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# yasyf/Mr. Warner.py

Created Apr 3, 2013
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 #!/usr/bin/env python """ Yasyf Mohamedali (http://www.yasyf.com) First published at https://gist.github.com/yasyf/5299301 A quick python script to solve the math riddle involving an insurance broker and a woman's daughters. This script will take a variable amount of daughters, whose ages multiply to any number, and present you with all possible solutions, if any. Quickstart: To run, save the script, make sure you have python installed, and simply run the following from the command line: python 'Mr. Warner.py' """ import math, itertools, collections def get_int(prompt): print prompt return input("> ") def find_factors(n): sqrt = math.sqrt(n) i = 1 factors = [] while i <= sqrt: if n % i == 0: factors.append(i) if i != sqrt: factors.append(n/i) i += 1 return sorted(factors) def product_combination(combination): product = 1 for i in range(0,len(combination)): product *= combination[i] return product def sum_combination(combination): sum = 0 for i in range(0,len(combination)): sum += combination[i] return sum def possible_combinations(n,product,factors): return [x for x in itertools.combinations_with_replacement(factors, n) if product_combination(x) == product] def combinations_with_repeated_sums(combinations): sums = dict(zip([x for x in combinations],[sum_combination(x) for x in combinations])) repeated_sums = [x for x,y in collections.Counter(sums.values()).items() if y > 1] return [x for x in sums.keys() if sums[x] in repeated_sums] def filter_single_oldest(combinations): return [x for x in combinations if x[-1] != x[-2]] def format_children(combination): response = "The ages of the %s children are " % len(combination) for x in combination: if x == 1: addition = "1 year old" else: addition = "%s years old" % x if x == combination[len(combination)-1]: response += "and %s." % addition else: response += "%s, " % addition return response def print_answers(combinations): if len(combinations) == 0: print "There is no solution to this question." elif len(combinations) == 1: print format_children(combinations[0]) else: print "There are %s possible answers:" % len(combinations) for x in combinations: print format_children(x) def run(number_children=None,age_product=None): if number_children is None: number_children = get_int("How Many Children?") if age_product is None: age_product = get_int("What Is The Product Of Their Ages?") #each age must be a factor of (the product of the ages) #find all the factors of (the product of the ages) factors = find_factors(age_product) #find every possible combination of choosing the a certain number of ages (the number of children) from all the factors of (the product of the ages) #for example, if there are three children, find all the possible combinations of choosing three ages from the list of all the factors combinations = possible_combinations(number_children,age_product,factors) #since the man didn't have enough information after the first clue, there must be multiple combinations that add up to the same number (the house next door) #for each combination, sum up the ages, and find the combinations that correspond with sums that are repeated at least twice repeats = combinations_with_repeated_sums(combinations) #since there is only one eldest child, the two eldest children cannot have the same age #filter out any answers where the two eldest children have the same age valid_answers = filter_single_oldest(repeats) #print the answers, if any print_answers(valid_answers) return len(valid_answers) def find_result(n,strict=True,x=1,y=1): #find numbers that give n results found = False while found is False: if x >= 5: x = 1 y += 1 else: x += 1 print "Testing with %s children whose ages multiply to %s." % (x,y) result = run(x,y) if (result == n) or (result > n and strict is False): print "There are %s solutions with %s children whose ages multiply to %s." % (result,x,y) found = True def main(): run() #find_result(3) if __name__ == '__main__': main()

### yasyf commented Apr 4, 2013

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