Airbus crypto challenge write-up posted December 2014
Airbus made a "private" challenge called « Trust the future » and accessible only by some selected schools (epitech, insa, and others). I wasn't invited to participate but there was a "crypto" challenge I thought was interesting. Since the challenge just finished I'm posting the write up.
Crypto challenge #1
We have 4 certificates and a
challenge1 file that seems to be a s/mime file of a pkcs#7 enveloped-data object.
2.4.3 EnvelopedData Content Type
This content type is used to apply privacy protection to a message. A sender needs to have access to a public key for each intended message recipient to use this service. This content type does not provide authentication.
3.2 The application/pkcs7-mime Type
The application/pkcs7-mime type is used to carry CMS objects of several types including envelopedData and signedData. The details of constructing these entities is described in subsequent sections. This section describes the general characteristics of the application/pkcs7-mime type.
We dump the info of each certificates in human readable format, openssl has commands for that (I think certtool does as well, but I'm on windows using cmder and openssl is the one included).
openssl x509 -in alice.crt -text -noout -out alice.crt.txt
We see that alice, bob and charly use the same rsa exponent (3).
If you're familiar with RSA (and it's highly probable you are if you read this blog) you can skip this section.
RSA is an asymmetric encryption scheme (also used as a signature). It works by generating a set of private key/public key, the private key is of course kept private and the public key is publicly disclosed. If someone wants to send us a private message he can encrypt it with our public key and we will be able to decrypt it with the private key. The public key is the pair of number (n, e) where n is called the modulus and e is called the exponent. If we want to encrypt a message m with the public key we "basically" do
c = m^e modulo n and send c. To decrypt it we use our private key d like this:
m = c^d modulo n.
The math behind this is that n is generated from two secret primes p and q (big enough)
n = p x q and
d = e^-1 modulo (p-1)(q-1),
phi(n) being the order of the multiplicative group
Z/nZ. The security comes from the fact that it's Computationally Hard to find the inverse of e if we don't know p and q. By the way, Heartbleed (a recent attack on openssl) led to finding one of the prime, thus the entire decomposition of n.
Textbook RSA vs real life RSA
This is all theory. And in practice we have to go through several steps to encrypt an ascii message, make sure it is of length lesser than the modulus, make sure the modulus is big enough, etc...
Textbook RSA is also deterministic and thus not semantically secure (see my previous post) + it is malleable: imagine you intercept c, and of course you know (n, e) (the public key). You could compute
c' = 2^e * c = 2^e * m^e = (2m)^e modulo n, this would correctly decrypt as
Thus, to counter those in practice, RSA Encrytion uses padding (usually OAEP) to make it probabilist and not malleable.
Let's go back to our challenge
We open our
MIME-Version: 1.0 Content-Disposition: attachment; filename="smime.p7m" Content-Type: application/x-pkcs7-mime; smime-type=enveloped-data; name="smime.p7m" Content-Transfer-Encoding: base64 MIIy1wYJKoZIhvcNAQcDoIIyyDCCMsQCAQAxggQ0MIIBYgIBADBKMDYxCzAJBgNV BAYTAkZSMQ4wDAYDVQQHEwVQYXJpczEXMBUGA1UEAxQOY2FAZXhhbXBsZS5jb20C EGOE4rIYS8v1jszxDKemVjwwDQYJKoZIhvcNAQEBBQAEggEAweI1fG/FPxzF4Odu sSJL6PJOiDklHPlUqYCQxFSfG6+3vLEAbdKpgtVsHS0+a0IhItAfeNoAmXdreJFi 6M6U7j0ee4iqgXXbuG8vSsZTYbyUmzuQRgdByu5vGr3FvWxSlvvI8tr/d/cRDqMt
To read that we need to extract the pkcs7 object and parse it. Openssl allows us to do this:
openssl smime -in challenge1 -pk7out -out challenge1.p7m openssl asn1parse -text -in challenge1.p7m
We get an annoying dump of info to read. With three of those things:
95:d=6 hl=2 l= 16 prim: INTEGER :6384E2B2184BCBF58ECCF10CA7A6563C 113:d=5 hl=2 l= 13 cons: SEQUENCE 115:d=6 hl=2 l= 9 prim: OBJECT :rsaEncryption 126:d=6 hl=2 l= 0 prim: NULL 128:d=5 hl=4 l= 256 prim: OCTET STRING [HEX DUMP]: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
Which means the same message was sent to three recipients, identified by their serial number which we recognize as being our alice, bob and charly.
