Category Archives: Cryptography

Modulus Switching in LWE

Posted on by

The Learning With Errors problem is the basis of a few cryptosystems, and a foundation for many fully homomorphic encryption (FHE) schemes. In this article I’ll describe a technique used in some of these schemes called modulus switching.

In brief, an LWE sample is a vector of values in $\mathbb{Z}/q\mathbb{Z}$ for some $q$, and in LWE cryptosystems an LWE sample can be modified so that it hides a secret message $m$. Modulus switching allows one to convert an LWE encryption from having entries in $\mathbb{Z}/q{Z}$ to entries in some other $\mathbb{Z}/q'{Z}$, i.e., change the modulus from $q$ to $q’ < q$.

The reason you’d want to do this are a bit involved, so I won’t get into them here and instead back-reference this article in the future.

LWE encryption

Briefly, the LWE encryption scheme I’ll use has the following parameters:

  • A plaintext space $ \mathbb{Z}/q\mathbb{Z}$, where $ q \geq 2$ is a positive integer. This is the space that the underlying message comes from.
  • An LWE dimension $ n \in \mathbb{N}$.
  • A discrete Gaussian error distribution $ D$ with a mean of zero and a fixed standard deviation.

An LWE secret key is defined as a vector in $ \{0, 1\}^n$ (uniformly sampled). An LWE ciphertext is defined as a vector $ a = (a_1, \dots, a_n)$, sampled uniformly over $ (\mathbb{Z} / q\mathbb{Z})^n$, and a scalar $ b = \langle a, s \rangle + m + e$, where $ e$ is drawn from $ D$ and all arithmetic is done modulo $ q$.

Without the error term, an attacker could determine the secret key from a polynomial-sized collection of LWE ciphertexts with something like Gaussian elimination. The set of samples looks like a linear (or affine) system, where the secret key entries are the unknown variables. With an error term, the problem of solving the system is believed to be hard, and only exponential time/space algorithms are known.

However, the error term in an LWE encryption encompasses all of the obstacles to FHE. For starters, if your message is $ m=1$ and the error distribution is wide (say, a standard deviation of 10), then the error will completely obscure the message from the start. You can’t decrypt the LWE ciphertext because you can’t tell if the error generated in a particular instance was 9 or 10. So one thing people do is have a much smaller cleartext space (actual messages) and encode cleartexts as plaintexts by putting the messages in the higher-order bits of the plaintext space. E.g., you can encode 10-bit messages in the top 10 bits of a 32-bit integer, and leave the remaining 22 bits of the plaintext for the error distribution.

Moreover, for FHE you need to be able to add and multiply ciphertexts to get the corresponding sum/product of the underlying plaintexts. One can easily see that adding two LWE ciphertexts produces an LWE ciphertext of the sum of the plaintexts (multiplication is harder and beyond the scope of this article). Summing ciphertexts also sums the error terms together. So the error grows with each homomorphic operation, and eventually the error may overtake the message, at which point decryption fails. How to deal with this error accumulation is 99% of the difficulty of FHE.

Finally, because the error can be negative, even if you store a message in the high-order bits of the plaintext, you can’t decrypt by simply clearing the low order error bits. In that case an error of -1 would result in a corrupted message. Instead, to decrypt, we round the value $ b – \langle a, s \rangle = m + e$ to the nearest multiple of $ 2^k$, where $k$ is the number of bits “reserved” for error, as described above. In particular, decryption will only succeed if the error is small enough in absolute value. So to make this work in practice, one must coordinate the encoding scheme (how many bits to reserve for error), the dimension of the vector $ a$, and the standard deviation of the error distribution.

Modulus switching

With a basic outline of an LWE ciphertext, we can talk about modulus switching.

Start with an LWE ciphertext for the plaintext $ m$. Call it $ (a_1, \dots, a_n, b) \in (\mathbb{Z}/q\mathbb{Z})^{n+1}$, where

$ \displaystyle b = \left ( \sum_{i=1}^n a_i s_i \right ) + m + e_{\textup{original}}$

Given $ q’ < q$, we would like to produce a vector $ (a’_1, \dots, a’_n, b’) \in (\mathbb{Z}/q’\mathbb{Z})^{n+1}$ (all that has changed is I’ve put a prime on all the terms to indicate which are changing, most notably the new modulus $ q’$) that also encrypts $ m$, without knowing $ m$ or $ e_{\textup{original}}$, i.e., without access to the secret key.

Failed attempt: why not simply reduce each entry in the ciphertext vector modulo $ q’$? That would set $ a’_i = a_i \mod q’$ and $ b’ = b \mod q’$. Despite the fact that this operation produces a perfectly valid equation, it won’t work. The problem is that taking $ m \mod q’$ destroys part or all of the underlying message. For example, say $ x$ is a 12-bit number stored in the top 12 bits of the plaintext, i.e., $ m = x \cdot 2^{20}$. If $ q’ = 2^{15}$, then the message is a multiple of $ q’$ already, so the proposed modulus produces zero.

For this reason, we can’t hope to perfectly encrypt $ m$, as the output ciphertext entries may not have a modulus large enough to represent $ m$ at all. Rather, we can only hope to encrypt something like “the message $ x$ that’s encoded in $ m$, but instead with $ x$ stored in lower order bits than $ m$ originally used.” In more succinct terms, we can hope to encrypt $ m’ = m q’ / q$. Indeed, the operation of $ m \mapsto m q’ / q$ shifts up by $ \log_2(q’)$ many bits (temporarily exceeding the maximum allowable bit length), and then shifting down by $ \log_2(q)$ many bits.

For example, say the number $ x=7$ is stored in the top 3 bits of a 32-bit unsigned integer ($ q = 2^{32}$), i.e., $ m = 7 \cdot 2^{29}$ and $ q’ = 2^{10}$. Then $ m q’ / q = 7 \cdot 2^{29} \cdot 2^{10} / 2^{32} = 7 \cdot 2^{29+10 – 32} = 7 \cdot 2^7$, which stores the same underlying number $ x=7$, but in the top three bits of a 10-bit message. In particular, $ x$ is in the same “position” in the plaintext space, while the plaintext space has shrunk around it.

Side note: because of this change to the cleartext-to-plaintext encoding, the decryption/decoding steps before and after a modulus switch are slightly different. In decryption you use different moduli, and in decoding you round to different powers of 2.

So the trick is instead to apply $ z \mapsto z q’ / q$ to all the entries of the LWE ciphertext vector. However, because the entries like $ a_i$ use the entire space of bits in the plaintext, this transformation will not necessarily result in an integer. So we can round the result to an integer and analyze that. The final proposal for a modulus switch is

$ \displaystyle a’_i = \textup{round}(a_i q’ / q)$

$ \displaystyle b’ = \textup{round}(b q’ / q)$

Because the error growth of LWE ciphertexts permeates everything, in addition to proving this transformation produces a valid ciphertext, we also have to understand how it impacts the error term.

Analyzing the modulus switch

The statement summarizing the last section:

Theorem: Let $ \mathbf{c} = (a_1, \dots, a_n, b) \in (\mathbb{Z}/q\mathbb{Z})^{n+1}$ be an LWE ciphertext encrypting plaintext $ m$ with error term $ e_\textup{original}$. Let $ q’ < q$. Then $ c’ = \textup{round}(\mathbf{c} q’ / q)$ (where rounding is performed entrywise) is an LWE encryption of $ m’ = m q’ / q$, provided $ m’$ is an integer.

Proof. The only substantial idea is that $ \textup{round}(x) = x + \varepsilon$, where $ |\varepsilon| \leq 0.5$. This is true by the definition of rounding, but that particular way to express it allows us to group the error terms across a sum-of-rounded-things in isolation, and then everything else has a factor of $ q’/q$ that can be factored out. Let’s proceed.

Let $ c’ = (a’_1, \dots, a’_n, b’)$, where $ a’_i = \textup{round}(a_i q’ / q)$ and likewise for $ b’$. need to show that $ b’ = \left ( \sum_{i=1}^n a’_i s_i \right ) + m q’ / q + e_{\textup{new}}$, where $ e_{\textup{new}}$ is a soon-to-be-derived error term.

