david wong

Hey ! I'm David, a security consultant at Cryptography Services, the crypto team of NCC Group . This is my blog about cryptography and security and other related topics that I find interesting.

Git client vulnerability December 2014

A new vulnerability has been discovered on the git client. See Github's announcement

Repositories hosted on github.com cannot contain any of the malicious trees that trigger the vulnerability because we now verify and block these trees on push.

The official announcement and the updated and fixed version of git is here.

We used to allow committing a path ".Git/config" with Git that is running on a case sensitive filesystem, but an attempt to check out such a path with Git that runs on a case insensitive filesystem would have clobbered ".git/config", which is definitely not what the user would have expected. Git now prevents you from tracking a path with ".Git" (in any case combination) as a path component.

More information about the vulnerability here

Git maintains various meta-information for its repository in files in .git/ directory located at the root of the working tree. The system does not allow a file in that directory (e.g. .git/config) to be committed in the history of the project, or checked out to the working tree from the project. Otherwise, an unsuspecting user can run git pull from an innocuous-looking-but-malicious repository and have the meta-information in her repository overwritten, or executable hooks installed by the owner of that repository she pulled from (i.e. an attacker).

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Last exam of my life December 2014

And I just passed the last exam of this semester, which should be the last exam of my life =) Now is time to take a few days to relax and eat nice food because it will soon be christmas ^^ (or holidays, as I heard some american say to avoid saying christmas).

A few interesting things I had to do during my exams these last few days:

  • Simple Power Analysis (SPA). Guess what algorithm is used from smartcards' traces and calculate the exponent if it's a binary exponentiation

spa

In the picture you can see two patterns, "1" is represented by two operations in the algorithm, and one of them is squaring which happens also when you have a "0" in your exponent's binary representation. So following the computations revealed by the power trace you can guess the binary representation of the exponent.

I had to read this article explaining two malloc implementations and their vulnerabilities. GNU Lib C (used in Linux) and System V AT&T (used in Solaris, IRIX). I knew the double chained list system but System V uses a different approach: binary tree and also a realfree function that completes the free function.

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Bruce Schneier December 2014

Schneier just gave a talk on security at Qcon in San Francisco. It was recorded and you can watch that here.

It's a high level talk that brings a lot of interesting points, like how much do we trust our devices, how companies are often doing very bad things in term of security, ...

The psychologist he's talking about is Daniel Kahneman, who won the nobel prize in economics for his work on Prospect Theory.

Prospect theory is a behavioral economic theory that describes the way people choose between probabilistic alternatives that involve risk, where the probabilities of outcomes are known. The theory states that people make decisions based on the potential value of losses and gains rather than the final outcome, and that people evaluate these losses and gains using certain heuristics.

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What might have been going on at Mtgox December 2014

I ran into an old post from nullc (Greg Maxwell one of the core Bitcoin developer) and it's interesting how small details might have been the fall of Mtgox.

First. You can't spend bitcoins you just mined.

Freshly generated Bitcoins (from mining) can not be spend until they are at least 100 blocks deep in the blockchain. This prevents the funds from vanishing forever if the chain reorgs.

see chain reorganization.

The term "blockchain reorganization" is used to refer to the situation where a client discovers a new difficultywise-longest well-formed blockchain which excludes one or more blocks that the client previously thought were part of the difficultywise-longest well-formed blockchain. These excluded blocks become orphans.
Chain reorganization is a client-local phenomenon; the entire bitcoin network doesn't "reorganize" simultaneously.

see orphan block.

An orphan block is a well-formed block which is no longer part of the difficultywise-longest well-formed blockchain.
The block reward in an orphaned block is no longer spendable on the difficultywise-longest well-formed blockchain; therefore whoever mined that block does not actually get the reward (or the transaction fees). This phenomenon must be taken into account by mining pools that use any payout strategy other than "proportional".

And here is a misunderstand of the padding of ECDSA (Elliptic Curve version of the Signature Scheme DSA) that might have be the problem:

This issue arises from several sources, one of them being OpenSSL's willingness to accept and make sense of signatures with invalid encodings. A normal ECDSA signature encodes two large integers, the encoding isn't constant length— if there are leading zeros you are supposed to drop them.
It's easy to write software that assumes the signature will be a constant length and then leave extra leading zeros in them.