We also get this at the end:
1110:d=4 hl=2 l= 9 prim: OBJECT :pkcs7-data 1121:d=4 hl=2 l= 20 cons: SEQUENCE 1123:d=5 hl=2 l= 8 prim: OBJECT :des-ede3-cbc 1133:d=5 hl=2 l= 8 prim: OCTET STRING [HEX DUMP]:01D4CE3AF4D17ABB
Which means that the data sent (after this dump) is encrypted by 3DES version 3 (three different keys) in CBC mode with an IV
I like to put reminders like this so you don't have to switch to Wikipedia if you don't remember what are those letters.
DES (Data Encryption Standard) is the famous no-longer-used block cipher (because it was broken ages ago). EDE3 short for the third version of the Triple DES block cipher (that is still considered secure today, it was a response to DES no longer being secure) which uses 3 different keys. Encrypting is done like this:
- we encrypt with key1
- then we decrypt with key2
- then we encrypt again with key3
E(k3, D(k2, E(k1, M)))
Hence the triple DES.
CBC is a mode of operation. A block cipher can only encrypt/decrypt blocks of a certain size (64bits with DES). If you want to do more (or less) you have to use a mode of operation (and a padding system).
Chinese Remainder Theorem
Here the interesting thing is that the same message was sent to three different recipients, encrypted with the same exponent (3). Let's write down the informations we have:
c1 = m^3 modulo n1 c2 = m^3 modulo n2 c3 = m^3 modulo n3
c1 being the encrypted message sent to Alice, n1 being Alice's modulus, and so on...
We have a system with one unknown: the message. The Chinese Remainder Theorem works in a similar fashion to Lagrange Interpolation (anecdote time: it is used in Shamir's Secret Sharing).
So that we have:
m^3 = c1 * n2 * n3 * ((n2 * n3)^-1 [n1]) + c2 * n1 * n3 * ((n1 * n3)^-1 [n2]) + c3 * n1 * n2 * ((n1 * n2)^-1 [n3]) modulo n1 * n2 * n3
A brief explanation:
`c1 = m^3 modulo n1, to place it in a formula modulo
n1 * n2 * n3 we have to cancel it when it's modulo n2 or modulo n3. How to make something congruent to zero when its modulo n2 or n3 ? Make it a multiple of n2 or n3. So we multiply c1 with n2 and n3. But then when it will be modulo n1 we will have the value
c1 * n2 * n3 which is not correct (
c1 = m^3 modulo n1 !). So let's cancel the n2 and n3 with their inverse modulo n1. We then have
c1 * n2 * n3 * ((n2 * n3)^-1 [n1]). We do this with all the equations to find the bigger equation. This is the Chinese Remainder Theorem. Simple no?
And this result is even more useful since we know that:
m < n1 m < n2 m < n3 => m^3 < n1*n2*n3
Of course if m was greater than one of the modulus then it would decrypt incorrectly. So what we have is:
m^3 = something modulo n1*n2*n3 => m^3 = something
That's right, we can get rid of the modulo. We then do a normal cubic root and we find m.
