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Co-authored-by: Sean Bowe <ewillbefull@gmail.com>
Co-authored-by: Daira Hopwood <daira@jacaranda.org>
Co-authored-by: ying tong <yingtong@z.cash>
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- [Gadgets](design/gadgets.md)
- [SHA-256](design/gadgets/sha256.md)
- [16-bit table chip](design/gadgets/sha256/table16.md)
- [Background Material](background.md)
- [Fields](background/fields.md)
- [Polynomials](background/polynomials.md)
- [Cryptographic groups](background/groups.md)
- [Elliptic curves](background/curves.md)
- [UltraPLONK arithmetisation](background/upa.md)
- [Polynomial commitment using inner product argument](background/pc-ipa.md)
- [Recursion](background/recursion.md)

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# Background Material
This section covers the background material required to understand the Halo 2 proving
system. It is targeted at an ELI15 (Explain It Like I'm 15) level; if you think anything
could do with additional explanation, let us know!

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# Elliptic curves
Elliptic curves constructed over finite fields are another important cryptographic tool.
There are several ways to define the curve equation, but for our purposes, let
$\mathbb{F}_p$ be a large (255-bit) field, and then let the set of solutions $(x, y)$ to
$y^2 = x^3 + b$ for some constant $b$ define the $\mathbb{F}_p$-rational points on an
elliptic curve $E(\mathbb{F}_p)$. These $(x, y)$ coordinates are called "affine
coordinates". Each of the $\mathbb{F}_p$-rational points, together with a "point at
infinity" $\mathcal{O}$ that serves as the group identity, can be interpreted as an
element of a group. By convention, elliptic curve groups are written additively.
![](https://i.imgur.com/JvLS6yE.png)
*"Three points on a line sum to zero, which is the point at infinity."*
The group addition law is simple: to add two points together, find the line that
intersects both points and obtain the third point, and then negate its $y$-coordinate. The
case that a point is being added to itself, called point doubling, requires special
handling: we find the line tangent to the point, and then find the single other point that
intersects this line and then negate. Otherwise, in the event that a point is being
"added" to its negation, the result is the point at infinity.
The ability to add and double points naturally gives us a way to scale them by integers.
The number of points on the curve is known as the "group order". If this number is prime
$q$, we call the numbers that we scale curve points by "scalars" and consider that they
are all elements of a scalar field $\mathbb{F}_q$.
Elliptic curves, when properly designed, have an important security property. Given two
random elements $G, H \in E(\mathbb{F}_p)$ finding $a$ such that $[a] G = H$, otherwise
known as the discrete log of $H$ with respect to $G$, is considered computationally
infeasible with classical computers. This is called the elliptic curve discrete log
assumption.
## Curve arithmetic
### Point doubling
The simplest situation is doubling a point $(x_0, y_0)$. Continuing with our example
$y^2 = x^3 + b$, this is done first by computing the derivative
$$
\lambda = \frac{dy}{dx} = \frac{3x^2}{2y}.
$$
To obtain expressions for $(x_1, y_1) = (x_0, y_0) + (x_0, y_0),$ we consider
$$
\begin{aligned}
\frac{-y_1 - y_0}{x_1 - x_0} = \lambda &\implies -y_1 = \lambda(x_1 - x_0) + y_0 \\
&\implies \boxed{y_1 = \lambda(x_0 - x_1) - y_0}.
\end{aligned}
$$
To get the expression for $x_1,$ we substitute $y = \lambda(x_0 - x) - y_0$ into the
elliptic curve equation:
$$
\begin{aligned}
y^2 = x^3 + b &\implies (\lambda(x_0 - x) - y_0)^2 = x^3 + b \\
&\implies x^3 - \lambda^2 x^2 + \cdots = 0 \leftarrow\text{(rearranging terms)} \\
&= (x - x_0)(x - x_0)(x - x_1) \leftarrow\text{(known roots $x_0, x_0, x_1$)} \\
&= x^3 - (x_0 + x_0 + x_1)x^2 + \cdots.
\end{aligned}
$$
Comparing coefficients for the $x^2$ term gives us
$\lambda^2 = x_0 + x_0 + x_1 \implies \boxed{x_1 = \lambda^2 - 2x_0}.$
### Projective coordinates
This unfortunately requires an expensive inversion of $2y$. We can avoid this by arranging
our equations to "defer" the computation of the inverse, since we often do not need the
actual affine $(x', y')$ coordinate of the resulting point immediately after an individual
curve operation. Let's introduce a third coordinate $Z$ and scale our curve equation by
$Z^3$ like so:
$$
Z^3 y^2 = Z^3 x^3 + Z^3 b
$$
Our original curve is just this curve at the restriction $Z = 1$. If we allow the affine
point $(x, y)$ to be represented by $X = xZ$, $Y = yZ$ and $Z \neq 0$ then we have the
[homogenous projective curve](https://en.wikipedia.org/wiki/Homogeneous_coordinates)
$$
Y^2 Z = X^3 + Z^3 b.
$$
Obtaining $(x, y)$ from $(X, Y, Z)$ is as simple as computing $(X/Z, Y/Z)$ when
$Z \neq 0$. (When $Z = 0,$ we are dealing with the point at infinity $O := (0:1:0)$.) In
this form, we now have a convenient way to defer the inversion required by doubling a
point. The general strategy is to express $x', y'$ as rational functions using $x = X/Z$
and $y = Y/Z$, rearrange to make their denominators the same, and then take the resulting
point $(X, Y, Z)$ to have $Z$ be the shared denominator and $X = x'Z, Y = y'Z$.
> Projective coordinates are often, but not always, more efficient than affine
> coordinates. There may be exceptions to this when either we have a different way to
> apply Montgomery's trick, or when we're in the circuit setting where multiplications and
> inversions are about equally as expensive (at least in terms of circuit size).
