Random point in Miller's Weil Pairing algorithm

Random point in Miller's Weil Pairing algorithm - python

I'm trying to implement the Miller's Weil Pairing algorithm but I have a problem.
I used the book "An introduction to mathematical cryptography" by Hoffstein, Pipher and Silverman and tried my implementation with the example given in the book :
I obtain the same response so the computations should be good but I have a problem choosing a random point S. In the book, it is stated that we must choose a point S not in {O,P,-Q,P-Q}.
But with this equation :
I try to compute e(P,Q) with P = (1,5) and Q = (12,2) of order 9 but there is always an error at one moment because the result of the Miller's function is 0/0. -Q = (12,11) and P-Q = (9,6) so I tried with all the others points and it does not work with points of the same order which is normal. (9,6) and (9,7) are the only ones that have a different order : 3. As (9,6) cannot be chosen I took (9,7) but I still get some 0 that create problems.
Can someone tell me what I did wrong or what I didn't understood ?
Thank you
Here is the code :
def g(p, q, r, curve):
modulo = curve.p
if p.x is None and p.y is None:
return (r.x - q.x) % modulo
if q.x is None and q.y is None:
return (r.x - p.x) % modulo
if p != q and p.x == q.x:
return (r.x - p.x) % modulo
l = slope(p, q, curve)
num = (r.y - p.y - l*(r.x - p.x)) % modulo
denom = (r.x + p.x + q.x - l**2) % modulo
return (num * mult_inverse(denom, modulo)) % modulo
def miller_algorithm(p, q, m, curve):
f = 1
t = copy(p)
n = list(bin(m)[2:])
for i in range(1,len(n)):
f = (f**2) * g(t, t, q, curve)
t = t.double_and_add(2)
if n[i] == '1':
f = f * g(t, p, q, curve)
t = t.add(p)
return f % curve.p
def weil_pairing(p, q, curve):
if p == q or (p.x is None and p.y is None) or (q.x is None and q.y is None):
return 1
m = q.order()
s = curve.random_point()
while s == p or s == q.neg() or s == (p.sub(q)) or s.order() == m:
s = curve.random_point()
fpnum = miller_algorithm(p, q.add(s), m, curve)
fpdenom = miller_algorithm(p, s, m, curve)
fqnum = miller_algorithm(q, p.sub(s), m, curve)
fqdenom = miller_algorithm(q, s.neg(), m, curve)
modulo = curve.p
num = (fpnum * mult_inverse(fpdenom, modulo)) % modulo
denom = (fqnum * mult_inverse(fqdenom, modulo)) % modulo
return (num * mult_inverse(denom, modulo)) % modulo

I can't tell you what you are doing incorrectly. But, the python program sagemath has an implementation of miller's algorithm, and all the pairings, weil, tate, and ate. If you install it, you can find the source code for miller and weil in the file ell_point.py. The algorithm is more general, and so more complex, than your implementation, but you should be able to translate to get what you want. Or, you can just import sagemath and use its implementations.

Related

Memory limit exceeded in Python

I am solving a problem that needs either a list of integers or a dictionary of size 10^18. Upon running the code the compiler throws an error message saying "Memory Limit Exceeded".
Here is my code:
def fun(l, r, p):
#f = [None, 1, 1]
f = {0:0, 1:1, 2:1}
su = 0
for i in range(1, r):
if i%2 == 0:
f[i+2] = 2*f[i+1] - f[i] + 2
#f.append(2*f[i+1] - f[i] + 2)
else:
f[i+2] = 3*f[i]
#f.append(3*f[i])
for k in range(l, r):
su = su + f[k]
su = (su + f[r]) % p
print(su)
t, p = input().split()
p = int(p)
t = int(t)
#t = 3
#p = 100000007
for i in range(t):
l , r = input().split()
l = int(l)
r = int(r)
fun(l, r, p)
It is showing memory limit exceeded with a maximum memory usage of 306612 KiB.

