def is_square_free(n):
prime_list = list(pyecm.factors(n, False, True, 8, 1))
print(prime_list)
for i in range(len(prime_list) - 1):
if prime_list[i] == prime_list[i + 1]:
return False
return True
def generator(dimensionPrimeNumber, startPrimeNumber):
pq = []
for i in range(2):
TMP = startPrimeNumber
while len(str(TMP)) < dimensionPrimeNumber:
N = 4 * TMP + 2
U = 0
while True:
candidate = (N + U) * TMP + 1
a = randint(2, 5)
if pow(a, int(candidate - 1), int(candidate)) == 1 and pow(
a, int(N + U),
int(candidate)) != 1 and candidate % 4 == 3:
TMP = candidate
break
else:
U = U - 2
pq.append(TMP)
startPrimeNumber = startPrimeNumber + 2
while len(list(pyecm.factors(startPrimeNumber, False, True, 8,
1))) != 1:
startPrimeNumber = startPrimeNumber + 2
TMP = startPrimeNumber
p = pq[0]
q = pq[1]
print("\np = {}, длина = {}".format(p, len(str(p))))
print("q = {}, длина = {}".format(q, len(str(q))))
return p, q
def generate_p_q(prime_number, start_prime):
tmp = start_prime
pq = []
for i in range(4):
while len(str(tmp)) < prime_number:
N = 4 * tmp + 2
U = 0
candidate = 0
while True:
candidate = (N + U) * tmp + 1
if pow(2, int(candidate - 1), int(candidate)) == 1 and pow(
2, int(N + U), int(candidate)) != 1:
tmp = candidate
break
else:
U = U - 2
pq.append(tmp)
start_prime = start_prime + 2
while len(list(pyecm.factors(start_prime, False, True, 8, 1))) != 1:
start_prime = start_prime + 2
tmp = start_prime
Pa = pq[0]
Qa = pq[1]
Pb = pq[2]
Qb = pq[3]
return Pa, Qa, Pb, Qb
def find_p():
a = 3
q = generator(100, a)
p = 2 * q + 1
while isprime(p) == False:
a = a + 2
while len(list(pyecm.factors(a, False, True, 8, 1))) != 1:
a += 2
#print('a', a)
q = generator(100, a)
#print('q', q)
p = 2 * q + 1
#print('p', p)
if len(list(pyecm.factors(p, False, True, 8, 1))) == 1:
#print('len', len(list(pyecm.factors(p, False, True, 8, 1))))
return p, q
def generate_p_q():
prime_number = 82
start_prime = 367
tmp = start_prime
pq = []
for i in range(2):
while len(str(tmp)) < prime_number:
N = 4 * tmp + 2
U = 0
candidate = 0
while True:
candidate = (N + U) * tmp + 1
if pow(2, int(candidate - 1), int(candidate)) == 1 and pow(
2, int(N + U), int(candidate)) != 1:
tmp = candidate
break
else:
U = U - 2
pq.append(tmp)
start_prime = start_prime + 2
while len(list(pyecm.factors(start_prime, False, True, 8, 1))) != 1:
start_prime = start_prime + 2
tmp = start_prime
p = pq[0]
q = pq[1]
print("\np = {}, длина = {}".format(p, len(str(p))))
print("\nq = {}, длина = {}".format(q, len(str(q))))
return p, q
def ecmtestcomps():
rows = [['Magnitude', 'Time (sec)']]
path = r'C:\Users\Kenny\Dropbox\BCA\Senior Year\ToK\ecm_composite.csv'
for i in range(len(primes)):
start = time.clock()
for j in range(100):
len(list(factors(composites[i][0], False, True, 10, 1))) == 1
len(list(factors(composites[i][1], False, True, 10, 1))) == 1
t = time.clock() - start
row = [i, t]
print(row)
rows.append(row)
with open(path, 'w', newline='') as f:
writer = csv.writer(f)
writer.writerows(rows)
def ecmtestprimes():
rows = [['Magnitude', 'Time (sec)']]
path = r'C:\Users\Kenny\Dropbox\BCA\Senior Year\ToK\ecm_prime.csv'
for i in range(len(primes)):
start = time.clock()
for j in range(100):
len(list(factors(primes[i][0], False, True, 10, 1))) == 1
len(list(factors(primes[i][1], False, True, 10, 1))) == 1
t = time.clock() - start
row = [i, t]
print(row)
rows.append(row)
with open(path, 'w', newline='') as f:
writer = csv.writer(f)
writer.writerows(rows)
def is_square_free(n):
multipliers_list = list(pyecm.factors(
n, False, True, 8, 1)) # данная функция факторизирует число n
# описание библиотеки pyecm https://github.com/martingkelly/pyecm/blob/master/pyecm.py
for i in range(len(multipliers_list) - 1): # просматриваем уже с
if multipliers_list[i] == multipliers_list[i + 1]:
return False # смотрим, если два числа в массиве равны - то это квадрат
return True
def getECMFactors( target ):
from pyecm import factors
n = int( floor( target ) )
verbose = g.verbose
randomSigma = True
asymptoticSpeed = 10
