43 lines
981 B
Markdown
43 lines
981 B
Markdown
## Difficulty
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50/100 | MEDIUM
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They need to figure out that the first part of the message is a
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substitution alphabet, then decrypt the code, and then finally
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figure out the flag is embedded in the last message they sent.
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## Category
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Crypto
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## How To Solve
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The first part of every message line was a substitution table,
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the second part the message. Just substitute the message and the
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last message will contain the flag.
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```python3
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from pwn import *
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import string
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p = remote("localhost", 3006)
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p.recvline()
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while True:
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msg = p.recvline().decode()
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p.recvuntil(b"> ")
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print(msg)
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alphabet = msg.split(" | ")[0]
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message = msg.split(" | ")[1][:-1]
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alphabet_map = dict(zip(alphabet, string.ascii_letters))
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decoded = "".join([alphabet_map[c] if c in alphabet_map else c for c in message])
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print(f"Sending back decoded: '{decoded}'")
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p.sendline(decoded.encode())
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p.recvline()
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```
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## Flag
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IGCTF{Some-wicked-story-unfolded-here-great-work} |