Here is my take in 12 lines of Python code:
1 p = 7 2 g = [x for x in range(1, p) if len(set([x**i % p for i in range(1, p)])) == p-1] 3 k = 4 4 t = g**k % p 5 m = 6 6 import random 7 r = random.randint(1, p-1) 8 c1 = g**r % p 9 c2 = (t**r * m) % p 10 inv_k = p - 1 - k 11 inv_c1 = c1**inv_k % p 12 message = c2 * inv_c1 % p 13 print "YOU ARE A CRYPTOSTAR!" if message == m else "YOU SUCK" YOU ARE A CRYPTOSTAR!
Besides line #2 that involves an algorithm to find the generator `g` and line #10 that calculates the additive inverse of private key `k`, everything else is just modular multiplication and exponentiation.
Interested in more details and line by line explanations? please read on.
Discrete logarithm trapdoor
First thing first, lets pick a prime number:
1 p = 7
For the prime number above we need to find the generator `g` that defines a cyclic group which in theory is a little bit involved but in simple terms the generator is a specific number that returns unique values for each exponent between `1` and `p-1` where `p-1` is called the order of the cyclic group.
Lets take a few examples and see how it looks like: pick `2` as generator:
print [2**i % p for i in range(1, p)] [2, 4, 1, 2, 4, 1]
We can easily see that returning values are not unique, lets pick `3` as generator:
print [3**i % p for i in range(1, p)] [3, 2, 6, 4, 5, 1]
BINGO! returning values are unique and we’ve found our generator.
The line below is just a brute force algorithm that finds multiple generators of order `p-1` but only takes the first one that is found.
2 g = [x for x in range(1, p) if len(set([x**i % p for i in range(1, p)])) == p-1]
Now is time to pick a private key `k` between `1` and `p-1`, execute the trapdoor function and calculate `t` value:
3 k = 4 4 t = g**k % p
The line #4 is the secret sauce of ElGamal cryptography, the trapdoor function called the discrete logarithm which says that for given `g` and `t` values it is infeasible to calculate the private key `k` when working with very large numbers.
The `p, g and t` form the public key and will be used in encryption below, while `k` is kept secret.
`m` is the message that we want to encrypt and `r` is a random number between `1` and `p-1`:
5 m = 6 6 import random 7 r = random.randint(1, p-1) 8 c1 = g**r % p 9 c2 = (t**r * m) % p
`c1, c2` are the encrypted ciphers that will be used to decrypt the original message.
To decrypt the original message we need apply the following formula:
`message = c2 / c1k mod p`
The problem is that division is not an operation defined for finite fields but multiplication is and we can re-write it using multiplication / exponentiation operations only:
`message = c2 * c1invk mod p`
`invk` is called additive inverse of `k` over finite field ‘p’ where `k + invk mod p = 0` is valid. We can brute force and find it but there is a simpler way: if `k < p` (which it is) then `invk = p - 1 - k`.
10 inv_k = p - 1 - k 11 inv_c1 = c1**inv_k % p 12 message = c2 * inv_c1 % p 13 print "YOU ARE A CRYPTOSTAR!" if message == m else "YOU SUCK" YOU ARE A CRYPTOSTAR!
Why this magic works the way it works?
1 message == (c2 * inv_c1) % p 2 message == (c2 * c1**(p-1-k)) % p 3 message == (t**r * m * (g**r)**(p-1-k)) % p 4 message == ((g**k)**r * m * (g**r)**(p-1-k)) % p 5 message == (g**(k*r) * m * (g**(r*(p-1-k)))) % p 6 message == (g**(k*r) * m * g**(r*(p-1)) * g**(-r*k)) % p
Lets do the math backwards:
- start with last decryption formula and substitute inverses `invc1` and `invk`
- substitute ciphers `c1` and `c2`
- substitute trapdoor `t`
- apply exponentiation power rule `(ab)c = a(b*c)`
- distribute the exponent `r`
Check if the intuition is valid by substituting with numbers: generator `g` is 3, private key `k` is 4, random number `r` is 5 and message `m` is 6:
print "MAGIC" if 6 == (3**(4*5) * 6 * 3**(5*(7-1)) * 3**(-5*4)) % 7 else "ERROR" MAGIC