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tylerl/rsa.py

Created Sep 24, 2011
RSA Explained in Python
 #!/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. def isPrime(x): if x%2==0 and x>2: return False # False for all even numbers i=3 # we don't divide by 1 or 2 sqrt=x**.5 while i T: raise Exception("E must be > 1 and < T") if T%E==0: raise Exception("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. import sys print "Enter \">NUMBER\" to apply private key and \"': key = D else: print "Must start with either < or >" print "Enter \">NUMBER\" to apply private key and \"NUMBER\" to apply private key and \"= 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. if key == D: print "PRIVATE(%i) >> %i" %(before,after) else: print "PUBLIC(%i) >> %i" %(before,after)

MichelPoulain 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?

jdavid54 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%E==0: raise Exception("E is not coprime with T") must be if gcd(E,T)!=1: raise Exception("E is not coprime with T")