HackThisSite - Steganography Mission 9

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Challenge

Download the song Here kiss me alone

Solution

Analysis

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curl -sL -b "$HTS_COOKIE" -o Stego9.zip \
  'https://www.hackthissite.org/missions/stego/lvl/Stego9.zip'
7z l Stego9.zip
7z x Stego9.zip
soxi ecstasy-atb-music.wav
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Type = zip
   Date      Time    Attr         Size   Compressed  Name
2008-06-24 08:51:34 ....A      1784818      1675128  ecstasy-atb-music.wav
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Channels       : 2
Sample Rate    : 11025
Precision      : 16-bit
Duration       : 00:00:40.47 = 446190 samples
File Size      : 1.78M
Sample Encoding: 16-bit Signed Integer PCM

ZIP 里只有一个 40 秒、11025 Hz、16 bit 立体声的 WAV。单看左右声道各自都是一首连续的 电子舞曲，波形和频谱都很正常，秘密只在混音层面才会显形。

Step 1: WAV chunks

先按 WAV 的块结构逐块核对尾部附加数据，以及头部声明长度与实际文件长度的一致性：

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import struct

d = open("ecstasy-atb-music.wav", "rb").read()
print("riff", d[:4], "file_size", len(d))
pos = 12
while pos + 8 <= len(d):
    cid = d[pos:pos + 4].decode("ascii", "replace")
    size = struct.unpack_from("<I", d, pos + 4)[0]
    print("%-4s offset=%d size=%d" % (cid, pos + 8, size))
    pos += 8 + size + (size & 1)          # 块按 2 字节对齐
print("end_of_chunks", pos, "== file size", pos == len(d))
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riff b'RIFF' file_size 1784818
fmt  offset=20 size=18
fact offset=46 size=4
data offset=58 size=1784760
end_of_chunks 1784818 == file size True

data 块正好覆盖到文件结尾（58 + 1784760 = 1784818），没有任何尾部附加数据； 头部声明的长度也和真实文件一致。信息只能在 PCM 采样里。

Step 2: Channel inversion

分别看左右声道的能量和相关性：

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import wave
import numpy as np

with wave.open("ecstasy-atb-music.wav", "rb") as w:
    sr = w.getframerate()
    pcm = np.frombuffer(w.readframes(w.getnframes()), dtype="<i2").astype(np.float64)
L, R = pcm[0::2], pcm[1::2]
print("corr(L,R) = %.6f" % np.corrcoef(L, R)[0, 1])
print("std L=%.1f R=%.1f  L-R=%.1f  L+R=%.1f" % (
    L.std(), R.std(), (L - R).std(), (L + R).std()))
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corr(L,R) = -0.999697
std L=18641.6 R=18642.0  L-R=37280.8  L+R=458.9

这一步就是全部关键。两个声道的相关系数是 -0.9997：右声道是左声道的反相副本。 所以：

  • 现场听感正常，是因为播放器把两声道混合后音乐照旧；
  • L - R 的幅度几乎是被乘了 2 倍的歌曲本身（std 37281 ≈ 2×18641）；
  • L + R 却塌缩成 std 只有 459 的极小残差。

音乐在求和时被抵消，留下的 L + R 才是承载信息的信号。把它单独导出成单声道：

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import wave
import numpy as np

with wave.open("ecstasy-atb-music.wav", "rb") as w:
    sr, nch, sw = w.getframerate(), w.getnchannels(), w.getsampwidth()
    pcm = np.frombuffer(w.readframes(w.getnframes()), dtype="<i2").astype(np.float64)
residual = (pcm[0::2] + pcm[1::2]).astype(np.int16)
with wave.open("residual_sum.wav", "wb") as o:
    o.setnchannels(1)
    o.setsampwidth(sw)
    o.setframerate(sr)
    o.writeframes(residual.tobytes())
print("residual std=%.1f absmax=%d" % (residual.std(), np.abs(residual).max()))
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residual std=458.9 absmax=1044

这个残差和歌曲本身几乎不相关（corr(L+R, L) = 0.0115），是一个独立的信号。

Step 3: Carrier

对残差做精细 FFT，看能量集中在哪：

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import wave
import numpy as np

with wave.open("ecstasy-atb-music.wav", "rb") as w:
    sr = w.getframerate()
    pcm = np.frombuffer(w.readframes(w.getnframes()), dtype="<i2").astype(np.float64)
res = pcm[0::2] + pcm[1::2]

chunk = res[100000:100000 + 8192] * np.hanning(8192)
spec = np.abs(np.fft.rfft(chunk))
freqs = np.fft.rfftfreq(8192, 1.0 / sr)
band = (freqs > 500) & (freqs < 2000)
top = band.nonzero()[0][np.argsort(spec[band])[-5:]][::-1]
for i in top:
    print("%7.1f Hz  mag %8.1f" % (freqs[i], spec[i]))
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 998.6 Hz  mag 963327.3
 999.9 Hz  mag 917071.2
1002.6 Hz  mag 903649.9
1001.3 Hz  mag 665056.8
1004.0 Hz  mag 341981.2

残差里几乎只有一个 约 1000 Hz 的纯载波（相邻几条谱线都挤在 1 kHz 附近， 是同一根载波在 FFT 分辨率上的展宽）。再用时间轴看它，按 100 ms 一格量载波的能量， 会发现它一段一段地通断：

