# Zero Knowledge Proofs for NP

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 -> int
hardcorePredicate: int -> {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!

# Zero Knowledge Proofs — A Primer

In this post we’ll get a strong taste for zero knowledge proofs by exploring the graph isomorphism problem in detail. In the next post, we’ll see how this relates to cryptography and the bigger picture. The goal of this post is to get a strong understanding of the terms “prover,” “verifier,” and “simulator,” and “zero knowledge” in the context of a specific zero-knowledge proof. Then next time we’ll see how the same concepts (though not the same proof) generalizes to a cryptographically interesting setting.

## Graph isomorphism

Let’s start with an extended example. We are given two graphs $G_1, G_2$, and we’d like to know whether they’re isomorphic, meaning they’re the same graph, but “drawn” different ways.

The problem of telling if two graphs are isomorphic seems hard. The pictures above, which are all different drawings of the same graph (or are they?), should give you pause if you thought it was easy.

To add a tiny bit of formalism, a graph $G$ is a list of edges, and each edge $(u,v)$ is a pair of integers between 1 and the total number of vertices of the graph, say $n$. Using this representation, an isomorphism between $G_1$ and $G_2$ is a permutation $\pi$ of the numbers $\{1, 2, \dots, n \}$ with the property that $(i,j)$ is an edge in $G_1$ if and only if $(\pi(i), \pi(j))$ is an edge of $G_2$. You swap around the labels on the vertices, and that’s how you get from one graph to another isomorphic one.

Given two arbitrary graphs as input on a large number of vertices $n$, nobody knows of an efficient—i.e., polynomial time in $n$—algorithm that can always decide whether the input graphs are isomorphic. Even if you promise me that the inputs are isomorphic, nobody knows of an algorithm that could construct an isomorphism. (If you think about it, such an algorithm could be used to solve the decision problem!)

## A game

Now let’s play a game. In this game, we’re given two enormous graphs on a billion nodes. I claim they’re isomorphic, and I want to prove it to you. However, my life’s fortune is locked behind these particular graphs (somehow), and if you actually had an isomorphism between these two graphs you could use it to steal all my money. But I still want to convince you that I do, in fact, own all of this money, because we’re about to start a business and you need to know I’m not broke.

Is there a way for me to convince you beyond a reasonable doubt that these two graphs are indeed isomorphic? And moreover, could I do so without you gaining access to my secret isomorphism? It would be even better if I could guarantee you learn nothing about my isomorphism or any isomorphism, because even the slightest chance that you can steal my money is out of the question.

Zero knowledge proofs have exactly those properties, and here’s a zero knowledge proof for graph isomorphism. For the record, $G_1$ and $G_2$ are public knowledge, (common inputs to our protocol for the sake of tracking runtime), and the protocol itself is common knowledge. However, I have an isomorphism $f: G_1 \to G_2$ that you don’t know.

Step 1: I will start by picking one of my two graphs, say $G_1$, mixing up the vertices, and sending you the resulting graph. In other words, I send you a graph $H$ which is chosen uniformly at random from all isomorphic copies of $G_1$. I will save the permutation $\pi$ that I used to generate $H$ for later use.

Step 2: You receive a graph $H$ which you save for later, and then you randomly pick an integer $t$ which is either 1 or 2, with equal probability on each. The number $t$ corresponds to your challenge for me to prove $H$ is isomorphic to $G_1$ or $G_2$. You send me back $t$, with the expectation that I will provide you with an isomorphism between $H$ and $G_t$.

Step 3: Indeed, I faithfully provide you such an isomorphism. If I you send me $t=1$, I’ll give you back $\pi^{-1} : H \to G_1$, and otherwise I’ll give you back $f \circ \pi^{-1}: H \to G_2$. Because composing a fixed permutation with a uniformly random permutation is again a uniformly random permutation, in either case I’m sending you a uniformly random permutation.

Step 4: You receive a permutation $g$, and you can use it to verify that $H$ is isomorphic to $G_t$. If the permutation I sent you doesn’t work, you’ll reject my claim, and if it does, you’ll accept my claim.

Before we analyze, here’s some Python code that implements the above scheme. You can find the full, working example in a repository on this blog’s Github page.

