def remove_digits(number):
from maths.misc import get_digits, digit_list_to_num
digits = get_digits(number)
num_digits = len(digits)
for i in xrange(1, num_digits):
# Remove digits from the right
yield digit_list_to_num(digits[:i])
# Remove digits from the left
yield digit_list_to_num(digits[i:])
def list_to_num(digits):
length = len(digits)
num = 0
for i in xrange(length):
power = length - 1 - i
num += digits[i] * 10**power
return num
if __name__ == "__main__":
from maths.misc import get_digits, is_prime
circular_primes = []
for i in xrange(2, 1000000):
if not is_prime(i):
continue
prime = True
digits = get_digits(i)
for offset in xrange(1,len(digits)):
shifted = shift_digits(digits, offset)
num = list_to_num(shifted)
if not is_prime(num):
prime = False
break
if prime:
circular_primes.append(i)
print('The number of circular primes below 1,000,000 is %d'%len(circular_primes))
# We don't want to "over-reduce"
if num % i == 0 and denom % i == 0:
num = num / i
denom = denom / i
return num, denom
if __name__ == "__main__":
from maths.misc import get_digits
numerators = []
denominators = []
for i in xrange(10,100):
for j in xrange(i + 1,100):
# Get digits of i,j
digits_i, digits_j = get_digits(i), get_digits(j)
# Check for overlap
overlap = filter(set(digits_i).__contains__, digits_j)
if len(overlap) > 0:
# Pull first char from overlap
char_to_remove = overlap[0]
# Check for trivial case where digits to be removed are in the same power of 10s position
index_i = digits_i.index(char_to_remove)
index_j = digits_j.index(char_to_remove)
if index_i == index_j:
# Trivial case
continue