Two observations here:
You don't need to store all numbers simultaneously. You can use the deque and generator functions to generate the numbers by keeping track of only the last three digits generated instead of the entire sequence.
import itertools
from collections import deque
def infinite_fun_generator():
seed = [0, 1, 1]
dq = deque(maxlen=2)
dq.extend(seed)
yield from seed
for i in itertools.count(1):
if i % 2 == 0:
dq.append(2 * dq[-1] - dq[-2] + 2)
else:
dq.append(3 * dq[-2])
yield dq[-1]
def fun(l, r, p):
funs = itertools.islice(infinite_fun_generator(), l, r + 1)
summed_funs = itertools.accumulate(funs, lambda a, b: (a + b) % p)
return deque(summed_funs, maxlen=1)[-1]
You might have a better chance asking this in Math.SE since I don't want to do the math right now, but just like with the Fibonacci sequence there's likely an analytic solution that you can use to compute the nth member of the sequence analytically without having to iteratively compute the intermediate numbers and it may even be possible to analytically derive a formula to compute the sums in constant time.

(RSA) This script I got off of stackoverflow returns a negative d value

So I stumbled upon this thread on here with this script and it returns a negative d value and my p and q values are both prime. Any reason for this? Possibly just a faulty script?
def egcd(a, b):
x,y, u,v = 0,1, 1,0
while a != 0:
q, r = b//a, b%a
m, n = x-u*q, y-v*q
b,a, x,y, u,v = a,r, u,v, m,n
gcd = b
return gcd, x, y
def main():
p = 153143042272527868798412612417204434156935146874282990942386694020462861918068684561281763577034706600608387699148071015194725533394126069826857182428660427818277378724977554365910231524827258160904493774748749088477328204812171935987088715261127321911849092207070653272176072509933245978935455542420691737433
q = 156408916769576372285319235535320446340733908943564048157238512311891352879208957302116527435165097143521156600690562005797819820759620198602417583539668686152735534648541252847927334505648478214810780526425005943955838623325525300844493280040860604499838598837599791480284496210333200247148213274376422459183
e = 65537
ct = 313988037963374298820978547334691775209030794488153797919908078268748481143989264914905339615142922814128844328634563572589348152033399603422391976806881268233227257794938078078328711322137471700521343697410517378556947578179313088971194144321604618116160929667545497531855177496472117286033893354292910116962836092382600437895778451279347150269487601855438439995904578842465409043702035314087803621608887259671021452664437398875243519136039772309162874333619819693154364159330510837267059503793075233800618970190874388025990206963764588045741047395830966876247164745591863323438401959588889139372816750244127256609
# compute n
n = p * q
# Compute phi(n)
phi = (p - 1) * (q - 1)
# Compute modular inverse of e
gcd, a, b = egcd(e, phi)
d = a
print( "n: " + str(d) );
# Decrypt ciphertext
pt = pow(ct,d,n)
print( "pt: " + str(pt) )
if __name__ == "__main__":
main()

This can happen, I'll explain why below, but for practical purposes you'll want to know how to fix it. The answer to that is to add phi to d and use that value instead: everything will work as RSA should.
So why does it happen? The algorithm computes the extended gcd. The result of egcd is a*e + b*phi = gcd, and in the case of RSA, we have gcd = 1 so a*e + b*phi = 1.
If you look at this equation modulo phi (which is the order of the multiplicative group), then a*e == 1 mod phi which is what you need to make RSA work. In fact, by the same congruence, you can add or subtract any multiple of phi to a and the congruence still holds.
Now look at the equation again: a*e + b*phi = 1. We know e and phi are positive integers. You can't have all positive integers in this equation or else no way would it add up to 1 (it would be much larger than 1). So that means either a or b is going to be negative. Sometimes it will be a that is negative, other times it will be b. When it is b, then your a comes out as you would expect: a positive integer that you then assign to the value d. But the other times, you get a negative value for a. We don't want that, so simply add phi to it and make that your value of d.

