23 Oct 2016

The broken one-time pad was solved. On reddit a user called svvw 1 claimed to be the one who did it.

At least I was able to recover a QR code, and transfer some bitcoins to myself… I think, I’m very much a noob wrt. to this cryptocurrency stuff. The basic idea is this:

• The cipher is essentially a Vigenerer cipher
• The key length is provided (1024*100)
• The plaintext space is very limited (only valid base64 chars) So what one can do, is split the ciphertext into 1024*100 blocks such that each character in block i was encrypted with key-byte i. This means we can simply look for each key-byte separately. When assessing whether or not some byte k could be a key-byte, one simply checks that only valid base64 characters is the result, when calculating “k xor c” for every c in some block. Putting it all together (obtaining the QR code and so on) was done with python + some command line tools. I’ll try and write a little more detailed description later, a long with the code I wrote to solve it.

A more detailed write up can be found on github. Thank you for sharing the idea to your solution svvw.

In this post I’ll first try to reconstruct svvw’s solution, then present my own. At the end we’ll fix the found issues in the broken one-time pad so that we can restart the challenge in [broken one-time pad 2].

Notation and visualization

The message stream $$m$$ equals the repeated base64 encoded PNG image. The pad stream $$k$$ reassembles a key stream. The bitwise exclusive OR $$\oplus$$ of both streams is the resulting ciphertext stream $$c = m \oplus p$$. The streams are all chunked and indexed in bytes, e.g. $$m_{3}$$ is the third byte in the message stream. The streams repeat themselves after $$\vert m \vert$$, respectively $$\vert k \vert$$ bytes. In general $$\vert m \vert \neq \vert k \vert$$ which means after every repetition of $$k$$, $$m$$ is shifted by $$s \equiv \vert k \vert \mod{\vert m \vert}$$ bytes. Important to notice is that the two streams always line up after $$\DeclareMathOperator{\lcm}{lcm} \vert c \vert = \lcm(\vert m \vert, \vert k \vert)$$ bytes.

Variants of the following graphic are used to visualize the two streams. The outer ring represents the key stream $$k$$. The inner the message stream $$m$$. The upper right corner magnifies a section of the two streams.

The two streams aligned

svvw’s solution

First things first, broken one-time pad should implement a one-time pad. 2 Due to the broken nature of the implementation it becomes similar to a Vigenerer cipher. Cryptanalysis techniques working on Vigenerer ciphers will mostly work on the broken one-time pad. 3

But now let’s get started reconstructing svvw’s solution. First we’ve to find out $$\vert c \vert$$. From the visualization we can derive that $$\DeclareMathOperator{\lcm}{lcm} \vert c \vert = \lcm(\vert m \vert, \vert k \vert)$$, which is exactly as many bytes as we’ll get from the server. 4 Thus, obtaining $$\vert c \vert$$ is as simple as

\$ nc duckpond.ch 8888 | wc -c


Which tells us $$\vert c \vert = 14028800$$ bytes (13.37890625 MiB).

So what one can do, is split the ciphertext into 1024*100 blocks …

$$\vert k \vert = 1024 \cdot 100$$, this number comes directly from the broken one-time pad.

Known $$\vert k \vert$$

… such that each character in block i was encrypted with key-byte i. This means we can simply look for each key-byte separately.

For $$i = 1$$ this would correspond to the red lines in the following visualization:

e.g. $$i = 1$$

When assessing whether or not some byte k could be a key-byte, one simply checks that only valid base64 characters …

With $$M$$ as the set of valid base64 characters,

$M = \{ \text{'a'}, \text{'b'}, \text{'c'}, \dots, \text{'A'}, \text{'B'}, \text{'C'}, \dots, \text{'+'}, \text{'/'}, \text{'='}\}$

every byte in $$m$$ has to be a character from $$M$$.

$\forall_{j \in \mathbb{N}} m_{j} \in M$

… is the result, when calculating “k xor c” for every c in some block.