Expanding $ b’$ and using the “only substantial idea” above, we get

$ \displaystyle b’ = \textup{round}(b q’ / q) = bq’/q + \varepsilon_b$

For some $ \varepsilon_b$ with magnitude at most $ 1/2$. Continuing to expand, and noting that $ b$ is related to the $ a_i$ only modulo $ q$, we have

$ \displaystyle \begin{aligned} b’ &= bq’/q + \varepsilon_b \\ b’ &= \left ( \left ( \sum_{i=1}^n a_i s_i \right ) + m + e_{\textup{original}} \right ) \frac{q’}{q} + \varepsilon_b \mod q \end{aligned}$

Because we’re switching moduli, it makes sense to rewrite this over the integers, which means we add a term $ Mq$ for some integer $ M$ and continue to expand

$ \displaystyle \begin{aligned} b’ &= \left ( \left ( \sum_{i=1}^n a_i s_i \right ) + m + e_{\textup{original}} + Mq \right ) \frac{q’}{q} + \varepsilon_b \\ &= \left ( \sum_{i=1}^n \left ( a_i \frac{q’}{q} \right) s_i \right ) + m \frac{q’}{q} + e_{\textup{original}}\frac{q’}{q} + Mq \frac{q’}{q} + \varepsilon_b \\ &= \left ( \sum_{i=1}^n \left ( a_i \frac{q’}{q} \right) s_i \right ) + m’ + e_{\textup{original}}\frac{q’}{q} + Mq’ + \varepsilon_b \end{aligned}$

The terms with $ a_i$ are still missing their rounding, so, just like $ b’$, rewrite $ a’_i = a_i q’/q + \varepsilon_i$ as $ a_i q’/q = a’_i – \varepsilon_i$, expanding, simplifying, and finally reducing modulo $ q’$ to get

$ \displaystyle \begin{aligned} b’ &= \left ( \sum_{i=1}^n \left ( a’_i – \varepsilon_i \right) s_i \right ) + m’ + e_{\textup{original}}\frac{q’}{q} + Mq’ + \varepsilon_b \\ &= \left ( \sum_{i=1}^n a’_i s_i \right ) – \left ( \sum_{i=1}^n \varepsilon_i s_i \right) + m’ + e_{\textup{original}}\frac{q’}{q} + Mq’ + \varepsilon_b \\ &= \left ( \sum_{i=1}^n a’_i s_i \right ) + m’ + Mq’ + \left [ e_{\textup{original}}\frac{q’}{q} – \left ( \sum_{i=1}^n \varepsilon_i s_i \right) + \varepsilon_b \right ] \\ &= \left ( \sum_{i=1}^n a’_i s_i \right ) + m’ + \left [ e_{\textup{original}}\frac{q’}{q} – \left ( \sum_{i=1}^n \varepsilon_i s_i \right) + \varepsilon_b \right ] \mod q’ \end{aligned}$

Define the square bracketed term as $ e_{\textup{new}}$, and we have proved the theorem.

$ \square$

The error after modulus switching is laid out. It’s the original error scaled, plus at most $ n+1$ terms, each of which is at most $ 1/2$. However, note that this is larger than it appears. If the new modulus is, say, $ q’=1024$, and the dimension is $ n = 512$, then in the worst case the error right after modulus switching will leave us only $ 1$ bit left for the message. This is not altogether unrealistic, as production (128-bit) security parameters for LWE put $ n$ around 600. But it is compensated for by the fact that the secret $ s$ is chosen uniformly at random, and the errors are symmetric around zero. So in expectation only half the bits will be set, and half of the set bits will have a positive error, and half a negative error. Using these facts, you can bound the probability that the error exceeds, say, $ \sqrt{n \log n}$ using a standard Hoeffding bound argument. I further believe that the error is bounded by $ \sqrt{n}$. I have verified it empirically, but haven’t been able to quite nail down a proof.

Until next time!

Zero Knowledge Proofs for NP

Posted on by

Last time, we saw a specific zero-knowledge proof for graph isomorphism. This introduced us to the concept of an interactive proof, where you have a prover and a verifier sending messages back and forth, and the prover is trying to prove a specific claim to the verifier.

A zero-knowledge proof is a special kind of interactive proof in which the prover has some secret piece of knowledge that makes it very easy to verify a disputed claim is true. The prover’s goal, then, is to convince the verifier (a polynomial-time algorithm) that the claim is true without revealing any knowledge at all about the secret.

In this post we’ll see that, using a bit of cryptography, zero-knowledge proofs capture a much wider class of problems than graph isomorphism. Basically, if you believe that cryptography exists, every problem whose answers can be easily verified have zero-knowledge proofs (i.e., all of the class NP). Here are a bunch of examples. For each I’ll phrase the problem as a question, and then say what sort of data the prover’s secret could be.

  • Given a boolean formula, is there an assignment of variables making it true? Secret: a satisfying assignment to the variables.
  • Given a set of integers, is there a subset whose sum is zero? Secret: such a subset.
  • Given a graph, does it have a 3-coloring? Secret: a valid 3-coloring.
  • Given a boolean circuit, can it produce a specific output? Secret: a choice of inputs that produces the output.

The common link among all of these problems is that they are NP-hard (graph isomorphism isn’t known to be NP-hard). For us this means two things: (1) we think these problems are actually hard, so the verifier can’t solve them, and (2) if you show that one of them has a zero-knowledge proof, then they all have zero-knowledge proofs.

We’re going to describe and implement a zero-knowledge proof for graph 3-colorability, and in the next post we’ll dive into the theoretical definitions and talk about the proof that the scheme we present is zero-knowledge. As usual, all of the code used in making this post is available in a repository on this blog’s Github page. In the follow up to this post, we’ll dive into more nitty gritty details about the proof that this works, and study different kinds of zero-knowledge.

One-way permutations

In a recent program gallery post we introduced the Blum-Blum-Shub pseudorandom generator. A pseudorandom generator is simply an algorithm that takes as input a short random string of length $ s$ and produces as output a longer string, say, of length $ 3s$. This output string should not be random, but rather “indistinguishable” from random in a sense we’ll make clear next time. The underlying function for this generator is the “modular squaring” function $ x \mapsto x^2 \mod M$, for some cleverly chosen $ M$. The $ M$ is chosen in such a way that makes this mapping a permutation. So this function is more than just a pseudorandom generator, it’s a one-way permutation.

If you have a primality-checking algorithm on hand (we do), then preparing the Blum-Blum-Shub algorithm is only about 15 lines of code.

def goodPrime(p):
    return p % 4 == 3 and probablyPrime(p, accuracy=100)

def findGoodPrime(numBits=512):
    candidate = 1

    while not goodPrime(candidate):
        candidate = random.getrandbits(numBits)

    return candidate

def makeModulus(numBits=512):
    return findGoodPrime(numBits) * findGoodPrime(numBits)

def blum_blum_shub(modulusLength=512):
    modulus = makeModulus(numBits=modulusLength)

    def f(inputInt):
        return pow(inputInt, 2, modulus)

    return f

The interested reader should check out the proof gallery post for more details about this generator. For us, having a one-way permutation is the important part (and we’re going to defer the formal definition of “one-way” until next time, just think “hard to get inputs from outputs”).

The other concept we need, which is related to a one-way permutation, is the notion of a hardcore predicate. Let $ G(x)$ be a one-way permutation, and let $ f(x) = b$ be a function that produces a single bit from a string. We say that $ f$ is a hardcore predicate for $ G$ if you can’t reliably compute $ f(x)$ when given only $ G(x)$.

Hardcore predicates are important because there are many one-way functions for which, when given the output, you can guess part of the input very reliably, but not the rest (e.g., if $ g$ is a one-way function, $ (x, y) \mapsto (x, g(y))$ is also one-way, but the $ x$ part is trivially guessable). So a hardcore predicate formally measures, when given the output of a one-way function, what information derived from the input is hard to compute.

In the case of Blum-Blum-Shub, one hardcore predicate is simply the parity of the input bits.

def parity(n):
    return sum(int(x) for x in bin(n)[2:]) % 2

Bit Commitment Schemes

A core idea that will makes zero-knowledge proofs work for NP is the ability for the prover to publicly “commit” to a choice, and later reveal that choice in a way that makes it infeasible to fake their commitment. This will involve not just the commitment to a single bit of information, but also the transmission of auxiliary data that is provably infeasible to fake.

