ACTF2022 - FFSK - Official WriteUp
FFSK - WriteUp
众所周知歪歪歪英文很烂求轻喷
0. Intro
In the game period, only one team had solved this problem: MapleBacon, a genius team at the University of British Columbia. I’m happy about their praise, but after checking their solution I think what truly “impressive” is their creativity and persistence.
Strongly recommend reading their awesome solution: https://maplebacon.org/2022/06/actf-ffsk/
FSK = Frequency-shift keying.
FFSK = Double FSK or Fast FSK, whatever.
This problem is designed to invite participants to have a look at the principle of communication.
1. Description
A journey to solve a misc problem always begins from a problem description. Here’s it:
I’ve bought the second commercial modem for computers in a big city of the UK.
激情澎湃的球迷迷恋这个地方。遇上球赛季，酒吧里的热情、呐喊、啤酒、摇滚，足球让这个城市充满活力和希望。
从三万英尺的云端望去，往日的生活成了一个遥远微小的地图。
阳光明媚的日子，开始出发，北京时间00:50 开始起飞，一个梦的距离，就可以到达荷兰阿姆斯特丹，短暂停留之后，然后转机飞往英国
南航的飞机配置完备，全程可以充电，还有wifi，影视屏有面前最新的电影。睡睡醒醒，在飞机上觅到一部《北京爱情故事》，让我在三万英尺的空中哭的稀里哗啦。
Just Google it, and you’ll realize what it means:
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second commercial modem→Bell 103, corresponds with the file name “modem.wav”
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a big city in the UK: Manchester, which refers to the famous coding method.
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The source of the long Chinese paragraph: https://kknews.cc/zh-hk/travel/e6yjp34.html
It describes a trip to Manchester, which is indeed a big city in the UK.
2. Bell 103
Here’s an article that shows how the Bell 103 protocol works: https://vigrey.com/blog/emulating-bell-103-modem
So two key points need your attention. First, characters are stored in ASCII code and are little-endian; second, it has 2 channels for communication: one for the server-side(2025/2225 Hz), and another for the client-side(1070/1270 Hz).
You can also find it from the spectrogram of the .wav file.
2. Server channel
Using the minimodemtool (See MapleBacon’s write-up) is functional.
Also, you can find some useful tools in GitHub: https://github.com/laurenschneider/audiodecoder
It may be a faster way. In fact, the solve.py is based on its code.
After all, you’ll see this on the server channel:
HINT_Hamming@ddddPdddddddPdddPdPP(20).ECCode; Content: Why do you use such
a slow method with a high Bit Error Ratio for communication? It took me a lot of
effort to correct bit-flips, making sure the communication was less
error-prone...that is 2 say, THE ORIGINAL PROTOCOL IS WRAPPED BY SOME OTHER
TRANSFORMATIONS! Fortunately, we can now communicate properly on another channel
while enjoying a vacation in this BIG CITY--I mean, IEEE 802.3.....Wait, what is
the new protocol? Guess by yourself!
3. Client channel
We can extract the bit string on this channel using the same method but just make some tweaks of frequency. You’ll get a bit string of 53640 bits.
Notice that the bit string contains only “01” “10”, that is what Manchester is all about. The IEEE 802.3mentioned in the server channel message is actually to make sure you decode in the right way: there are 2 opposite ways to map 01/10 to 1/0, but what is widely used is defined in IEEE 802.3, which says “01”→1 and “10”→0
Then the key problem is to solve Hamming code. From the given information, you’ll realize the block size is 20bits. Implement it by yourself or just Google/GitHub/StackOverflow it.
Find every “1” bit in a block, XOR their positions, and magically you got the error bit position(0 if no error) which is a well-designed feature of Hamming code, then just flip the bit.
Actually, every block has, and only has an error bit: that’s an intended design to notify you that you’re on the right way 😃
4. Final Step
Now you’ve got the cipher bit string: just applied Bell 103 decoder to it once again.
It’ll yield a string that starts with data:image/png;base64,
An experienced CTFer will immediately put it into the browser (like what MapleBacon did). Or you can find a random online converter to recover this Base64-encoded image. It’s a QR Code. Scan it, and got the flag.
5. Hints Explanation
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所有人都认为，吃鸡蛋前，原始的方法是打破鸡蛋较大的一端。可是当今皇帝的祖父 时候吃鸡蛋，一次按古法打鸡蛋时碰巧将一个手指弄破了，因此他的父亲，当时的皇帝， 就下了一道敕令，命令全体臣民吃鸡蛋时打破鸡蛋较小的一端，违令者重罚。 老百姓们 对这项命令极为反感。历史告诉我们，由此曾发生过六次叛乱，其中一个皇帝送了命，另 一个丢了王位…关于这一争端，曾出版过几百本大部著作，不过大端派的书一直是受禁的 ，法律也规定该派的任何人不得做官。 ——乔纳森·斯威夫特，《格列佛游记》It is a quote from Gulliver's Travels. Fun fact: this paragraph is exactly the original source of the 2 words used in modern computer science: “big-endian” & “little-endian”.
This hint is intended to guide those who are stuck because of their ignorance of the contents(especially the coding method) of the Bell 103 protocol.
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Hamming code block size: 20bitsNoticed that minimodem may yield partly corrupt text and mislead participants. This hint is to make sure they see the hint hidden at the beginning of the service-side channel message.
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Bell 103For those who ignored the problem description.
