def keygen(n):
"""
Benaloh kriptosistemi ile public, private anahtarlari olusturup
sirasiyla publickey.txt, privatekey.txt dosyalarina yazar.
Arguments:
n (int) : anahtar bit uzunlugu
"""
p = q = y = x = None
while True:
p = generate_random(n // 2)
if (p - 1) % r == 0 and isprime(p) == 1 and math.gcd(
r, int((p - 1) / r)) == 1:
break
while True:
q = generate_random(n // 2)
if isprime(q) == 1 and math.gcd(r, int(q - 1)) == 1:
break
n_ = p * q
phi = (p - 1) * (q - 1)
while True:
y = randrange(1, n_)
if math.gcd(y, n_) == 1:
x = pow(y, phi * pow(r, -1, n_) % n_, n_)
if x != 1:
break
with open("publickey.txt", "w+") as publickey_file:
publickey_file.write(str(n_) + '\n' + str(y))
with open("privatekey.txt", "w+") as privatekey_file:
privatekey_file.write(str(n_) + '\n' + str(phi) + '\n' + str(x))
def problem_10():
running_sum = 2
for value in xrange(3, int(2E6)+1, 2):
if functions.isprime(value):
running_sum += value
return running_sum
def quadratic(a,b):
primecount = 0
for n in range(100):
if(isprime(n**2+a*n+b)==1):
primecount += 1
else:
break
return(primecount)
def problem_3(N=600851475143):
""" prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
"""
all_factors = list(functions.factors(N))
max_prime = 1
for factor in all_factors:
if factor > max_prime and functions.isprime(factor):
max_prime = factor
return max_prime
def problem_7(N=10001):
"""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?
"""
count = 0
value = 0
while count < N:
value += 1
if (not value % 2) and (value != 2):
continue
if functions.isprime(value):
count += 1
return value
def keygen(n):
w=[];wtotal=0;rcontrol=0;controlPrime=0
y=secrets.randbits(32)
w.append(y)
while len(w)!=8:
total=0
total=sum(w)
w.append(total+1)
wtotal=sum(w)
while controlPrime==0:
qNumber= random.randint(wtotal,wtotal+1000)
primeNumber=functions.isprime(qNumber)
if(primeNumber==0):
controlPrime=0
else:
controlPrime=1
while rcontrol==0:
rNumber= random.randint(1,1000)
gcd_q_r=math.gcd(qNumber,rNumber)
if(gcd_q_r==1):
rcontrol=1
else:
rcontrol=0
BArray=[]
for i in w:
b=(rNumber*i)%qNumber
BArray.append(b)
with open("publickey.txt","w") as public:
for i in BArray:
public.write(str(i)+" ")
private_key=(w,qNumber,rNumber)
with open("privatekey.txt","w") as private:
for i in w:
private.write(str(i)+" ")
private.write("-"+str(qNumber))
private.write("-"+str(rNumber))
def generatePrimeNumber(n):
p = 42
while not functions.isprime(p):
p = generateBigNumber(n)
return p
#!/usr/bin/python3
# -*- coding: utf-8 -*-
from functions import isprime
asal, n, p = 3, 9, 4
while (asal / (2 * p + 1) > 0.1):
for m in range(1, 5):
n += p
if (isprime(n) == 1):
asal += 1
p += 2
print(p + 1) # p kadar ekleyince karenin bir kenarı p+1 oluyor
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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 functions import isprime
n,p = 0,1
while n<10001:
p += 1
if(isprime(p)==1):
n += 1
print(p)
import functions as func
import numpy as np
primes = [2, 3]
value = 5
for i in np.arange(5, 2000000, 2):
if func.isprime(i):
primes.append(i)
value += float(i)
print(primes)
print(value)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# The prime 41, can be written as the sum of six consecutive primes:
# 41 = 2 + 3 + 5 + 7 + 11 + 13
# This is the longest sum of consecutive primes that adds to a prime below one-hundred.
# The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
# Which prime, below one-million, can be written as the sum of the most consecutive primes?
from functions import isprime
maks=1
primes=[]
for i in range(2,1000000):
if isprime(i)==1:
primes.append(i)
for p in range(100,600):
for i in range(len(primes)-p-1):
sums = 0
for j in range(p):
sums += primes[i+j]
if (i==1) and sums>primes[-1]:
break
if(primes.count(sums)==1):
maks=sums
print(maks)
from functions import allprimes
from functions import isprime
import random
deneme = 0
koltuklar=allprimes(100)
while(len(koltuklar)>0):
tercih = random.randint(1,200)%98
deneme += 1
if(isprime(tercih)==1 and koltuklar.count(tercih)==1):
koltuklar.remove(tercih)
print(deneme)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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 functions import isprime
sonuc = 13
for n in range(101, 1000000, 2):
if (isprime(n) == 0):
continue
for p in range(len(str(n))):
if (isprime(int(n)) == 0):
break
n = list(str(n))
n.append(n[0])
n.remove(n[0])
n = ''.join(str(i) for i in n)
else:
sonuc += 1
print(sonuc)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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 functions import isprime
sonuc = 13
for n in range(101,1000000,2):
if(isprime(n)==0):
continue
for p in range(len(str(n))):
if(isprime(int(n))==0):
break
n=list(str(n))
n.append(n[0])
n.remove(n[0])
n=''.join(str(i) for i in n)
else:
sonuc +=1
print(sonuc)
# Eoin Lees
# Computing the primes.
from functions import isprime
# My list of Primes - TBD
P = []
# Loop through all of the numbers we're checking for Primality.
for i in range(2, 100):
# If i is prime, then append to P.
if isprime(i):
P.append(i)
# Print out our list
print(P)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# Find the sum of all the primes below two million.
from functions import isprime
toplam = 0
for i in range(2, 2000000):
if isprime(i) == 1:
toplam += i
print(toplam)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# The prime 41, can be written as the sum of six consecutive primes:
# 41 = 2 + 3 + 5 + 7 + 11 + 13
# This is the longest sum of consecutive primes that adds to a prime below one-hundred.
# The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
# Which prime, below one-million, can be written as the sum of the most consecutive primes?
from functions import isprime
from functions import allprimes
asal = allprimes(4000)
toplam = []
for p in range(5,len(asal)):
for a in range(0,len(asal)-p):
if(isprime(sum(asal[a:a+p]))==1 and sum(asal[a:a+p])<1000000):
toplam.append(sum(asal[a:a+p]))
print(toplam[-1])
# 10001 inci asal sayıyı bulan kod.
# 10001. prime number
from functions import isprime
n, p = 0, 1
while n < 10001:
p += 1
if (isprime(p) == 1):
n += 1
print(p)
from functions import allprimes
from functions import isprime
import random
deneme = 0
koltuklar = allprimes(100)
while (len(koltuklar) > 0):
tercih = random.randint(1, 200) % 98
deneme += 1
if (isprime(tercih) == 1 and koltuklar.count(tercih) == 1):
koltuklar.remove(tercih)
print(deneme)
import numpy as np
import scipy as sp
import pandas as pd
import functions as func
import matplotlib.pyplot as plt
primes = [2,3]
number = 3
while len(primes) < 10001:
if func.isprime(number):
primes.append(number)
number+=2
print(primes[-1])
#!/usr/bin/python3
# -*- coding: utf-8 -*-
from functions import isprime
asal,n,p = 3,9,4
while (asal/(2*p+1)>0.1):
for m in range(1,5):
n += p
if(isprime(n)==1):
asal += 1
p += 2
print(p+1)# p kadar ekleyince karenin bir kenarı p+1 oluyor