I am currently working on a project replicating RSA key generation and testing using euclidean algorithm, extended euclidean algorithm to find the modular inverse of the value.
I used the Miller-Rabin test to choose two prime numbers, p and q.
After running the code, I am able to obtain Kpub and e, however Kpr returns as nan.
Please help!
#Euclidean Algorithm func
def EucAlgo(a, b):
if a==0:
return b
return EucAlgo(b % a,a)
def ExEucAlgo(a,b):
if a==0:
return b,0,1
gcd, s1, t1 = ExEucAlgo(b%a,a)
#gcd of a,b
s = t1 - (b/a) * s1
t = s1
return gcd, s, t
def ExEucAlgo_modInverse(a,b):
gcd, s, t = ExEucAlgo(b,a)
if (gcd == 1):
i = t % a
elif (gcd !=1):
print("There is no inverse modulo for the input.")
return i
def SqMul_ModularExpo(b, exp, n):
bin_exp = bin(exp)
base = b
for i in range (3, len(bin_exp)):
base = (base ** 2) % n
if(bin_exp[i]=='1'):
i+=1
base = (base * b) %n
return base
#RSA Key generation
p=9054583561027584891319616491815785011595937977633787663340258672121877196627062461308487615739189212918799813327175451021729047602129396754172486202100997
q=10115395220079214686776355235686624745626962891667413288473649946208213820942557513105240135405981494333016032659525466362014175268953946332375459648688023
n= p * q
phi_n= (p-1) * (q-1)
e= randint(1, phi_n - 1)
while((EucAlgo(e,phi_n)) !=1):
e = randint(1, (phi_n-1))
d = ExEucAlgo_modInverse(e,phi_n)
print(f"\nKpr={d}")
print(f"\nKpub=(n={n})\n \ne={e}")
The problem is that you are using float point division which will result in returning float a point which when dealing with large int can result in very large floats which python can't handle so the solution is to use integer division which means 5//2=2 not 2.5. The problem is that Now encrypting and decrypting data would result in wrong decryption. (You wont get 2 again) because of some bugs in your functions.
FIRST: use public exponent pf 65537(prime number) which is the default for all RSA implementations(see your browser certificates) rather than finding a random one. Then after calculating the extended Euclidean algorithm which is used to find modulo inverse you dont have to make any more calculations(just return this value if GCD is 1 otherwise raise an error or whatever).
Here is the complete code that works after removing some unneeded (functions, imports, and random public exponent) READ comments.
def EucAlgo(a, b):
if a == 0:
return b
return EucAlgo(b % a, a)
def ExEucAlgo(a,b):
if a==0:
return b, 0, 1
gcd, s1, t1 = ExEucAlgo(b%a, a)
# You dont use / use // to make integer division
s = t1 - (b//a) * s1
t = s1
return gcd, s, t
def ExEucAlgo_modInverse(a,b):
gcd, s, t = ExEucAlgo(a, b)
if (gcd == 1):
# Just return s which is the inverse of public exponent
return s
elif (gcd != 1):
# I think it's better to raise an error but it's up to you
print("There is no inverse modulo for the input.")
#RSA Key generation
p = 9054583561027584891319616491815785011595937977633787663340258672121877196627062461308487615739189212918799813327175451021729047602129396754172486202100997
q = 10115395220079214686776355235686624745626962891667413288473649946208213820942557513105240135405981494333016032659525466362014175268953946332375459648688023
n = p * q
phi_n = (p-1) * (q-1)
# Just use fixed prime public exponent rather than trying fixed ones
e = 65537
d = ExEucAlgo_modInverse(e, phi_n)
print(f"\nKpr={d}")
print(f"\nKpub=(n={n})\n \ne={e}")
# Try to encrypt and decrypt 36
ciphertext = pow(36, e, n)
print("Encrypted data {}".format(ciphertext))
print("Decrypted data is {}".format(pow(ciphertext, d, n)))
Here's the very dumb way:
def divisorGenerator(n):
for i in xrange(1,n/2+1):
if n%i == 0: yield i
yield n
The result I'd like to get is similar to this one, but I'd like a smarter algorithm (this one it's too much slow and dumb :-)
I can find prime factors and their multiplicity fast enough.
I've an generator that generates factor in this way:
(factor1, multiplicity1)
(factor2, multiplicity2)
(factor3, multiplicity3)
and so on...
i.e. the output of
for i in factorGenerator(100):
print i
is:
(2, 2)
(5, 2)
I don't know how much is this useful for what I want to do (I coded it for other problems), anyway I'd like a smarter way to make
for i in divisorGen(100):
print i
output this:
1
2
4
5
10
20
25
50
100
UPDATE: Many thanks to Greg Hewgill and his "smart way" :)
Calculating all divisors of 100000000 took 0.01s with his way against the 39s that the dumb way took on my machine, very cool :D
UPDATE 2: Stop saying this is a duplicate of this post. Calculating the number of divisor of a given number doesn't need to calculate all the divisors. It's a different problem, if you think it's not then look for "Divisor function" on wikipedia. Read the questions and the answer before posting, if you do not understand what is the topic just don't add not useful and already given answers.