Our pair of one-way permutation $ G$ and hardcore predicate $ f$ comes in very handy. Let’s say I want to commit to a bit $ b \in \{ 0,1 \}$. Let’s fix a security parameter that will measure how hard it is to change my commitment post-hoc, say $ n = 512$. My process for committing is to draw a random string $ x$ of length $ n$, and send you the pair $ (G(x), f(x) \oplus b)$, where $ \oplus$ is the XOR operator on two bits.

The guarantee of a one-way permutation with a hardcore predicate is that if you only see $ G(x)$, you can’t guess $ f(x)$ with any reasonable edge over random guessing. Moreover, if you fix a bit $ b$, and take an unpredictably random bit $ y$, the XOR $ b \oplus y$ is also unpredictably random. In other words, if $ f(x)$ is hardcore, then so is $ x \mapsto f(x) \oplus b$ for a fixed bit $ b$. Finally, to reveal my commitment, I just send the string $ x$ and let you independently compute $ (G(x), f(x) \oplus b)$. Since $ G$ is a permutation, that $ x$ is the only $ x$ that could have produced the commitment I sent you earlier.

Here’s a Python implementation of this scheme. We start with a generic base class for a commitment scheme.

class CommitmentScheme(object):
    def __init__(self, oneWayPermutation, hardcorePredicate, securityParameter):
        '''
            oneWayPermutation: int -&gt; int
            hardcorePredicate: int -&gt; {0, 1}
        '''
        self.oneWayPermutation = oneWayPermutation
        self.hardcorePredicate = hardcorePredicate
        self.securityParameter = securityParameter

        # a random string of length `self.securityParameter` used only once per commitment
        self.secret = self.generateSecret()

    def generateSecret(self):
        raise NotImplemented

    def commit(self, x):
        raise NotImplemented

    def reveal(self):
        return self.secret

Note that the “reveal” step is always simply to reveal the secret. Here’s the implementation subclass. We should also note that the security string should be chosen at random anew for every bit you wish to commit to. In this post we won’t reuse CommitmentScheme objects anyway.

class BBSBitCommitmentScheme(CommitmentScheme):
    def generateSecret(self):
        # the secret is a random quadratic residue
        self.secret = self.oneWayPermutation(random.getrandbits(self.securityParameter))
        return self.secret

    def commit(self, bit):
        unguessableBit = self.hardcorePredicate(self.secret)
        return (
            self.oneWayPermutation(self.secret),
            unguessableBit ^ bit,  # python xor
        )

One important detail is that the Blum-Blum-Shub one-way permutation is only a permutation when restricted to quadratic residues. As such, we generate our secret by shooting a random string through the one-way permutation to get a random residue. In fact this produces a uniform random residue, since the Blum-Blum-Shub modulus is chosen in such a way that ensures every residue has exactly four square roots.

Here’s code to check the verification is correct.

class BBSBitCommitmentVerifier(object):
    def __init__(self, oneWayPermutation, hardcorePredicate):
        self.oneWayPermutation = oneWayPermutation
        self.hardcorePredicate = hardcorePredicate

    def verify(self, securityString, claimedCommitment):
        trueBit = self.decode(securityString, claimedCommitment)
        unguessableBit = self.hardcorePredicate(securityString)  # wasteful, whatever
        return claimedCommitment == (
            self.oneWayPermutation(securityString),
            unguessableBit ^ trueBit,  # python xor
        )

    def decode(self, securityString, claimedCommitment):
        unguessableBit = self.hardcorePredicate(securityString)
        return claimedCommitment[1] ^ unguessableBit

and an example of using it

if __name__ == "__main__":
    import blum_blum_shub
    securityParameter = 10
    oneWayPerm = blum_blum_shub.blum_blum_shub(securityParameter)
    hardcorePred = blum_blum_shub.parity

    print('Bit commitment')
    scheme = BBSBitCommitmentScheme(oneWayPerm, hardcorePred, securityParameter)
    verifier = BBSBitCommitmentVerifier(oneWayPerm, hardcorePred)

    for _ in range(10):
        bit = random.choice([0, 1])
        commitment = scheme.commit(bit)
        secret = scheme.reveal()
        trueBit = verifier.decode(secret, commitment)
        valid = verifier.verify(secret, commitment)

        print('{} == {}? {}; {} {}'.format(bit, trueBit, valid, secret, commitment))

Example output:

1 == 1? True; 524 (5685, 0)
1 == 1? True; 149 (22201, 1)
1 == 1? True; 476 (34511, 1)
1 == 1? True; 927 (14243, 1)
1 == 1? True; 608 (23947, 0)
0 == 0? True; 964 (7384, 1)
0 == 0? True; 373 (23890, 0)
0 == 0? True; 620 (270, 1)
1 == 1? True; 926 (12390, 0)
0 == 0? True; 708 (1895, 0)

As an exercise, write a program to verify that no other input to the Blum-Blum-Shub one-way permutation gives a valid verification. Test it on a small security parameter like $ n=10$.

It’s also important to point out that the verifier needs to do some additional validation that we left out. For example, how does the verifier know that the revealed secret actually is a quadratic residue? In fact, detecting quadratic residues is believed to be hard! To get around this, we could change the commitment scheme reveal step to reveal the random string that was used as input to the permutation to get the residue (cf. BBSCommitmentScheme.generateSecret for the random string that needs to be saved/revealed). Then the verifier could generate the residue in the same way. As an exercise, upgrade the bit commitment an verifier classes to reflect this.

In order to get a zero-knowledge proof for 3-coloring, we need to be able to commit to one of three colors, which requires two bits. So let’s go overkill and write a generic integer commitment scheme. It’s simple enough: specify a bound on the size of the integers, and then do an independent bit commitment for every bit.

class BBSIntCommitmentScheme(CommitmentScheme):
    def __init__(self, numBits, oneWayPermutation, hardcorePredicate, securityParameter=512):
        '''
            A commitment scheme for integers of a prespecified length `numBits`. Applies the
            Blum-Blum-Shub bit commitment scheme to each bit independently.
        '''
        self.schemes = [BBSBitCommitmentScheme(oneWayPermutation, hardcorePredicate, securityParameter)
                        for _ in range(numBits)]
        super().__init__(oneWayPermutation, hardcorePredicate, securityParameter)

    def generateSecret(self):
        self.secret = [x.secret for x in self.schemes]
        return self.secret

    def commit(self, integer):
        # first pad bits to desired length
        integer = bin(integer)[2:].zfill(len(self.schemes))
        bits = [int(bit) for bit in integer]
        return [scheme.commit(bit) for scheme, bit in zip(self.schemes, bits)]

And the corresponding verifier

class BBSIntCommitmentVerifier(object):
    def __init__(self, numBits, oneWayPermutation, hardcorePredicate):
        self.verifiers = [BBSBitCommitmentVerifier(oneWayPermutation, hardcorePredicate)
                          for _ in range(numBits)]

    def decodeBits(self, secrets, bitCommitments):
        return [v.decode(secret, commitment) for (v, secret, commitment) in
                zip(self.verifiers, secrets, bitCommitments)]

    def verify(self, secrets, bitCommitments):
        return all(
            bitVerifier.verify(secret, commitment)
            for (bitVerifier, secret, commitment) in
            zip(self.verifiers, secrets, bitCommitments)
        )

    def decode(self, secrets, bitCommitments):
        decodedBits = self.decodeBits(secrets, bitCommitments)
        return int(''.join(str(bit) for bit in decodedBits))

A sample usage:

if __name__ == "__main__":
    import blum_blum_shub
    securityParameter = 10
    oneWayPerm = blum_blum_shub.blum_blum_shub(securityParameter)
    hardcorePred = blum_blum_shub.parity

    print('Int commitment')
    scheme = BBSIntCommitmentScheme(10, oneWayPerm, hardcorePred)
    verifier = BBSIntCommitmentVerifier(10, oneWayPerm, hardcorePred)
    choices = list(range(1024))
    for _ in range(10):
        theInt = random.choice(choices)
        commitments = scheme.commit(theInt)
        secrets = scheme.reveal()
        trueInt = verifier.decode(secrets, commitments)
        valid = verifier.verify(secrets, commitments)

        print('{} == {}? {}; {} {}'.format(theInt, trueInt, valid, secrets, commitments))

