## Schnorr basics

If you're having a difficult time wrapping your head around Schnorr signatures, you're not alone. In this post, I make an attempt at explaining Schnorr signatures at a level that I myself appreciate, and hopefully, you'll find it valuable too.

## What's Schnorr?

Schnorr is a new signature scheme in Bitcoin that got activated in the taproot upgrade. Schnorr has some nice properties listed in BIP340, but I won't reiterate those here.

## Signing and verifying

A signature scheme consists of two actions, signing and verification as the following diagram shows.

The signer has a private key and a message to sign (for example, a Bitcoin transaction or a cat picture), and produces a signature. The verifier has the public key corresponding to the private key that the signer used, the message, and the signature. The verification process makes sure that the signature was created with the correct private key without knowing the private key.

If you want a more in-depth explanation, please visit the second half of chapter 2 of Grokking Bitcoin.

### Signer

We're now going to see how a Schnorr signature is created. It represents the left side of the diagram above. I assume that you're familiar with how a key pair is created using a random number generator and an elliptic curve. If not, I suggest you read section 4.8 of Grokking Bitcoin.

Suppose that you want to make a signature for the cat picture. The picture is the message, $$m$$, to sign. Your private key is $$p$$ which belongs to your public key $$P$$.

The first thing you'll do is to draw a random number, $$r$$, that we'll call the nonce (for "Number Once"). Then you'll treat $$r$$ as if it was a private key, which means that you can generate the corresponding public key by multiplying it with $$G$$, which is the generator point. At this stage, you have the following:

\begin{align} p & : \text{private key} \\ P = pG & : \text{public key} \\ m & : \text{message} \\ r & : \text{random nonce} \\ R = rG & : \text{nonce commitment} \end{align}

$$R$$ is your nonce commitment which will become the first part of the final signature, and $$r$$ must remain secret (explained later) and never be reused (also explained later). You've prepared everything you need to make the signature. You'll do it in two steps. Step 1 is to calculate a so-called challenge hash, $$e$$:

$e = H(R || P || m)$

The challenge hash is the hash of the challenge, which is the concatenation of $$R$$, $$P$$, and $$m$$. These components will all be available to the verifier. Using the challenge hash, you can now do step 2: Calculate the challenge response, or simply response, $$s$$, which is the second part of the signature (Thanks to Anthony Towns, Ruben Somsen, and nothingmuch on Twitter for their help with the name response.):

$s = r + ep \tag{1} \label{respeqn}$

Finally, your signature is

$(R,s)$

You send the cat picture, $$m$$, and your signature $$(R,s)$$ to your friend, Fred.

### Verifier

Fred wants to make sure the cat picture hasn't been compromised during transfer and that it really originates from you, the only one with access to your private key $$p$$. He has access to $$P$$, $$m$$, $$R$$, and $$s$$. Of course, he also has access to $$G$$ because that's a widely known constant. From this information, he can calculate the challenge hash and verify that the verification equation balances:

\begin{align} e & = H(R || P || m) \\ sG & = R + eP \tag{2} \label{vereqn} \end{align}

If this equation balances, Fred can be sure that the signature was made with $$p$$. Note that the verification equation, equation $$(\ref{vereqn})$$, is the response equation, $$(\ref{respeqn})$$, where both sides are multiplied by $$G$$. Starting with the response equation, we get

\begin{align} &s = r + ep \iff sG = (r+ep)G \\ &\iff sG = rG + epG \iff sG = R + eP \end{align}

As you can see: If the response equation holds, the verification equation holds. Likewise, if the verification equation holds, the response equation also holds. Thus, when Fred verifies the equation based on points on the elliptic curve, the verification equation, he also implicitly verifies that the response equation, based on scalars, holds.

When Fred has verified the signature, he can enjoy the cat picture, fully confident that it's actually the same picture as you sent him.

## Why is $$r$$ secret?

You might wonder why the nonce $$r$$ must be kept secret. You might even wonder why it's needed at all? Let's start with the latter. Let's remove $$r$$ from the process and see what happens.

You would create the signature as follows:

$\begin{array}{} e = H(P || m) \\ s = ep \end{array}$

The signature would consist only of $$s$$. Fred would then verify your signature as:

$\begin{array}{} e = H(P || m) \\ sG = eP \end{array}$

That equation holds, but it would also allow Fred to extract the private key $$p$$ since he knows both $$s$$ and $$e$$. He'll take the response equation and solve it for p:

$s = ep \iff p = \frac{s}{e}$

OK, we need the nonce to prevent Fred from figuring out your private key, but why must we keep the nonce secret? Why the hassle of using the nonce commitment, instead of the nonce itself?

