def find_solution():
start_time = datetime.now()
answer = 0
primes = []
progress = ProgressBar(1000000)
for n in xrange(2, progress.max):
progress.set_value(n)
if is_prime(n, primes):
primes.append(n)
#~ digits = map(lambda x: int(x), str(n))
try:
permutate(map(lambda x: x, str(n)), [], raise_if_not_prime)
print "Found palindrome: %s" % n
answer += 1
except IsNotPrimeError:
pass
#~ all_prime = True
#~ for item in permutate(map(lambda x: x, str(n)), ""):
#~ if not is_prime(int(item)):
#~ all_prime = False
#~ break
#~ if all_prime:
end_time = datetime.now()
progress.finished()
print "answer: %s" % answer
print "number of primes: %s" % len(primes)
print "first 10 primes: %s" % primes[:10]
print end_time - start_time
def find_solution():
start_time = datetime.now()
answer = 0
primes = []
progress = ProgressBar(1000000)
for n in xrange(2, progress.max):
progress.set_value(n)
if is_prime(n, primes):
primes.append(n)
all_prime = True
for item in rotate(n):
if not is_prime(int(item)):
all_prime = False
break
if all_prime:
print "Found palindrome: %s" % n
answer += 1
end_time = datetime.now()
progress.finished()
print "answer: %s" % answer
print "number of primes: %s" % len(primes)
print "first 10 primes: %s" % primes[:10]
print end_time - start_time
Find the sum of all the primes below two million.
"""
if __name__ == "__main__":
import sys, os
sys.path.append( os.path.join( os.getcwd(), '..' ) )
from datetime import datetime
from euler_tools.prime import is_prime
from euler_tools.progress_bar import ProgressBar
start_time = datetime.now()
sum = 0
primes = []
progress = ProgressBar(2000000)
for n in xrange(2, progress.max):
progress.set_value(n)
if is_prime(n, primes):
sum += n
primes.append(n)
end_time = datetime.now()
progress.finished()
print "sum: %s" % sum
print "number of primes: %s" % len(primes)
print end_time - start_time
#2,000,000: 142913828922
Triangle T(n)=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal P(n)=n(3n-1)/2 1, 5, 12, 22, 35, ...
Hexagonal H(n)=n(2n-1) 1, 6, 15, 28, 45, ...
It can be verified that T_(285) = P_(165) = H_(143) = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
"""
if __name__ == "__main__":
import sys, os
sys.path.append( os.path.join( os.getcwd(), '..' ) )
from euler_tools.progress_bar import ProgressBar
progress = ProgressBar(1000000)
max_iter = progress.max
for n_t in range(max_iter):
progress.set_value(n_t)
#Triangle
result_t = n_t * (n_t + 1) / 2
# Pentagonal
success = False
for n_p in range(max_iter):
result_p = n_p * (3 * n_p - 1) / 2
if result_p == result_t:
success = True