Living with Elegance

6 minutes to read

We are given the following Python script that encrypts the flag:

from secrets import token_bytes, randbelow
from Crypto.Util.number import bytes_to_long as b2l

class ElegantCryptosystem:
    def __init__(self):
        self.d = 16
        self.n = 256
        self.S = token_bytes(self.d)

    def noise_prod(self):
        return randbelow(2*self.n//3) - self.n//2

    def get_encryption(self, bit):
        A = token_bytes(self.d)
        b = self.punc_prod(A, self.S) % self.n
        e = self.noise_prod()
        if bit == 1:
            return A, b + e
        else:
            return A, randbelow(self.n)
    
    def punc_prod(self, x, y):
        return sum(_x * _y for _x, _y in zip(x, y))

def main():
    FLAGBIN = bin(b2l(open('flag.txt', 'rb').read()))[2:]
    crypto = ElegantCryptosystem()

    while True:
        idx = input('Specify the index of the bit you want to get an encryption for : ')  
        if not idx.isnumeric():
            print('The index must be an integer.')
            continue
        idx = int(idx)
        if idx < 0 or idx >= len(FLAGBIN):
            print(f'The index must lie in the interval [0, {len(FLAGBIN)-1}]')
            continue
        
        bit = int(FLAGBIN[idx])
        A, b = crypto.get_encryption(bit)
        print('Here is your ciphertext: ')
        print(f'A = {b2l(A)}')
        print(f'b = {b}')


if __name__ == '__main__':
    main()

Source code analysis

The server allows us to get the encryption result for each bit we indicate from the flag:

    while True:
        idx = input('Specify the index of the bit you want to get an encryption for : ')  
        if not idx.isnumeric():
            print('The index must be an integer.')
            continue
        idx = int(idx)
        if idx < 0 or idx >= len(FLAGBIN):
            print(f'The index must lie in the interval [0, {len(FLAGBIN)-1}]')
            continue
        
        bit = int(FLAGBIN[idx])
        A, b = crypto.get_encryption(bit)
        print('Here is your ciphertext: ')
        print(f'A = {b2l(A)}')
        print(f'b = {b}')

The ciphertext is composed by two values: A and b. Let’s see how they are computed:

class ElegantCryptosystem:
    def __init__(self):
        self.d = 16
        self.n = 256
        self.S = token_bytes(self.d)

    def noise_prod(self):
        return randbelow(2*self.n//3) - self.n//2

    def get_encryption(self, bit):
        A = token_bytes(self.d)
        b = self.punc_prod(A, self.S) % self.n
        e = self.noise_prod()
        if bit == 1:
            return A, b + e
        else:
            return A, randbelow(self.n)
    
    def punc_prod(self, x, y):
        return sum(_x * _y for _x, _y in zip(x, y))

When the class ElegantCryptosystem is initialized, a value S is computed, as a 16-byte number
Then, on each encryption, the server computes a random value A
Then multiplies A and S using a scalar product (punc_prod) to get b
e is some random noise
If the bit is 1, the result is A and b + e; otherwise it is A and a random number below n = 256

Learning With Errors

In math terms, we can express A and S as vectors of 8-bit integers:

$$ A = (a_0, a_1, \dots, a_{15}) \qquad S = (s_0, s_1, \dots, s_{15}) $$

Then, b is just their scalar product:

$$ b = A \circ S = \sum_{i = 0}^{15} a_i \cdot s_i $$

And if the bit to encrypt is a 1, then we receive the following ciphertext:

$$ \begin{cases} A = (a_0, a_1, \dots, a_{15}) \\ c = A \circ S + e = b + e \end{cases} $$

This cryptosystem is known as Learning With Errors (LWE), and the name of the challenge is kind of a hint. The idea behind this cryptosystem is basically that it hard to recover the secret key $S$ from $A$ and $c$ because $c$ is just $b$ corrupted with some random noise $e$.

