Easy Peasy
A one-time pad is unbreakable, but can you manage to recover the flag? (Wrap with picoCTF{}) nc mercury.picoctf.net 20266 otp.py
- 1.As stated in the description, this is a one-time pad challenge. One of the criteria of a one-time pad is that the key is never reused in part or in whole. We can modify the program so that it does not meet this requirement.
- 2.if stop >= KEY_LEN:stop = stop % KEY_LENkey = kf[start:] + kf[:stop]
stop
equals the previous ending point of the key plus the length of the new user input. However, ifstop
is greater than the key length,stop
is set tostop % KEY_LEN
. Thus, inputting just enough text to get to the end of the keyfile will setstop
to 0 because40000 % 40000
is0
.key_location
is then set tostop
andkey_location
is returned. So, when we input the next string to be encrypted theencrypt
function will receivekey_location=0
, thus allowing us to use the same key that was used to encrypt the flag. - 3.Generate a
pwntools
template withpwn template --host mercury.picoctf.net --port 20266 otp.py
. - 4.The solve script is commented and explains the solution. In brief, since we know a clear text message and an encrypted message, we can find the key (as explained in this Computer Science StackExchange answer) and then decrypt the flag.
picoCTF{99072996e6f7d397f6ea0128b4517c23}
Last modified 6mo ago