def find_solution(self):
"""
This method contains the guts of finding the solution.
The timer starts just prior to calling this method
and stops just after returning the solution's value.
"""
up_to = 0
delta = 5000
while len(Arithmetic.PRIMES_LIST) < 10002:
up_to += delta
Arithmetic.primes_up_to(up_to)
return Arithmetic.PRIMES_LIST[10000]
def find_solution(self):
"""
This method contains the guts of finding the solution.
The timer starts just prior to calling this method
and stops just after returning the solution's value.
"""
up_to = 0
delta = 50000
while up_to < 2000000:
up_to += delta
Arithmetic.primes_up_to(up_to)
p_list = Arithmetic.primes_up_to(2000000)
return "%d" % np.sum(p_list)
def test_set_position(self):
values = Arithmetic.primes_up_to(2)
msg = 'Primes through 2'
exp = np.array([2])
self.assertListEqual(list(exp), list(values), msg)
values = Arithmetic.primes_up_to(6)
msg = 'Primes through 6'
exp = np.array([2, 3, 5])
self.assertListEqual(list(exp), list(values), msg)
values = Arithmetic.primes_up_to(20)
msg = 'Primes through 6'
exp = np.array([2, 3, 5, 7, 11, 13, 17, 19])
self.assertListEqual(list(exp), list(values), msg)
def find_solution(self):
"""
This method contains the guts of finding the solution.
The timer starts just prior to calling this method
and stops just after returning the solution's value.
"""
# Initialize the palindrome instance with a palindrome just
# above the largest possible and walk backwards.
#
# 999 * 999 = 998001 so 998899 is too large
p_value = 998899
p_seq = Palindrome(p_value)
p_list = Arithmetic.primes_up_to(1000)
while p_value >= 10000:
p_value = p_seq.previous()
factors = np.array([])
value = p_value
for prime in p_list:
# If the value is 1, we've got it all
if value == 1:
break
while value % prime == 0:
# this prime is a factor of value so pull it out
factors = np.append(factors, prime)
value = value / prime
if value != 1:
# If the value is not 1 by now, a larger than 3 digit
# prime factor exists so this is not a solution.
continue
# Try to construct two 3-digit factors. Start at the highest
# prime factor, and multiply it in to the lowest of the two
# factors as you go trying to keep them even so both end up
# under the 1000 mark.
factor_1 = 1
factor_2 = 1
for index in range(len(factors), 0, -1):
factor = factors[index-1]
if factor_1 <= factor_2:
factor_1 *= factor
else:
factor_2 *= factor
if factor_1 < 1000 and factor_2 < 1000:
break
return '%d x %d = %d' % (factor_1, factor_2, factor_1 * factor_2)