Given your factorGenerator function, here is a divisorGen that should work:
def divisorGen(n):
factors = list(factorGenerator(n))
nfactors = len(factors)
f = [0] * nfactors
while True:
yield reduce(lambda x, y: x*y, [factors[x][0]**f[x] for x in range(nfactors)], 1)
i = 0
while True:
f[i] += 1
if f[i] <= factors[i][1]:
break
f[i] = 0
i += 1
if i >= nfactors:
return
The overall efficiency of this algorithm will depend entirely on the efficiency of the factorGenerator.
To expand on what Shimi has said, you should only be running your loop from 1 to the square root of n. Then to find the pair, do n / i, and this will cover the whole problem space.
As was also noted, this is a NP, or 'difficult' problem. Exhaustive search, the way you are doing it, is about as good as it gets for guaranteed answers. This fact is used by encryption algorithms and the like to help secure them. If someone were to solve this problem, most if not all of our current 'secure' communication would be rendered insecure.
Python code:
import math
def divisorGenerator(n):
large_divisors = []
for i in xrange(1, int(math.sqrt(n) + 1)):
if n % i == 0:
yield i
if i*i != n:
large_divisors.append(n / i)
for divisor in reversed(large_divisors):
yield divisor
print list(divisorGenerator(100))
Which should output a list like:
[1, 2, 4, 5, 10, 20, 25, 50, 100]
I think you can stop at math.sqrt(n) instead of n/2.
I will give you example so that you can understand it easily. Now the sqrt(28) is 5.29 so ceil(5.29) will be 6. So I if I will stop at 6 then I will can get all the divisors. How?
First see the code and then see image:
import math
def divisors(n):
divs = [1]
for i in xrange(2,int(math.sqrt(n))+1):
if n%i == 0:
divs.extend([i,n/i])
divs.extend([n])
return list(set(divs))
Now, See the image below:
Lets say I have already added 1 to my divisors list and I start with i=2 so
So at the end of all the iterations as I have added the quotient and the divisor to my list all the divisors of 28 are populated.
Source: How to determine the divisors of a number
Although there are already many solutions to this, I really have to post this :)
This one is:
readable
short
self contained, copy & paste ready
quick (in cases with a lot of prime factors and divisors, > 10 times faster than the accepted solution)
python3, python2 and pypy compliant
Code:
def divisors(n):
# get factors and their counts
factors = {}
nn = n
i = 2
while i*i <= nn:
while nn % i == 0:
factors[i] = factors.get(i, 0) + 1
nn //= i
i += 1
if nn > 1:
factors[nn] = factors.get(nn, 0) + 1
primes = list(factors.keys())
# generates factors from primes[k:] subset
def generate(k):
if k == len(primes):
yield 1
else:
rest = generate(k+1)
prime = primes[k]
for factor in rest:
prime_to_i = 1
# prime_to_i iterates prime**i values, i being all possible exponents
for _ in range(factors[prime] + 1):
yield factor * prime_to_i
prime_to_i *= prime
# in python3, `yield from generate(0)` would also work
for factor in generate(0):
yield factor
An illustrative Pythonic one-liner:
from itertools import chain
from math import sqrt
def divisors(n):
return set(chain.from_iterable((i,n//i) for i in range(1,int(sqrt(n))+1) if n%i == 0))
But better yet, just use sympy:
from sympy import divisors
I like Greg solution, but I wish it was more python like.
I feel it would be faster and more readable;
so after some time of coding I came out with this.
The first two functions are needed to make the cartesian product of lists.
And can be reused whnever this problem arises.
By the way, I had to program this myself, if anyone knows of a standard solution for this problem, please feel free to contact me.
"Factorgenerator" now returns a dictionary. And then the dictionary is fed into "divisors", who uses it to generate first a list of lists, where each list is the list of the factors of the form p^n with p prime.
Then we make the cartesian product of those lists, and we finally use Greg' solution to generate the divisor.
We sort them, and return them.
I tested it and it seem to be a bit faster than the previous version. I tested it as part of a bigger program, so I can't really say how much is it faster though.
Pietro Speroni (pietrosperoni dot it)
from math import sqrt
##############################################################
### cartesian product of lists ##################################
##############################################################
def appendEs2Sequences(sequences,es):
result=[]
if not sequences:
for e in es:
result.append([e])
else:
for e in es:
result+=[seq+[e] for seq in sequences]
return result
def cartesianproduct(lists):
"""
given a list of lists,
returns all the possible combinations taking one element from each list
The list does not have to be of equal length
"""
return reduce(appendEs2Sequences,lists,[])
##############################################################
### prime factors of a natural ##################################
##############################################################
def primefactors(n):
'''lists prime factors, from greatest to smallest'''
i = 2
while i<=sqrt(n):
if n%i==0:
l = primefactors(n/i)
l.append(i)
return l
i+=1
return [n] # n is prime
##############################################################
### factorization of a natural ##################################
##############################################################
def factorGenerator(n):
p = primefactors(n)
factors={}
for p1 in p:
try:
factors[p1]+=1
except KeyError:
factors[p1]=1
return factors
def divisors(n):
factors = factorGenerator(n)
divisors=[]
listexponents=[map(lambda x:k**x,range(0,factors[k]+1)) for k in factors.keys()]
listfactors=cartesianproduct(listexponents)
for f in listfactors:
divisors.append(reduce(lambda x, y: x*y, f, 1))
divisors.sort()
return divisors
print divisors(60668796879)
P.S.
it is the first time I am posting to stackoverflow.
