Generate Private Key Rsa Euclidian

• Public Private Key Encryption
• Generate Private Key Rsa Euclidean Property

RSA encryption and decryption in Python (3) I need help using RSA encryption and decryption in Python. I am creating a private/public key pair, encrypting a message with keys and writing message to a file. Then I am reading ciphertext from file and decrypting text using key. I understand that finding the multiplicative inverse in that way is computationally difficult. If this weren't the case, then any attacker could determine the private key from the public key. What am I misunderstanding here? What does the key generator know that an attacker doesn't, and how is that used to determine the private key?

rsa.py

#!/usr/bin/env python

# This example demonstrates RSA public-key cryptography in an

# easy-to-follow manner. It works on integers alone, and uses much smaller numbers

# for the sake of clarity.

#####################################################################

# First we pick our primes. These will determine our keys.

#####################################################################

# Pick P,Q,and E such that:

# 1: P and Q are prime; picked at random.

# 2: 1 < E < (P-1)*(Q-1) and E is co-prime with (P-1)*(Q-1)

P=97# First prime

Q=83# Second prime

E=53# usually a constant; 0x10001 is common, prime is best

#####################################################################

# Next, some functions we'll need in a moment:

#####################################################################

# Note on what these operators do:

# % is the modulus (remainder) operator: 10 % 3 is 1

# // is integer (round-down) division: 10 // 3 is 3

# ** is exponent (2**3 is 2 to the 3rd power)

# Brute-force (i.e. try every possibility) primality test.

defisPrime(x):

ifx%20andx>2: returnFalse# False for all even numbers

i=3# we don't divide by 1 or 2

sqrt=x**.5

whilei<sqrt:

ifx%i0: returnFalse

i+=2

returnTrue

# Part of find_inverse below

# See: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

defeea(a,b):

ifb0:return (1,0)

(q,r) = (a//b,a%b)

(s,t) =eea(b,r)

return (t, s-(q*t) )

# Find the multiplicative inverse of x (mod y)

# see: http://en.wikipedia.org/wiki/Modular_multiplicative_inverse

deffind_inverse(x,y):

inv=eea(x,y)[0]

ifinv<1: inv+=y#we only want positive values

returninv

#####################################################################

# Make sure the numbers we picked above are valid.

#####################################################################

ifnotisPrime(P): raiseException('P (%i) is not prime'% (P,))

ifnotisPrime(Q): raiseException('Q (%i) is not prime'% (Q,))

T=(P-1)*(Q-1) # Euler's totient (intermediate result)

# Assuming E is prime, we just have to check against T

ifE<1orE>T: raiseException('E must be > 1 and < T')

ifT%E0: raiseException('E is not coprime with T')

#####################################################################

# Now that we've validated our random numbers, we derive our keys.

#####################################################################

# Product of P and Q is our modulus; the part determines as the 'key size'.

MOD=P*Q

# Private exponent is inverse of public exponent with respect to (mod T)

D=find_inverse(E,T)

# The modulus is always needed, while either E or D is the exponent, depending on

# which key we're using. D is much harder for an adversary to derive, so we call

# that one the 'private' key.

print'public key: (MOD: %i, E: %i)'% (MOD,E)

print'private key: (MOD: %i, D: %i)'% (MOD,D)

# Note that P, Q, and T can now be discarded, but they're usually

# kept around so that a more efficient encryption algorithm can be used.

# http://en.wikipedia.org/wiki/RSA#Using_the_Chinese_remainder_algorithm

#####################################################################

# We have our keys, let's do some encryption

#####################################################################

# Here I only focus on whether you're applying the private key or

# applying the public key, since either one will reverse the other.

importsys

print'Enter '>NUMBER' to apply private key and '<NUMBER' to apply public key; 'Q' to quit.'

whileTrue:

sys.stdout.write('? ')

line=sys.stdin.readline().strip()

ifnotline: break

ifline'q'orline'Q': break

ifline[0]'<': key=E

elifline[0]'>': key=D

else:

print'Must start with either < or >'

print'Enter '>NUMBER' to apply private key and '<NUMBER' to apply public key; 'Q' to quit.'

continue

line=line[1:]

try: before=int(line)

exceptValueError:

print'not a number: '%s''% (line)

print'Enter '>NUMBER' to apply private key and '<NUMBER' to apply public key; 'Q' to quit.'

continue

ifbefore>=MOD:

print'Only values up to %i can be encoded with this key (choose bigger primes next time)'% (MOD,)

continue

# Note that the pow() built-in does modulo exponentation. That's handy, since it saves us having to

# implement that ablity.

# http://en.wikipedia.org/wiki/Modular_exponentiation

after=pow(before,key,MOD) #encrypt/decrypt using this ONE command. Surprisingly simple.

ifkeyD: print'PRIVATE(%i) >> %i'%(before,after)

else: print'PUBLIC(%i) >> %i'%(before,after)

commented Mar 1, 2017

Perfect explanation! Thanks for your answer to «Is there a simple example of an Asymmetric encryption/decryption routine?» I was looking for this kind of routine to encrypt numbers inferiors to 1 billion with results inferiors to 1 billion. (I'm limited in string length). Is this routine safe for that task? Can we computed (by brute force?) the private key from and only the public key and the modulus?

Public Private Key Encryption

commented Feb 6, 2018

The condition to have an inverse in line 63 is wrong ! E and T must be coprime then their gcd must be 1 and not T%E!=0 as given for the raise condition if T%E0: raise Exception('E is not coprime with T') must be if gcd(E,T)!=1: raise Exception('E is not coprime with T')

Generate Private Key Rsa Euclidean Property

Sign up for freeto join this conversation on GitHub. Already have an account? Sign in to comment