How to generate a list of all distinct 3x3 Latin Square in Python

Vince :

Latin Square is an nxn array filled with n different symbols, each occurring exactly once in each row and exactly once in each column (like sudoku). An example of Latin Square:

1 2 3
2 3 1
3 1 2

This is what I've tried, but it is still not all distinct

grid = []
temp = []
block = [[1,2,3],
         [2,3,1],
         [3,1,2]]
perm = permutations(block)
for i in perm:  #row permutations
    temp.extend(i)
    if len(temp)==3:
        grid.extend([temp])
        temp = []
for perm in zip(permutations(block[0]), permutations(block[1]), permutations(block[2])): #column permutations
    temp.extend([perm])
for i in range(len(temp)):  #convert to list
    temp[i] = list(temp[i])
    for j in range(len(temp[0])):
        temp[i][j] = list(temp[i][j])
grid.extend(temp)
for i in grid:
    for j in i:
        print(j)
    print()

Output is:

[1, 2, 3]
[2, 3, 1]
[3, 1, 2]

[1, 2, 3]
[3, 1, 2]
[2, 3, 1]

[2, 3, 1]
[1, 2, 3]
[3, 1, 2]

[2, 3, 1]
[3, 1, 2]
[1, 2, 3]

[3, 1, 2]
[1, 2, 3]
[2, 3, 1]

[3, 1, 2]
[2, 3, 1]
[1, 2, 3]

[3, 1, 2]
[2, 3, 1]
[1, 2, 3]

[3, 2, 1]
[2, 1, 3]
[1, 3, 2]

[1, 3, 2]
[3, 2, 1]
[2, 1, 3]

[1, 2, 3]
[3, 1, 2]
[2, 3, 1]

[2, 3, 1]
[1, 2, 3]
[3, 1, 2]

[2, 1, 3]
[1, 3, 2]
[3, 2, 1]

The result is supposed to be like this (order doesn't matter): Latin Square

[1, 2, 3]
[2, 3, 1]
[3, 1, 2]

[1, 2, 3]
[3, 1, 2]
[2, 3, 1]

[1, 3, 2]
[2, 1, 3]
[3, 2, 1]

[1, 3, 2]
[3, 2, 1]
[2, 1, 3]

[2, 1, 3]
[1, 3, 2]
[3, 2, 1]

[2, 1, 3]
[3, 2, 1]
[1, 3, 2]

[2, 3, 1]
[1, 2, 3]
[3, 1, 2]

[2, 3, 1]
[3, 1, 2]
[1, 2, 3]

[3, 2, 1]
[1, 3, 2]
[2, 1, 3]

[3, 2, 1]
[2, 1, 3]
[1, 3, 2]

[3, 1, 2]
[1, 2, 3]
[2, 3, 1]

[3, 1, 2]
[2, 3, 1]
[1, 2, 3]

Ajax1234 :

You can use recursion with a generator:

def row(n, r, c = []):
   if len(c) == n:
      yield c
   for i in range(1, n+1):
      if i not in c and i not in r[len(c)]:
         yield from row(n, r, c+[i])

def to_latin(n, c = []):
  if len(c) == n:
     yield c
  else:
     for i in row(n, [[]]*n if not c else list(zip(*c))):
        yield from to_latin(n, c+[i])

for i in to_latin(3):
  for b in i:
    print(b)
  print('-'*9)

Output:

[1, 2, 3]
[2, 3, 1]
[3, 1, 2]
---------
[1, 2, 3]
[3, 1, 2]
[2, 3, 1]
---------
[1, 3, 2]
[2, 1, 3]
[3, 2, 1]
---------
[1, 3, 2]
[3, 2, 1]
[2, 1, 3]
---------
[2, 1, 3]
[1, 3, 2]
[3, 2, 1]
---------
[2, 1, 3]
[3, 2, 1]
[1, 3, 2]
---------
[2, 3, 1]
[1, 2, 3]
[3, 1, 2]
---------
[2, 3, 1]
[3, 1, 2]
[1, 2, 3]
---------
[3, 1, 2]
[1, 2, 3]
[2, 3, 1]
---------
[3, 1, 2]
[2, 3, 1]
[1, 2, 3]
---------
[3, 2, 1]
[1, 3, 2]
[2, 1, 3]
---------
[3, 2, 1]
[2, 1, 3]
[1, 3, 2]
---------