Overview Of The Challenge
The challenge involved an encryption program which performed the following encryption to genereate the cipher text:
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for i in range(0, len(message)):
encrypted += chr((ord(message[i]) ^ ord(key[i % len(key)]) ^ ord(encrypted[i]))%128)
encrypted is the ciphertext and a random ascii character was added to it before the encryption loop.
The goal was to find the flag from a message which consisted essentially three parts
- The flag
- A predefined character
| - And a key
An output of the encryption program was given which when hex decoded gave a ciphertext of length 36. Consequently I found out the Length of the flag part by subtracting ciphertext length from key lentgth plus 2 (taking into account the predefined chracter |. and the randomly generated ascii character)
The Approach
Since in the encryption equation there are two unknowns it seems impossible to find the key and hence the flag. The idea here is that the program already gives us a character in the message namel the predefined character | whose position we know. taking the position of that character as i value we can get the correspnding value of the character at index i%13 in key. On the index counter of the list message the corresponding vlaue of key[i%13] will be at the position message[22+i%13] (as Key comes after flag text and the predefined character)
Now I have got message[22+i%13].Now i can assign i to i=22+i%13. If I put these three instructions:
- Finding value of key[i%13]
- Assigning that value to message[22+i%13]
- Changing value of i to 22+i%13
In a loop that repets 12 times what we surprisingly get is nothing other than value of key in all its thirteen characters!!!
The core logic behind this piece of code is pretty easy eliminate one unkown out of the two which essentially solves the challenge to a very large extent . Here the variable being elimnated is messge[i]. We do this by assignig message[index] the value of the key which we find in the first loop instruction and consequently assigning that i with value index in the third loop instruction so that when the loop iterates the next time the value for message[i] is no longer an unknown thus we find out the entire key (but not the full message).
Now with only one unknown finding the complete message is childs play (just solve the encryption equation with message[i] as the unknown)
In the end we get the key as: ENJ0YHOLIDAY!
Flag as: TWCTF{Crypto-is-fun!}