And a sample output:

527 == 527? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 0), (342, 1), (54363, 1), (63975, 0), (5426, 0), (9124, 1), (23973, 0), (44832, 0), (33044, 0), (68501, 0)]
67 == 67? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 1), (342, 1), (54363, 1), (63975, 1), (5426, 0), (9124, 1), (23973, 1), (44832, 1), (33044, 0), (68501, 0)]
729 == 729? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 0), (342, 1), (54363, 0), (63975, 1), (5426, 0), (9124, 0), (23973, 0), (44832, 1), (33044, 1), (68501, 0)]
441 == 441? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 1), (342, 0), (54363, 0), (63975, 0), (5426, 1), (9124, 0), (23973, 0), (44832, 1), (33044, 1), (68501, 0)]
614 == 614? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 0), (342, 1), (54363, 1), (63975, 1), (5426, 1), (9124, 1), (23973, 1), (44832, 0), (33044, 0), (68501, 1)]
696 == 696? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 0), (342, 1), (54363, 0), (63975, 0), (5426, 1), (9124, 0), (23973, 0), (44832, 1), (33044, 1), (68501, 1)]
974 == 974? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 0), (342, 0), (54363, 0), (63975, 1), (5426, 0), (9124, 1), (23973, 0), (44832, 0), (33044, 0), (68501, 1)]
184 == 184? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 1), (342, 1), (54363, 0), (63975, 0), (5426, 1), (9124, 0), (23973, 0), (44832, 1), (33044, 1), (68501, 1)]
136 == 136? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 1), (342, 1), (54363, 0), (63975, 0), (5426, 0), (9124, 1), (23973, 0), (44832, 1), (33044, 1), (68501, 1)]
632 == 632? True; [25461, 56722, 25739, 2268, 1185, 18226, 46375, 8907, 54979, 23095] [(29616, 0), (342, 1), (54363, 1), (63975, 1), (5426, 1), (9124, 0), (23973, 0), (44832, 1), (33044, 1), (68501, 1)]

Before we move on, we should note that this integer commitment scheme “blows up” the secret by quite a bit. If you have a security parameter $ s$ and an integer with $ n$ bits, then the commitment uses roughly $ sn$ bits. A more efficient method would be to simply use a good public-key encryption scheme, and then reveal the secret key used to encrypt the message. While we implemented such schemes previously on this blog, I thought it would be more fun to do something new.

A zero-knowledge proof for 3-coloring

First, a high-level description of the protocol. The setup: the prover has a graph $ G$ with $ n$ vertices $ V$ and $ m$ edges $ E$, and also has a secret 3-coloring of the vertices $ \varphi: V \to \{ 0, 1, 2 \}$. Recall, a 3-coloring is just an assignment of colors to vertices (in this case the colors are 0,1,2) so that no two adjacent vertices have the same color.

So the prover has a coloring $ \varphi$ to be kept secret, but wants to prove that $ G$ is 3-colorable. The idea is for the verifier to pick a random edge $ (u,v)$, and have the prover reveal the colors of $ u$ and $ v$. However, if we run this protocol only once, there’s nothing to stop the prover from just lying and picking two distinct colors. If we allow the verifier to run the protocol many times, and the prover actually reveals the colors from their secret coloring, then after roughly $ |V|$ rounds the verifier will know the entire coloring. Each step reveals more knowledge.

We can fix this with two modifications.

  1. The prover first publicly commits to the coloring using a commitment scheme. Then when the verifier asks for the colors of the two vertices of a random edge, he can rest assured that the prover fixed a coloring that does not depend on the verifier’s choice of edge.
  2. The prover doesn’t reveal colors from their secret coloring, but rather from a random permutation of the secret coloring. This way, when the verifier sees colors, they’re equally likely to see any two colors, and all the verifier will know is that those two colors are different.

So the scheme is: prover commits to a random permutation of the true coloring and sends it to the verifier; the verifier asks for the true colors of a given edge; the prover provides those colors and the secrets to their commitment scheme so the verifier can check.

The key point is that now the verifier has to commit to a coloring, and if the coloring isn’t a proper 3-coloring the verifier has a reasonable chance of picking an improperly colored edge (a one-in-$ |E|$ chance, which is at least $ 1/|V|^2$). On the other hand, if the coloring is proper, then the verifier will always query a properly colored edge, and it’s zero-knowledge because the verifier is equally likely to see every pair of colors. So the verifier will always accept, but won’t know anything more than that the edge it chose is properly colored. Repeating this $ |V|^2$-ish times, with high probability it’ll have queried every edge and be certain the coloring is legitimate.

Let’s implement this scheme. First the data types. As in the previous post, graphs are represented by edge lists, and a coloring is represented by a dictionary mapping a vertex to 0, 1, or 2 (the “colors”).

# a graph is a list of edges, and for simplicity we'll say
# every vertex shows up in some edge
exampleGraph = [
    (1, 2),
    (1, 4),
    (1, 3),
    (2, 5),
    (2, 5),
    (3, 6),
    (5, 6)
]

exampleColoring = {
    1: 0,
    2: 1,
    3: 2,
    4: 1,
    5: 2,
    6: 0,
}

Next, the Prover class that implements that half of the protocol. We store a list of integer commitment schemes for each vertex whose color we need to commit to, and send out those commitments.

class Prover(object):
    def __init__(self, graph, coloring, oneWayPermutation=ONE_WAY_PERMUTATION, hardcorePredicate=HARDCORE_PREDICATE):
        self.graph = [tuple(sorted(e)) for e in graph]
        self.coloring = coloring
        self.vertices = list(range(1, numVertices(graph) + 1))
        self.oneWayPermutation = oneWayPermutation
        self.hardcorePredicate = hardcorePredicate
        self.vertexToScheme = None

    def commitToColoring(self):
        self.vertexToScheme = {
            v: commitment.BBSIntCommitmentScheme(
                2, self.oneWayPermutation, self.hardcorePredicate
            ) for v in self.vertices
        }

        permutation = randomPermutation(3)
        permutedColoring = {
            v: permutation[self.coloring[v]] for v in self.vertices
        }

        return {v: s.commit(permutedColoring[v])
                for (v, s) in self.vertexToScheme.items()}

    def revealColors(self, u, v):
        u, v = min(u, v), max(u, v)
        if not (u, v) in self.graph:
            raise Exception('Must query an edge!')

        return (
            self.vertexToScheme[u].reveal(),
            self.vertexToScheme[v].reveal(),
        )

In commitToColoring we randomly permute the underlying colors, and then compose that permutation with the secret coloring, committing to each resulting color independently. In revealColors we reveal only those colors for a queried edge. Note that we don’t actually need to store the permuted coloring, because it’s implicitly stored in the commitments.

It’s crucial that we reject any query that doesn’t correspond to an edge. If we don’t reject such queries then the verifier can break the protocol! In particular, by querying non-edges you can determine which pairs of nodes have the same color in the secret coloring. You can then chain these together to partition the nodes into color classes, and so color the graph. (After seeing the Verifier class below, implement this attack as an exercise).

Here’s the corresponding Verifier:

class Verifier(object):
    def __init__(self, graph, oneWayPermutation, hardcorePredicate):
        self.graph = [tuple(sorted(e)) for e in graph]
        self.oneWayPermutation = oneWayPermutation
        self.hardcorePredicate = hardcorePredicate
        self.committedColoring = None
        self.verifier = commitment.BBSIntCommitmentVerifier(2, oneWayPermutation, hardcorePredicate)

    def chooseEdge(self, committedColoring):
        self.committedColoring = committedColoring
        self.chosenEdge = random.choice(self.graph)
        return self.chosenEdge

    def accepts(self, revealed):
        revealedColors = []

        for (w, bitSecrets) in zip(self.chosenEdge, revealed):
            trueColor = self.verifier.decode(bitSecrets, self.committedColoring[w])
            revealedColors.append(trueColor)
            if not self.verifier.verify(bitSecrets, self.committedColoring[w]):
                return False

        return revealedColors[0] != revealedColors[1]

As expected, in the acceptance step the verifier decodes the true color of the edge it queried, and accepts if and only if the commitment was valid and the edge is properly colored.

