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If you are a student looking for an internship in Cryptography, that might be the right post for you.
Those past few months, before landing an internship at Cryptography Services, I applied in many places around the world and went through several interviews. Be it irl interviews, phone interviews, video interviews... I traveled to some remote places and even applied to companies I really did not want to work at. But those are important to get you the experience. You will be bad at your first interviews, and you don't want your first interviews to be the important ones.
In this post I'll talk about how I got interviews, how I prepared and dealt with them, how I got propositions. I might be wrong on some stuff but I'm sure this will help some students anyway :)
Think about them
Put yourself in the employer's shoes, he doesn't want to read a cover letter, he doesn't want to spend more than a few seconds on a resume, he doesn't want to hurt his eyes on horrific fonts, too many colors and typos.
- Ditch all those premade themes.
- Learn how to do something pretty with Word/LaTeX/Indesign/CSS...
- Okay, if you're really bad at building a nice layout, maybe you should use a premade theme, but then use a very simple/plain one.
- No more than one page (you're a student, don't be cocky).
- Use keywords.
- Be simple. Don't write something that belongs in the cover letter.
There is a lot of theory on C.V. making. I'm sure you can find better resources on how fabricate your resume.
- Be concise (I use bullet points for my cover letters and it saves the reader time)
- Write well, make your friends read it, take a break and re-read it.
- Be formal, polite, etc...
- Don't hesitate to contact them again if they don't answer
- Use my application 3pages to write your cover letter (shameless plug :D)
Side projects are important, like really important. If you don't have anything to show you're just as good as the next student, (almost) no one will ask for your grades.
If you don't know what side project to work on, check Cryptopals (I'll be working with the folks who made this by the way :P)
Also if you're looking for something in development, applied crypto, get a github. Many employers will check for your github.
And, you know, you could... you could write a blog. That's a nice way to write down what you're learning, to motivate you into reading more about crypto and to be able to show what you're doing.
Where to apply?
Check with your professors, they usually know a list of known companies that do crypto. In France there are Thales, Morpho, Airbus and many big companies that you might want to avoid, and also very good companies/start-ups like Cryptoexpert, Quarkslab, ... that you should aim for.
You can also ping universities. They will usually accept you without an interview but will rarely pay you. If you really want to do academic researches, you should check some good universities in Finland, France, California, Sydney, India... wherever you want to go.
If you want to do crypto or applied crypto you might want to look for a start-up/smaller companies as they are not too big and you might learn more with them. But even in big companies you might find sizable crypto teams (rarely more than 2 or 3), check Rootlabs, Cloudflare, Matasano...
Preparing for an interview is a really good opportunity to learn many things! If you know what the subject is about read more about it beforehand. If you don't know anything about the internship subject, read more about what your interviewers have done. If you don't know anything because you have been talking to a PR all along then you might want to rethink applying there.
In all the interviews I've done I just had one PR interview. I would advise you to refuse/avoid those, unless you really want to work for the company, as it's most always a waste of time /rant.
Most interviews in France were focusing on my side projects, whereas the ones I had in the US were way more technical. Younger and smaller companies did ask me things about me and my hobbies (one even asked me if I was doing some sport! I thought that was a neat question :))
Now, here's an unsorted list of questions I got asked:
- Explain whitebox cryptography
- How do you obfuscate a program?
- How can you tell a cipher is secure?
- How do stream ciphers work?
- What is a LFSR?
- name different stream/block ciphers
- Explain Simple Power Analysis
- Explain Differential Power Analysis
- Explain Differential Fault Analysis
- Any counter measures?
- Explain the Chinese Remainder Theorem
- How are points on an Elliptic Curve represented?
- What prevents me from signing a bad certificate?
- Imagine a system to send encrypted messages between two persons
- How to do it with Perfect Forward Secrecy?
- How does a compiler works?
- What is XSS?
- Explain Zero-knowledge with the discrete logarithm example
- You have a smartcard you can inject code on, what do you do to perform a DPA?
- You've recorded traces from the smartcard, do you have to do some precomputations on those traces before doing a DPA?
- Techniques to multiply points on an Elliptic Curve
- Example of homomorphic encryption
- Knapsack problem
An interview should be an interaction. You can cheat your way through by trying not to talk too much, but it still should be a conversation because eventually, if you get the job, you guys will be having real work conversations.
You should be a nice dude. Because nobody wants to work with a boring, elitist dude.
