shown above, and i want to find a solution to this array. It should satisfy the following conditions:

The first element 0:

it should have atleast one solution from [0,1,11],[0,2,11],[0,3,10],[0,4,10],[0,5,9],[0,6,9]

The first element 1:

this : [1,7,9],[1,5,11],[1,6,11]

The first element 2:

and this : [2,7,11],[2,8,10]

Such that the second element and 3rd element is unique for each solution(where 1st element=0,second element=1 and 3rd element=2)

o/p can be : [0,1,11] and [1,7,9] and [2,8,10]

wrong output : [0,1,11], [1,6,11] ,[2,8,10] here parameter 3 of the first and the second are same.

Answer 1

If I understand correctly, you want to produce triplets from the given x array so that the first, second and third element are all unique in one triplet. Code to do that:

import itertools

x = [[0,1,11],[0,2,11],[0,3,10],[0,4,10],[0,5,9],[0,6,9],[1,7,9],
              [1,5,11],[1,6,11],[2,7,11],[2,8,10]]
triplets = itertools.combinations(x,3)
for t in triplets:
  isGood = True
  for pos in range(3):
    if (t[0][pos] == t[1][pos] or t[0][pos] == t[2][pos] or t[1][pos] == t[2][pos]):
      isGood = False
  if (isGood):
    print(repr(t))

This procudes the following output:

([0, 1, 11], [1, 7, 9], [2, 8, 10])
([0, 2, 11], [1, 7, 9], [2, 8, 10])
([0, 5, 9], [1, 6, 11], [2, 8, 10])
([0, 6, 9], [1, 5, 11], [2, 8, 10])

---

A more pythonic solution which does the same in only 3 lines

for t in itertools.combinations(x,3):
  if all(len(col) == len(set(col)) for col in zip(*t)):
    print(repr(t))

---

Insane one-liner:

print(''.join(repr(t) + '\n' for t in itertools.combinations(x,3) if all(len(col) == len(set(col)) for col in zip(*t))))