After having great fun last year in Google CTF with a nice RSA challenge, and a couple of strange crypto schemes, and despite the lack of enthusiasm of my fellow team members, I decided to play again this year. And the first crypto challenge I solved was also about RSA, it said:

Perfect Secrecy

nc perfect-secrecy.ctfcompetition.com 1337


And it provided us an attachment, which contained a file called flag.txt (probably encrypted, since it contained seemingly random bytes), a key_pub.pem file, containing a RSA public key and a python file challenge.py.

Let us first take a look at the key:

\$> openssl rsa -pubin -text -modulus -noout < key_pub.pem
Public-Key: (1024 bit)
Modulus:
00:da:53:a8:99:d5:57:30:91:af:6c:c9:c9:a9:fc:
31:5f:76:40:2c:89:70:bb:b1:98:6b:fe:8e:29:ce:
d1:2d:0a:df:61:b2:1d:6c:28:1c:cb:f2:ef:ed:79:
aa:7d:d2:3a:27:76:b0:35:03:b1:af:35:4e:35:bf:
58:c9:1d:b7:d7:c6:2f:6b:92:c9:18:c9:0b:68:85:
9c:77:ca:e9:fd:b3:14:f8:24:90:a0:d6:b5:0c:5d:
c8:5f:5c:92:a6:fd:f1:97:16:ac:84:51:ef:e8:bb:
df:48:8a:e0:98:a7:c7:6a:dd:25:99:f2:ca:64:20:
73:af:a2:0d:14:3a:f4:03:d1
Exponent: 65537 (0x10001)


So, we have a 1024 bit RSA key, which is a bit low, but not necessarily weak and still out of reach of a brute force factorization. It features a common exponent: 65537, so far nothing fishy.

We can now look into the challenge.py file. It has very few functions:

def ReadPrivateKey(filename): ...



And a rather simple main function:

def main():
return Challenge(private_key, sys.stdin.buffer, sys.stdout.buffer)


The ReadPrivateKey function is using the Cryptography.io library to read a private key file in a standard way, however the RsaDecrypt function is a bit more interesting:

def RsaDecrypt(private_key, ciphertext):
assert (len(ciphertext) <=
(private_key.public_key().key_size // 8)), 'Ciphertext too large'
return pow(
int.from_bytes(ciphertext, 'big'),
private_key.private_numbers().d,
private_key.public_key().public_numbers().n)


As you can see, it is performing the RSA decryption function manually by taking a ciphertext as bytes in big endian format, converting it into an integer $$c$$ and computing using pow the RSA decryption $$c^d \bmod{n}$$, for $$d$$ the private exponent and $$n$$ the RSA modulus.

So, this tells us two things:

1. the ciphertext is in big endian.
2. no padding was used and it is textbook RSA.

As you all know textbook RSA without any padding has a lot of shortcomings. It is now time to dig into the bigger Challenge function called by the main:

def Challenge(private_key, reader, writer):
try:
for rounds in range(100):
p = [m0, m1][dice & 1]
k = random.randint(0, 2)
c = (ord(p) + k) % 2
writer.write(bytes((c,)))
writer.flush()
return 0

except Exception as e:
return 1


So, it first reads 2 bytes from the reader (which is sys.stdin.buffer in the main) and stores them as m0 and m1 respectively. It then proceeds to read the ciphertext from the same reader as being $$\frac{1024}{8}=128$$ bytes. That ciphertext gets decrypted into the dice variable and then for 100 rounds, it generates a random value $$k\in\{0,1,2\}$$, adds it to the Unicode code point of the byte selected by [m0, m1][dice & 1], reduces it modulo $$2$$, and sends it back on the writer (which is sys.stdout.buffer in the main).

That's definitively fishy. So what's really going on there?

We get 100 times the value $$(\operatorname{ord}(p) + k) \bmod{2}$$, with a different, random $$k\in\{0,1,2\}$$... But this actually leaks information about the value of $$\operatorname{ord}(p)$$, because the operation $$k \bmod{2}$$ does not result in a uniform distribution: indeed, we are expecting twice as many $$0$$s as $$1$$s, because when $$k$$ is equal to both $$0$$ and $$2$$, it is congruent to $$0 \bmod 2$$.
So, if $$(\operatorname{ord}(p) + k) \bmod{2}$$ gives us more 0s than 1s, we can be pretty sure that $$\operatorname{ord}(p)\equiv 0 \bmod{2}$$. On the other hand, if $$\operatorname{ord}(p)\equiv 1 \bmod{2}$$, the addition with $$k$$ would shift the value and we would expect to get twice as many $$1$$s as $$0$$s on average.

