def stage19():
calculator = calc666.Calc666([".", "1", "2"],
binary_inverse=stage19_inverse)
factors = helper.get_factors(666)
nodes = [calculator.root]
for i in range(1, 3):
next_level = []
for node in nodes:
print(node)
factors = helper.get_factors(node.value)
for val in factors:
calculator.add_nodes_binary(node, val)
for n in node.binary_children:
next_level.append(n[0])
next_level.append(n[1])
nodes = list(next_level)
def problem_12():
"""What is the value of the first triangle number to have over five hundred
divisors?"""
# Closed form: \sum_i=1^n i = (n * (n + 1)) / 2
for n in xrange(7, 100000): # Don't loop forever.
triangle = (n * (n + 1)) / 2
if len(helper.get_factors(triangle)) >= 250:
return triangle
raise Exception('No solution to problem 12 found!')
def problem_9():
"""There exists exactly one Pythagorean triplet for which
a + b + c = 1000. Find the product abc."""
# Using Euclid's formula:
# http://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
# 1000 = a + b + c = (m**2 - n**2) + 2mn + (m**2 + n**2) = 2m**2 + 2mn =>
# 500 = m**2 + mn = m (m + n)
for f1, f2 in helper.get_factors(500):
m, n = (f1, f2 - f1) if f2 > f1 else (f2, f1 - f2) # Avoid negatives.
m, n = (m, n) if m > n else (n, m) # Make sure m > n.
a, b, c = (m ** 2 - n ** 2, 2 * m * n, m ** 2 + n ** 2) # Euclid's formula.
if a + b + c == 1000:
return a * b * c
raise error("No solution to problem 9 found!")
