def isASquare(a,p,d=0,log=False):
space = " " * d
if log: print(f"{space}({a}/{p})",end = " => ")
if a == 0:
if log: print("0")
return 0
if a == 1:
if log: print(1)
return 1
if a == -1:
if log: print(-1)
return (-1)**(p-1 // 2)
if a == 2:
res = (-1)**((p**2 - 1) // 8)
if log: print(res)
return res
if a >= p:
if log: print()
return isASquare(a % p, p,d+1, log)
if not(isPrime(a)):
print(f"factoring {a}")
aFactored = pFactor(a)
res = 1
for factor in aFactored:
for _ in range(aFactored[factor]):
if log: print()
res *= isASquare(int(factor),p,d+1, log)
return res
if log: print(f"* {((-1)**(((p-1)//2) * ((a-1)//2)))}")
return isASquare(p,a,d+1,log) * ((-1)**(((p-1)//2) * ((a-1)//2)))
def characteristic(base):
if isPrime(base):
return base
factors = pFactor(base)
if len(factors) != 1:
return None
return int(next(iter(factors)))
def nthPrime(n):
found = 0
guess = 0
while found <= n:
guess += 1
if isPrime(guess):
found += 1
return guess
def read_file_text():
data = []
with open("primeNumber.txt", "r") as f:
line = f.readline()
while line:
line = line.strip()
data.append(isPrime(int(line)))
line = f.readline()
f.close()
return data
def isTrunctPrime(n):
number = []
for i in str(n):
number.append(int(i))
test_num = number[:]
while len(test_num) > 0:
if isPrime(int(''.join(map(str, test_num)))) == False:
return False
else:
test_num.pop()
test_num = number[:]
test_num.pop(0)
while len(test_num) > 0:
if isPrime(int(''.join(map(str, test_num)))) == False:
return False
else:
test_num.pop(0)
return True
def factor(n):
'''pollard's rho algorithm'''
if n < 1:
raise Exception('[Error]: {} is less than 1'.format(n))
if n == 1:
return []
if isPrime(n):
return [n]
fact = 1
cycle_size = 2
x = x_fixed = 2
c = randint(1, n)
while fact == 1:
for i in range(cycle_size):
if fact > 1:
break
x = (x * x + c) % n
if x == x_fixed:
c = randint(1, n)
continue
fact = gcd(x - x_fixed, n)
cycle_size *= 2
x_fixed = x
return factor(fact) + factor(n // fact)
from isPrime import isPrime
'''
for n in range(40):
p = n**2 + n + 41
print(p, isPrime(p))
for n in range(80):
p = n**2 - 79*n + 1601
print(p, isPrime(p))
'''
A = 0
B = 0
N = 0
for a in range(-1000, 1001):
for b in range(-1000, 1001):
n = 0
p = n**2 + a * n + b
while isPrime(p):
n += 1
p = n**2 + a * n + b
if n > N:
A = a
B = b
N = n
print(A * B)
from isPrime import isPrime
for x in range(__ITERS__):
isPrime(x)
print('end')
def test_prime(self):
self.assertEqual(isPrime(19), True)
from isPrime import isPrime
limit = 2000000
sumPrime = 5
n = 5
##for n in range(5, limit,2):
while n <= limit:
if isPrime(n):
sumPrime = sumPrime + n
n = n + 2
elif n <= limit and isPrime(n):
sumPrime = sumPrime + n
n = n + 4
else:
n = n + 2
print sumPrime
def main():
num = int(input("Enter a number: "))
if isPrime.isPrime(num):
print("This number is prime")
else:
print("This is not prime number")
def test_nonprime(self):
self.assertEqual(isPrime(12), False)
def test_is_a_prime(self):
a = [2, 3, 5, 7, 11, 31, 59, 15485863]
for i in a:
self.assertTrue(isPrime(i))
#Problem 10
from isPrime import isPrime
u=input('Upper limit: ')
s=0
for i in xrange(1,u+1):
if isPrime(i)==True:
s=s+i
print s
import isPrime
while True:
print('Please enter the number of digits your prime should have.')
length = int(input())
num = '9' * length
while len(str(num)) != length - 1:
if isPrime.isPrime(num) == True and length == len(str(num)):
f = open('prime.txt', 'w')
f.write(str(num))
f.close()
print(num)
break
else:
num = int(num)
num -= 1
'''
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
Execution time = 0.232 secs
'''
from isPrime import isPrime
import time
n = int(input("Enter number to find largest prime factor: "))
start = time.time()
i = 2
largest = 0
while(i*i<n):
if not (n%i):
if(isPrime(i)):
largest = i
if(isPrime(n/i)):
largest = n/i
break
i += 1
print(largest)
end = time.time()
print("Executed in {} secs".format(end - start))
#Problem 7
#By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13,
#we can see that the 6th prime is 13.
