# The missing explanation of zk-SNARKs: Part 2 posted November 2020

**warning: some errors got into this post. If you're following it to the letter they might be confusing to you. Note that they have been fixed in the book.**

In part 1 of this series I explained what exactly zk-SNARKs attempt to prove. The answer was confusing: zk-SNARKs are surely efficient at proving things, but they’re only useful at proving that you know a polynomial constrained by some known roots. How is that useful for us? How does one uses such a system to prove more general statements? This is what I will answer in the second part of this series. Make sure that you read part 1 before getting into this, as this won’t make too much sense otherwise :)

## From programs to polynomials

So far the constraints on the polynomial that the prover must find is that it has some roots (some values that evaluate to 0 with our polynomial). But how do we translate a more general statement into a polynomial knowledge proof? Typical statements in cryptocurrencies (which are the applications currently making the most use of zk-SNARKs these days) are of the form:

- prove that a value is in the range $[0, 2^64)$ (this is called a range proof)
- prove that a (secret) value is included in some given (public) Merkle tree
- prove that the sum of some values is equal to the sum of some other values
- etc.

And here is the difficult part… as I said in part 1 of this series, converting a program execution into the knowledge of a polynomial “sorts of requires a graduate course into the subject”. The good news is that I’m not going to tell you all about the details, but give you enough to give you a sense of how things work. From there, you should be able to understand what are the parts that are missing from my explanation and fill in the gaps as you wish.

### A program to an arithmetic circuits

First, let’s acknowledge that any program can be re-written (more or less easily) in math. The reason why we would want to do that should be obvious: we can’t prove code, but we can prove math.

For example, let’s take the following function where every input is public except for `a`

which is our secret input:

```
fn my_function(w, a, b) {
if w == true {
return a + b;
} else {
return a x b;
}
}
```

In this simple example, if every input and output is public except for `a`

, one can still deduce what `a`

is. So this example also serves as an example of what you shouldn’t try to prove in zero-knowledge. Anyway, the program can be re-written in math with this equation:

$$ w(a+b) + (1-w)(ab) = v $$

Where $v$ is the output and $w$ is either $0$ (`false`

) or $1$ (`true`

). Notice that we this equation is not really a program or a circuit, it just looks like a constraint: if you execute the program correctly, and then fill in the different inputs given and outputs obtained in the equation, the **equality** should be correct.

That’s one mental step we need to take: instead of executing a function in zero-knowledge (which doesn’t mean much really), what we can do is use zk-SNARKs to prove that some given inputs and outputs (secret or public) correctly match the execution of a program.

In any case, we’re only one step into the process of converting our execution to something we can prove with zk-SNARKs, the next step is to convert that in a series of constraints that can be converted in the knowledge of a polynomial. What zk-SNARKs want is a **rank-1 constraint system (R1CS)**. R1CS are really just a series of equations that are of the form $L \times R = O$ where $L, R, O$ can only be the addition of some variables, thus the only multiplication is between $L$ and $R$. It really doesn’t matter why we need to transform our arithmetic circuit into such a system of equations, besides that it helps doing the conversion to the final stuff we can prove.

Try to do this with the equation we have, and we can obtain something like

- $a \times b = m$
- $w \times (m - a - b) = v - a - b$

We actually forgot the constraint that $w$ is either $0$ or $1$, which we can add to our system via a clever trick:

- $a \times b = m$
- $w \times (m - a - b) = v - a - b$
- $w \times w = w$

Does that make sense? You should really see this system as a set of constraints: if you give me a set of values that you claim match the inputs and outputs of the execution of my program, then I should be able to verify that they also correctly verify these $equalities$. If one of the equality is wrong, then it must mean that the program does not output the value you gave me for the inputs you gave me.

Another way to think about it is that zk-SNARKs allow you to verifiably remove inputs or outputs of the transcript of a correct execution of a program.

### R1CS to a polynomial

The question now is: how do we transform this system into a polynomial? With a series of tricks!

Since we have three different equations in our system, the first step is to agree on three roots for our polynomial. We can simply choose $1, 2, 3$ as roots. Meaning that our polynomial solves $f(x) = 0$ for $x = 1$, $x = 2$, and $x = 3$. Why do that? So that we can make our polynomial represent all the equations in our system at the same time, by representing the first equation when evaluated at $1$, and representing the second equation when evaluated at $2$, and so on. So the prover’s job is now to create a polynomial $f(x)$ such that:

- $f(1) = a \times b - m$
- $f(2) = w \times (m - a - b) - (v - a -b)$
- $f(3) = w \times w - w$

And notice that all these equations should evaluate to $0$ if the values correctly match an execution of our original program. In other words, our polynomial $f(x)$ has roots $1, 2, 3$ only if we created it correctly. And this is all what zk-SNARKs are about if you remember part 1 of this blog series, we have the protocol to prove that our polynomial $f(x)$ indeed has these roots (that were decided during the set up phase of our protocol).