Testing RSA in Python

I created a small code for testing RSA, but when I try to decrypt a message
with keys that are 6-7 digit long, it takes a while and gives me the wrong
result.
from math import sqrt
def isPrime(n):
x = int(sqrt(n)) + 1
if n < 2:
return False`
for i in range(2, x):
if (n / i).is_integer():
return (i, False
return True
def factor(num):
hold = list()
inum = int(sqrt(num) + 1)
hold.append((1, num))
if num % 2 == 0: hold.append((2, int(num / 2)))
for i in range(3, inum, 2):
x = num / i
if x.is_integer():
hold.append((i, int(x)))
return hold
def egcd(a, b):
#Extended Euclidean Algorithm
x,y, u,v = 0,1, 1,0
while a != 0:
q, r = b//a, b%a
m, n = x-u*q, y-v*q
b,a, x,y, u,v = a,r, u,v, m,n
gcd = b
return y
def fastMod(n, e):
if e == 0:
return 1
if e % 2 == 1:
return n * fastMod(n, e - 1)
p = fastMod(n, e / 2)
return p * p
def decrypt(p, q, em):
#Uses CRT for decrypting
mp = em % p; mq = em % q;
dp = d % (p-1); dq = d % (q-1);
xp = fastMod(mp, dp) % p; xq = fastMod(mq, dq) % q
log = egcd(p, q)
cp = (p-log) if log > 0 else (p+log)
cq = cp
m = (((q*cp)*xp) + ((p*cq)*xq)) % n
return m
def encrypt(pm):
return fastMod(pm, e) % n
Is there any way to improve speed or fix any errors?
I try to decrypt a few messages I made with a key 9-10 digits long, but it takes
too long.

A lot of things need improvement, but most notably:
For RSA encryption/decryption: fastMod( ) should take the modulus as an input parameter, and reduce by the modulus each iteration. I found this code which illustrates the right way to do it.
For parameter generation: In practice, one could never use a function like your isPrime( ) to determine primality because it runs in exponential time. Instead, you should be doing Miller-Rabin / Strong pseudo prime tests, which can use fastMod( ) as a sub-routine.
By the way, you are implementing textbook RSA here, which is hugely insecure. You would need to use padding such as OAEP to have security, but you need to be very careful on how you implement that to prevent various forms of attacks (such as side channel attacks).
As for why you are getting the wrong result, it is hard to tell without seeing all of your code. Maybe you want to include a main function that generates params and tries to use them for encryption and decryption.
EDIT: I did notice this which looks suspicious: log = egcd(p, q). Not sure what you are doing here. I suggest you first compute d as the inverse of e mod (p-1)*(q-1) and verify that you are getting that correct (ie multiply d*e mod (p-1)*(q-1) and make sure the result is 1). If so, then do a fastMod( ) with d to see if it decrypts (it should). Once you get that working, then move on to making CRT work.

How to check if the number can be represented prime power (nth root is prime or not)

