hack.lu CTF 2011 Simplexor (200)

Category: crypto

To get a better security we deceided to encrypt our most secret document with the secure xor-algorithm. Unfortunately we lost the key. Now we are sad. Can you help us recovering the key?

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Summary: recovering multibyte xor-key, using autocorrelation

As the text says – the file was xored so we should use my xortool or Cryptool or whatever else.

First, we’ll decode base64:

$ base64 -d simplexor.txt >ciphertext.bin

Now, let’s try xortool:

$ xortool ciphertext.bin 
Probable key lengths:
   2:   4.9 %
   4:   7.3 %
   6:   4.8 %
   8:   9.5 %
  10:   4.8 %
  12:   7.1 %
  14:   4.9 %
  16:   14.1 %
  18:   4.8 %
  20:   7.1 %
  22:   4.9 %
  24:   9.2 %
  26:   4.8 %
  28:   7.0 %
  30:   4.8 %
Key-length can be 4*n
Most possible char is needed to guess the key!

Ok, we see that 16 is the most possible length of the key. By checking it we get pretty key but a wasty output:

$ xortool ciphertext.bin -c 20
Probable key lengths:
...
1 possible key(s) of length 16:
WklF6e5TEc5XmEG8

$ xxd xortool_out/0_WklF6e5TEc5XmEG8 | head
0000000: 2e72 4b08 367f 3e03 1646 4700 6054 7f32  .rK.6.>..FG.`T.2
0000010: 2e38 512f 6320 7435 2020 2020 2020 2020  .8Q/c t5        
0000020: 2d04 2020 200f 5623 0829 2d10 2f65 450d  -.   .V#.)-./eE.

The problem is that key looks to be longer. xortool by default tries only values, smaller than 32. We should force it to, say, 257:

$ xortool ciphertext.bin -m 257 -c 20
...
Key-length can be 4*n
1 possible key(s) of length 64:
WvhnPry60NRl41weWY7IueaAEc5XmEG8ZOlF6JCWmj8hbvmYkkwFox5Tz1HLvdKl

We see, it choose 64 from 257, which means we are on the right way. Let’s check output:

$ head xortool_out/0_*
.oO Phrack 49 Oo.

                          Volume Seven, Issue Forty-Nine
                                     
                                  File 14 of 16

                      BugTraq, r00t, and Underground.Org
                                   bring you

                     XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Here it is! Obviously we should try xor key as flag.

The flag: WvhnPry60NRl41weWY7IueaAEc5XmEG8ZOlF6JCWmj8hbvmYkkwFox5Tz1HLvdKl