Hey! I'm David, the author of the Real-World Cryptography book. I'm a crypto engineer at O(1) Labs on the Mina cryptocurrency, previously I was the security lead for Diem (formerly Libra) at Novi (Facebook), and a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.

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# OneRNG posted December 2014

I like how people make an extreme effort to create "sure" source of random numbers.

OneRNG has released a new usb source. Everything is opensource (open hardware, open software), you can even create your own by following instructions on their websites.

OneRNG collects entropy from an avalanche diode circuit, and from a channel-hopping RF receiver. It even has a “tinfoil hat” to prevent RF interference — you can remove the hat in order to visually verify the components being used.

Now I'm wondering who is using that and for what

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# Awesome nmap cheatsheet posted December 2014

High on Coffee has released a cheatsheet on nmap. Full of examples and tips. You can find it here

EDIT: There are also Linux Commands for Penetration Testers there. The blog seems pretty new and it already has really good content :)

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# Git client vulnerability posted 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.

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 posted 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

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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Josip Franjković found a vulnerability in one of the file uploader of facebook.

He described what he did here

basically he uploaded a zipped file of a symbolic link to /etc/passwd

ln -s /etc/passwd link
zip --symlinks test.zip link

And since uploaders are always a mess to secure. Facebook just replied displaying the content of what he thought was the unzipped resume.

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# Bruce Schneier posted 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 posted 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.

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 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.

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

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

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]:C1E2357C6FC53F1CC5E0E76EB1224BE8F24E8839251CF954A98090C4549F1BAFB7BCB1006DD2A982D56C1D2D3E6B422122D01F78DA0099776B789162E8CE94EE3D1E7B88AA8175DBB86F2F4AC65361BC949B3B90460741CAEE6F1ABDC5BD6C5296FBC8F2DAFF77F7110EA32D330D38DD2CA2FE13E785C86FE2210B58074C2DA5F440794BA023FC98B3D1E7DC979DBAC6672B5C19ABF4A91E21D5E474475BC09B78910D1F8E0290B38AE8D756E04D7F5EFBA64BFB5A0E96CD3DE1D82F609544A423F666D08B63262229687E1982BC8E424C7B5266B11A59036625F8E92C06740A3C9D8F3CE87FEB1F4444BC2039C8C6FF0AB9457D8AA63851ECF3C4AF1A2328FD

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).

### 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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

## 9F9D51BC70EF21CA5C14F307980A29D8 (Bob)

## A6D4EF4DD38B1BB016D250C16A680470 (Charly)

## 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)

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 :)