Which means for a given $$c$$ and $$k$$ the following condition must hold:

$\forall_{i \in \{1, \dots, \vert k \vert \}} \forall_{j \in \{1, \dots, \vert c \vert \}}\{ x \mid x = k_{i} \oplus c_{i + j \vert k \vert} \} \subset M$

$$M$$ contains few elements, so this comes in handy.

With all this information it is straight forward to write a guessing algorithm in python3 for $$k_{i}$$.

with open('data', 'rb') as fd:

M = set('abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/=')
K_REPEAT = 1024 * 100
C_REPEAT = 14028800

def guess_k_i(i):
for candidate in range(256):
if all(
[
chr(candidate ^ C[i + j * K_REPEAT]) in M
for j in range(int(C_REPEAT / K_REPEAT))
]
):
yield candidate


We could now get $$k$$

K = [ [ k_i for k_i in guess_k_i(i) ] for i in range(K_REPEAT) ]


but this takes ages to complete. So we’ve to parallelize the algorithm.

from multiprocessing import Pool

def k_i(i):
return [ k for k in guess_k_i(i) ]

K = Pool().map(k_i, range(K_REPEAT))


It can happen that this finds multiple possible solutions for a certain $$k_{i}$$. Next we verify that this is not the case.

all([
len(k_i) == 1
for k_i in K
])


Finally, we decrypt the whole $$m$$.

M = ''.join([
chr(K[i % len(K)][0] ^ C[i])
for i in range(len(C))
])


And what we’ll get is our base64 encoded PNG, repeated several times.

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iVBORw0KGgoAAAANSUhEUgAAAG8AAABvAQMAAADYCwwjAAAABlBMVEUAAAD///+l2Z/dAAAAAnRSTlP//8i138cAAAAJcEhZcwAACxIAAAsSAdLdfvwAAAErSURBVDiN1dQxbsQgEAXQsSjo7AsgcQ06rrRcYL2+wPpKdFwDiQvgbgrkyXhjaZMiHoooUpBl+RUW/GEA6NuA/8EKEFJZyQBYkRu1OeUQ20wdjObu84Dtprs4J1rR9BLLElsXqd0d7fq9yAu+8hp+vsT/kTyqV1t6F/aC1XFe ...


In order to isolate the image we can either look for the base64 padding ==, or the string iVBORw0KGgo which stands for the base64 encoded PNG magic bytes.

Discussion

A brute-force attack on the keystream $$k$$ is possible because $$M$$ contains only $$65$$ out of $$2^8 = 256$$ possible elements. This means with every $$c_{i+j \vert k \vert}$$ tested, we can exclude $$1 - \frac{65}{256} = 0.74609375 \rightarrow ~ 74.6 \%$$ of the possible $$k_{i}$$. The huge amount of data $$\vert c \vert = 14028800$$ allows for a total of $$\frac{\vert c \vert}{\vert k \vert} = 137$$ tests, which reduces the chance for an ambiguity of a certain $$k_{i}$$ down to $$(\frac{65}{256})^{137} = 2.7558438890447563 \cdot 10^{-82}$$.

“Go towards the light”

Ente’s solution

My solution is based on the observation that $$\oplus : \{0,1\}^{n} \rightarrow \{0,1\}^{n}$$ is a linear and commutative function. So we can use the additivity property from linearity $$f(x) + f(y) = f(x + y)$$

$(k_{i} \oplus m_{i}) \oplus (k_{i} \oplus m_{i + s}) = k_{i} \oplus m_{j} \oplus k_{i} \oplus m_{j+ s}$

Commutativity tells us that we are allowed to change the order of the operands

$k_{i} \oplus m_{j} \oplus k_{i} \oplus m_{j+ s} = k_{i} \oplus k_{i} \oplus m_{i} \oplus m_{i + s}$

From the $$\oplus$$ truth table

pqp $$\oplus$$ q
000
011
101
110

we see that $$k_{i} \oplus k_{i}$$ always results in a byte with all bits set to zero $$(0,0,0,0,0,0,0,0)$$. Furthermore, we find that

$(0,0,0,0,0,0,0,0) \oplus m_{i} \oplus m_{i + s} = m_{i} \oplus m_{i + s}$

The keystream $$k$$ cancels out.

