# Oil Circuit Breaker (Crypto) \[714\]
## __Description__

## __Solution__
It's an OCB2 cipher, and its vulnerability can be found [here](https://eprint.iacr.org/2018/1040.pdf).
First encryption, send in: (+ is concatenation and (n) is 16 bytes expansion to n)
```nonce = 00000000000000000000000000000000plain = 000000000000000000000000000000010000000000000000000000000000008000000000000000000000000000000000      = (1) + (128) + nonce```Where plain = M\[0\]+M\[1\]+M\[2\].
Get C\[0\]+C\[1\]+C\[2\] and T.
In descryption, send in:```nonce  = 00000000000000000000000000000000cipher = C[0] + C[1] ^ (M[0] ^ 128)       = C'[0] + C'[1]tag    = C[2] ^ M[2]       = C[2] ^ nonce```This will reply auth = True because:```T' = e(S' ^ 8L ^ 4L)   = e((M'[1] ^ C'[1] ^ pad') ^ 8L ^ 4L)   = e(M[1] ^ C'[1] ^ C[1] ^ 4L ^ 8L ^ 4L)   = e(M[1] ^ C'[1] ^ C[1] ^ 8L)   = e(M[1] ^ (M[1] ^ C[1] ^ (128)) ^ C[1] ^ 8L)   = e((128) ^ 8L)   = pad   = C[2] ^ nonce```Where T' is new tag, e stands for aes encryption, and L = e(nonce).
So T' is profit. We also have information about L because:```M'[2] = C'[2] ^ pad'      = C[1] ^ (M[0] ^ 128) ^ C[1] ^ 4L      = M[0] ^ (128) ^ 4L```So we can recover L for the encryption.
Now that we have L, we also need to know other components for encryption of 'giveme flag.txt', but we can only send the same nonce once.
That's ok because if we let nonce = M\[0\] ^ L, then L\'\' = e(M\[0\] ^ L) = C\[0\] ^ L in the first encryption!
In the second encryption, I send in:
```nonce'' = L'' = C[0] ^ Lplain'' = (120)for i in range (256):      s = b'giveme flag.txt' + (i).to_bytes(1)      s = s ^ 2L'' ^ 4L''      plain'' += s ^ pow(2,i+2)L''```Where plain'' = M''\[0\] ~ M''\[256\].
From the return C''\[0\] ~ C''\[256\] and T'', we can recove the disired cipher C\* and T\* by:```pad'' = e(120 ^ 2L'')      = C''[0] ^ 2L''C* = pad''[0:15] ^ b'giveme flag.txt'i = pad''[16]T* = c2[16*(i+1):16*(i+2)] ^ pow(2, i+2)L''```This is true because ```T* = e(S* ^ 2L'' ^ 4L'')   = e(pad'' ^ (C* + '00') ^ 4L'')   = e((b'giveme flag.txt' + pad[16]) ^ 2L'' ^ 4L'')   = e(M''[i+1] ^ pow(2, i+2)L'') ^ pow(2, i+2)L''    = c2[16*(i+1):16*(i+2)] ^ pow(2, i+2)L''``````BAMBOOFOX{IThOUgHtitWAspRoVaBles3cuRE}```
