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@yasyf 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()
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commented Apr 4, 2013

Related Blog Post: A Welcome Back Challenge

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