24puzzle.py

from __future__ import absolute_import, division, print_function
import itertools
import operator

ops = {operator.add: '+', operator.sub: '-', operator.mul: '*', operator.truediv: '/'}

def run(a, b, c, d, op1, op2, op3, order):
  try:
    if order == 1:
      return op1(a, op2(b, op3(c, d)))
    elif order == 2:
      return op1(a, op3(op2(b, c), d))
    elif order == 3:
      return op3(op2(op1(a, b), c), d)
    elif order == 4:
      return op3(op1(a, op2(b, c)), d)
    elif order == 5:
      return op2(op1(a, b), op3(c, d))
  except ZeroDivisionError:
    return None
brackets = {
  1 : ('  ', ' ', ' (', ' ', ' (', '  ', ' ', '))'),
  2 : ('  ', ' ', ' ((', ' ', ' ', ') ', ' ', ')'),
  3 : ('((', ' ', ' ', ') ', ' ', ') ', ' ', ''),
  4 : (' (', ' ', ' (', ' ', ' ', ')) ', ' ', ''),
  5 : ('((', ' ', ' ', ') ', ' (', ' ', ' ', '))'),
}

def find_all(target):
  count = 0
  for a, b, c, d in itertools.permutations((1,3,4,6)):
    for op1, op2, op3 in itertools.product(ops, ops, ops):
      for order in (1,2,3,4,5):
        result = run(a, b, c, d, op1, op2, op3, order)
        if result is not None and abs(result - target) < 0.01:
          count = count + 1
          print("{count:3d}. {p[0]}{a}{p[1]}{op1}{p[2]}{b}{p[3]}{op2}{p[4]}{c}{p[5]}{op3}{p[6]}{d}{p[7]} = {result}".format(count=count, result=result, a=a, b=b, c=c, d=d, op1=ops[op1], op2=ops[op2], op3=ops[op3], p=brackets[order]))

find_all(24)