I am looking forward for any feedback.
Here is a smart and fast way to do it for numbers up to and around 10**16 in pure Python 3.6,
from itertools import compress
def primes(n):
""" Returns a list of primes < n for n > 2 """
sieve = bytearray([True]) * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = bytearray((n-i*i-1)//(2*i)+1)
return [2,*compress(range(3,n,2), sieve[1:])]
def factorization(n):
""" Returns a list of the prime factorization of n """
pf = []
for p in primeslist:
if p*p > n : break
count = 0
while not n % p:
n //= p
count += 1
if count > 0: pf.append((p, count))
if n > 1: pf.append((n, 1))
return pf
def divisors(n):
""" Returns an unsorted list of the divisors of n """
divs = [1]
for p, e in factorization(n):
divs += [x*p**k for k in range(1,e+1) for x in divs]
return divs
n = 600851475143
primeslist = primes(int(n**0.5)+1)
print(divisors(n))
If your PC has tons of memory, a brute single line can be fast enough with numpy:
N = 10000000; tst = np.arange(1, N); tst[np.mod(N, tst) == 0]
Out:
array([ 1, 2, 4, 5, 8, 10, 16,
20, 25, 32, 40, 50, 64, 80,
100, 125, 128, 160, 200, 250, 320,
400, 500, 625, 640, 800, 1000, 1250,
1600, 2000, 2500, 3125, 3200, 4000, 5000,
6250, 8000, 10000, 12500, 15625, 16000, 20000,
25000, 31250, 40000, 50000, 62500, 78125, 80000,
100000, 125000, 156250, 200000, 250000, 312500, 400000,
500000, 625000, 1000000, 1250000, 2000000, 2500000, 5000000])
Takes less than 1s on my slow PC.
Adapted from CodeReview, here is a variant which works with num=1 !
from itertools import product
import operator
def prod(ls):
return reduce(operator.mul, ls, 1)
def powered(factors, powers):
return prod(f**p for (f,p) in zip(factors, powers))
def divisors(num) :
pf = dict(prime_factors(num))
primes = pf.keys()
#For each prime, possible exponents
exponents = [range(i+1) for i in pf.values()]
return (powered(primes,es) for es in product(*exponents))
Old question, but here is my take:
def divs(n, m):
if m == 1: return [1]
if n % m == 0: return [m] + divs(n, m - 1)
return divs(n, m - 1)
You can proxy with:
def divisorGenerator(n):
for x in reversed(divs(n, n)):
yield x
NOTE: For languages that support, this could be tail recursive.
Assuming that the factors function returns the factors of n (for instance, factors(60) returns the list [2, 2, 3, 5]), here is a function to compute the divisors of n:
function divisors(n)
divs := [1]
for fact in factors(n)
temp := []
for div in divs
if fact * div not in divs
append fact * div to temp
divs := divs + temp
return divs
Here's my solution. It seems to be dumb but works well...and I was trying to find all proper divisors so the loop started from i = 2.
import math as m
def findfac(n):
faclist = [1]
for i in range(2, int(m.sqrt(n) + 2)):
if n%i == 0:
if i not in faclist:
faclist.append(i)
if n/i not in faclist:
faclist.append(n/i)
return facts
If you only care about using list comprehensions and nothing else matters to you!
from itertools import combinations
from functools import reduce
def get_devisors(n):
f = [f for f,e in list(factorGenerator(n)) for i in range(e)]
fc = [x for l in range(len(f)+1) for x in combinations(f, l)]
devisors = [1 if c==() else reduce((lambda x, y: x * y), c) for c in set(fc)]
return sorted(devisors)
My solution via generator function is:
def divisor(num):
for x in range(1, num + 1):
if num % x == 0:
yield x
while True:
yield None
Try to calculate square root a given number and then loop range(1,square_root+1).
number = int(input("Enter a Number: "))
square_root = round(number ** (1.0 / 2))
print(square_root)
divisor_list = []
for i in range(1,square_root+1):
if number % i == 0: # Check if mod return 0 if yes then append i and number/i in the list
divisor_list.append(i)
divisor_list.append(int(number/i))
print(divisor_list)
def divisorGen(n): v = n last = [] for i in range(1, v+1) : if n % i == 0 : last.append(i)
I don´t understand why there are so many complicated solutions to this problem.
Here is my take on it:
def divisors(n):
lis =[1]
s = math.ceil(math.sqrt(n))
for g in range(s,1, -1):
if n % g == 0:
lis.append(g)
lis.append(int(n / g))
return (set(lis))
return [x for x in range(n+1) if n/x==int(n/x)]