I am trying this problem for a while but getting wrong answer again and again.
number can be very large <=2^2014.
22086. Prime Power Test
Explanation about my algorithm:
For a Given number I am checking if the number can be represented as form of prime power or not.
So the the maximum limit to check for prime power is log n base 2.
Finally problem reduced to finding nth root of a number and if it is prime we have our answer else check for all i till log (n base 2) and exit.
I have used all sort of optimizations and have tested enormous test-cases and for all my algorithm gives correct answer
but Judge says wrong answer.
Spoj have another similar problem with small constraints n<=10^18 for which I already got accepted with Python and C++(Best solver in c++)
Here is My python code Please suggest me if I am doing something wrong I am not very proficient in python so my algorithm is a bit lengthy. Thanks in advance.
My Algorithm:
import math
import sys
import fractions
import random
import decimal
write = sys.stdout.write
def sieve(n):
sqrtn = int(n**0.5)
sieve = [True] * (n+1)
sieve[0] = False
sieve[1] = False
for i in range(2, sqrtn+1):
if sieve[i]:
m = n//i - i
sieve[i*i:n+1:i] = [False] * (m+1)
return sieve
def gcd(a, b):
while b:
a, b = b, a%b
return a
def mr_pass(a, s, d, n):
a_to_power = pow(a, d, n)
if a_to_power == 1:
return True
for i in range(s-1):
if a_to_power == n - 1:
return True
a_to_power = (a_to_power * a_to_power) % n
return a_to_power == n - 1
isprime=sieve(1000000)
sprime= [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997]
def smooth_num(n):
c=0
for a in sprime:
if(n%a==0):
c+=1
if(c>=2):
return True;
return False
def is_prime(n):
if(n<1000000):
return isprime[n]
if any((n % p) == 0 for p in sprime):
return False
if n==2:
return True
d = n - 1
s = 0
while d % 2 == 0:
d >>= 1
s += 1
for repeat in range(10):
a=random.randint(1,n-1)
if not mr_pass(a, s, d, n):
return False
return True
def iroot(n,k):
hi = 1
while pow(hi, k) < n:
hi *= 2
lo = hi // 2
while hi - lo > 1:
mid = (lo + hi) // 2
midToK = (mid**k)
if midToK < n:
lo = mid
elif n < midToK:
hi = mid
else:
return mid
if (hi**k) == n:
return hi
else:
return lo
def isqrt(x):
n = int(x)
if n == 0:
return 0
a, b = divmod(n.bit_length(), 2)
x = pow(2,(a+b))
while True:
y = (x + n//x)>>1
if y >= x:
return x
x = y
maxx=2**1024;minn=2**64
def nth_rootp(n,k):
return int(round(math.exp(math.log(n)/k),0))
def main():
for cs in range(int(input())):
n=int(sys.stdin.readline().strip())
if(smooth_num(n)):
write("Invalid order\n")
continue;
order = 0;m=0
power =int(math.log(n,2))
for i in range(1,power+1):
if(n<=maxx):
if i==1:m=n
elif(i==2):m=isqrt(n)
elif(i==4):m=isqrt(isqrt(n))
elif(i==8):m=isqrt(isqrt(isqrt(n)))
elif(i==16):m=isqrt(isqrt(isqrt(isqrt(n))))
elif(i==32):m=isqrt(isqrt(isqrt(isqrt(isqrt(n)))))
elif(i==64):m=isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(n))))))
elif(i==128):m=isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(n)))))))
elif(i==256):m=isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(n))))))))
else:m=int(nth_rootp(n,i))
else:
if i==1:m=n
elif i==2:m=isqrt(n)
elif(i==4):m=isqrt(isqrt(n))
elif(i==8):m=isqrt(isqrt(isqrt(n)))
elif(i==16):m=isqrt(isqrt(isqrt(isqrt(n))))
elif(i==32):m=isqrt(isqrt(isqrt(isqrt(isqrt(n)))))
elif(i==64):m=isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(n))))))
elif(i==128):m=isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(n)))))))
elif(i==256):m=isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(isqrt(n))))))))
else:m=iroot(n,i)
if m<2:
order=0
break
if(is_prime(m) and n==(m**i)):
write("%d %d\n"%(m,i))
order = 1
break
if(order==0):
write("Invalid order\n")
main()

I'm not going to read all that code, though I suspect the problem is floating-point inaccuracy. Here is my program to determine if a number n is a prime power; it returns the prime p and the power k:
# prime power predicate
from random import randint
from fractions import gcd
def findWitness(n, k=5): # miller-rabin
s, d = 0, n-1
while d % 2 == 0:
s, d = s+1, d/2
for i in range(k):
a = randint(2, n-1)
x = pow(a, d, n)
if x == 1 or x == n-1: continue
for r in range(1, s):
x = (x * x) % n
if x == 1: return a
if x == n-1: break
else: return a
return 0
# returns p,k such that n=p**k, or 0,0
# assumes n is an integer greater than 1
def primePower(n):
def checkP(n, p):
k = 0
while n > 1 and n % p == 0:
n, k = n / p, k + 1
if n == 1: return p, k
else: return 0, 0
if n % 2 == 0: return checkP(n, 2)
q = n
while True:
a = findWitness(q)
if a == 0: return checkP(n, q)
d = gcd(pow(a,q,n)-a, q)
if d == 1 or d == q: return 0, 0
q = d
The program uses Fermat's Little Theorem and exploits the witness a to the compositeness of n that is found by the Miller-Rabin algorithm. It is given as Algorithm 1.7.5 in Henri Cohen's book A Course in Computational Algebraic Number Theory. You can see the program in action at http://ideone.com/cNzQYr.

