Question

So I have to decrypt a .txt file that is crypted with XOR code and with a repeated password that is unknown, and the goal is to discover the message.

Here are the things that I already know because of the professor:

1. First I need to find the length of the unknown password

2. The message has been altered and it doesn’t have spaces (this may add a bit more difficulty because the space character has the highest frequency in a message)

Any ideas on how to solve this?

thx in advanced 🙂

Answer 1

First you need to find out the length of the password. You do this by assessing the Index of Coincidence or Kappa-test. XOR the ciphertext with itself shifted 1 step and count the number of characters that are the same (value 0). You get the Kappa value by dividing the result with the total number of characters minus 1. Shift one more time and again calculate the Kappa value. Shift the ciphertext as many times as needed until you discover the password length. If the length is 4 you should see something similar to this:

Offset             Hits
-------------------------
  1              2.68695%
  2              2.36399%
  3              3.79009%
  4              6.74012%
  5              3.6953%
  6              1.81582%
  7              3.82744%
  8              6.03504%
  9              3.60273%
 10              1.98052%
 11              3.83241%
 12              6.5627%

As you see the Kappa value is significantly higher on multiples of 4 (4, 8 and 12) than the others. This suggests that the length of the password is 4.

Now that you have the password length you should again XOR the cipher text with itself but now you shift by multiples of the length. Why? Since the ciphertext looks like this:

THISISTHEPLAINTEXT    <- Plaintext
PASSPASSPASSPASSPA    <- Password
------------------
EJKELDOSOSKDOWQLAG    <- Ciphertext

When two values which are the same are XOR:ed the result is 0:

EJKELDOSOSKDOWQLAG        <- Ciphertext
    EJKELDOSOSKDOWQLAG    <- Ciphertext shifted 4.

Is in reality:

THISISTHEPLAINTEXT        <- Plaintext
PASSPASSPASSPASSPA        <- Password
    THISISTHEPLAINTEXT    <- Plaintext
    PASSPASSPASSPASSPA    <- Password

Which is:

THISISTHEPLAINTEXT        <- Plaintext
    THISISTHEPLAINTEXT    <- Plaintext

As you see the password “disappears” and the plaintext is XOR:ed with itself.

So what can we do now then? You wrote that the spaces are removed. This makes it a bit harder to get the plaintext or password. But not at all impossible.

The following table shows the ciphertext values for all english characters:

   A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z
A  0                                                                           
B  3  0                                                                        
C  2  1  0                                                                     
D  5  6  7  0                                                                  
E  4  7  6  1  0                                                               
F  7  4  5  2  3  0                                                            
G  6  5  4  3  2  1  0                                                         
H  9 10 11 12 13 14 15  0                                                      
I  8 11 10 13 12 15 14  1  0                                                   
J 11  8  9 14 15 12 13  2  3  0                                                
K 10  9  8 15 14 13 12  3  2  1  0                                             
L 13 14 15  8  9 10 11  4  5  6  7  0                                          
M 12 15 14  9  8 11 10  5  4  7  6  1  0                                       
N 15 12 13 10 11  8  9  6  7  4  5  2  3  0                                    
O 14 13 12 11 10  9  8  7  6  5  4  3  2  1  0                                 
P 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31  0                              
Q 16 19 18 21 20 23 22 25 24 27 26 29 28 31 30  1  0                           
R 19 16 17 22 23 20 21 26 27 24 25 30 31 28 29  2  3  0                        
S 18 17 16 23 22 21 20 27 26 25 24 31 30 29 28  3  2  1  0                     
T 21 22 23 16 17 18 19 28 29 30 31 24 25 26 27  4  5  6  7  0                  
U 20 23 22 17 16 19 18 29 28 31 30 25 24 27 26  5  4  7  6  1  0               
V 23 20 21 18 19 16 17 30 31 28 29 26 27 24 25  6  7  4  5  2  3  0            
W 22 21 20 19 18 17 16 31 30 29 28 27 26 25 24  7  6  5  4  3  2  1  0         
X 25 26 27 28 29 30 31 16 17 18 19 20 21 22 23  8  9 10 11 12 13 14 15  0      
Y 24 27 26 29 28 31 30 17 16 19 18 21 20 23 22  9  8 11 10 13 12 15 14  1  0   
Z 27 24 25 30 31 28 29 18 19 16 17 22 23 20 21 10 11  8  9 14 15 12 13  2  3  0

What does this mean then? If an A and a B is XOR:ed then the resulting value is 3. E and P will result in 21. Etc. OK but how will this help you?

Remember that the plaintext is XOR:ed with itself shifted by multiples of the password length. For each value you can check the above table and determine what combinations that position could have. Lets say the value is 25 then the two characters that resulted in the value 25 could be one of the following combinations:(I-P), (H-Q), (K-R), (J-S), (M-T), (L-U), (O-V), (N-W), (A-X) or (C-Z). But which one? Now you do more shifts and look up the corresponding values in the table again for each position. Next time the value might be 7 and since you already have a list of possible character combinations you only check against them. At the next two shifts the values are 3 and 1. Now you can determine that the character is W since that is the only common character in each shift, (N-W), (P-W), (T-W), (V-W). You can do this for most positions.

You will not get all the plaintext but you will get enough characters to discover the password. Take the known characters and XOR them in the correct position in the ciphertext. This will yield the password. The number of known characters you need atleast is the number of characters in the password if they are at the “correct” positions in regards to the password.

Good luck!