You might ask yourself now, what is this guy trying to tell me?

Theorem: If you XOR every byte of the cipher stream $$c$$ with a byte $$\vert k \vert$$ bytes further down the same stream, you’ll end up with a new stream $$m^{'} = m_{i} \oplus m_{i + s}$$.

Proof: From $$c_i = (k_{i} \oplus m_{i})$$ we get

$$c_{i} \oplus c_{i + \vert k \vert} = (k_{i} \oplus m_{i}) \oplus (k_{i + \vert k \vert} \oplus m_{i + \vert k \vert})$$

The two stream $$k$$ and $$m$$ are periodic, thus $$k_{i} = k_{i + \vert k \vert}$$ and $$m_{i + \vert k \vert} = m_{i + s}$$

$$(k_{i} \oplus m_{i}) \oplus (k_{i + \vert k \vert} \oplus m_{i + \vert k \vert}) = (k_{i} \oplus m_{i}) \oplus (k_{i} \oplus m_{i + s})$$

and with the observation from above

$$(k_{i} \oplus m_{i}) \oplus (k_{i} \oplus m_{i + s}) = m_{i} \oplus m_{i + s} = m^{‘} \qquad \Box$$

e.g. $$m^{'}_{1} = (k_{1} \oplus m_{1}) \oplus (k_{1} \oplus m_{1+s})$$

We lack only one last piece of information: $$\vert m \vert$$. Once we notice that $$\vert m \vert = \vert m^{'} \vert$$, we can write a python script which searches for repeating sub lists in $$m^{'}$$ and derive $$\vert m \vert$$ accordingly.

from math import floor

def chunks(l, n):
"""Yield successive n-sized chunks from l."""
for i in range(0, len(l), n):
ret = l[i:i+n]
if(len(ret) == n):
yield ret

def same(ll):
"""Checks if all the lists in ll are the same. same([[1], [1]]) == True ; same([[1], [1,2]]) == False"""
return ll.count(ll[0]) == len(ll)

def sublists(l):
"""Yields all repeating sublists of l: [1,2,3,1,2,3,1,2,3,1,2,3] -> [ [1,2,3], [1,2,3,1,2,3] ]"""
for i in range(1, floor(len(l) / 2) + 1):
if same(list(chunks(l, i))):
yield l[:i]

with open('data', 'rb') as fd:

K_REPEAT = 1024 * 100
C_REPEAT = 14028800
M_XOR_M = [c[0] ^ c[1] for c in zip(C[:K_REPEAT], C[K_REPEAT:2*K_REPEAT])]
M_REPEAT = next(len(i) for i in sublists(M_XOR_M))
SHIFT = K_REPEAT % M_REPEAT


There’s a PNG inside the stream, which starts with 0x89504e470d0a1a0a, this means we can break the cipher with a known-plaintext attack on $$m^{'}$$.

from base64 import b64encode

# PNG magic bytes
WORD = bytearray(b64encode(b"\x89PNG\r\n\x1a\n").rstrip(b"="))

# known plaintext attack
M = list(WORD) + [0] * (M_REPEAT - len(WORD))
o = (-SHIFT) % M_REPEAT
while o:
for i in range(o, o + len(WORD)):
M[i % M_REPEAT] = M_XOR_M[i % M_REPEAT] ^ M[(i + SHIFT) % M_REPEAT]
o = ( o - SHIFT ) % M_REPEAT

M = bytearray(M).decode("ascii")


Mitigation

The cipher is called ONE-time pad. Which means the pad should only be used once 5! Mitigation of all these problems is as simple as not reusing the key stream.

1. aka. potater on github

2. Captain obvious speaking

3. That makes me wonder if somebody already sampled relative character frequencies of base64 encoded PNG images.

4. It’s getting late..