Here’s the whole protocol, which is syntactically very similar to the one for graph isomorphism.

def runProtocol(G, coloring, securityParameter=512):
    oneWayPermutation = blum_blum_shub.blum_blum_shub(securityParameter)
    hardcorePredicate = blum_blum_shub.parity

    prover = Prover(G, coloring, oneWayPermutation, hardcorePredicate)
    verifier = Verifier(G, oneWayPermutation, hardcorePredicate)

    committedColoring = prover.commitToColoring()
    chosenEdge = verifier.chooseEdge(committedColoring)

    revealed = prover.revealColors(*chosenEdge)
    revealedColors = (
        verifier.verifier.decode(revealed[0], committedColoring[chosenEdge[0]]),
        verifier.verifier.decode(revealed[1], committedColoring[chosenEdge[1]]),
    )
    isValid = verifier.accepts(revealed)

    print("{} != {} and commitment is valid? {}".format(
        revealedColors[0], revealedColors[1], isValid
    ))

    return isValid

And an example of running it

if __name__ == "__main__":
    for _ in range(30):
        runProtocol(exampleGraph, exampleColoring, securityParameter=10)

Here’s the output

0 != 2 and commitment is valid? True
1 != 0 and commitment is valid? True
1 != 2 and commitment is valid? True
2 != 0 and commitment is valid? True
1 != 2 and commitment is valid? True
2 != 0 and commitment is valid? True
0 != 2 and commitment is valid? True
0 != 2 and commitment is valid? True
0 != 1 and commitment is valid? True
0 != 1 and commitment is valid? True
2 != 1 and commitment is valid? True
0 != 2 and commitment is valid? True
2 != 0 and commitment is valid? True
2 != 0 and commitment is valid? True
1 != 0 and commitment is valid? True
1 != 0 and commitment is valid? True
0 != 2 and commitment is valid? True
2 != 1 and commitment is valid? True
0 != 2 and commitment is valid? True
0 != 2 and commitment is valid? True
2 != 1 and commitment is valid? True
1 != 0 and commitment is valid? True
1 != 0 and commitment is valid? True
2 != 1 and commitment is valid? True
2 != 1 and commitment is valid? True
1 != 0 and commitment is valid? True
0 != 2 and commitment is valid? True
1 != 2 and commitment is valid? True
1 != 2 and commitment is valid? True
0 != 1 and commitment is valid? True

So while we haven’t proved it rigorously, we’ve seen the zero-knowledge proof for graph 3-coloring. This automatically gives us a zero-knowledge proof for all of NP, because given any NP problem you can just convert it to the equivalent 3-coloring problem and solve that. Of course, the blowup required to convert a random NP problem to 3-coloring can be polynomially large, which makes it unsuitable for practice. But the point is that this gives us a theoretical justification for which problems have zero-knowledge proofs in principle. Now that we’ve established that you can go about trying to find the most efficient protocol for your favorite problem.

Anticipatory notes

When we covered graph isomorphism last time, we said that a simulator could, without participating in the zero-knowledge protocol or knowing the secret isomorphism, produce a transcript that was drawn from the same distribution of messages as the protocol produced. That was all that it needed to be “zero-knowledge,” because anything the verifier could do with its protocol transcript, the simulator could do too.

We can do exactly the same thing for 3-coloring, exploiting the same “reverse order” trick where the simulator picks the random edge first, then chooses the color commitment post-hoc.

Unfortunately, both there and here I’m short-changing you, dear reader. The elephant in the room is that our naive simulator assumes the verifier is playing by the rules! If you want to define security, you have to define it against a verifier who breaks the protocol in an arbitrary way. For example, the simulator should be able to produce an equivalent transcript even if the verifier deterministically picks an edge, or tries to pick a non-edge, or tries to send gibberish. It takes a lot more work to prove security against an arbitrary verifier, but the basic setup is that the simulator can no longer make choices for the verifier, but rather has to invoke the verifier subroutine as a black box. (To compensate, the requirements on the simulator are relaxed quite a bit; more on that next time)

Because an implementation of such a scheme would involve a lot of validation, we’re going to defer the discussion to next time. We also need to be more specific about the different kinds of zero-knowledge, since we won’t be able to achieve perfect zero-knowledge with the simulator drawing from an identical distribution, but rather a computationally indistinguishable distribution.

We’ll define all this rigorously next time, and discuss the known theoretical implications and limitations. Next time will be cuffs-off theory, baby!

Until then!

The Mathematics of Secret Sharing

Posted on by

Here’s a simple puzzle with a neat story. A rich old woman is drafting her will and wants to distribute her expansive estate equally amongst her five children. But her children are very greedy, and the woman knows that if he leaves her will unprotected her children will resort to nefarious measures to try to get more than their fair share. In one fearful scenario, she worries that the older four children will team up to bully the youngest child entirely out of his claim! She desperately wants them to cooperate, so she decides to lock the will away, and the key is a secret integer $ N$. The question is, how can she distribute this secret number to her children so that the only way they can open the safe is if they are all present and willing?

estateA mathematical way to say this is: how can she distribute some information to her children so that, given all of their separate pieces of information, they can reconstruct the key, but for every choice of fewer than 5 children, there is no way to reliably recover the key? This is called the secret sharing problem. More generally, say we have an integer $ N$ called the secret, a number of participants $ k$, and a number required for reconstruction $ r$. Then a secret sharing protocol is the data of a method for distributing information and a method for reconstructing the secret. The distributing method is an algorithm $ D$ that accepts as input $ N, k, r$ and produces as output a list of $ k$ numbers $ D(N, k) = (x_1, x_2, \dots, x_k)$. These are the numbers distributed to the $ k$ participants. Then the reconstruction method is a function $ R$ which accepts as input $ r$ numbers $ (y_1, \dots, y_r)$ and outputs a number $ M$. We want two properties to hold :

  • The reconstruction function $ R$ outputs $ N$ when given any $ r$ of the numbers output by $ D$.
  • One cannot reliably reconstruct $ N$ with fewer than $ r$ of the numbers output by $ D$.

The question is: does an efficient secret sharing protocol exist for every possible choice of $ r \leq k$? In fact it does, and the one we’ll describe in this post is far more secure than the word “reliable” suggests. It will be so hard as to be mathematically impossible to reconstruct the secret from fewer than the desired number of pieces. Independently discovered by Adi Shamir in 1979, the protocol we’ll see in this post is wonderfully simple, and as we describe it we’ll build up a program to implement it. This time we’ll work in the Haskell programming language, and you can download the program from this blog’s Github page. And finally, a shout out to my friend Karishma Chadha who worked together with me on this post. She knows Haskell a lot better than I do.

Polynomial Interpolation

The key to the secret sharing protocol is a beautiful fact about polynomials. Specifically, if you give me $ k+1$ points in the plane with distinct $ x$ values, then there is a unique degree $ k$ polynomial that passes through the points. Just as importantly (and as a byproduct of this fact), there are infinitely many degree $ k+1$ polynomials that pass through the same points. For example, if I give you the points $ (1,2), (2,4), (-2,2)$, the only quadratic (degree 2) polynomial that passes through all of them is $ 1 + \frac{1}{2}x + \frac{1}{2} x^2$.interpolating polynomial example The proof that you can always find such a polynomial is pretty painless, so let’s take it slowly and write a program as we go. Suppose you give me some list of $ k+1$ points $ (x_0, y_0), \dots, (x_k, y_k)$ and no two $ x$ values are the same. The proof has two parts. First we have to prove existence, that some degree $ k$ polynomial passes through the points, and then we have to prove that the polynomial is unique. The uniqueness part is easier, so let’s do the existence part first. Let’s start with just one point $ (x_0, y_0)$. What’s a degree zero polynomial that passes through it? Just the constant function $ f(x) = y_0$. For two points $ (x_0, y_0), (x_1, y_1)$ it’s similarly easy, since we all probably remember from basic geometry that there’s a unique line passing through any two points. But let’s write the line in a slightly different way:

$ \displaystyle f(x) = \frac{(x-x_1)}{x_0-x_1}y_0 + \frac{(x-x_0)}{x_1-x_0} y_1$

Why write it this way? Because now it should be obvious that the polynomial passes through our two points: if I plug in $ x_0$ then the second term is zero and the first term is just $ y_0(x_0 – x_1) / (x_0 – x_1) = y_0$, and likewise for $ x_1$.