You should never correct the interviewer. This feels like a stupid advice but correcting someone that knows more than you, even if you are right, might lead to bad things... Wait to be hired for that.
You should never let the interviewer tell you something you already know. This is an occasion for you to shine.
If you can't answer a question, ask the interviewer how he would have answer, this is a good opportunity to learn something.
Eventually, ask questions about the job, the workplace, the city. Not only they will appreciate it, it is showing that you are interested in their job, but it's also nice for you to see if the company is the kind of place you would like to work at (it also makes the table turns).
I don't know if any of you were planning on bullshiting but we're in a technical field, avoid bullshiting!
I finally chose where I'll be doing my internship to finish my master of Cryptography: it will be at Cryptography Services (a new crypto team, part of NCC Group (iSEC Partners, Matasano, IntrepidusGroup)). I will be conducting researches and audits in the offices of Matasano in Chicago. I'm super excited and the next 6 months of my life should be full of surprise!
Cryptography Services is a dedicated team of consultants from iSEC Partners, Matasano, Intrepidus Group, and NCC Group focused on cryptographic security assessments, protocol and design reviews, and tracking impactful developments in the space of academia and industry.
I guess I'll soon have to create a "life in Chicago" category =)
I was trying to access the Journal of Cryptology on Springer but I had to pay. Thanks to __x86 I realized I had free access to Springer thanks to my university!
So this post is oriented to my fellow classmates. If any of you want to check something there, it's free for us! (Well until we graduate).
I was trying to check the papers that got Dan Boneh and Antoine Joux their Gödel prize:
I'm reading through A Key Recovery Attack on Discrete Log-based Schemes Using a Prime Order Subgoup which is a Small subgroup confinement attack.
It deals with stuff I had no knowledge of, like Schnorr's Signature that I talk about in a previous post, or like what I'm going to talk about now:
The Pohlig-Hellman Algorithm is a method to compute a Discrete Logarithm (which is a difficult problem) on a multiplicative group whose order is a smooth number (also called friable). Meaning its order can be factorized into small primes.
y = g^x mod p
ord_p(g) = p - 1
p - 1 = q_1^(i_1) * ... * q_j^(i_j)
Here y is the public key,
x is the secret key we're trying to compute.
The order of
g, our generator, is
p - 1 since p is prime.
p - 1 is smooth so it can be factorized into something like 2^2 * 3 * 7 (extremely convenient example!)
Following is an overview of the method, if you read an equation and feel like it comes from nowhere (and it should feel like that), I posted a very short paper containing the simple proofs of those bellow.
The idea that should come to your mind, if you're used to seeing that kind of problem, is that there might be a way to use the Chinese Remainder Theorem (abbreviated CRT) to our advantage. What if we could write
x modulo the factors of
p - 1 and then reconstruct the real
x with the CRT? Well, let's do just that!
x modulo a factor
p - 1 we can write
y^((p-1) / q) which we know and can compute, and is also equal to
g^(x_q * (p-1) / q)
(If you have a factor that appears multiple times in the prime decomposition of
p - 1 (for example
p - 1 = 2^5, then there is also a way to ease the computations by finding multiple unknowns (5 unknowns in our example))
We then have a discrete logarithm to compute, but a small one, that we can compute efficiently thanks to Shanks' method (baby-step giant-step) or Pollard's rho algorithm.
Then we have multiple ways of writing our
x (modulo the factors of
p - 1) and we find out what is
x with CRT (I explained this step in my airbus write-up here).
Proofs + Video
You can read through the Algorithm (or watch the video bellow), but I don't really like to do that since I'm not really good at memorizing something if I don't understand the nut and bolts of it. So here's a really good paper written by D. R. Stinson that demonstrate where the equations of the algorithm come from. And here's an explanation + example of the algorithm:
Interactive Protocols are basically a discussion between a Prover and a Verifier where the Prover has some piece of information he wants to prove, without giving out the information.
It is often illustrated with Peggy and Victor and their super tunnel.
Usualy it takes 3 steps:
- Prover sends a fixed value.
- Verifier sends a challenge.
- Prover answers the challenge.
The Verifier can then verify the answer based on the fixed value. If the answer is correct, the Verifier can assume the Prover knows what he's trying to prove. Sometimes you have to repeat the protocols multiple time to be sure, and not all problems have an Interactive Proof.
Classic examples of such proofs can be found on the Discrete Logarithm problem (we'll talk about that later) and the Hamiltonian Cycle problem.
