IvB+AsSA+9xiVMDm/3YABBBM/R7wDcciTmKucZoEuDeWJg+cwLur8kKSVNwwkwCT Np2qVHo5CFvSTAgrZKBCmKD2c2AGD9O+TSOrT0RDzvJosW4P1QJ9qsCjKFSPW/8K
S5B/QBzmSeNYlYYyGnO9GxyuHyR7P5xqF9AcpSskn9gjYy0koTLmvg/bwF圆jnci OuGgVJLJGPTw/NXAnLq6XlwB3C+zQMWpxvcMub4D6/IltP/PUpKNj9QzKKGhZF+6 MIIEvgIBADANBgkqhkiG9w0BAQEFAASCBKgwggSkAgEAAoIBAQCTiP1x58DboFjSĤWgzMm8tuY9VoHwANbOHNuuiElsZ4xIVFS+ZG7lu7Iz9gzmsno/YiqQXms8hXmUH Our strategy will be to take a string and break it up into individualĬharacters and encrypt each character, just as we did with the CaesarĬipher.I tried to decrypt rsa using a private key that exists as a string, but it failed. How to adapt the RSA encryption and decryption algorithms to
With the Caesar cipher and One-Time Pad cryptosystem. Unsatisfying because it encrypts numbers instead of strings, like we saw The above implementation of RSA is correct, but is a little Preconditions: - private_key is a valid RSA private key (p, q, d) - 0 < ciphertext < private_key * private_key """ p, q, d = private_key, public_key, public_key n = p * q decrypted = (ciphertext ** d) % n return decrypted Encrypting and This mathematicalĭefinition translates directly into Python:ĭef rsa_decrypt(private_key: tuple ciphertext: int) -> int: """Decrypt the given ciphertext using the recipient's private key. Recall that the plaintext here is a number \(m\) between \(1\) and \(n -ġ\) inclusive, and the ciphertext is another number \(c = m^e ~\%~ n\). Next, let’s look at RSA encryption, which only uses the public key. (which, in turn, used the extended_euclidean_gcd
Modular_inverse function we defined in the last chapter Once we haveĮ, we can finally calculate the last part of our private Phi_n have a greatest common divisor of 1. Generate an e, but the while loop ensures that theĮ we finally choose is valid. The algorithm makes use of both a while loop and the # Notice that we're using our modular_inverse from our work in the last chapter! d = modular_inverse(e, phi_n) return ((p, q, d), (n, e)) e = random.randint( 2, phi_n - 1) while math.gcd(e, phi_n) != 1: e = random.randint( 2, phi_n - 1) # Choose d such that e * d % phi_n = 1. phi_n = (p - 1) * (q - 1) # Since e is chosen randomly, we repeat the random choice # until e is coprime to phi_n. Preconditions: - p and q are prime - p != q """ # Compute the product of p and q n = p * q # Choose e such that gcd(e, phi_n) = 1. The second tuple is the public key, containing (n, e). The first tuple is the private key, containing (p, q, d). The return value is a tuple containing two tuples: 1. \newcommandĭef rsa_generate_key(p: int, q: int) -> \ tuple, tuple]: """Return an RSA key pair generated using primes p and q.
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