For example, if we’re given $ (1, 3), (2, 5)$ we get:

$ \displaystyle f(x) = \frac{(x – 2)}{(1-2)} \cdot 3 + \frac{(x-1)}{(2-1)} \cdot 5 $

Plugging in $ x = 1$ cancels the second term out, leaving $ f(1) = \frac{1-2}{1-2} \cdot 3 = 3$, and plugging in $ x = 2$ cancels the first term, leaving $ f(2) = \frac{(2-1)}{(2-1)} \cdot 5 = 5$.

Now the hard step is generalizing this to three points. But the suggestive form above gives us a hint on how to continue.

$ \displaystyle f(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}y_1+ \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}y_2$

Notice that the numerators of the terms take on the form $ y_j \prod_{i \ne j} (x-x_i)$, that is, a product $ (x-x_0)(x-x_1), \dots, (x-x_n) y_j$ excluding $ (x – x_j)$. Thus, all terms will cancel out to 0 if we plug in $ x_i$, except one term, which has the form

$ \displaystyle y_i \cdot \frac{\prod_{j \neq i} (x-x_j)}{\prod_{j \neq i} (x_i – x_j)}$

Here, the fraction on the right side of the term cancels out to 1 when $ x_i$ is plugged in, leaving only $ y_i$, the desired result. Now that we’ve written the terms in this general product form, we can easily construct examples for any number of points. We just do a sum of terms that look like this, one for each $ y$ value. Try writing this out as a summation, if you feel comfortable with notation.

Let’s go further and write an algorithm to construct the polynomial for us. Some preliminaries: we encode a polynomial as a list of coefficients in degree-increasing order, so that $ 1 + 3x + 5x^3$ is represented by [1,3,0,5].

type Point = (Rational, Rational)
type Polynomial = [Rational] --Polynomials are represented in ascending degree order

Then we can write some simple functions for adding and multiplying polynomials

addPoly :: Polynomial -&gt; Polynomial -&gt; Polynomial
addPoly [] [] = []
addPoly [] xs = xs
addPoly xs [] = xs
addPoly (x:xs) (y:ys) = (x+y) : (addPoly xs ys)

multNShift :: Polynomial -&gt; (Rational, Int) -&gt; Polynomial
multNShift xs (y, shift) =
    (replicate shift 0) ++ ( map ((*) y) xs)

multPoly :: Polynomial -&gt; Polynomial -&gt; Polynomial
multPoly [] [] = []
multPoly [] _ = []
multPoly _ [] = []
multPoly xs ys = foldr addPoly [] $ map (multNShift ys) $ zip xs [0..]

In short, multNShift multiplies a polynomial by a monomial (like $ 3x^2 (1 + 7x + 2x^4)$), and multPoly does the usual distribution of terms, using multNShift to do most of the hard work. Then to construct the polynomial we need one more helper function to extract all elements of a list except a specific entry:

allBut :: Integer -&gt; [a] -&gt; [a]
allBut i list = snd $ unzip $ filter (\ (index,_) -&gt; i /= index) $ zip [0..] list

And now we can construct a polynomial from a list of points in the same way we did mathematically.

findPolynomial :: [Point] -&gt; Polynomial
findPolynomial points =
   let term (i, (xi,yi)) =
          let prodTerms = map (\ (xj, _) -&gt; [-xj/(xi - xj), 1/(xi - xj)]) $ allBut i points
          in multPoly [yi] $ foldl multPoly [1] prodTerms
   in foldl addPoly [] $ map term $ zip [0..] points

Here the sub-function term constructs the $ i$-th term of the polynomial, and the remaining expression adds up all the terms. Remember that due to our choice of representation the awkward 1 sitting in the formula signifies the presence of $ x$. And that’s it! An example of it’s use to construct $ 3x – 1$:

*Main&gt; findPolynomial [(1,2), (2,5)]
[(-1) % 1,3 % 1]

Now the last thing we need to do is show that the polynomial we constructed in this way is unique. Here’s a proof.

Suppose there are two degree $ n$ polynomials $ f(x)$  and $ g(x)$ that pass through the $ n+1$ given data points $ (x_0, y_0), (x_1, y_1), \dots , (x_n, y_n)$. Let $ h(x) = p(x) – q(x)$, and we want to show that $ h(x)$ is the zero polynomial. This proves that $ f(x)$ is unique because the only assumptions we made at the beginning were that $ f,g$ both passed through the given points. Now since both $ f$ and $ g$ are degree $ n$ polynomials, $ h$ is a polynomial of degree at most $ n$. It is also true that $ h(x_i) = p(x_i) – q(x_i) = y_i – y_i = 0$ where $ 0\leq i\leq n$. Thus, we have (at least) $ n+1$ roots of this degree $ n$ polynomial. But this can’t happen by the fundamental theorem of algebra! In more detail: if a nonzero degree $ \leq n$ polynomial really could have $ n+1$ distinct roots, then you could factor it into at least $ n+1$ linear terms like $ h(x) = (x – x_0)(x – x_1) \dots (x – x_n)$. But since there are $ n+1$ copies of $ x$, $ h$ would need to be a degree $ n+1$ polynomial! The only way to resolve this contradiction is if $ h$ is actually the zero polynomial, and thus $ h(x) = f(x) – g(x) = 0$, $ f(x) = g(x)$.

This completes the proof. Now that we know these polynomials exist and are unique, it makes sense to give them a name. So for a given set of $ k+1$ points, call the unique degree $ k$ polynomial that passes through them the interpolating polynomial for those points.

Secret Sharing with Interpolating Polynomials

Once you think to use interpolating polynomials, the connection to secret sharing seems almost obvious. If you want to distribute a secret to $ k$ people so that $ r$ of them can reconstruct it here’s what you do:

  1. Pick a random polynomial $ p$ of degree $ r-1$ so that the secret is $ p(0)$.
  2. Distribute the points $ (1, p(1)), (2, p(2)), \dots, (k, p(k))$.

Then the reconstruction function is: take the points provided by at least $ r$ participants, use them to reconstruct $ p$, and output $ p(0)$. That’s it! Step 1 might seem hard at first, but you can just notice that $ p(0)$ is equivalent to the constant term of the polynomial, so you can pick $ r-1$ random numbers for the other coefficients of $ p$ and output them. In Haskell,

makePolynomial :: Rational -&gt; Int -&gt; StdGen -&gt; Polynomial
makePolynomial secret r generator =
  secret : map toRational (take (r-1) $ randomRs (1, (numerator(2*secret))) generator)

share :: Rational -&gt; Integer -&gt; Int -&gt; IO [Point]
share secret k r = do
  generator &lt;- getStdGen
  let poly = makePolynomial secret r generator
      ys = map (eval poly) $ map toRational [1..k]
  return $ zip [1..] ys

In words, we initialize the Haskell standard generator (which wraps the results inside an IO monad), then we construct a polynomial by letting the first coefficient be the secret and choosing random coefficients for the rest. And findPolynomial is the reconstruction function.

Finally, just to flush the program out a little more, we write a function that encodes or decodes a string as an integer.

encode :: String -&gt; Integer
encode str = let nums = zip [0..] $ map (toInteger . ord) str
                 integers = map (\(i, n) -&gt; shift n (i*8)) nums
             in foldl (+) 0 integers

decode :: Integer -&gt; String
decode 0 = &quot;&quot;
decode num = if num &lt; 0
             then error &quot;Can't decode a negative number&quot;
             else chr (fromInteger (num .&amp;. 127)) : (decode $ shift num (-8))

And then we have a function that shows the whole process in action.

example msg k r =
   let secret = toRational $ encode msg
   in do points  (numerator x, numerator y)) points
      let subset = take r points
          encodedSecret = eval (findPolynomial subset) 0
      putStrLn $ show $ numerator encodedSecret
      putStrLn $ decode $ numerator encodedSecret

And a function call:

*Main&gt; example &quot;Hello world!&quot; 10 5
10334410032606748633331426632
[(1,34613972928232668944107982702),(2,142596447049264820443250256658),(3,406048862884360219576198642966),(4,916237517700482382735379150124),(5,1783927975542901326260203400662),(6,3139385067235193566437068631142),(7,5132372890379242119499357692158),(8,7932154809355236501627439048336),(9,11727493455321672728948666778334),(10,16726650726215353317537380574842)]
10334410032606748633331426632
Hello world!