this is not really an answer, but I don't have enough space to write it as a comment.
So, if the problem still not solved, you may try the following function for nth_rootp, though it is a bit ugly (it is just a binary search to find the precise value of the function):
def nth_rootp(n,k):
r = int(round(math.log(n,2)/k))
left = 2**(r-1)
right = 2**(r+1)
if left**k == n:
return left
if right**k == n:
return right
while left**k < n and right**k > n:
tmp = (left + right)/2
if tmp**k == n:
return tmp
if tmp == left or tmp == right:
return tmp
if tmp**k < n:
left = tmp
else:
if tmp**k > n:
right = tmp

your code look like a little overcomplicated for this task, I will not bother to check it, but the thing you need are the following
is_prime, naturally
a prime generator, optional
calculate the nth root of a number in a precise way
for the first one I recommend the deterministic form of the Miller-Rabin test with a appropriate set of witness to guaranty a exact result until 1543267864443420616877677640751301 (1.543 x 1033) for even bigger numbers you can use the probabilistic one or use a bigger list of witness chosen at your criteria
with all that a template for the solution is as follow
import math
def is_prime(n):
...
def sieve(n):
"list of all primes p such that p<n"
...
def inthroot(x,n):
"calculate floor(x**(1/n))"
...
def is_a_power(n):
"return (a,b) if n=a**b otherwise throw ValueError"
for b in sieve( math.log2(n) +1 ):
a = inthroot(n,b)
if a**b == n:
return a,b
raise ValueError("is not a power")
def smooth_factorization(n):
"return (p,e) where p is prime and n = p**e if such value exists, otherwise throw ValueError"
e=1
p=n
while True:
try:
p,n = is_a_power(p)
e = e*n
except ValueError:
break
if is_prime(p):
return p,e
raise ValueError
def main():
for test in range( int(input()) ):
try:
p,e = smooth_factorization( int(input()) )
print(p,e)
except ValueError:
print("Invalid order")
main()
And the code above should be self explanatory
Filling the blacks
As you are familiar with Miller-Rabin test, I will only mention that if you are interested you can find a implementation of the determinist version here just update the list of witness and you are ready to go.
For the sieve, just change the one you are using to return a list with primes number like this for instance [ p for p,is_p in enumerate(sieve) if is_p ]
With those out of the way, the only thing left is calculate the nth root of the number and to do that in a precise way we need to get rip of that pesky floating point arithmetic that only produce headaches, and the answer is implement the Nth root algorithm using only integer arithmetic, which is pretty similar to the one of isqrt that you already use, I guide myself with the one made by Mark Dickinson for cube root and generalize it and I get this
def inthroot(A, n) :
"calculate floor( A**(1/n) )"
#https://en.wikipedia.org/wiki/Nth_root_algorithm
#https://en.wikipedia.org/wiki/Nth_root#nth_root_algorithm
#https://stackoverflow.com/questions/35254566/wrong-answer-in-spoj-cubert/35276426#35276426
#https://stackoverflow.com/questions/39560902/imprecise-results-of-logarithm-and-power-functions-in-python/39561633#39561633
if A<0:
if n%2 == 0:
raise ValueError
return - inthroot(-A,n)
if A==0:
return 0
n1 = n-1
if A.bit_length() < 1024: # float(n) safe from overflow
xk = int( round( pow(A,1.0/n) ) )
xk = ( n1*xk + A//pow(xk,n1) )//n # Ensure xk >= floor(nthroot(A)).
else:
xk = 1 << -(-A.bit_length()//n) # 1 << sum(divmod(A.bit_length(),n))
# power of 2 closer but greater than the nth root of A
while True:
sig = A // pow(xk,n1)
if xk <= sig:
return xk
xk = ( n1*xk + sig )//n
and with all the above you can solve the problem without inconvenient

from sympy.ntheory import factorint
q=int(input("Give me the number q="))
fact=factorint(q) #We factor the number q=p_1^{n_1}*p_2^{n_2}*...
p_1=list(fact.keys()) #We create a list from keys to be the the numbers p_1,p_2,...
n_1=list(fact.values()) #We create a list from values to be the the numbers n_1,n_2,...
p=int(p_1[0])
n=int(n_1[0])
if q!=p**n: #Check if the number q=p_{1}[0]**n_{1}[0]=p**n.
print("The number "+str(q)+" is not a prime power")
else:
print("The number "+str(q)+" is a prime power")
print("The prime number p="+str(p))
print("The natural number n="+str(n))