So, we have an information leak depending on the number of $$0$$s and $$1$$s that this Challenge function writes to the writer. But what information do we actually get? We learn the parity of the value $$\operatorname{ord}(p)$$, but what's $$p$$ again? Well, it is set just above as being p = [m0, m1][dice & 1], so it builds a list made of m0 and m1 (which are both inputs we provide in the first part of the Challenge function), and it sets $$p$$ as being the element at index dice & 1. And as mentioned above, dice is the integer resulting from the RSA decryption operation. So, dice & 1 is equal to the bitwise AND of the plaintext and $$1$$, which is basically equal to $$0$$ if the least significant bit (LSB) of the plaintext is $$0$$ and is equal to $$1$$ if the LSB of the plaintext is equal to $$1$$ as well, since the AND truth table goes as follows:

$\begin{array}{c|c||c} x & y & x\land y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \end{array}$

Now, this means that we have in the end an oracle telling us what's the value of the LSB of the decrypted ciphertext!

And while we all know the famous padding oracle attacks against RSA PKCS and RSA OAEP, which are relying on the most significant bit of the padded plaintext to decrypt a ciphertext, there also exists an attack exploiting the LSB. This is done by iteratively narrowing down the range of possible plaintext down to just one. At each iteration, we cut down the remaining possibilities, and while we quickly converge to the corrects most significant bytes, we still have to perform all the 1024 iterations to be able to pinpoint just one plaintext.

So, we now have all the building blocks we need to build an attack against this service.

Firstly, we have a decryption oracle that we can query.

The oracle will then decrypt the ciphertext and give us a lot of "somewhat" randomized answers, based on values we chose, e.g. 0 and 1.
And we can deduce the right value out of the oracle's answers, allowing us to conduct the "LSB oracle" attack against RSA.

In the end, the following code, inspired from a previous CTF and from the above linked page explaining this LSB-oracle, allows to query the netcat server, provide it with two consecutive bytes (we need to have a different parity betwen m0 and m1, otherwise we don't leak any information) and interpret the oracle's answer in order to rebuild the actual plaintext out of its answers.

And guess what? The flag was at the very end of the plaintext, so that it wouldn't leak before we had recovered the whole plaintext. ;)

All right, time to go to bed, now. I'll be following tomorrow with another write-up for the "DM collision" challenge.

b"\x00\x02a\xb4\x02\x0c3\xd2.:\xe7\x9eB'\xf2\xd5\x1c7\xe4\r\xcd\xac\xd8\x7f\x02_L\x84q\xea\x8c\xb4\x1d}\x82\x90g\x170\x00.oO CTF{h3ll0__17_5_m3_1_w45_w0nd3r1n6_1f_4f73r_4ll_7h353_y34r5_y0u_d_l1k3_70_m337} Oo."

import decimal
from socket import socket
import sys

keysize =1024

# from the flag.txt file read in big endian
c  = 119217309829194046348522363146747304937295954356236084387775000685800908617629318417110348204755477939424239274691807984378678076883364880141274514261178345172539063655348321636037215361924332171794520555913635132454086644353250127828076172493408546050292999559913199868787754769594591462488323013485213560406
# extracted with openssl rsa -pubin -text -modulus -noout < key_pub.pem
e = 65537

def oracle(c,i):
# I'm opening a new socket each time
r = socket()
r.connect(('perfect-secrecy.ctfcompetition.com', 1337))
# we first send our two bytes m0 and m1
tosend = b'\x00'
if r.send(tosend) == 0:
print("borken socket")
sys.exit()
tosend = b'\x01'
if r.send(tosend) == 0:
print("borken socket")
sys.exit()
# we now send the ciphertext c
tosend = c.to_bytes(keysize//8,'big')
print("step", i,"sending:", tosend)
if r.send(tosend) == 0:
print("borken socket")
sys.exit()
# and now we listen to our oracle
zero,one = 0,0
for rounds in range(100):
a= r.recv(1)
#print ("> ", a)
if a == b'\x00':
zero += 1
if a == b'\x01':
one+=1

print("zero:",zero, "\n one:", one)
if zero > one:
return 0
return 1

# Encrypt the integer 2 to exploint RSA's malleability
c_of_2 = pow(2,e,n)

def partial(c,n):
k = n.bit_length() # that is 1024
decimal.getcontext().prec = k # allows for 'precise enough' floats
lower = decimal.Decimal(0)
upper = decimal.Decimal(n)
for i in range(k): # will take a while
possible_plaintext = (lower + upper)/2
if not oracle(c, i):
upper = possible_plaintext # plaintext is in the lower half
else:
lower = possible_plaintext # plaintext is in the upper half
c=(c*c_of_2) % n # multiply y by the encryption of 2 again, thanks to RSA's malleability
# in the end we got the plaintext!
return int(upper).to_bytes(128,'big')

# and we exploit it
print ("\nAnd we got:", partial((c*c_of_2)%n,n))
`