Security

The final question to really close this problem with a nice solution is, “How secure is this protocol?” That is, if you didn’t know the secret but you had $ r-1$ numbers, could you find a way to recover the secret, oh, say, 0.01% of the time?

Pleasingly, the answer is a solid no. This protocol has something way stronger, what’s called information-theoretic security. In layman’s terms, this means it cannot possibly be broken, period. That is, without taking advantage of some aspect of the random number generator, which we assume is a secure random number generator. But with that assumption the security proof is trivial. Here it goes.

Pick a number $ M$ that isn’t the secret $ N$. It’s any number you want. And say you only have $ r-1$ of the correct numbers $ y_1, \dots, y_{r-1}$. Then there is a final number $ y_r$ so that the protocol reconstructs $ M$ instead of $ N$. This is no matter which of the unused $ x$-values you pick, no matter what $ M$ and $ r-1$ numbers you started with. This is simply because adding in $ (0, M)$ defines a new polynomial $ q$, and you can use any point on $ q$ as your $ r$-th number.

Here’s what this means. A person trying to break the secret sharing protocol would have no way to tell if they did it correctly! If the secret is a message, then a bad reconstruction could produce any message. In information theory terms, knowing $ r-1$ of the numbers provides no information about the actual message. In our story from the beginning of the post, no matter how much computing power one of the greedy children may have, the only algorithm they have to open the safe is to try every combination. The mother could make the combination have length in the millions of digits, or even better, the mother could encode the will as an integer and distribute that as the secret. I imagine there are some authenticity issues there, since one could claim to have reconstructed a false will, signatures and all, but there appear to be measures to account for this.

One might wonder if this is the only known secret sharing protocol, and the answer is no. Essentially, any time you have an existence and uniqueness theorem in mathematics, and the objects you’re working with are efficiently constructible, then you have the potential for a secret sharing protocol. There are two more on Wikipedia. But people don’t really care to find new ones anymore because the known protocols are as good as it gets.

On a broader level, the existence of efficient secret sharing protocols is an important fact used in the field of secure multiparty computation. Here the goal is for a group of individuals to compute a function depending on secret information from all of them, without revealing their secret information to anyone. A classic example of this is to compute the average of seven salaries without revealing any of the salaries. This was a puzzle featured on Car Talk, and it has a cute answer. See if you can figure it out.

Until next time!

Sending and Authenticating Messages with Elliptic Curves

Posted on by

Last time we saw the Diffie-Hellman key exchange protocol, and discussed the discrete logarithm problem and the related Diffie-Hellman problem, which form the foundation for the security of most protocols that use elliptic curves. Let’s continue our journey to investigate some more protocols.

Just as a reminder, the Python implementations of these protocols are not at all meant for practical use, but for learning purposes. We provide the code on this blog’s Github page, but for the love of security don’t actually use them.

Shamir-Massey-Omura

Recall that there are lots of ways to send encrypted messages if you and your recipient share some piece of secret information, and the Diffie-Hellman scheme allows one to securely generate a piece of shared secret information. Now we’ll shift gears and assume you don’t have a shared secret, nor any way to acquire one. The first cryptosystem in that vein is called the Shamir-Massey-Omura protocol. It’s only slightly more complicated to understand than Diffie-Hellman, and it turns out to be equivalently difficult to break.

The idea is best explained by metaphor. Alice wants to send a message to Bob, but all she has is a box and a lock for which she has the only key. She puts the message in the box and locks it with her lock, and sends it to Bob. Bob can’t open the box, but he can send it back with a second lock on it for which Bob has the only key. Upon receiving it, Alice unlocks her lock, sends the box back to Bob, and Bob can now open the box and retrieve the message.

To celebrate the return of Game of Thrones, we’ll demonstrate this protocol with an original Lannister Infographic™.

Assuming the box and locks are made of magical unbreakable Valyrian steel, nobody but Jamie will be able to read the message.

Assuming the box and locks are made of magically unbreakable Valyrian steel, nobody but Bob (also known as Jamie) will be able to read the message.

Now fast forward through the enlightenment, industrial revolution, and into the age of information. The same idea works, and it’s significantly faster over long distances. Let $ C$ be an elliptic curve over a finite field $ k$ (we’ll fix $ k = \mathbb{Z}/p$ for some prime $ p$, though it works for general fields too). Let $ n$ be the number of points on $ C$.

Alice’s message is going to be in the form of a point $ M$ on $ C$. She’ll then choose her secret integer $ 0 < s_A < p$ and compute $ s_AM$ (locking the secret in the box), sending the result to Bob. Bob will likewise pick a secret integer $ s_B$, and send $ s_Bs_AM$ back to Alice.

Now the unlocking part: since $ s_A \in \mathbb{Z}/p$ is a field, Alice can “unlock the box” by computing the inverse $ s_A^{-1}$ and computing $ s_BM = s_A^{-1}s_Bs_AM$. Now the “box” just has Bob’s lock on it. So Alice sends $ s_BM$ back to Bob, and Bob performs the same process to evaluate $ s_B^{-1}s_BM = M$, thus receiving the message.

Like we said earlier, the security of this protocol is equivalent to the security of the Diffie-Hellman problem. In this case, if we call $ z = s_A^{-1}$ and $ y = s_B^{-1}$, and $ P = s_As_BM$, then it’s clear that any eavesdropper would have access to $ P, zP$, and $ yP$, and they would be tasked with determining $ zyP$, which is exactly the Diffie-Hellman problem.

Now Alice’s secret message comes in the form of a point on an elliptic curve, so how might one translate part of a message (which is usually represented as an integer) into a point? This problem seems to be difficult in general, and there’s no easy answer. Here’s one method originally proposed by Neal Koblitz that uses a bit of number theory trickery.

Let $ C$ be given by the equation $ y^2 = x^3 + ax + b$, again over $ \mathbb{Z}/p$. Suppose $ 0 \leq m < p/100$ is our message. Define for any $ 0 \leq j < 100$ the candidate $ x$-points $ x_j = 100m + j$. Then call our candidate $ y^2$-values $ s_j = x_j^3 + ax_j + b$. Now for each $ j$ we can compute $ x_j, s_j$, and so we’ll pick the first one for which $ s_j$ is a square in $ \mathbb{Z}/p$ and we’ll get a point on the curve. How can we tell if $ s_j$ is a square? One condition is that $ s_j^{(p-1)/2} \equiv 1 \mod p$. This is a basic fact about quadratic residues modulo primes; see these notes for an introduction and this Wikipedia section for a dense summary.

Once we know it’s a square, we can compute the square root depending on whether $ p \equiv 1 \mod 4$ or $ p \equiv 3 \mod 4$. In the latter case, it’s just $ s_j^{(p+1)/4} \mod p$. Unfortunately the former case is more difficult (really, the difficult part is $ p \equiv 1 \mod 8$). You can see Section 1.5 of this textbook for more details and three algorithms, or you could just pick primes congruent to 3 mod 4.

I have struggled to find information about the history of the Shamir-Massey-Omura protocol; every author claims it’s not widely used in practice, and the only reason seems to be that this protocol doesn’t include a suitable method for authenticating the validity of a message. In other words, some “man in the middle” could be intercepting messages and tricking you into thinking he is your intended recipient. Coupling this with the difficulty of encoding a message as a point seems to be enough to make cryptographers look for other methods. Another reason could be that the system was patented in 1982 and is currently held by SafeNet, one of the US’s largest security providers. All of their products have generic names so it’s impossible to tell if they’re actually using Shamir-Massey-Omura. I’m no patent lawyer, but it could simply be that nobody else is allowed to implement the scheme.

Digital Signatures

Indeed, the discussion above raises the question: how does one authenticate a message? The standard technique is called a digital signature, and we can implement those using elliptic curve techniques as well. To debunk the naive idea, one cannot simply attach some static piece of extra information to the message. An attacker could just copy that information and replicate it to forge your signature on another, potentially malicious document. In other words, a signature should only work for the message it was used to sign. The technique we’ll implement was originally proposed by Taher Elgamal, and is called the ElGamal signature algorithm. We’re going to look at a special case of it.