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Airbus crypto challenge write-up 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

crypto1

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.

and

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.

rfc2633

Certificates

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

crypto2

We see that alice, bob and charly use the same rsa exponent (3).

Reminder: RSA

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), (p-1)(q-1) being 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 2m.

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 challenge1 file:

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 01D4CE3AF4D17ABB.

Reminder: DES-EDE3-CBC

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).

cbc

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: We have `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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

## 9F9D51BC70EF21CA5C14F307980A29D8 (Bob)
c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
n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

## A6D4EF4DD38B1BB016D250C16A680470 (Charly)
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)[1]
n1n3_ = xgcd(n1n3, n2)[1]
n1n2_ = xgcd(n1n2, n3)[1]

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)

Note:

  • 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:

0001ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff004f8957408f0ea202c785b95e206b3ba8da3dba7aea08dca1

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.

rfc 2315

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 :)

comments (6)

Is there any particular reason to use Diffie-Hellman over RSA for key exchange? December 2014

I was wondering why RSA was used in the SSL handshake, and why Diffie-Hellman was used instead in a Perfect Forward Secrecy scheme.

http://security.stackexchange.com/questions/35471/is-there-any-particular-reason-to-use-diffie-hellman-over-rsa-for-key-exchange

There is, however, an advantage of DH over RSA for generating ephemeral keys: producing a new DH key pair is extremely fast (provided that some "DH parameters", i.e. the group into which DH is computed, are reused, which does not entail extra risks, as far as we know). This is not a really strong issue for big servers, because a very busy SSL server could generate a new "ephemeral" RSA key pair every ten seconds for a very small fraction of his computing power, and keep it in RAM only, and for only ten seconds, which would be PFSish enough.

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Transform your messages into spam! December 2014

When you encrypt your mail through PGP or GPG it's great. But people can tell you're sending an important email. What if you could encrypt your message to something innocent? This is what spammimic does. It transforms your message into a spam message so no one can guess it's a legit message! This idea is so neat.

There is tons of spam flying around the Internet. Most people can't delete it fast enough. It's virtually invisible. This site gives you access to a program that will encrypt a short message into spam. Basically, the sentences it outputs vary depending on the message you are encoding. Real spam is so stupidly written it's sometimes hard to tell the machine written spam from the genuine article.

The encrypted messages look like that:

Dear Friend ; Thank-you for your interest in our publication 
  . If you no longer wish to receive our publications 
  simply reply with a Subject: of "REMOVE" and you will 
  immediately be removed from our club ! This mail is 
  being sent in compliance with Senate bill 1626 ; Title 
  3 , Section 308 . THIS IS NOT MULTI-LEVEL MARKETING 
  . Why work for somebody else when you can become rich 
  as few as 10 WEEKS ! Have you ever noticed more people 
  than ever are surfing the web plus nearly every commercial 
  on television has a .com on in it ! Well, now is your 
  chance to capitalize on this . We will help you process 
  your orders within seconds and deliver goods right 
  to the customer's doorstep ! You are guaranteed to 
  succeed because we take all the risk . But don't believe 
  us ! Prof Simpson who resides in Illinois tried us 
  and says "Now I'm rich, Rich, RICH" . This offer is 
  100% legal ! We BESEECH you - act now . Sign up a friend 
  and you'll get a discount of 20% . God Bless ! Dear 
  Friend , Especially for you - this amazing news ! We 
  will comply with all removal requests . This mail is 
  being sent in compliance with Senate bill 1618 ; Title 
  2 , Section 301 . This is not multi-level marketing 
  ! Why work for somebody else when you can become rich 
  in 58 weeks ! Have you ever noticed people will do 
  almost anything to avoid mailing their bills plus most 
  everyone has a cellphone ! Well, now is your chance 
  to capitalize on this ! We will help you SELL MORE 
  and increase customer response by 170% ! You are guaranteed 
  to succeed because we take all the risk . But don't 
  believe us . Mr Jones of Georgia tried us and says 
  "Now I'm rich many more things are possible" ! This 
  offer is 100% legal ! So make yourself rich now by 
  ordering immediately ! Sign up a friend and you'll 
  get a discount of 60% . Best regards !
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Real Life side-channels attacks December 2014

Some funny slides from Vitaly Shmatikov on side channels attacks: http://www.cs.utexas.edu/~shmat/courses/cs361s/sidechannels.pdf

So you can tell what someone is typing just by analyzing the sound of the fingers on the keyboard, from a certain distance.

