# Aarhus CTF challenge: Indistiguishable

I recently attended a capture the flag event at Aarhus University and decided to write a couple of paragraphs on a particular challenge I found interesting.

In this flag we were given the task to determine if a bunch of text has been encrypted with RC4 or AES. This is a variation of the classic cryptographic problem: Is a given text an encrypted message or random data? We therefore have to perform a distinguishing attack on the text at hand.

## The challenge

We are given access to a server and the source code for said server. The server produces 32 lines of text and asks if it is AES or RC4. Each line is encrypted with a random key and for each line the plaintext is 16 bytes long containing only zeroes. You then type aes or rc4 to the server and it responds with Correct! or Wrong. You can ask the server for 32 more lines by typing more. Guessing randomly will not help us since the probability of success for guessing 42 successive rounds correctly is: $$\frac{1}{2^{42}} \approx \frac{1}{4.4 \cdot 10^{12}} \approx 0.00000000000023$$. Given our time constraints and guessing over the network: zero.

## The cryptography

AES is a very good cipher that produces ciphertexts that look VERY much like random, uniform data. The best attacks against AES1 require a lot of ciphertext and have time complexities in the order of ~$2^{40}$. We do not have a lot of ciphertext (only 16 byte per key) and not a enough time for brute force.

RC4 on the other hand has some flaws. It works by taking your key as input and creating a keystream. The keystream is XORed with the plaintext and you have your ciphertext. Decrypting is simple, you just encrypt the ciphertext and due to XOR magic you have your original plaintext!

There are a couple of known attacks that work against RC4, but most of them require lots of ciphertext. Since the plaintext is only zeroes the cipher actually leaks it’s internal state, but since the ciphertexts are only 16 bytes there’s not enough data to derive anything of value. In 2001 Shamir showed that the second output byte of the cipher is biased towards zero with a probability of 1128 instead of 1256.2 The second byte of the output has double the probability of being zero than any other byte. If we observe 256 lines there will, on average, be 1 occurrence of 0 when it’s AES and 2 occurrences when it’s RC4. Now we’re talking! How much data do we actually need to observe before we can be certain? Let’s run some tests and find out.

## Distinguishing RC4 with Python

Let’s start by writing the code that does the distinguishing of the bytes. First we generate some testdata. This can be done by modifying indistinguishable.py to dump the lines to a file. The code needs look at the distribution of the bytes on every position in the ciphertext. The ciphertext is encoded as HEX.

The first version iterates all the lines and calculates the number of zeroes for every byte. It then prints the number of zeroes at the given byte position.

def distinguish_rc4(lines):
max = 0
sums = []
for start in range(0, 31, 2):
zeroes = 0
for line in lines:
# Get the second byte
char = line.strip()[start:start+2]
if char == "00":
zeroes += 1

sums.append(zeroes)
if(zeroes > max):
max = zeroes

# Print the average number of zeroes per position.
for sum in sums:
print(sum / len(lines))

# If the max was at index 1, we assume it is RC4
return sums.index(max) == 1

Let’s look at some output. For 1 round of 32 lines we get basically all zeroes. For 320 lines around half of the lines are zero. Bumping this to 3200 we get (we’re showing the first 8 bytes):

ByteAESRC4
10.00500000.0028125
20.00531250.0103125
30.00531250.0031250
40.00375000.0043750
50.00187500.0046875
60.00562500.0037500
70.00468750.0056250
80.00312500.0062500

In the AES column the numbers differ at the third decimal whereas in RC4 the occurrences of zero in the second byte is noticably larger. You can see that while the values are not 1/128 and 1/256, they are close to 2x. The more data we analyze the closer the frequencies will resemble 1/128 and 1/256 respectively. For demonstration purposes let’s try with a million lines:

ByteAESRC4
10.00383200.0038190
20.00404200.0077910
30.00393000.0037490
40.00392200.0039170
50.00386500.0039120
60.00393200.0039700
70.00393700.0037490
80.00386500.0039240

As you can see the numbers are much closer to the ideals of 1/128 = 0.0078125 and 1/256 = 0.00390625.

## Getting the flag

Now that we have an algorithm to distinguish RC4 from randomness, we can now attack the server!

The server starts by sending a header that looks like this:

Welcome to the distinguishing game!


Every round starts with the following text (n is round number, starting from 0):

===== Round n =====
Is this AES or RC4?


Following directly after you see 32 lines of hex-encoded ciphertext. After the ciphertext there is a prompt. A whole round looks like this:

===== Round n =====
Is this AES or RC4?
[ 32 lines go here]
>


This means we can simply use a blocking socket to get the data we need. First we need to read one line to get past the initial header. For every round we read two lines which we ignore. We store the next 32 lines and then read another two bytes that is the prompt. We send more to the server until we have enough data and finally we send our guess. Let’s have a look at the code:

#!/usr/bin/python3
import sys
import socket
from time import sleep

HOST = sys.argv
PORT = int(sys.argv)
MORE = int(sys.argv)

print("Attacking RC4 on " + sys.argv + " at port "+sys.argv)

def distinguish_rc4(lines):
max = 0
sums = []
for start in range(0, 31, 2):
zeroes = 0
for line in lines:
# Get the second byte
char = line.strip()[start:start+2]
if char == "00":
zeroes += 1

sums.append(zeroes)
if(zeroes > max):
max = zeroes

print()
print("Average number of zeroes:")
for sum in sums:
print(sum / len(lines))

# If the max was at index 1, we assume it is RC4
rc4 = sums.index(max) == 1
if rc4:
return "rc4"
else:
return "aes"

lines = []
for i in range(n):
lines.append(line.strip())

return lines

with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as s:
s.connect((HOST, PORT))
input = s.makefile("r")

for round in range(42):
print()
print("Attacking round "+str(round+1))

lines = []
for more in range(MORE):
if (more + 1) % int(MORE/10) == 0:
print(" getting data {}/{}".format(more+1, MORE))

# Read the 32 lines of ciphertext
sleep(0.01)

if more < MORE-1:
s.sendall(b'more\n')
sleep(0.01)

guess = distinguish_rc4(lines)

print()
print(" Guess: {}".format(guess))
s.sendall(bytes("{}\n".format(guess), "utf-8"))

sleep(0.01)

print("Result: {}".format(result))

if result == "Wrong":
# If the result was wrong, exit.
exit(1)

if round == 41:
# If this was the last round, print out the result.
print(next)

Let’s take it from the top. This script takes a couple of commandline parameters: the hostname, port and number of more calls to send. Assuming your server is localhost on port 8080 and you want to call more 200 times, you run the distinguish.py script:

\$ ./distinguish.py localhost 8080 200


Our function distinguish_rc4 has changed a little and now returns either rc4 or aes for convenience. The next function is called read_n which simply reads n lines from a given file. Next up we have the actual network code.

First we create a socket and connect to it. Then we use the socket.makefile function to be able to read lines from the socket instead of raw bytes. We then iterate through a range of our rounds (0 to 41). For every round we read the lines and ask for more data the number of times you specified. There are a couple of calls to sleep to avoid buffering issues. After we have enough lines we run our previously explained algorithm on said lines and send the result to the server. If the server responds with Wrong we exit the program.

If all goes well and the 41st round was correct, we print the last line containing the flag.

## Conclusion

This challenge demonstrates just how even the smallest leak of information from a cipher can lead to an attacker gaining an advantage. RC4 is a little dated but was used in something as common as the WiFi security protocol WEP and ultimately led to WEP being a security swiss cheese. Crypto is very hard to get right and the takeaway here is: Use AES.

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