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
def test(a, b):
n = 0
while 1:
term = f(n, a, b)
p = primes.get(term, None)
if p is None:
p = euler.is_prime(term)
primes[term] = p
if p:
n+=1
else: break
return n
"""
http://projecteuler.net/index.php?section=problems&id=58
"""
from lib import spiraldiagonals
from lib import euler
"""
this is pretty easy - we can generate the diagonals indefinitely from the
formula we came up with in q28 and simply check the prime ratio
it's a bit slow, probably the prime check, but still less than a minute
"""
primes = []
ds = [1]
for i, diagonals in enumerate(spiraldiagonals.diagonals()):
for d in diagonals:
ds += [d]
if euler.is_prime(d): primes += [d]
ratio = float(len(primes)) / len(ds)
if ratio <= 0.1:
print (i+1)*2 + 1 # this is the spiral width
break
"""
What is the largest n-digit pandigital prime that exists?
"""
from lib import euler
import math
digits = '123456789'
while digits:
n = digits
p = 0
while 1:
if int(n) > p and euler.is_prime(int(n)):
p = int(n)
try:
n = ''.join(euler.lexicographic_permutation(n))
except: break
if p: break
digits = digits[:-1]
print p
