Boring Assignment

April 18 2017

# # # #

There was once a guy, who hadn't anything to do. So he made haiku.

CTF question, related to cryptography, he presents to you.

Solve you can or not, you must at least try or else, you disappoint him.

  • Flag Format : /FLAG[A-Z]+/

Provided boring-assignment

Hint: Cipher used is a really popular cipher.

Solution

Starting this with the usual.

$ file boring-assignment
boring-assignment: data

Huh… Nothing. Ok… Let us open this file and see what it is.

$ cat boring-assignment

<Redacted Binary>
./a.py
<listcomp>s
dSdS)Nr)
plain_textarr<module>
<Redacted Binary>

Ok… So now we are getting somewhere. This has some interesting characters. The ./a.py catches my eye. As does the , tags. This is quite plainly a python object compiled file. These can be decompiled.

$ cp boring-assignment boring-assignment.pyc

$ uncompyle6 boring-assignment.pyc
# uncompyle6 version 2.9.10
# Python bytecode 3.5 (3350)
# Decompiled from: Python 3.5.2 (default, Nov 17 2016, 17:05:23) 
# [GCC 5.4.0 20160609]
# Embedded file name: ./a.py
# Compiled at: 2017-03-26 02:56:19
# Size of source mod 2**32: 298 bytes
cipher_text = ''.join([chr(x) for x in [80, 80, 89, 81, 71, 80, 73, 84, 82, 89, 86, 67, 65, 89, 71, 66, 73, 81, 68, 76, 67, 69, 87, 67, 89, 74, 75, 75, 88, 70, 79, 81, 89, 68, 77, 65, 67]])

def decrypt(cipher_text=None):
    if not cipher_text:
        return


plain_text = decrypt(cipher_text)
# okay decompiling boring-assignment.pyc

Woohoo! We now have a source. And there seems to be a python function here. That gives us a cipher text. And a template for a decrypt function, which gives us a nothing.

Let's see what the cipher text is. The program seems kind enough to join it into a string for us.

    In [1]: cipher_text
    Out[1]: 'PPYQGPITRYVCAYGBIQDLCEWCYJKKXFOQYDMAC'

So… Weird cipher_text. It is too readable. Seems like a simple shift cipher or a substitution cipher. Substitution ciphers generally provide more text for frequency analysis, so we can try a shift cipher.

I quickly whip up a code for shift cipher, and fire it up.

    In [9]: def shift_cipher(n, cipher):
       ...:     d = {}
       ...:     for i in range(0, len(string.ascii_uppercase)):
       ...:         d[string.ascii_uppercase[i]] = string.ascii_uppercase[(i + n) % len(string.ascii_uppercase)]
       ...:     print("".join([d[x] for x in cipher]))
       ...:     
       ...:     
       ...:     

    In [10]: for i in range(26):
        ...:     shift_cipher(i, cipher_text)
        ...:     
    PPYQGPITRYVCAYGBIQDLCEWCYJKKXFOQYDMAC
    QQZRHQJUSZWDBZHCJREMDFXDZKLLYGPRZENBD
    RRASIRKVTAXECAIDKSFNEGYEALMMZHQSAFOCE
    SSBTJSLWUBYFDBJELTGOFHZFBMNNAIRTBGPDF
    TTCUKTMXVCZGECKFMUHPGIAGCNOOBJSUCHQEG
    UUDVLUNYWDAHFDLGNVIQHJBHDOPPCKTVDIRFH
    VVEWMVOZXEBIGEMHOWJRIKCIEPQQDLUWEJSGI
    WWFXNWPAYFCJHFNIPXKSJLDJFQRREMVXFKTHJ
    XXGYOXQBZGDKIGOJQYLTKMEKGRSSFNWYGLUIK
    YYHZPYRCAHELJHPKRZMULNFLHSTTGOXZHMVJL
    ZZIAQZSDBIFMKIQLSANVMOGMITUUHPYAINWKM
    AAJBRATECJGNLJRMTBOWNPHNJUVVIQZBJOXLN
    BBKCSBUFDKHOMKSNUCPXOQIOKVWWJRACKPYMO
    CCLDTCVGELIPNLTOVDQYPRJPLWXXKSBDLQZNP
    DDMEUDWHFMJQOMUPWERZQSKQMXYYLTCEMRAOQ
    EENFVEXIGNKRPNVQXFSARTLRNYZZMUDFNSBPR
    FFOGWFYJHOLSQOWRYGTBSUMSOZAANVEGOTCQS
    GGPHXGZKIPMTRPXSZHUCTVNTPABBOWFHPUDRT
    HHQIYHALJQNUSQYTAIVDUWOUQBCCPXGIQVESU
    IIRJZIBMKROVTRZUBJWEVXPVRCDDQYHJRWFTV
    JJSKAJCNLSPWUSAVCKXFWYQWSDEERZIKSXGUW
    KKTLBKDOMTQXVTBWDLYGXZRXTEFFSAJLTYHVX
    LLUMCLEPNURYWUCXEMZHYASYUFGGTBKMUZIWY
    MMVNDMFQOVSZXVDYFNAIZBTZVGHHUCLNVAJXZ
    NNWOENGRPWTAYWEZGOBJACUAWHIIVDMOWBKYA
    OOXPFOHSQXUBZXFAHPCKBDVBXIJJWENPXCLZB

So… Nothing. However if we now see the flag format, and take a look at the hint provided, two things become clear.

  • It is a historically used cipher
  • The plaintext begins with FLAG

This rules out mono-aliphatic substitution ciphers (because cipher text begins with two P's). And also reduces the chance of a poly-aliphatic substitution cipher. Because we don't have any more information, and this given information is incomplete.

One cipher comes to mind which maps each character to different characters. It is also quite well known. The Vigenère cipher. A pretty nice website for decoding such a cipher is here.

It even gives us an exact solution for our case, where we presumably know the first word of the series. We feed it the values, and it spits out the following.

Plain Text: FLAGCRYPTOREQUIRESTHEUSEOFMATHEMATICS

Cipher Key : KEY

Alphabet: ABCDEFGHIJKLMNOPQRSTUVWXYZ

The flag is CRYPTOREQUIRESTHEUSEOFMATHEMATICS

Flag

CRYPTOREQUIRESTHEUSEOFMATHEMATICS

Recommended Reading

Foren-Steg

# # # #

Find the flag, flag finder.

  • Flag Format: /flag{.+}/

Provided foren-steg.docx

Solution

The file provided to us appears to be a docx file. Let us see what happens when we try to open it.

There appears to be nothing in the file but gibberish...

...

Recommended Reading

Macbeth

# # # # #

Find the flag.

  • Flag Format /flag:[a-zA-Z]+/

Provided Macbeth.docx

Hint : Not all characters are created equal.

Solution

Opening the file we can quickly tell that the entire file has been written in two fonts. We unzip the docx file, and grep...

...