6. Script to Solve
# goertzel.py
"""
Module to create a Goertzel filter
Original source: https://github.com/laurenschneider/audiodecoder
"""
import numpy as np
class Goertzel():
def __init__(self, rate, freq):
self.normalize = 0
self.coeffs = 0
self.sample_rate = rate
self.target_freq = freq
def calculate_coeff(self):
"""
Precompute coefficients needed for filter equation.
Coeff formulas courtesy of Prof. Massey
"""
n = 160
w0 = (2 * np.pi * self.target_freq) / self.sample_rate
self.normalize = np.exp(1j * w0 * n)
self.coeffs = np.array([np.exp((-1j) * w0 * k) for k in range(n)])
def filter(self, samples):
"""
Goertzel filter equation
:param samples: array of samples
:returns: amplitude
"""
y = self.normalize * 160 * np.dot(self.coeffs, samples)
ampl = np.abs(y)
return ampl
# decode.py
"""
Decode a wav file using a Goertzel filter.
Modified from https://github.com/laurenschneider/audiodecoder
"""
from goertzel import Goertzel
import numpy as np
import os
from scipy.io import wavfile
DATA = os.path.dirname(os.path.abspath(__file__))
filepath = os.path.join(DATA, "filename.wav")
# Read sample rate and data from audio file
rate, data = wavfile.read(filepath)
message = ''
bit_string = ''
mark_freq = 2225
space_freq = 2025
mark_filter = Goertzel(rate, mark_freq)
space_filter = Goertzel(rate, space_freq)
# calculate coefficients for each filter
mark_filter.calculate_coeff()
space_filter.calculate_coeff()
for i in range(data.size + 1):
# for each chunk of 160 samples
if i%160 == 0 and i != 0:
start = i - 160
end = i
samples = data[start:end]
# get amplitutes of sample set
mark_amp = mark_filter.filter(samples)
space_amp = space_filter.filter(samples)
if mark_amp > space_amp:
# bit is 1
to_add = '1'
else:
# bit is zero
to_add = '0'
bit_string = to_add + bit_string
for x in range(10, len(bit_string)+10):
if x%10 == 0:
start = x - 9
end = x - 1
message = chr(int(bit_string[start:end],2)) + message
print(message)
# set target frequencies
mark_freq = 1270
space_freq = 1070
# create two filters
mark_filter = Goertzel(rate, mark_freq)
space_filter = Goertzel(rate, space_freq)
# calculate coefficients for each filter
mark_filter.calculate_coeff()
space_filter.calculate_coeff()
bit_string = ''
message = ''
for i in range(data.size + 1):
# for each chunk of 160 samples
if i%160 == 0 and i != 0:
start = i - 160
end = i
samples = data[start:end]
# get amplitutes of sample set
mark_amp = mark_filter.filter(samples)
space_amp = space_filter.filter(samples)
if mark_amp > space_amp:
to_add = '1'
else:
to_add = '0'
bit_string = to_add + bit_string
def HammingBolck(message):
assert(len(message)==15)
message=message[::-1]
code = 0
m_pos = 0
for ind in range(20):
if (ind+1)&(ind): # Not parity check bit
if int(message[m_pos],2):
code = code ^ (1<<ind)
code = code ^ (((ind+1)&0b1)<<0)
code = code ^ (((ind+1)&0b10)<<0)
code = code ^ (((ind+1)&0b100)<<1)
code = code ^ (((ind+1)&0b1000)<<4)
m_pos = m_pos + 1
code = code ^ (1<<random.randint(0,19))
retStr = "{0:020b}".format(code)
print("from "+message+" to "+retStr)
return retStr
def HammingBolckInv(message):
assert(len(message)==20)
code = int(message,2)
wrong = 0
for ind in range(20):
if (1<<ind)&code:
wrong = wrong ^ (ind+1)
if wrong:
code = code ^ (1<<(wrong-1))
retStr = ""
for ind in range(20):
if (ind+1)&(ind): # Not parity check bit
retStr = retStr + ("1" if code&(1<<ind) else "0")
return retStr[::-1]
def Hamming(message):
retStr = ""
for ind in range(0,len(message),15):
retStr = retStr + HammingBolck(message[ind:ind+15])
return retStr
def HammingInv(message):
retStr = ""
for ind in range(0,len(message),20):
retStr = retStr + HammingBolckInv(message[ind:ind+20])
return retStr
def manchester(message):
retStr = ""
for char in message:
retStr += "01" if char=='1' else "10"
return retStr
def manchesterInv(message):
retStr = ""
for ind in range(0, len(message), 2):
char = message[ind]
retStr += "0" if char=='1' else "1"
return retStr
bit_string=bit_string[::-1]
bit_string = HammingInv(manchesterInv(bit_string))[::-1]
for x in range(10, len(bit_string)+10):
if x%10 == 0:
start = x - 9
end = x - 1
message = chr(int(bit_string[start:end],2)) + message
print(message)
7. Final Words
In my eyes, a good misc problem should not be an annoying puzzle. Steps to solve a misc problem have to be reasonable. For example, in this problem, Manchester coding is applied after the application of Hamming coding: that’s because the former is channel coding, and the latter is source coding. We shouldn’t just pick some random encryption and apply it to plaintext.
I have made my best to make the solving process more natural. I hope you enjoy digging deep into the problem. You’ll earn much more fun than those who can just use tools written by others without understanding fundamental principles (like me).
ご武運を！