Simple RSA code

Hi I am trying to create a working RSA program, but on a very small level, I am having problems encrypting and decrypting with this code, can someone help me figure out what is wrong? I have tried doing this many different ways, but this way seems to be the right math, so I believe it might just be my lack of coding skills? Thanks
import random, math
def RandomPrime():
prime = False
while prime == False:
n = 2
while n % 2 == 0:
n = random.randint(10000, 100000)
s = math.trunc(n**0.5)
s = int(s)
x = 3
# While n doesn't exactly divide to equal 0, and x is less then the sqrt of n
while ( n % x != 0 ) and (x <= s):
x = x + 2
# if n is greater than s, it means it has run out of numbers to test, so is prime
if x > s:
prime = True
return n
def Modulus(p, q):
M = p * q
return M
def Totient(p, q):
T = ((p-1) * (q-1))
return T
def Pubkey(T):
prime = False
while prime == False:
n = 2
while n % 2 == 0:
n = random.randint(3, T)
s = math.trunc(n**0.5)
s = int(s)
x = 3
# While
while ( n % x != 0 ) and (x <= s):
x = x + 2
if x > s:
prime = True
return n
def privkey( T, n):
y = math.fmod(1, T)
d = float((y / n))
return d
# z is my encyption in this scenario
z = 8
# I generate p and q, using my random prime generator, i used low primes in
# this example just to see if it would work but it is still not showing reults
p = RandomPrime()
q = RandomPrime()
print(p, q)
#This creates the modulus
M = Modulus(p, q)
print(M)
# Eulier's totient
T = Totient(p, q)
print(T)
#Pub key creation
n = Pubkey(T)
print(n)
#Priv key creation
d = privkey(n, T)
print(d)
enc = (pow(z, n)) % M
print('enc: ', enc)
dec = (pow(enc, d)) % M
print('dec: ', dec)

Your privkey function appears wrong - I'm guessing you saw the definition of RSA's private key value as something like:
the value "e" such that e * d = 1 mod Phi(N)
However in this case, 1 mod Phi(N) does not mean The remainder when 1 is divided by Phi(N) (which appears to be the way you have translated it into code, based on your use of math.fmod(1, T), but in fact should be read more like:
the value "e" such that (e * d) mod Phi(N) = 1
This value is generally calculated using the Extended Euclidean Algorithm. An example Python implementation is here.
It's also worth noting that you seem to be defining privkey(T, n) but calling it as privkey(n, T).

Check my blog which in detail contains the implementation of the following using python:
MD5 Secure hash Algorithm RFC 1321, RSA public Key cryptography RFC 3447, OpenPGP RFC 4880
def keyGen():
''' Generate Keypair '''
i_p=randint(0,20)
i_q=randint(0,20)
# Instead of Asking the user for the prime Number which in case is not feasible,
# generate two numbers which is much highly secure as it chooses higher primes
while i_p==i_q:
continue
primes=PrimeGen(100)
p=primes[i_p]
q=primes[i_q]
#computing n=p*q as a part of the RSA Algorithm
n=p*q
#Computing lamda(n), the Carmichael's totient Function.
# In this case, the totient function is the LCM(lamda(p),lamda(q))=lamda(p-1,q-1)
# On the Contrary We can also apply the Euler's totient's Function phi(n)
# which sometimes may result larger than expected
lamda_n=int(lcm(p-1,q-1))
e=randint(1,lamda_n)
#checking the Following : whether e and lamda(n) are co-prime
while math.gcd(e,lamda_n)!=1:
e=randint(1,lamda_n)
#Determine the modular Multiplicative Inverse
d=modinv(e,lamda_n)
#return the Key Pairs
# Public Key pair : (e,n), private key pair:(d,n)
return ((e,n),(d,n))
Blog Link :Python Cryptography
Github Link : Python Cryptography

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