If you observe someone typing at his computer from an outside window, you can analyze the reflections in many objects (glass teapots, plastic bottles, spoons!!! and even eyes).

Like we weren't worried enough.

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Network/Software Security December 2014

I was trying to find some info about the Heap and malloc (for the level 14 of microcorruption) when I ran into some very good videos from the Infosec Institute. I cannot find the name of the speaker but damn he's so good I just lost 2 hours of my life just watching his videos about nmap, pentesting, metaspoilt, and so on...

Here is his video on Heap Overflow:

And his talks are on several different youtube channels. I don't know how legit this is, and if someone can find the name of that guy I would love to know more about him. More about him: Advanced Recon, Advanced Exploitation, and so on...

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Pandoc : Markdown -> LaTeX December 2014

I just turned in my cryptanalysis project: A Linear Cryptanalysis of A5/2 with Sage. Had to write a rapport in LaTeX.

Now I have to finish the challenges over at Microcorruption and produce a write up in LaTeX as well.

Well LaTeX is awful as a writing syntax. I'd rather focus on writing with John Gruber's excellent markdown syntax and then later convert it to LaTeX. And Pandoc does just that! It's magic. Now that I have all my .md files I concat them with a quick python script

output = ""
for ii in range(1,15):
    markdown = open(str(ii) + ".md", "r")
    output += markdown.read()
    output += "\n\n"
    markdown.close()

output_file = open("rapport.md", "w")
output_file.write(output)
output_file.close()

A bit of Pandoc magic and voilà ! I have a beautiful .tex

Now let's finish Microcorruption (or at least try :D it's getting pretty hard).

PS: I use Markdown for everything. This blog is written in Markdown and then converted to HTML. I'm also writing a book in Markdown. Well Markdown is awesome.

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Forward Secrecy December 2014

I was asked during an interview how to build a system where Alice could send encrypted messages to Bob. And I was asked to think outloud.

So I imagined a system where both Alice and Bob would have a set of (public key, private key). I thought of RSA as they all use RSA but ECIES (Elliptic Curve Integrated Encryption Scheme) would be more secure for smaller keys. Although here ECIES is not a "pure" asymmetric encryption scheme and Elgamal with ECs might be better.

Once one wants to communicate he could send a "hello" request and a handshake protocol could take place (to generate a symmetric encryption key (called a session key in this case)).

I imagined that two session keys would be generated by each end. Different set of keys depending on the direction. One for encrypting the messages and one for making the MAC (that would be then appended to the encrypted message. So we EtM (Encrypt-then-Mac)).

Then those keys would be encrypted with the public signature of the other one and sent over the wire like this. And Let's add a signature so we can get authentication and they also won't get tampered. Let's use ECDSA (Elliptic Curve Digital Signature Algorithm) for that.

Although I'm wondering if two symmetric keys for encrypting messages according to the direction is really useful.

I was then asked to think about renewal of keys. Well, the public keys of Alice and Bob are long term keys so I didn't bother with that. About the symmetric keys? What about their TTL (Time To Live)?

My interviewer was nice enough to give me some clues: "It depends on the quantity of messages you encrypt in that time also."

So I thought. Why not using AES in CTR mode. So that after a certain number of iteration we would just regenerate the symmetric keys.

I was then asked to think about Forward Secrecy.

Well I knew the problem it was solving but didn't know it was solving it.

Mark Stamp (a professor of cryptography in California (how many awesome cryptography professors are in California? Seriously?)) kindly uploaded this video of one of his class on "Perfect Forward Secrecy".

So here the trick would be to do an Ephemeral Diffie-Hellman to use a different session key everytime we send something.

This EDH would of course be encrypted as well by our public key system so the attacker would first need to get the private keys to see that EDH.

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