Zebra Puzzle

#reference: https://en.wikipedia.org/wiki/Zebra_Puzzle
# keypoint in python here: 
#1. generator expression
#2. *args to function calls
#3. itertools module 
#4. is (if two variables point to the same object)
#5. rearrage for and if clause to stop program earlier (if not this program may run estimately an hour)
#5. time module

import itertools
import time

def imright (h1, h2):
  "House h1 is immediately right of h2 if h1 - h2 == 1"
  return h1 - h2 == 1
  
def nextto(h1, h2):
  "Two houses are next to each other if they differ by 1"
  return abs(h1 - h2) == 1

def zebra_puzzle():
  "Return a tuple (WATER, ZEBRA) indicating their house numbers."
  houses = first, _, middle, _, _ = [1, 2, 3, 4, 5]
  orderings = list(itertools.permutations(houses)) #1
  
  return next((WATER, ZEBRA)
          for (red, green, ivory, yellow, blue) in orderings
            if imright(green, ivory) #6
          for (Englishman, Spaniard, Ukranian, Japanese, Norwegian) in orderings
            if (Englishman is red) #2
            if Norwegian is first #10
            if nextto(Norwegian, blue) #15
          for (coffee, tea, milk, oj, WATER) in orderings
            if coffee is green #4
            if Ukranian is tea #5
            if milk is middle #9
          for (OldGold, Kools, Chesterfields, LuckyStrike, Parliaments) in orderings
            if Kools is yellow #8
            if LuckyStrike is oj #13
            if Japanese is Parliaments #14      
          for (dog, snails, fox, horse, ZEBRA) in orderings
            if (Spaniard is dog) #3
            if OldGold is snails #7
            if nextto(Chesterfields, fox) #11
            if nextto(Kools, horse) #12
          )
          
def timedcall(fn, *args):
  "Call function with args; return the time in seconds and results."
  t0 = time.clock()
  result = fn(*args)
  t1 = time.clock()
  return t1-t0, result
  
print(timedcall(zebra_puzzle)) # time cost: less than 0.001 second