Rsa 2049 Key Generation 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. |
| 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) |
Rsa 2048 Bits
Ssh-keygen -t rsa -b 2048 -f key The -t option specifies the key generation algorithm (RSA in this case), while the -b option specifies the length of the key in bits. The -f option sets the name of the output file. If not present, ssh-keygen will ask the name of the file, offering to save it to the default file /.ssh/idrsa. That generates a 2048-bit RSA key pair, encrypts them with a password you provide and writes them to a file. You need to next extract the public key file. You will use this, for instance, on your web server to encrypt content so that it can only be read with the private key. Users must generate a public/private key pair when their site implements host-based authentication or user public-key authentication. For additional options, see the ssh-keygen(1) man page. Before You Begin. Determine from your system administrator if host-based authentication is configured. Start the key generation program. There isn't too much to see here because the key generation simply relies on RSA.generate(2048), but I wonder why you would need this code as it is exceedingly shallow. Regenerating key pairs for signing at startup is utter nonsense because a key pair is next to useless if the public key.
Rsa 2048 Bit Encryption
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. Adobe pagemaker 7.0 crack download. (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? |
2048 Bit Rsa Key
commented Feb 6, 2018
The condition to have an inverse in line 63 is wrong !
|
