def decrypt(data, number):
prime_num = prime.get_prime(number)
cur_data = ord(data) - prime_num
new_data = chr(cur_data % 128)
return new_data
def num_of_divisors(n):
i = 0
div = 1
while n > 1:
p = get_prime(i)
j = 1
while n % p == 0:
j += 1
n = n / p
i += 1
div *= j
return div
def weak_forge(l, attacks):
if '(p-1)-метод' or 'итерация' in attacks:
p = weak_low_p_1(l)
q = weak_low_p_1(l)
n = p * q
else:
p = get_prime(l)
q = get_prime(l)
n = p * q
if 'атака-Винера' in attacks:
if 'малый-модуль-шифрования' in attacks:
e = 2
while True:
if math.gcd(e, (p - 1) * (q - 1)) != 1:
e += 1
continue
d = reverse(e, (p - 1) * (q - 1))
if d < (1 / 3 * (n**0.25)):
print("Weak parameters: ({}, {})".format(e, n))
return 0
e += 1
else:
d = 2
while True:
if math.gcd(d, (p - 1) * (q - 1)) != 1:
d += 1
continue
e = reverse(d, (p - 1) * (q - 1))
print("Weak parameters: ({}, {})".format(e, n))
return 0
elif 'малый-модуль-шифрования' in attacks:
e = 3
while math.gcd(e, (p - 1) * (q - 1)) != 1:
e += 1
print("Weak parameters: ({}, {})".format(e, n))
return 0
def prime_permutations(digit): candidate = digit_filter(get_prime(pow(10,digit)-1), digit) i = 0 do_nothing = 0 while i < len(candidate): j = i while j < len(candidate): diff = [] diff.append(candidate[j]-candidate[i]) if not has_diff_number(candidate[i]): j+=1 continue elif diff[0] == 0: j+=1 continue if not has_same_number(candidate[i],candidate[j]): j+=1 continue # elif not digit_filter(diff,digit-1): # j+=1 # continue try: candidate.index(candidate[i] + (diff[0] * 2)) except ValueError: j+=1 continue if not has_same_number(candidate[i],candidate[j]+ diff[0]): j+=1 continue print candidate[i],candidate[j],diff,candidate[i] + (diff[0] * 2) j+=1 i+=1
def encrypt(data, number):
prime_num = prime.get_prime(number)
new_data = chr(ord(data) + prime_num)
return new_data
'''
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
'''
from prime import get_prime, is_prime
result, i = {}, 0
p = get_prime(i)
def get_next_number(x):
return (x % 10) * (10 ** (len(str(x)) - 1)) + x / 10
while p < 1000000:
if result.get(p, 0):
continue;
else:
t = get_next_number(p)
while t != p:
if is_prime(t):
t = get_next_number(t)
else:
break
if t == p:
result[p] = 1
i += 1
p = get_prime(i)
print len(result)