from sys import exit
"""
What we're going to do here is iterate over odd integers indefinitely,
filling up an array of primes and squares up to the current integer as we go.
Then we're going to iterate over the primes, subtracting the prime from the
integer, and seeing if the difference can be made up in the list of squares
HURRAH FOR GENERATORS
"""
def squares():
for x in euler.integers(1):
yield x**2
prime_generator = euler.primes()
square_generator = squares()
"""Using dicts for speedy lookup"""
primes = {}
squares = {}
for i in euler.integers(9, 2):
while 1:
p = prime_generator.next()
primes[p] = True
if p > i: break
while 1:
s = square_generator.next()
squares[s] = True
if s > i: break
import atexit
from lib import euler
existing_primes = []
try:
with open('data/primes') as f:
l = f.read()
existing_primes = [int(p) for p in l.splitlines()]
except IOError: pass
def write():
global new_primes
if new_primes:
to_write = '\n'.join([str(n) for n in new_primes]) + '\n'
with open('data/primes', 'a') as f:
f.write(to_write)
new_primes = []
i = len(existing_primes)
generator = euler.primes(i, existing_primes)
new_primes = []
while 1:
new_primes += [generator.next()]
if len(new_primes) > 1000: write()
"""
Find the sum of the only eleven primes that are both truncatable from
left to right and right to left.
"""
# we don't know the upper bound, but since there's only 11 of them we don't
# need to, I guess.
from lib import euler
truncatable = []
primes = {} # we'll want to look up numbers we've already generated for primality
for p in euler.primes():
primes[p] = True
if p < 10: continue
# truncate
p_ltr = str(p)
p_rtl = str(p)
t = True
while p_ltr:
if not primes.get(int(p_ltr), False) or not primes.get(int(p_rtl), False):
t = False
break
p_ltr = p_ltr[1:]
p_rtl = p_rtl[:-1]
if t:
truncatable += [p]
if len(truncatable) == 11:
"""Find the sum of all the primes below two million.""" from lib import euler primes = euler.primes() s = 0 for p in primes: if p >= 2*10**6: break s += p print s
def concat(a, b):
return int(str(a) + str(b))
def are_prime(l):
for n in l:
is_p = primes_map.get(n, None)
if is_p is None:
is_p = euler.is_prime(n)
primes_map[n] = is_p
if not is_p: return False
return True
for i, p1 in enumerate(euler.primes()):
#print p1
primes += [p1]
primes_map[p1] = True
for j, p2 in enumerate(primes):
c = [concat(p1, p2),
concat(p2, p1)]
if not are_prime(c): continue
for k, p3 in enumerate(primes[j+1:]):
c = [
concat(p1, p3),
concat(p3, p1),
concat(p2, p3),
concat(p3, p2),
]
"""
Which prime, below one-million, can be written as the sum of the most
consecutive primes?
"""
# seems easy: generate list of primes up to 1m and have a look
from lib import euler
generator = euler.primes()
primes = [] # list of primes, as an optimisation, this only goes up to
#the last two primes p1 and p2 where p1 + p2 < 1m
primes_map = {} # lookup, this goes up to 1m
last_p = 0
for p in euler.primes():
if p > 10**6: break
if p + last_p < 10**6:
primes += [p]
last_p = p
primes_map[p] = True
print 'generated primes', len(primes)
max_ = (0, 0)
for i, p in enumerate(primes):
#print p