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import wave
import numpy as np
from scipy.signal import butter, hilbert, sosfiltfilt

with wave.open("ecstasy-atb-music.wav", "rb") as w:
    sr = w.getframerate()
    pcm = np.frombuffer(w.readframes(w.getnframes()), dtype="<i2").astype(np.float64)
res = pcm[0::2] + pcm[1::2]

sos = butter(4, [800 / (sr / 2), 1200 / (sr / 2)], btype="band", output="sos")
env = np.abs(hilbert(sosfiltfilt(sos, res)))
env = np.convolve(env, np.ones(32) / 32, mode="same")
step = sr // 10
print(" ".join("%4d" % env[i * step:(i + 1) * step].mean()
               for i in range(30)))
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517  249 1034  781    4 1019 1027   12  747 1017  271  495 1035  482    0    1  997 1006   23  708  994  283  394  856  513  222 1033  676    1  947

每 100 ms 的能量在接近 1000 和接近 0 两档之间来回跳，这正是通断键控（OOK）， 把通断的时间长度读出来就是莫尔斯电码。

Step 4: Morse

对上一步的包络取阈值、做游程编码，按运行长度区分点 / 划和字符内间隔 / 字符间间隔 / 词间间隔，然后查表：

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import wave
import numpy as np
from scipy.signal import butter, hilbert, sosfiltfilt

with wave.open("ecstasy-atb-music.wav", "rb") as w:
    sr = w.getframerate()
    pcm = np.frombuffer(w.readframes(w.getnframes()), dtype="<i2").astype(np.float64)
res = pcm[0::2] + pcm[1::2]

sos = butter(4, [800 / (sr / 2), 1200 / (sr / 2)], btype="band", output="sos")
env = np.abs(hilbert(sosfiltfilt(sos, res)))
env = np.convolve(env, np.ones(32) / 32, mode="same")
on = env > env.max() * 0.25
first, last = np.flatnonzero(on)[[0, -1]]
on = on[first:last + 1]

runs = []
cur, start = bool(on[0]), 0
for i in range(1, len(on)):
    if bool(on[i]) != cur:
        runs.append((cur, i - start))
        cur, start = bool(on[i]), i
runs.append((cur, len(on) - start))

ms = lambda n: round(n / sr * 1000)
print("on  ms:", [ms(n) for c, n in runs if c][:24])
print("off ms:", [ms(n) for c, n in runs if not c][:24])
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on  ms: [51, 201, 202, 202, 201, 202, 202, 202, 201, 202, 201, 201, 52, 50, 52, 52, 51, 52, 51, 51, 202, 52, 51, 52]
off ms: [124, 124, 124, 124, 249, 124, 124, 124, 124, 249, 124, 124, 124, 125, 499, 124, 124, 124, 124, 249, 124, 124, 124, 124]

时长分成非常干净的四档：

  • 通 ≈ 51 ms（点）与 ≈ 201 ms（划），约 1:4；
  • 断 ≈ 124 ms（字符内部间隔）、≈ 249 ms（字符之间）、≈ 499 ms（词之间），约 1:2:4。

按这个比例把游程翻译成 ./-，用 124/249/499 三档间隔切分字符和词：

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import wave
import numpy as np
from scipy.signal import butter, hilbert, sosfiltfilt

MORSE = {
    "-----": "0", ".----": "1", "..---": "2", "...--": "3", "....-": "4",
    ".....": "5", "-....": "6", "--...": "7", "---..": "8", "----.": "9",
}

with wave.open("ecstasy-atb-music.wav", "rb") as w:
    sr = w.getframerate()
    pcm = np.frombuffer(w.readframes(w.getnframes()), dtype="<i2").astype(np.float64)
res = pcm[0::2] + pcm[1::2]

sos = butter(4, [800 / (sr / 2), 1200 / (sr / 2)], btype="band", output="sos")
env = np.abs(hilbert(sosfiltfilt(sos, res)))
env = np.convolve(env, np.ones(32) / 32, mode="same")
on = env > env.max() * 0.25
first, last = np.flatnonzero(on)[[0, -1]]
on = on[first:last + 1]

runs = []
cur, start = bool(on[0]), 0
for i in range(1, len(on)):
    if bool(on[i]) != cur:
        runs.append((cur, i - start))
        cur, start = bool(on[i]), i
runs.append((cur, len(on) - start))

unit = min(n for c, n in runs if c)         # 最短的通段 = 一个点
words, word, sym = [], [], ""
for is_on, n in runs:
    if is_on:
        sym += "." if n < unit * 2.5 else "-"
    elif n < unit * 4:                      # 字符内间隔，忽略
        continue
    elif n < unit * 7:                      # 字符间间隔
        word.append(sym)
        sym = ""
    else:                                   # 词间间隔
        word.append(sym)
        words.append(word)
        word, sym = [], ""
if sym:
    word.append(sym)
if word:
    words.append(word)

digits = ["".join(MORSE[s] for s in w) for w in words]
print("morse digits:", " ".join(digits))
print("ascii       :", " ".join(chr(int(d)) for d in digits))
k8x0w5bbuq