Digital Media Concepts/RSA (cryptosystem)
What is RSA[edit | edit source]
It is one of the first asymmetric cryptography algorithms which uses a pair of keys to encrypt and decrypt data. This algorithm is now widely used to transmit sensitive data over insecure network like the Internet.
History[edit | edit source]
Before mid-1970s, people used symmetric cryptography algorithms to encrypt data. With these algorithms, data encrypted with a particular cryptographic key could only be decrypted with the same key. Anyone without the key cannot decrypt the data. Therefore people could safely send sensitive information through insecure communication channel. However, people could not find a way to safely exchange their keys between them. Asymmetric cryptography algorithms were invented to solve this problem, and RSA is one of them. In April 1977, RSA was invented by three people from Massachusetts Institute of Technology, including two computer scientists, Ron Rivest, Adi Shamir, and a mathematician Leonard Adleman, and was later publicized in August of the same year.
RSA is now in public domain, and can be freely implemented and used by anyone.
Steps[edit | edit source]
Generate the keys[edit | edit source]
As an asymmetric cryptography algorithm, RSA cryptosystem involves two keys, the public key and the private key. Data encrypted by one key can only be decrypted with the other.
- Randomly choose two prime numbers, and .
- Multiply them together:
- Get the least common multiple of and :
- Randomly choose a positive integer which is less than and coprime to .
- Calculate modular multiplicative inverse of modulo . Which means, find a positive integer which is less than that satisfies:
- Now the key pair is generated. is the private key, and is the public key
Sample Code[edit | edit source]
Here's a sample implementation of generating RSA keys written in Python 3.6
import secrets def gcd(x: int, y: int) -> tuple: ''' Extended Euclidean Algorithm return: a, b, gcd ax + by = gcd ''' def f(x: int, y: int) -> tuple: l =  while y != 0: q, r = divmod(x, y) l.append(q) x, y = y, r x, y, g = (1, 0, x) if x >= 0 else (-1, 0, -x) while l: x, y = y, x - l.pop() * y return x, y, g # below is a recursive approach. easier to understand # but it may exceeds python recursion limit with large numbers # if y == 0: # return (1, 0, x) if x >= 0 else (-1, 0, -x) # quot, rem = divmod(x, y) # rem + quot * y = x . . . . . (1) # _a, _b, _gcd = f(y, rem) # _a * y + _b * rem = _gcd . . (2) # return _b, _a - quot * _b, _gcd # <- plug rem from (1) into (2) assert isinstance(x, int) and isinstance(y, int), 'Integers expected' return f(x, y) def miller_rabin(n: int, k: int=100): ''' Miller-Rabin primality test https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test : Input #1: n > 3, an odd integer to be tested for primality Input #2: k, the number of rounds of testing to perform Output: “composite” if n is found to be composite, “probably prime” otherwise write n as 2^r·d + 1 with d odd (by factoring out powers of 2 from n − 1) WitnessLoop: repeat k times: pick a random integer a in the range [2, n − 2] x ← a^d mod n if x = 1 or x = n − 1 then continue WitnessLoop repeat r − 1 times: x ← x^2 mod n if x = n − 1 then continue WitnessLoop return “composite” return “probably prime” ''' assert isinstance(n, int) and isinstance(k, int) and k > 0 if n < 4: return n > 1 if n & 1 == 0: # even numbers return False r = 0 n_1 = n - 1 d = n_1 while d & 1 == 0: d >>= 1 r += 1 for i in range(k): a = secrets.randbelow(n - 3) + 2 x = pow(a, d, n) if x == 1 or x == n_1: continue for j in range(r): x = pow(x, 2, n) if x == n_1: break else: return False return True def get_prime(bits: int) -> int: ''' Get a pseudo-prime with given bits This function randomly generates a number with given bits, and test its primality using the miller-rabin test. ''' assert isinstance(bits, int) and bits > 1 low = 1 << bits - 1 while True: r = secrets.randbits(bits - 1) + low if miller_rabin(r): return r def generate_rsa_keys(bits: int): ''' Generate RSA key pair It returns e, d, n, which makes (m^e mod n)^d mod n == m ''' def lcm(x: int, y: int) -> int: a, b, g = gcd(x, y) return x * y // g assert isinstance(bits, int) and bits > 5 p = q = 0 while p == q: p = get_prime(bits // 2) q = get_prime(bits // 2) n = p * q l = lcm(p - 1, q - 1) _g = 0 while _g != 1: e = secrets.randbelow(l - 3) + 3 # [3, l) d, _not_used, _g = gcd(e, l) if d < 0: # Got negative d d += l return e, d, n
Send the public key to others[edit | edit source]
Let's say Alice and Bob want to have a secure communication. They may exchange their public keys without encryption. After that, a sender should always encrypt the data with the receiver's public key before sending it.
For example, Bob's data is encrypted with Bob's public key, and only people who know Alice's private key can decrypt the data. So Alice is the only person that meets this requirement. The same works with Bob.
Encryption and Decryption[edit | edit source]
- Let's say the original data is an positive integer which should be less than . Encrypted data, positive integer , can be generated with a public key:
- The encrypted data can be decrypted with the corresponding private key:
Proof of correctness[edit | edit source]
Because the encrypted data is calculated as:
When decrypting the encrypted data , the result is:
If equals to the original unencrypted data , then this algorithm is correct.
It says when and are coprime (), if both of these statements are true:
Then this is also true:
As and are both prime numbers, they are obviously coprime. So if both of these statements could be proved to be true:
Then this is also true:
Then the correctness of RSA algorithm could be proved:
To accomplish this, Fermat's little theorem is necessary.
It says, if is a prime, and is not a multiple of (), then this statement is true:
As the relationship between and is unknown, the problem need to be divided into two cases.
As is a multiple of both and , both of these statements are true:
So can be expressed as:
Safety[edit | edit source]
The public key, which is and , can be known to everyone. As far as is kept private, no one except the private key owner is able to decrypt the data encrypted by the public key. However, number has two prime factors and . If one can find and by factoring , then this person can also find out , and eventually . So the problem of how safe RSA is, is equivalent to how hard it is to factor .
It is proved that, currently with traditional (non-quantum) computer, factoring a big number is an NP problem. It may or may not be an NP-complete problem. RSA remains safe when integer factorization can't be solved in polynomial time.
External Links[edit | edit source]
The original paper published by its inventors could be found here.
See also[edit | edit source]
- Cisco IOS/Configure public RSA key authentication
- OpenSSH/Configure OpenSSH to allow Public-key authentication:
- Digital Signature
- GNU Privacy Guard - Open Source Public-Private-Key Implementation
References[edit | edit source]
- Sheposh, Richard. 2017. “RSA (Cryptosystem).” Salem Press Encyclopedia of Science.
- "RSA Security Releases RSA Encryption Algorithm into Public Domain - RSA, The Security Division of EMC". web.archive.org. 2007-11-20. Retrieved 2019-09-30.
- "What is RSA algorithm (Rivest-Shamir-Adleman)? - Definition from WhatIs.com". SearchSecurity. Retrieved 2019-09-30.
- Rivest, R. L.; Shamir, A.; Adleman, L. (1978-2). "A Method for Obtaining Digital Signatures and Public-key Cryptosystems". Commun. ACM 21 (2): 120–126. doi:10.1145/359340.359342. ISSN 0001-0782. http://doi.acm.org/10.1145/359340.359342.
- "cryptography - How safe and secure is RSA?". Stack Overflow. Retrieved 2019-10-01.