Interactive Protocols are verified if they are :
- Complete: a Prover can successfully answer the challenge if he is honest.
- Sound : a dishonest Prover cannot convince the Verifier he knows the secret.
In the real definitions we use probabilities (an honest prover still has a small chance of making a mistake, a dishonest prover still has a small chance of convincing the Verifier).
We also often want a 3rd condition on our Interactive Protocols: we want it to be Zero-knowledge, no information about our secret should be leaked in this interaction.
Here are how you prove each one of them:
- Completeness: Can the Prover answer correctly thanks to his secret?
- Soundness: From the point of view of the Verifier. If the Prover can correctly answer two different challenges for the same fixed value (however he crafted the answers and the fixed value), does it mean that he must know the secret then?
- Zero-Knowledgeness: If you see a transcript of a recorded instance of this interaction, will you learn anything about the secret? (See if you can create fake transcripts)
There are also notions of weak Zero-knowledge, strong Zero-knowledge, dishonnest verifiers, etc...
But let's talk about something else.
Since we said that a recorded transcript of a past interaction has no value (if it is zero-knowledge), then we could assume that there is no way of proving something by showing an old transcript, or by showing a transcript with yourself.
Don't fool yourself! Yes we can. We do this by using hash functions that we deem random enough.
The idea is that, by replacing the Verifier by a random oracle, we cannot predict the challenges and we thus cannot craft a fake transcript (we like to use random oracles instead of hashes, to prove some scheme is secure).
a random oracle is an oracle (a theoretical black box) that responds to every unique query with a (truly) random response chosen uniformly from its output domain. If a query is repeated it responds the same way every time that query is submitted.
What is interesting is that this protocol was used in a Signature Scheme.
Interactive Proof of a Discrete Logarithm
The most famous academic example of Interactive Protocol is done using the Discrete Logarithm problem.
<g> = G, with
g of order
q. The Prover wants to show he knows
g^x = y.
- the Prover sends
t = g^e
- the Verifier sends a challenge
- the Prover sends
d = e + cx
The Verifier can then compute
y^c * t = g^(e + cx) and see if it equals
g^d = g^(e + cx)
A transcript would look like this:
(t, c, d)
Non-Interactive Proof of a Discrete Logarithm
Doing this with a non-interactive protocol, it would look like this:
(t, h(t), d) with
h a hash function.
This is what Schnorr's Signature is doing:
t = g^e
c = H(m || t)
d = e - x*c
he would then send
(c, d) as the signature, along the message
m. Which is basically a hash of the message with a proof that he knows the secret
To verify the signature you would use the public key
y = g^x to compute
y^c * g^d = t and then you would compute the hash. It would give you the proof that the signer knows
x (authentication, non-repudiation) and that the message hasn't been tampered (integrity).
So this is one way of using Non-interactive proofs!
A great FAQ, written by Marc Joye (Thomson R&D), on Whitebox Cryptography.
Thansk Tancrède for the link!
Q1: What is white-box cryptography?
A major issue when dealing with security programs is the protection of "sensitive" (secret, confidential or private) data embedded in the code. The usual solution consists in encrypting the data but the legitimate user needs to get access to the decryption key, which also needs to be protected. This is even more challenging in a software-only solution, running on a non-trusted host.
White-box cryptography is aimed at protecting secret keys from being disclosed in a software implementation. In such a context, it is assumed that the attacker (usually a "legitimate" user or malicious software) may also control the execution environment. This is in contrast with the more traditional security model where the attacker is only given a black-box access (i.e., inputs/outputs) to the cryptographic algorithm under consideration.
Q2: What is the difference with code obfuscation?
Related and complementary techniques for protecting software implementations but with different security goals include code obfuscation and software tamper-resistance. Code obfuscation is aimed at protecting against the reverse engineering of a (cryptographic) algorithm while
software tamper-resistance is aimed at protecting against modifications of the code.
All these techniques have however in common that the resulting implementation must remain directly executable.
Or as Francis Gabriel writes here
Code obfuscation means code protection. A piece of code which is obfuscated is modified in order to be harder to understand. As example, it is often used in DRM (Digital Rights Management) software to protect multimedia content by hiding secrets informations like algorithms and encryption keys.
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
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
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
Content-Disposition: attachment; filename="smime.p7m"
Content-Type: application/x-pkcs7-mime; smime-type=enveloped-data; name="smime.p7m"
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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
## 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
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
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)
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)
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
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 :)