Solver for "Passengers in a Railroad Compartment" puzzle.

railroad_puzzle.py

"""Passengers in a Railroad Compartment

Six passengers sharing a train compartment are from Moscow, Leningrad, Tula,
Kiev, Kharkov, and Odessa.

1. A and the man from Moscow are physicians
2. E and the Leningrader are teachers.
3. The man from Tula and C are engineers.
4. B and F are WWII veterans, but the man from Tula has never served in the
   armed forces.
5. The man from Kharkov is older than A.
6. The man from Odessa is older than C.
7. At Kiev, B and the man from Moscow got off.
8. At Vinnitsa, C and the man from Kharkov got off.

Match initials, professions, and cities. Also, are the facts both sufficient
and necessary to solve the problem?

from Boris A Kordemksy, The Moscow Puzzles: 359 Mathematical Recreations, p111.
"""

from itertools import permutations

def solve():
    """Search through enumerated solution space."""
    perms = list(permutations([1, 2, 3, 4, 5, 6]))
    a, b, c, d, e, f = (1, 2, 3, 4, 5, 6)

    return ((phys1, phys2, teach1, teach2, eng1, eng2,
             Moscow, Leningrad, Tula, Kiev, Kharkov, Odessa)
            for phys1, phys2, teach1, teach2, eng1, eng2 in perms
            if a is phys1 or a is phys2
            if e is teach1 or e is teach2
            if c is eng1 or c is eng2
            for Moscow, Leningrad, Tula, Kiev, Kharkov, Odessa in perms
            if Moscow is phys1 or Moscow is phys2
            if Leningrad is teach1 or Leningrad is teach2
            if Tula is eng1 or Tula is eng2
            if a is not Moscow
            if e is not Leningrad
            if c is not Tula
            if b is not Tula
            if f is not Tula
            if Kharkov is not a
            if Odessa is not c
            if Moscow is not b
            if Kharkov is not c)

if __name__ == '__main__':
    print list(solve())