Originally, I wanted to write up the write-up for the airport challenge, but since Niklas has already done that, I’m doing the Wood Island challenge instead. It was worth 150 points on the Boston Key Party 2015.

The goal was to break ElGamal Signatures. In short, the solution is based on exploiting “random” values that occur multiple times.

We are given access to a network server, and a tar archive with the server source code and a couple of signatures. From the server source code, we can see that the server will happily dump our flag if we manage to forge a signature for the message “There is no need to be upset”:

When viewing the signature file, we notice that we already have a signature for that message, but the server will not hand over the flag when we give it the exact same signature. So we need to somehow create a new signature for the message. This is a good time to have a look at the ElGamal signature scheme if you did not know it before:

## The ElGamal Signature Scheme

The ElGamal scheme works in the multiplicative group of integers modulo a prime number P with generator g. (In the task at hand, P even is a safe prime, i.e. a number of the form P = 2p + 1 where p is also prime.) The signing key is a secret, random exponent x. The corresponding verification key is y = gx.

To sign a message, firstly choose a random integer k between 1 and P - 1 (excluding both) with gcd(k, P - 1) = 1, and set r = gk. Then compute s = (H(m) - xr)k-1 (mod P - 1), where H is a cryptographic hash function and m is the message being signed. The final signature consists of r and s.

To verify such a signature, check that gH(m) = rs yr (mod P). This equation can be rewritten as gH(m) = (gk)s (gx)r (mod P) by filling in y and r. If we look purely at the exponent, this equation implicitly checks whether H(m) = ks + xr (mod P - 1).

The ElGamal scheme is similar to the DSA and the Schnorr signature scheme, and (like these two) breaks when the same random value is used for different signatures.

## The Attack

If we have two signatures (r, s1), (r, s2) with the same r value, which are valid for two message m1, m2, respectively, we know that the k values used for signing must also be equal. We can then subtract the s values to eliminate the unknown secret key x from the above equation:

s1 - s2 = (H(m1) - xr)k-1 - (H(m2) - xr)k-1 = (H(m1) - H(m2))k-1 (mod P - 1)

We can thus compute k-1 = (s1 - s2)/(H(m1) - H(m2)) (mod P - 1). (Doing this modular division is actually a little tricky, see below.) Once we have the correct k-1, we can create a forgery for any message m we like, by letting

s’ = s1 + (H(m) - H(m1))k-1 (mod P - 1)

This s’ together with r will be a valid signature for m, because

ks’ = k(s1 + (H(m) - H(m1))k-1) = ks1 + (H(m) - H(m1)) = H(m1) - xr + H(m) - H(m1) = H(m) - xr (mod P - 1)

and thus ks’ + xr = H(m) (mod P - 1), which is implicitly checked during verification.

My python code for this attack is:

Here, H was instantiated with the SHA-384 hash function.

When seeing the code above, you notice that I have multiple candidates for k-1. This is because the aforementioned difficulty dividing s1 - s2 by H(m1) - H(m2). Namely, both H(m1) - H(m2) and s1 - s2 are even and thus have a common factor with P - 1 = 2p. Thus, H(m1) - H(m2) is not invertible modulo 2p. Nonetheless, the division has two solutions, which are returned by the moddiv function.

I then filter for the correct one by checking that r(k-1) = (gk)k-1 = gkk-1 = g1 = g.

## Dividing modulo 2p

Now it is time to dive into the details of modular division. Our main tool is the Chinese Remainder Theorem. It allows us to do calculations modulo 2 and p distinctly, instead of doing calculations modulo 2p directly.

More precisely, if we have an integer h between 0 and 2p-1 (including both), we can represent h as the pair (h mod 2, h mod p). This representation is equivalent to the number h, i.e. we can recover h given just h mod 2 and h mod p. Moreover, this even is an isomorphism: If we are to multiply two numbers h, i modulo 2p, we can represent both as (h mod 2, h mod p) and (i mod 2, i mod p), and multiply componentwise. The result is (hi mod 2, hi mod p), which can be transformed back to hi mod 2p.

We can now use this knowledge to our advantage when doing modular division. Suppose we want to compute the modular division of s = s1-s2 by h = H(m1) - H(m2). Then, if s is even (as is the case in our scenario), then s mod 2 = 0, so s is equivalent to (0, s mod p). Likewise, h is even, so h can be represented as (0, h mod p).

As said before, multiplication is done componentwise. Suppose we know the modular inverse e of h modulo p. (This can be computed easily using the extended euclidean algorithm.) Then there are two possible numbers r1, r2 that map h to (0, s mod p) upon multiplication: (0, se mod p) and (1, se mod p). (Of course, (2, se mod p) would also work. But since we are computing modulo 2 in the first component, this is equivalent to (0, se mod p).)

These are the possible solutions to the division of s by h. We now only need to transform them back. There is a general method for this, but I decided to take a shortcut.

Basically, we are looking for two numbers r1, r2 that satisfy (r1 mod 2, r1 mod p) = (0, se mod p) and (r2 mod 2, r2 mod p) = (1, se mod p). In words, these numbers must be congruent to se when reduced modulo p, and one of them must be even, while the other must be uneven. It is easy to see that se mod p is one of them and (se mod p) + p is the other.

My code for this division is given below. It also implements a few other cases that I needed while toying around.

The code uses a few helper functions (like modinv, isUnit and is_probable_prime) I partially copied from websites, and partially wrote myself. You can find their implementation in the full source code in our github repository. (Note that my code is written for Python 3 and apparently not compatible with Python 2.)

## Putting the Pieces together

Above, I have covered the main concepts behind the solution. But the full solution must also accomplish some other stuff that is less exciting: For example, actually finding the signatures with repeated r values. This is done by the following code:

The findDoubles function mentioned here simply sorts the list of signatures by their r values and then does a linear scan over the result. This search returns 3 pairs of signatures. However, only one of these pairs is actually built from valid signatures. The other pairs contain at least one invalid signature. We therefore need to filter for the signature pairs:

For each of the remaining pairs (which is just one in our case), we can then attempt to forge a signature.

The submitSolution function takes care of the network communication with the server and solves the “captcha” that the server sends. It returns the answer received from the server, which is the flag

(Whatever that means.)