It would be too simple if this is was the end of my explanation, because now the problem is that the prover has way too much freedom in choosing their polynomial $f(x)$, they can simply find a polynomial that has roots $1, 2, 3$ without caring about the values $a, b, m, v, w$. They can do pretty much whatever they want. What we want is a system that locks every parts of the polynomial except for the secret values that the verifier must not learn about.

### It takes two to evaluate a polynomial hiding in the exponent

Let's recap:

- We want a prover that has to correctly execute the program with their secret value $a$ and the public values $b$ and $w$ and obtain the output $v$ that they can publish.
- The prover then must create a polynomial by only filling the parts that the verifier must not learn about: the values $a$ and $m$.

Thus, in a real zk-SNARK protocol you want the prover to have the least amount of freedom possible when they create their polynomials and then evaluate it to a random point (as seen in the first part of this blogpost series).

To do this, **the polynomial is created somewhat dynamically by having the prover only fill in their part, and having the verifier fill in the other parts**.

For example, let’s take the first equation $f(1) = a \times b - m$ and represent it as $f_1(x) = aL_1(x) \times bR_1(x) - mO_1(x)$ where $L_1(x), R_1(x), O_1(x)$ are polynomials that evaluate to $1$ for $x=1$ and to 0 for $x=2, 3$. This is necessary so that they only influence our first equation, and it is easy to use algorithms like Lagrange Interpolation to find such polynomials. Notice two things:

- We now have the inputs, intermediate values, and outputs, as coefficients of our polynomials.
- The polynomial $f(x)$ is the sum $f_1(x) + f_2(x) + f_3(x)$ where we can define $f_2(x)$ and $f_3(x)$ to represent equations 2 and 3, similarly to $f_1(x)$.

The prover can then:

- take the exponentiation of the random point $r$ hidden in the exponent to reconstruct the polynomials $L_1(r)$ and $O_1(r)$
- exponentiate $g^{L_1(r)}$ with the secret value $a$ to obtain $(g^{L_1(r)})^a=g^{aL_1(r)}$ which represents $a \times L_1(x)$ evaluated at the unknown and random point $x=r$ (and hidden in the exponent).
- Multiply $g^{O_1(r)}$ with $g^{m}$ to obtain $g^{O_1(r)}g^{m}=g^{O_1(r)+m}$ which represents the hidden evaluation of $O_1(x) + m$ at the random point $r$ (and hidden in the exponent).

And then the verifier can fill in the missing parts:

- reconstruct $g^{bR_1(x)}$ for some value $b$ and using the same techniques the prover used
- reconstruct $f_1(r)$ by using a bilinear pairing (seen in part 1 of this blog post series) as such: $f_1(r) = e(g^{aL(r)}, g^{bR(r)}) - e(g, g^{O(r) + m}) = e(g, g)^{aL(r) \times bR(r)} - e(g,g)^{O(r) + m}$

If you extrapolate these techniques to the whole polynomial $f(x)$ you can figure out the final protocol.

Of course, this is still a gross simplification of a real zk-SNARK protocol, my explanations still leave way too much power to the prover. All the other tricks that you can learn about are meant to restrict what the prover can do, ensuring that they correctly and consistently fill in the missing parts, and optimizing what can be optimized.

### I want to know more

I learned everything in the tutorial Why and How zk-SNARK Works: Definitive Explanation which goes much more in depth and explains all of parts that I’ve overlooked.

And if you like this content, be sure to check my book real-world cryptography.

## Comments

## Jude

I may be missing something but when you transform your equation into a R1CS, isn't your W now the opposite of what it was in the initial program and in the equation?

Like, w(m - a - b) = v - a - b evaluates to v equaling a + b when w is 0, and ab when w is 1. But our program was the other way round (v = ab when w is 0 and v= a + b when w is 1)

Sorry if I've misunderstood something!

## david

indeed I inverted the two. I made some mistakes in this post that I corrected in the chapter in the book[1], I'll see if I have the time to correct this post one day but I'll add a disclaimer in the meantime :o

thanks!

[1]: https://livebook.manning.com/book/real-world-cryptography/chapter-15?origin=product-toc

## Vlad Patryshev

Thank you. It's so beautiful. Never guessed number theory can make people very rich.

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