Here's the quick python code I hacked together for this:
(by the way we can quickly get the modulus of each recipients with openssl:
openssl x509 -in alice.crt -modulus)
## 6384E2B2184BCBF58ECCF10CA7A6563C (Alice) c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n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ob) c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n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harly) c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n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base16 -> base10 c1 = int(c1, 16) c2 = int(c2, 16) c3 = int(c3, 16) n1 = int(n1, 16) n2 = int(n2, 16) n3 = int(n3, 16) ## extended euclide algorithm def xgcd(a,b): """Extended GCD: Returns (gcd, x, y) where gcd is the greatest common divisor of a and b with the sign of b if b is nonzero, and with the sign of a if b is 0. The numbers x,y are such that gcd = ax+by.""" prevx, x = 1, 0; prevy, y = 0, 1 while b: q, r = divmod(a,b) x, prevx = prevx - q*x, x y, prevy = prevy - q*y, y a, b = b, r return a, prevx, prevy ## chinese remainder formula n2n3 = n2 * n3 n1n3 = n1 * n3 n1n2 = n1 * n2 n2n3_ = xgcd(n2n3, n1) n1n3_ = xgcd(n1n3, n2) n1n2_ = xgcd(n1n2, n3) m3 = c1 * n2n3 * n2n3_ + c2 * n1n3 * n1n3_ + c3 * n1n2 * n1n2_ m3 = m3 % (n1n2 * n3) print(m3) from decimal import * getcontext().prec = len(str(m3)) x = Decimal(m3) power = Decimal(1)/Decimal(3) answer = x**power ranswer = answer.quantize(Decimal('1.'), rounding=ROUND_UP) diff = x - ranswer**Decimal(3) if diff == Decimal(0): print("x is the cubic number of", ranswer) else: print("x has a cubic root of ", answer)
- The xgcd function is included in sage but here I use Python so I included it in the code.
- We need to use the decimal package to calculate the cubic root because our number is too big.
We then get this big ass number that we convert to hexadecimal (
hex(number) in python). This yields:
We refer once more to the RFCs
8.1 Encryption-block formatting
A block type BT, a padding string PS, and the data D shall be formatted into an octet string EB, the encryption block.
EB = 00 || BT || PS || 00 || D . (1)
The block type BT shall be a single octet indicating the structure of the encryption block. For this version of the document it shall have value 00, 01, or 02. For a private- key operation, the block type shall be 00 or 01. For a public-key operation, it shall be 02.
The padding string PS shall consist of k-3-||D|| octets. For block type 00, the octets shall have value 00; for block type 01, they shall have value FF; and for block type 02, they shall be pseudorandomly generated and nonzero. This makes the length of the encryption block EB equal to k.
We have our 3DES key:
4f8957408f0ea202c785b95e206b3ba8da3dba7aea08dca1 to use.
Let's get the hexdump the end of the file (you can use commandline utilities like base64, hexdump, dd and xdd):
openssl smime -in challenge1 -pk7out > b64file` base64 -d b64file > hexfile hexdump -s 1135 hexfile dd xdd
And finally decrypt our
encrypted file with openssl since it provides a command for that:
openssl des-ede3-cbc -d -iv 01D4CE3AF4D17ABB -K 4f8957408f0ea202c785b95e206b3ba8da3dba7aea08dca1 -in encrypted
Voila ! That was really fun :)
This is a Bleichenbacher-style attack that shows up in all sorts of competitions (like PicoCTF, Matasano). It also broke Firefox's certs a few years ago.
Aren't all the attacks from Bleinchenbacher anyway : )) ? How can we call one type of attack a Bleichenbacher-style attack.
I am aware of two attacks that are attributed to Bleichenbacher: one is an adaptive chosen-ciphertext attack on a padding oracle, the other is related to small e / small plaintext, in which case you can just extract the cubic root from the ciphertext.
The case discussed in this post, however, is Håstad's Broadcast Attack, demonstrated in 1988 (see http://www.csc.kth.se/~johanh/rsalowexponent.pdf). Nowadays it is prevented by random paddings.
To follow up on that: yes, there are more attacks from Bleichenbacher, and no, I don't why someone mentioned the cubic root thing being his idea. Still, his attacks do not apply here, and this attack is quite clearly Håstad.
My question if off-topic, but do you have the GPG key used to decrypt the "challenges.gpg" file?
I'm sorry but what challenges.gpg file?
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