processingPower = 1.0
if n < -1:
return [ ( -1, 1 ) ] + getECMFactors( fneg( n ) )
elif n == -1:
return [ ( -1, 1 ) ]
elif n == 0:
return [ ( 0, 1 ) ]
elif n == 1:
return [ ( 1, 1 ) ]
if verbose:
print( '\nfactoring', n, '(', int( floor( log10( n ) ) ), ' digits)...' )
if g.factorCache is None:
loadFactorCache( )
if n in g.factorCache:
if verbose and n != 1:
print( 'cache hit:', n )
print( )
return g.factorCache[ n ]
result = [ ]
for factor in factors( n, verbose, randomSigma, asymptoticSpeed, processingPower ):
result.append( factor )
result = [ int( i ) for i in result ]
largeFactors = list( collections.Counter( [ i for i in result if i > 65535 ] ).items( ) )
product = int( fprod( [ power( i[ 0 ], i[ 1 ] ) for i in largeFactors ] ) )
save = False
if product not in g.factorCache:
g.factorCache[ product ] = largeFactors
save = True
result = list( collections.Counter( result ).items( ) )
if n > g.minValueToCache and n not in g.factorCache:
g.factorCache[ n ] = result
g.factorCacheIsDirty = True
if verbose:
print( )
return result
def find_e(p, q, fi):
e = 17
#i = 10 ** 2
#while i < fi:
if isprime(e) and (gcd(e, p - 1) == gcd(e, q - 1) == 1):
if len(list(pyecm.factors(e, False, True, 8, 1))) == 1:
#e = i
#break
return e % fi
else:
#i += 1
e += 1
return e % fi
def validate(jamcoin, divisors):
outstrings = []
for i, j in zip(range(2, 11), divisors):
jamcoin_ten = int(jamcoin, i)
if gmpy2.is_prime(jamcoin_ten):
return False
outstrings.append(
pyecm.factors(jamcoin_ten, False, True, 10, 1).__next__())
# outstrings.append(factors.factors(jamcoin_ten).__next__()[0])
print(jamcoin, end='')
for i in outstrings:
print(' ', end='')
print(i, end='')
print()
return True
def get_primitive_root(prime):
# calculate phi(n)
# phi(prime) is (prime-1)
phi = prime - 1
prime_factors = list(pyecm.factors(phi, False, True, 10, 1))
# set range for primitive root
# so that we don't have to calculate all the roots
d_range = np.random.randint(20, 30)
possible_root = 5
while possible_root < prime:
temp = []
is_root = True
for i in range(len(prime_factors)):
remainder = gmpy2.powmod(possible_root, (phi // prime_factors[i]),
prime)
# if remainder is 1, then current number we're checking is not a primitive root
if remainder == 1:
is_root = False
break
if is_root and possible_root >= d_range:
return possible_root
possible_root = possible_root + 1
def findPrimitive(n):
s = set()
# находим функцию эйлера
phi = n - 1
# факторизуем функцию эйлера
s = set(pyecm.factors(phi, False, True, 8, 1))
for r in range(2, phi + 1):
#Проходимся по всем простым делителям phi
# и проверяем, если найдена степень равная 1
flag = False
for it in s:
# Проверяем если r^((phi)/primefactors)
# mod n сравнимо с 1 или нет
if (pow(r, phi // it, n) == 1):
flag = True
break
# если найден первообразный корень
if (flag == False):
return r
# если не найдено первообразного корня
return -1
def get_one_factor(n):
for x in factors(n, False, False, 2 * math.log(math.log(n)), 1.0):
return x
f = open('primes2011.txt', 'w')
start_time = time.time()
dimensionPrimeNumber = 500
startPrimeNumber = 2011
TMP = startPrimeNumber
for i in range(1000):
print(i + 1)
while len(str(TMP)) < dimensionPrimeNumber:
N = 4 * TMP + 2
U = 0
candidate = 0
while True:
candidate = (N + U) * TMP + 1
a = randint(2, 5)
if pow(a, int(candidate - 1), int(candidate)) == 1 and pow(
a, int(N + U), int(candidate)) != 1:
TMP = candidate
break
else:
U = U - 2
print(TMP)
f.write("Число№ " + str(i + 1) + "= " + str(TMP) + "\n")
startPrimeNumber = startPrimeNumber + 2
while len(list(pyecm.factors(startPrimeNumber, False, True, 8, 1))) != 1:
startPrimeNumber = startPrimeNumber + 2
TMP = startPrimeNumber
print("Время выполнения: ", time.time() - start_time)
f.write("Время выполнения: " + str(time.time() - start_time) + "\n")
def Func_isSquareFree(n):
multipliers_list = list(pyecm.factors(n, False, True, 8, 1))
for i in range(len(multipliers_list) - 1):
if multipliers_list[i] == multipliers_list[i + 1]:
return False
return True
