Hackfest 2022: CrunchyCiphers

Writeup for the “CrunchyCiphers” challenge created by @muemmelmoehre for the Hackfest CTF 2022.

01 - CrunchAsAService

The challenge consists of decrypting and decoding a flag provided to us in the following file:

---------------------------------------------------------------------------------------------

	  ___                       _     _  _         ___  _  _ __  _
	 / __| _ _  _  _  _ _   __ | |_  | || |       / __|(_)| '_ \| |_   ___  _ _  ___
	| (__ | '_|| || || ' \ / _||   \  \_. |      | (__ | || .__/|   \ / -_)| '_|(_-/
	 \___||_|   \_._||_||_|\__||_||_| |__/        \___||_||_|   |_||_|\___||_|  /__/
	
---------------------------------------------------------------------------------------------
Welcome to Crunchy Ciphers! I assume you are here for the flag. (y/n)
y

[*] Making your flag extra crunchy...

[*] Encrypting your flag...

[+] Here's your encrypted flag:
HycukjBYbDz3OVNbozbSY5XVCOH1puUWd+BD5iMcGhhXuniyclzuF2kYvjLGWpQgNKO6a6dZEDEU8NiZj+krbcGvI1zFmNfAgb8zAeY8RI5upSoBPyVqGAEL5G7PREab+iy0oCkUBOe2ul2sPJ3B+gx2j8QKMpHQ1vY7ieXsqymv7w0+00s59ZVQKoxb+pFR5/7VJK0LG3pY5UoKUAJKIc9UowCKRYC+lb2W2ns5NT5DN9Jq4YWC52ouZT1GOHVqSU5f9qZdAqoVG7vRV7dWbxwc9IY5zpzTvVDytqqGRT7V1sE+T0X1HD58OvgBSUTkXdDqaqWrY4Tf1QMrFEsTVw==

Would you like some keys with your flag? (y/n)
y

[*] Private key coming right up!
-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----

[*] Public key coming right up!
-----BEGIN RSA PUBLIC KEY-----
MIIBCgKCAQEAoDjQxbABCgWr3avBKcH3zM3zwDV4eqav3CARZmUwyP+u9BnITJ0u
dWb5EB++ahH/iDAp9XDP7DoT+bJZuoqZtwibDSLTXY5gFrNcnQfzVCG+gPqvP59n
GunrV7izjxYfQZ7XadFMPqKY8pv7hpiQKEcEx0SEGXeMrkmhpSups93shPIGs9vu
BIeQeEGEFzoyqleYVvuFR9H81hkLNwKehYXe8yY0SOJs6uHWSqzQQSrHsmhTpxIq
a4XADEKKP10j7HPWbzigZYq3dM5JyGjY+qHAFFsEasBD/TEAXfGbKyu/LzrMS18P
vbw8r7boxt7gqhY54by8LBquRlKcWjYIAQIDAQAB
-----END RSA PUBLIC KEY-----


[+] Enjoy! Come again! :)

With the provided RSA private key, we can decrypt the flag using CyberChef or the openssl command.

After decoding, we get the following message that starts to resemble a flag:

HF-'ÞOÜeCÂnRead.u9¦u8erSecäetHFFlag

It is possible to easily reconstruct certain parts of the flag manually. For example, ¦u8erSecäetHFFlag seems to mean SuperSecretHFFlag.

By analyzing the replaced characters, we can see a pattern emerge:

Encoded characters	Decoded characters
ord('¦') => 166	ord('S') => 83
ord('8') => 56	ord('p') => 112
ord('ä') => 228	ord('r') => 114

The value of each character is either doubled or halved.

The following script decodes almost all the characters in the flag:

import string

flag = 'HF-\'ÞOÜeCÂnRead.u9¦u8erSecäetHFFlag'

for letter in flag:
    if ord(letter) > 122:
        print(chr(int(ord(letter) / 2)), end='')
    elif ord(letter) < 65 and not letter == '-':
        print(chr(ord(letter) * 2), end='')
    else:
        print(letter, end='')

print()

HF-NoOneCanRead\urSuperSecretHFFlag

There is one last character that is not properly decoded, but we can easily deduce it and the final flag is:

HF-NoOneCanReadOurSuperSecretHFFlag