So Alice wants to send a message $ m$ with some extra information that is unique to the message and that can be used to verify that it was sent by Alice. She picks an elliptic curve $ E$ over $ \mathbb{F}_q$ in such a way that the number of points on $ E$ is $ br$, where $ b$ is a small integer and $ r$ is a large prime.

Then, as in Diffie-Hellman, she picks a base point $ Q$ that has order $ r$ and a secret integer $ s$ (which is permanent), and computes $ P = sQ$. Alice publishes everything except $ s$:

Public information: $ \mathbb{F}_q, E, b, r, Q, P &fg=000000$

Let Alice’s message $ m$ be represented as an integer at most $ r$ (there are a few ways to get around this if your message is too long). Now to sign $ m$ Alice picks a message specific $ k < r$ and computes what I’ll call the auxiliary point $ A = kQ$. Let $ A = (x, y)$. Alice then computes the signature $ g = k^{-1}(m + s x) \mod r$. The signed message is then $ (m, A, g)$, which Alice can safely send to Bob.

Before we see how Bob verifies the message, notice that the signature integer involves everything: Alice’s secret key, the message-specific secret integer $ k$, and most importantly the message. Remember that this is crucial: we want the signature to work only for the message that it was used to sign. If the same $ k$ is used for multiple messages then the attacker can find out your secret key! (And this has happened in practice; see the end of the post.)

So Bob receives $ (m, A, g)$, and also has access to all of the public information listed above. Bob authenticates the message by computing the auxiliary point via a different route. First, he computes $ c = g^{-1} m \mod r$ and $ d = g^{-1}x \mod r$, and then $ A’ = cQ + dP$. If the message was signed by Alice then $ A’ = A$, since we can just write out the definition of everything:

authentication-formula

Now to analyze the security. The attacker wants to be able to take any message $ m’$ and produce a signature $ A’, g’$ that will pass validation with Alice’s public information. If the attacker knew how to solve the discrete logarithm problem efficiently this would be trivial: compute $ s$ and then just sign like Alice does. Without that power there are still a few options. If the attacker can figure out the message-specific integer $ k$, then she can compute Alice’s secret key $ s$ as follows.

Given $ g = k^{-1}(m + sx) \mod r$, compute $ kg \equiv (m + sx) \mod r$. Compute $ d = gcd(x, r)$, and you know that this congruence has only $ d$ possible solutions modulo $ r$. Since $ s$ is less than $ r$, the attacker can just try all options until they find $ P = sQ$. So that’s bad, but in a properly implemented signature algorithm finding $ k$ is equivalently hard to solving the discrete logarithm problem, so we can assume we’re relatively safe from that.

On the other hand one could imagine being able to conjure the pieces of the signature $ A’, g’$ by some method that doesn’t involve directly finding Alice’s secret key. Indeed, this problem is less well-studied than the Diffie-Hellman problem, but most cryptographers believe it’s just as hard. For more information, this paper surveys the known attacks against this signature algorithm, including a successful attack for fields of characteristic two.

Signature Implementation

We can go ahead and implement the signature algorithm once we’ve picked a suitable elliptic curve. For the purpose of demonstration we’ll use a small curve, $ E: y^2 = x^3 + 3x + 181$ over $ F = \mathbb{Z}/1061$, whose number of points happens to have the a suitable prime factorization ($ 1047 = 3 \cdot 349$). If you’re interested in counting the number of points on an elliptic curve, there are many theorems and efficient algorithms to do this, and if you’ve been reading this whole series something then an algorithm based on the Baby-Step Giant-Step idea would be easy to implement. For the sake of brevity, we leave it as an exercise to the reader.

Note that the code we present is based on the elliptic curve and finite field code we’re been implementing as part of this series. All of the code used in this post is available on this blog’s Github page.

The basepoint we’ll pick has to have order 349, and $ E$ has plenty of candidates. We’ll use $ (2, 81)$, and we’ll randomly generate a secret key that’s less than $ 349$ (eight bits will do). So our setup looks like this:

if __name__ == &quot;__main__&quot;:
   F = FiniteField(1061, 1)

   # y^2 = x^3 + 3x + 181
   curve = EllipticCurve(a=F(3), b=F(181))
   basePoint = Point(curve, F(2), F(81))
   basePointOrder = 349
   secretKey = generateSecretKey(8)
   publicKey = secretKey * basePoint

Then so sign a message we generate a random key, construct the auxiliary point and the signature, and return:

def sign(message, basePoint, basePointOrder, secretKey):
   modR = FiniteField(basePointOrder, 1)
   oneTimeSecret = generateSecretKey(len(bin(basePointOrder)) - 3) # numbits(order) - 1

   auxiliaryPoint = oneTimeSecret * basePoint
   signature = modR(oneTimeSecret).inverse() *
         (modR(message) + modR(secretKey) * modR(auxiliaryPoint[0]))

   return (message, auxiliaryPoint, signature)

So far so good. Note that we generate the message-specific $ k$ at random, and this implies we need a high-quality source of randomness (what’s called a cryptographically-secure pseudorandom number generator). In absence of that there are proposed deterministic methods for doing it. See this draft proposal of Thomas Pornin, and this paper of Daniel Bernstein for another.

Now to authenticate, we follow the procedure from earlier.

def authentic(signedMessage, basePoint, basePointOrder, publicKey):
   modR = FiniteField(basePointOrder, 1)
   (message, auxiliary, signature) = signedMessage

   sigInverse = modR(signature).inverse() # sig can be an int or a modR already
   c, d = sigInverse * modR(message), sigInverse * modR(auxiliary[0])

   auxiliaryChecker = int(c) * basePoint + int(d) * publicKey
   return auxiliaryChecker == auxiliary

Continuing with our example, we pick a message represented as an integer smaller than $ r$, sign it, and validate it.

&gt;&gt;&gt; message = 123
&gt;&gt;&gt; signedMessage = sign(message, basePoint, basePointOrder, secretKey)
&gt;&gt;&gt; signedMessage
(123, (220 (mod 1061), 234 (mod 1061)), 88 (mod 349))
&gt;&gt;&gt; authentic(signedMessage, basePoint, basePointOrder, publicKey)
True

So there we have it, a nice implementation of the digital signature algorithm.

When Digital Signatures Fail

As we mentioned, it’s extremely important to avoid using the same $ k$ for two different messages. If you do, then you’ll get two signed messages $ (m_1, A_1, g_1), (m_2, A_2, g_2)$, but by definition the two $ g$’s have a ton of information in common! An attacker can recognize this immediately because $ A_1 = A_2$, and figure out the secret key $ s$ as follows. First write

$ \displaystyle g_1 – g_2 \equiv k^{-1}(m_1 + sx) – k^{-1}(m_2 + sx) \equiv k^{-1}(m_1 – m_2) \mod r$.

Now we have something of the form $ \text{known}_1 \equiv (k^{-1}) \text{known}_2 \mod r$, and similarly to the attack described earlier we can try all possibilities until we find a number that satisfies $ A = kQ$. Then once we have $ k$ we have already seen how to find $ s$. Indeed, it would be a good exercise for the reader to implement this attack.

The attack we just described it not an idle threat. Indeed, the Sony corporation, producers of the popular Playstation video game console, made this mistake in signing software for Playstation 3. A digital signature algorithm makes sense to validate software, because Sony wants to ensure that only Sony has the power to publish games. So Sony developers act as one party signing the data on a disc, and the console will only play a game with a valid signature. Note that the asymmetric setup is necessary because if the console had shared a secret with Sony (say, stored as plaintext within the hardware of the console), anyone with physical access to the machine could discover it.

Now here come the cringing part. Sony made the mistake of using the same $ k$ to sign every game! Their mistake was discovered in 2010 and made public at a cryptography conference. This video of the humorous talk includes a description of the variant Sony used and the attacker describe how the mistake should have been corrected. Without a firmware update (I believe Sony’s public key information was stored locally so that one could authenticate games without an internet connection), anyone could sign a piece of software and create games that are indistinguishable from something produced by Sony. That includes malicious content that, say, installs software that sends credit card information to the attacker.

So here we have a tidy story: a widely used cryptosystem with a scare story of what will go wrong when you misuse it. In the future of this series, we’ll look at other things you can do with elliptic curves, including factoring integers and testing for primality. We’ll also see some normal forms of elliptic curves that are used in place of the Weierstrass normal form for various reasons.

Until next time!