#What is the 10 001st prime number?
from isPrime import isPrime
n=input('Upper Limit: ')
i=0
j=0
while i!=n:
j=j+1
if isPrime(j):
i=i+1
print j
from isPrime import isPrime
limit = 2000000
sumPrime = 5
n = 5
##for n in range(5, limit,2):
while n <= limit:
if isPrime(n):
sumPrime = sumPrime + n
n = n + 2
elif n <= limit and isPrime(n):
sumPrime = sumPrime + n
n = n + 4
else:
n = n + 2
print sumPrime
#We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example,
#2143 is a 4-digit pandigital and is also prime.
#What is the largest n-digit pandigital prime that exists?
from itertools import permutations
from isPrime import isPrime
x = [1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789]
permut = []
for y in x:
perm = [''.join(p) for p in permutations(str(y))]
permut = permut + perm
prime = 0
for x in permut:
if isPrime(int(x)) == True and int(x) > int(prime):
prime = x
print(x)
print(prime)
def test_is_not_prime(self):
a = [9, 81, 140, 200, -2]
for i in a:
self.assertFalse(isPrime(i))
def test_input_is_not_number(self):
a = ["a", "1"]
for i in a:
self.assertFalse(isPrime(i))
def rwh_primes(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * int(n / 2)
for i in range(3, int(n**0.5) + 1, 2):
if sieve[int((i - 1) / 2)]:
sieve[int(i * i / 2)::i] = [False] * int((n - i * i - 1) /
(2 * i) + 1)
return [2] + [2 * i + 1 for i in range(1, int(n / 2)) if sieve[i]]
def rotate(s, d):
return s[-d:] + s[:-d]
rot = set()
primes = rwh_primes(10**6)
for prime in primes:
if prime not in rot:
add_prime = True
for d in range(len(str(prime))):
if not isPrime(int(rotate(str(prime), d))):
add_prime = False
if add_prime:
for d in range(len(str(prime))):
rot.add(int(rotate(str(prime), d)))
print(len(rot))
print(rot)
from isPalindrome import isPalindrome
from isPrime import isPrime
if __name__ == '__main__':
num = int(input('请输入正整数: '))
if (isPalindrome(num) and isPrime(num)):
print('%d是回文素数' % num)
else:
print('%d不是回文素数' % num)
#'Man kann die Fragestellung mit der Gleichung phi(x)=n*Produkt der Produkte aus (1-1/p) wobei p die Primfaktoren der Zahl n sind.
#So lässt sich das Problem verstehen, als die Suche nach dem der Zahl, bei der folgende Funktion maximal wird:
#1/Produkt aus (1-1/p)
# wobei P die Primfaktoren von der gesuchten Zahl sind. Die Funktion wird dann maximal, wenn die
# gesuchte Zahl n aus möglichst vielen Primfaktoren zusammengestzt ist.
#Man kann also die Fragestellung auch Formulieren, als die Suche nach der Zahl unter 1000000, die die meißten unterschiedlichen Primfaktoren enthält.
from isPrime import isPrime
primes = []
for x in range(2, 400):
if isPrime(x) == True:
primes.append(x)
product = 1
for x in primes:
if product * x < 1000000:
product = product * x
print(product)
def generatePrimes(limit):
prime_list = [2, 3, 5, 7, 11]
for n in range(13, (limit // 2), 2):
if isPrime(n):
prime_list.append(n)
return prime_list
def fermat(a, b, n):
b1 = 0
if isPrime.isPrime(n) and a > 0:
b1 = b % (n - 1)
return DownGradedModulo.power(a, b1, n)
return -1
from isPrime import isPrime
from fasterIsPrime import fasterIsPrime
import time
# Verify these are the same
for n in range(100):
assert(isPrime(n) == fasterIsPrime(n))
print("They seem to work the same!")
# Now let's see if we really sped things up
bigPrime = 499 # Try 1010809, or 10101023, or 102030407
print("Timing isPrime(", bigPrime, ")", end=" ")
time0 = time.time()
print(", returns ", isPrime(bigPrime), end=" ")
time1 = time.time()
print(", time = ", (time1-time0) * 1000, "ms")
print("Timing fasterIsPrime(", bigPrime, ")", end=" ")
time0 = time.time()
print(", returns ", fasterIsPrime(bigPrime), end=" ")
time1 = time.time()
print(", time = ", (time1-time0) * 1000, "ms")
def test_invalid(self):
self.assertEqual(isPrime(-1), False)
from isPrime import isPrime
from primes_to_n import list_primes
import itertools as it
#primes = list_primes(10**10)
#print("part 1")
def is_pandigital(s):
for d in range(1, len(s) + 1):
if s.count(str(d)) != 1:
return False
return True
cand = set()
for d in range(2, 9):
for i in it.permutations([str(i) for i in range(1, d + 1)]):
p = ''.join(i)
if isPrime(int(p)):
if is_pandigital(p):
cand.add(p)
print("count:", len(cand))
print("max:", max(cand))
from isPrime import isPrime print sum([x for x in xrange(2,2000000) if isPrime(x)])
