'''
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http://projecteuler.net/problem=118
Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different sets can be formed. Interestingly with the set {2,5,47,89,631}, all of the elements belonging to it are prime.
How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?
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'''
import itertools as it
from math import log10, ceil
from problem007 import primes
from problem027 import is_prime
P = primes('lt', int(98765432 ** 0.5) + 1)
is_p = lambda x: is_prime(x, P)
'''Digits = set of remaining digits. n_min = maximum number in set so far.'''
# num_prime_sets = lambda digits, n_min: sum(num_prime_sets(digits - set(map(int, str(x))), x) for x in
# it.ifilter(is_p, xrange(n_min + 1, min(100000000, int(''.join(sorted(map(str, digits), reverse=True))) + 1)))) if digits else 1
def num_prime_sets(digits, n_min, n_max=None):
if not digits: return 1
digits_str = map(str, digits)
eligible = (lambda x: x > n_min and x <= n_max and is_p(x)) if n_max else (lambda x: x > n_min and is_p(x))
max_digits = min(8, len(digits))
if n_max: max_digits = min(max_digits, int(ceil(log10(n_max))))
x_primes = it.ifilter(eligible, (int(''.join(x)) for r in xrange(1, max_digits + 1) for x in it.permutations(digits_str, r)))
return sum(num_prime_sets(digits - set(map(int, str(x))), x) for x in x_primes)
if __name__ == "__main__":
print num_prime_sets(set(range(1, 10)), 0, 9876)
9 3 7 48073
For d = 0 to 9, the sum of all S(4, d) is 273700.
Find the sum of all S(10, d).
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'''
import itertools as it, numpy as np
from problem007 import primes
from problem027 import is_prime
def repeated_digit_nums(n, d, m):
D = str(d)
for stars in it.combinations(xrange(n), m):
a = np.array(['*'] * n)
a[np.array(stars)] = D
free = np.array(list(set(range(n)) - set(stars)))
for values in it.product('0123456789', repeat=n - m):
a[free] = values
if a[0] != '0':
yield int(''.join(a))
S = lambda n, d, m, primes: sum(filter(lambda x: is_prime(x, primes), repeated_digit_nums(n, d, m)))
def sum_s(n):
p = primes('lt', 10 ** (0.5 * n))
return sum(it.dropwhile(lambda x: x[1] == 0, ((m, S(n, d, m, p)) for m in xrange(n - 1, 0, -1))).next()[1] for d in xrange(10))
if __name__ == "__main__":
print sum_s(4)
print sum_s(10)
The smallest positive integer n for which the numbers n2+1, n2+3, n2+7, n2+9, n2+13, and n2+27 are consecutive primes is 10. The sum of all such integers n below one-million is 1242490.
What is the sum of all such integers n below 150 million?
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'''
from problem007 import primes
from problem027 import is_prime
from random import choice
'''Fermat probable primality test'''
N_LIST = 10 ** 6
SQRT_N_LIST = int(N_LIST ** 0.5)
P = primes('lt', SQRT_N_LIST + 1)
is_probable_prime = lambda n, tries = 20: all(pow(int(choice(P)), n - 1, n) == 1 for _ in xrange(tries))
isp = lambda x, tries = 20: is_prime(x, P) if x <= N_LIST else is_probable_prime(x, tries=tries)
def consecutive_n(N, offset, complement, mods_primes=50):
'''Return n for which {n^2+x: x in offset} is prime and {n^2+x: x in complement} is not prime.'''
# Tabulate permissible remainders n mod p for the first mods_primes primes p
p_list = primes('first', mods_primes)
nc = int(p_list[-1] ** 0.5) + 1
R = [[all((i2 + x) % p for x in offsets) for i2 in (i * i for i in xrange(p))] for p in p_list]
# By the Chinese remainder theorem, n must be of the form 210q + [10|80|130|200]
for s in [10, 80, 130, 200]:
for n in xrange(s, N, 210):
n2 = n * n
# First check against remainder table; if passes, run the more expensive primality test
if (n <= nc or all(R[i][n % p] for i, p in enumerate(p_list))) and \
all(isp(n2 + x) for x in offsets) and all(not isp(n2 + x) for x in complement):
yield n