Solution

This time, we need to find a way to tell if we are been given b + e or randbelow(self.n). We can do this by looking at the boundaries that these values can take:

We know that b must be an integer value in $[0, 256)$ because it is reduced modulo n = 256
The value of e comes from noise_prod, which is computed as randbelow(2*self.n//3) - self.n//2. So, we have the following boundary for e:

$$ [0, 2 \cdot 256 / 3) - 256 / 2 = [0, 170) - 128 = [-128, 42) $$

Therefore, the boundary for b + e is:

$$ [0, 256) + [-128, 42) = [-128, 298) $$

On the other hand, randbelow(self.n) is an integer value in $[0, 256)$

As a result, we have a way to determine if the bit is 1: That is, if we receive a value $c$ such that $c < 0$ or $c \geqslant 256$, then we can be sure that it came from b + e, so the encrypted bit is a 1.

Since we have unlimited queries to the server, we can perform a probabilistic approach. For instance, we query 30 times for the same bit:

If any of these queries return $c < 0$ or $c \geqslant 256$, then we know for sure that it’s a 1 and we stop the process to continue
If none of the queries give a result outside of $[0, 256)$, then we can assume with high probability that the encrypted bit is a 0

We can repeat this process for all the bits until we have the flag.

Implementation

This time, I am using Go with my pwntools module. This is a helper function to query the server on a given bit index (it only returns the ciphertext value, because the A is not relevant):

func getEncryption(index int) int {
        io.SendLineAfter([]byte("Specify the index of the bit you want to get an encryption for : "), []byte(strconv.Itoa(index)))  
        io.RecvUntil([]byte("b = "))
        c, _ := strconv.Atoi(strings.TrimSpace(io.RecvLineS()))
        return c
}

We can find the bit length of the flag (the server will send an error if the queried index is not valid and will show the maximum index) and start finding bits with the above procedure:

func main() {
        io = getProcess()
        defer io.Close()

        io.SendLineAfter([]byte("Specify the index of the bit you want to get an encryption for : "), []byte("10000"))  
        io.RecvUntil([]byte("The index must lie in the interval [0, "))
        bitLength, _ := strconv.Atoi(io.RecvUntilS([]byte{']'}, true))
        bitLength++

        bits := make([]int, bitLength)
        prog := pwn.Progress("Bits")

        for i := 0; i < bitLength; i++ {
                prog.Status(fmt.Sprintf("%d / %d", i+1, bitLength))

                for range 30 {
                        c := getEncryption(i)

                        if c < 0 || 256 < c {
                                bits[i] = 1
                                break
                        }
                }
        }

        prog.Success(fmt.Sprintf("%[1]d / %[1]d", bitLength))

Next, we need to apply padding to fill 8-bit blocks (for later decoding):

        for len(bits)%8 != 0 {
                bits = append([]int{0}, bits...)  
        }

Finally, we decode each 8-bit block to a byte and show the flag as a string:

        flag := make([]byte, len(bits)/8)

        for i := 0; i < bitLength; i += 8 {
                for j, v := range bits[i : i+8] {
                        flag[i/8] |= byte(v << (7 - j))  
                }
        }

        pwn.Success(string(flag))
}

Flag

With this script, we get the flag:

$ echo 'HTB{f4k3_fl4g_f0r_t3st1ng}' > flag.txt

$ go run solve.go
[+] Starting local process 'python3': pid 91076
[+] Bits: 215 / 215
[+] HTB{f4k3_fl4g_f0r_t3st1ng}
[*] Stopped process 'python3' (pid 91076)

$ go run solve.go 94.237.52.200:59555
[+] Opening connection to 94.237.52.200 on port 59555: Done  
[+] Bits: 175 / 175
[+] HTB{s3cur3_cust0m_LW3}
[*] Closed connection to 94.237.52.200 port 59555

The full script can be found in here: solve.go.