First, a few helper functions for generating random permutations (and turning their list-of-zero-based-indices form into a function-of-positive-integers form)

import random

def randomPermutation(n):
L = list(range(n))
random.shuffle(L)
return L

def makePermutationFunction(L):
return lambda i: L[i - 1] + 1

def makeInversePermutationFunction(L):
return lambda i: 1 + L.index(i - 1)

def applyIsomorphism(G, f):
return [(f(i), f(j)) for (i, j) in G]


Here’s a class for the Prover, the one who knows the isomorphism and wants to prove it while keeping the isomorphism secret:

class Prover(object):
def __init__(self, G1, G2, isomorphism):
'''
isomomorphism is a list of integers representing
an isomoprhism from G1 to G2.
'''
self.G1 = G1
self.G2 = G2
self.n = numVertices(G1)
assert self.n == numVertices(G2)

self.isomorphism = isomorphism
self.state = None

def sendIsomorphicCopy(self):
isomorphism = randomPermutation(self.n)
pi = makePermutationFunction(isomorphism)

H = applyIsomorphism(self.G1, pi)

self.state = isomorphism
return H

def proveIsomorphicTo(self, graphChoice):
randomIsomorphism = self.state
piInverse = makeInversePermutationFunction(randomIsomorphism)

if graphChoice == 1:
return piInverse
else:
f = makePermutationFunction(self.isomorphism)
return lambda i: f(piInverse(i))


The prover has two methods, one for each round of the protocol. The first creates an isomorphic copy of $G_1$, and the second receives the challenge and produces the requested isomorphism.

And here’s the corresponding class for the verifier

class Verifier(object):
def __init__(self, G1, G2):
self.G1 = G1
self.G2 = G2
self.n = numVertices(G1)
assert self.n == numVertices(G2)

def chooseGraph(self, H):
choice = random.choice([1, 2])
self.state = H, choice
return choice

def accepts(self, isomorphism):
'''
Return True if and only if the given isomorphism
is a valid isomorphism between the randomly
chosen graph in the first step, and the H presented
by the Prover.
'''
H, choice = self.state
graphToCheck = [self.G1, self.G2][choice - 1]
f = isomorphism

isValidIsomorphism = (graphToCheck == applyIsomorphism(H, f))
return isValidIsomorphism


Then the protocol is as follows:

def runProtocol(G1, G2, isomorphism):
p = Prover(G1, G2, isomorphism)
v = Verifier(G1, G2)

H = p.sendIsomorphicCopy()
choice = v.chooseGraph(H)
witnessIsomorphism = p.proveIsomorphicTo(choice)

return v.accepts(witnessIsomorphism)


Analysis: Let’s suppose for a moment that everyone is honestly following the rules, and that $G_1, G_2$ are truly isomorphic. Then you’ll always accept my claim, because I can always provide you with an isomorphism. Now let’s suppose that, actually I’m lying, the two graphs aren’t isomorphic, and I’m trying to fool you into thinking they are. What’s the probability that you’ll rightfully reject my claim?

Well, regardless of what I do, I’m sending you a graph $H$ and you get to make a random choice of $t = 1, 2$ that I can’t control. If $H$ is only actually isomorphic to either $G_1$ or $G_2$ but not both, then so long as you make your choice uniformly at random, half of the time I won’t be able to produce a valid isomorphism and you’ll reject. And unless you can actually tell which graph $H$ is isomorphic to—an open problem, but let’s say you can’t—then probability 1/2 is the best you can do.

Maybe the probability 1/2 is a bit unsatisfying, but remember that we can amplify this probability by repeating the protocol over and over again. So if you want to be sure I didn’t cheat and get lucky to within a probability of one-in-one-trillion, you only need to repeat the protocol 30 times. To be surer than the chance of picking a specific atom at random from all atoms in the universe, only about 400 times.

If you want to feel small, think of the number of atoms in the universe. If you want to feel big, think of its logarithm.

Here’s the code that repeats the protocol for assurance.

def convinceBeyondDoubt(G1, G2, isomorphism, errorTolerance=1e-20):
probabilityFooled = 1

while probabilityFooled > errorTolerance:
result = runProtocol(G1, G2, isomorphism)
assert result
probabilityFooled *= 0.5
print(probabilityFooled)


Running it, we see it succeeds

\$ python graph-isomorphism.py
0.5
0.25
0.125
0.0625
0.03125
...
<SNIP>
...
1.3552527156068805e-20
6.776263578034403e-21


So it’s clear that this protocol is convincing.