It's for the same reason. Suppose that that the nonce, $$r$$, was made available to Fred, then he could figure out $$p$$ by:

$\begin{array}{} e = H(R || P || m) \\ s = r + ep \iff p = \frac{s-r}{e} \end{array}$

So Fred could calculate $$p$$ by subtracting $$r$$ from $$s$$ and dividing the result by $$e$$.

By revealing just the nonce commitment, $$R$$, to Fred, we make sure that Fred can't calculate $$p$$ while at the same time allowing him to verify that $$p$$ was used to generate the signature.

## Don't reuse nonces

Even if you keep the nonce secret, you may still leak your private key if you use the nonce twice for the same private key. Suppose that you make two signatures with the same nonce and private key as follows:

$\begin{array}{ll} e = H(R || P || m) & e' = H(R || P || m')\\ s = r + ep & s' = r + e'p \end{array}$

Then you give the signatures $$(R,s)$$ and $$(R,s')$$ to the verifier. The verifier can then use simple arithmetic to calculate your private key. He can set up an equation system with two equations and two unknown as follows:

$\left\{\begin{array}{@{}l@{}} s =r + ep\\ s' = r + e'p \end{array}\right.$

This is solvable for $$p$$ by subtracting $$s'$$ from $$s$$:

$s-s'=r+ep-r-e'p\\ =(e-e')p \implies p=\frac{s-s'}{e-e'}$

As you can see, the verifier will be able to extract the private key. Lesson learned: Don't reuse nonces.

If the nonce is reused, but for different private keys, $$p$$ and $$p'$$, the above equation system wouldn't be solvable because you'd have three unknowns, $$p$$, $$p'$$, and $$r$$, but only two equations.

## What's with the challenge?

The challenge hash, $$e$$, is the hash of the challenge $$R||P||m$$. Why do we use this particular challenge? Let's look at the three components separately.

### $$m$$

The message to sign is $$m$$, so it's really important that $$m$$ is somehow committed to by the signature. If we'd remove $$m$$ from the challenge, the "signature" would be valid for any message.

### $$R$$

(Thanks to waxwing and A J Towns for their help in sorting this out.)

To make sure that no one but the owner of the private key can create a signature, the challenge must contain the nonce commitment $$R$$. Suppose that the challenge didn't include the nonce commitment, then a signature can be trivially forged by anyone with access to the public key $$P$$. They can make up an arbitrary $$s$$ and do:

\begin{align} &e = H(P||m)\\ &s = \text{any number} \\ &sG=R+eP \implies R=sG-eP \end{align}

The last equation is the "verification equation", but solved for $$R$$. The right side of that equation contains only known variables, $$s$$, $$e$$, and $$P$$. Thus the signature $$(R,s)$$ is valid.

With $$R$$ in the challenge, it's impossible the solve the verification equation for $$R$$ because $$R$$ is part of the challenge $$e$$. It's hard to find an $$R$$ such that $$R=sG-H(R||P||m)P$$.

### $$P$$

Let's finally see what $$P$$ is doing in the challenge. Suppose that we didn't have $$P$$ in the challenge, $$e=H(R||m)$$, and that the signature $$(R,s)$$ is valid for public key $$P$$ and message $$m$$. Then the signature $$(R,s')=(R,s+ex)$$, where $$x$$ is an arbitrary number, would be valid for a public key $$P'=P+xG$$ and message $$m$$. Let's look at why:

$e = H(R || m)\\ s'G=R+eP' \iff (s+ex)G=R+e(P+xG) \iff \\ sG+exG=R+eP+exG \iff sG=R+eP+exG-exG \iff \\ sG=R+eP$

This is known as a related-key attack. If you're familiar with how extended public key derivation in BIP32 works, you might see the potential danger. For a refresher, here's the general idea:

This means that if an attacker knows the parent extended public key (xpub), and a valid signature for a child key, then the attacker can use this trick to forge signatures for the parent xpub, as well as any child xpubs that can be derived from the parent xpub. This is a problem not only for BIP32, but for many schemes using public key addition somehow, for example, Taproot (BIP341). For a bit more details, please visit the Design section of BIP340.

## Next steps

In the next post, I'll show how Schnorr signatures can be used in a multisignature setting to produce a signature that looks just like a normal single signature. This is very practical in Bitcoin because it reduces resource requirements for verifying the blockchain.