But how can we be sure that there’s no leakage of knowledge in the protocol? What does “leakage” even mean? That’s where this topic is the most difficult to nail down rigorously, in part because there are at least three a priori different definitions! The idea we want to capture is that anything that you can efficiently compute after the protocol finishes (i.e., you have the content of the messages sent to you by the prover) you could have computed efficiently given only the two graphs $G_1, G_2$, and the claim that they are isomorphic.

Another way to say it is that you may go through the verification process and feel happy and confident that the two graphs are isomorphic. But because it’s a zero-knowledge proof, you can’t do anything with that information more than you could have done if you just took the assertion on blind faith. I’m confident there’s a joke about religion lurking here somewhere, but I’ll just trust it’s funny and move on.

In the next post we’ll expand on this “leakage” notion, but before we get there it should be clear that the graph isomorphism protocol will have the strongest possible “no-leakage” property we can come up with. Indeed, in the first round the prover sends a uniform random isomorphic copy of $G_1$ to the verifier, but the verifier can compute such an isomorphism already without the help of the prover. The verifier can’t necessarily find the isomorphism that the prover used in retrospect, because the verifier can’t solve graph isomorphism. Instead, the point is that the probability space of “$G_1$ paired with an $H$ made by the prover” and the probability space of “$G_1$ paired with $H$ as made by the verifier” are equal. No information was leaked by the prover.

For the second round, again the permutation $\pi$ used by the prover to generate $H$ is uniformly random. Since composing a fixed permutation with a uniform random permutation also results in a uniform random permutation, the second message sent by the prover is uniformly random, and so again the verifier could have constructed a similarly random permutation alone.

Let’s make this explicit with a small program. We have the honest protocol from before, but now I’m returning the set of messages sent by the prover, which the verifier can use for additional computation.

def messagesFromProtocol(G1, G2, isomorphism):
p = Prover(G1, G2, isomorphism)
v = Verifier(G1, G2)

H = p.sendIsomorphicCopy()
choice = v.chooseGraph(H)
witnessIsomorphism = p.proveIsomorphicTo(choice)

return [H, choice, witnessIsomorphism]


To say that the protocol is zero-knowledge (again, this is still colloquial) is to say that anything that the verifier could compute, given as input the return value of this function along with $G_1, G_2$ and the claim that they’re isomorphic, the verifier could also compute given only $G_1, G_2$ and the claim that $G_1, G_2$ are isomorphic.

It’s easy to prove this, and we’ll do so with a python function called simulateProtocol.

def simulateProtocol(G1, G2):
# Construct data drawn from the same distribution as what is
# returned by messagesFromProtocol
choice = random.choice([1, 2])
G = [G1, G2][choice - 1]
n = numVertices(G)

isomorphism = randomPermutation(n)
pi = makePermutationFunction(isomorphism)
H = applyIsomorphism(G, pi)

return H, choice, pi


The claim is that the distribution of outputs to messagesFromProtocol and simulateProtocol are equal. But simulateProtocol will work regardless of whether $G_1, G_2$ are isomorphic. Of course, it’s not convincing to the verifier because the simulating function made the choices in the wrong order, choosing the graph index before making $H$. But the distribution that results is the same either way.

So if you were to use the actual Prover/Verifier protocol outputs as input to another algorithm (say, one which tries to compute an isomorphism of $G_1 \to G_2$), you might as well use the output of your simulator instead. You’d have no information beyond hard-coding the assumption that $G_1, G_2$ are isomorphic into your program. Which, as I mentioned earlier, is no help at all.

In this post we covered one detailed example of a zero-knowledge proof. Next time we’ll broaden our view and see the more general power of zero-knowledge (that it captures all of NP), and see some specific cryptographic applications. Keep in mind the preceding discussion, because we’re going to re-use the terms “prover,” “verifier,” and “simulator” to mean roughly the same things as the classes Prover, Verifier and the function simulateProtocol.

Until then!