The following shows an example of doubling a point $(X, Y, Z) = (xZ, yZ, Z)$ without an
inversion. Substituting with $X, Y, Z$ gives us
$$
\lambda = \frac{3x^2}{2y} = \frac{3(X/Z)^2}{2(Y/Z)} = \frac{3 X^2}{2YZ}
$$
and gives us
$$
\begin{aligned}
x' &= \lambda^2 - 2x \\
&= \lambda^2 - \frac{2X}{Z} \\
&= \frac{9 X^4}{4Y^2Z^2} - \frac{2X}{Z} \\
&= \frac{9 X^4 - 8XY^2Z}{4Y^2Z^2} \\
&= \frac{18 X^4 Y Z - 16XY^3Z^2}{8Y^3Z^3} \\
\\
y' &= \lambda (x - x') - y \\
&= \lambda (\frac{X}{Z} - \frac{9 X^4 - 8XY^2Z}{4Y^2Z^2}) - \frac{Y}{Z} \\
&= \frac{3 X^2}{2YZ} (\frac{X}{Z} - \frac{9 X^4 - 8XY^2Z}{4Y^2Z^2}) - \frac{Y}{Z} \\
&= \frac{3 X^3}{2YZ^2} - \frac{27 X^6 - 24X^3Y^2Z}{8Y^3Z^3} - \frac{Y}{Z} \\
&= \frac{12 X^3Y^2Z - 8Y^4Z^2 - 27 X^6 + 24X^3Y^2Z}{8Y^3Z^3}
\end{aligned}
$$
Notice how the denominators of $x'$ and $y'$ are the same. Thus, instead of computing
$(x', y')$ we can compute $(X, Y, Z)$ with $Z = 8Y^3Z^3$ and $X, Y$ set to the
corresponding numerators such that $X/Z = x'$ and $Y/Z = y'$. This completely avoids the
need to perform an inversion when doubling, and something analogous to this can be done
when adding two distinct points.
### TODO: Point addition
$$
\begin{aligned}
P + Q &= R\\
(x_p, y_p) + (x_q, y_q) &= (x_r, y_r) \\
\lambda &= \frac{y_q - y_p}{x_q - x_p} \\
x_r &= \lambda^2 - x_p - x_q \\
y_r &= \lambda(x_p - x_r) - y_p
\end{aligned}
$$
----------
Important notes:
* There exist efficient formulae[^complete-formulae] for point addition that do not have
edge cases (so-called "complete" formulae) and that unify the addition and doubling
cases together. The result of adding a point to its negation using those formulae
produces $Z = 0$, which represents the point at infinity.
* In addition, there are other models like the Jacobian representation where
$(x, y) = (xZ^2, yZ^3, Z)$ where the curve is rescaled by $Z^6$ instead of $Z^3$, and
this representation has even more efficient arithmetic but no unified/complete formulae.
* We can easily compare two curve points $(X_1, Y_1, Z_1)$ and $(X_2, Y_2, Z_2)$ for
equality in the homogenous projective coordinate space by "homogenizing" their
Z-coordinates; the checks become $X_1 Z_2 = X_2 Z_1$ and $Y_1 Z_2 = Y_2 Z_1$.
## Curve endomorphisms
Imagine that $\mathbb{F}_p$ has a primitive cube root of unity, or in other words that
$3 | p - 1$ and so an element $\zeta_p$ generates a $3$-order multiplicative subgroup.
Notice that a point $(x, y)$ on our example elliptic curve $y^2 = x^3 + b$ has two cousin
points: $(\zeta_p x, \zeta_p^2 x)$, because the computation $x^3$ effectively kills the
$\zeta$ component of the $x$-coordinate. Applying the map $(x, y) \mapsto (\zeta_p x, y)$
is an application of an endomorphism over the curve. The exact mechanics involved are
complicated, but when the curve has a prime $q$ number of points (and thus a prime
"order") the effect of the endomorphism is to multiply the point by a scalar in
$\mathbb{F}_q$ which is also a primitive cube root $\zeta_q$ in the scalar field.
## Curve point compression
TODO
## Cycles of curves
Let $E_p$ be an elliptic curve over a finite field $\mathbb{F}_p,$ where $p$ is a prime.
We denote this by $E_p/\mathbb{F}_p.$ and we denote the group of points of $E_p$ over
$\mathbb{F}_p,$ with order $q = \#E(\mathbb{F}_p).$ For this curve, we call $\mathbb{F}_p$
the "base field" and $\mathbb{F}_q$ the "scalar field".
We instantiate our proof system over the elliptic curve $E_p/\mathbb{F}_p$. This allows us
to prove statements about $\mathbb{F}_q$-arithmetic circuit satisfiability.
> **(aside) If our curve $E_p$ is over $\mathbb{F}_p,$ why is the arithmetic circuit instead in $\mathbb{F}_q$?**
> The proof system is basically working on encodings of the scalars in the circuit (or
> more precisely, commitments to polynomials whose coefficients are scalars). The scalars
> are in $\mathbb{F}_q$ when their encodings/commitments are elliptic curve points in
> $E_p/\mathbb{F}_p$.
However, most of the verifier's arithmetic computations are over the base field
$\mathbb{F}_p,$ and are thus efficiently expressed as an $\mathbb{F}_p$-arithmetic
circuit.
> **(aside) Why are the verifier's computations (mainly) over $\mathbb{F}_p$?**
> The Halo 2 verifier actually has to perform group operations using information output by
> the circuit. Group operations like point doubling and addition use arithmetic in
> $\mathbb{F}_p$, because the coordinates of points are in $\mathbb{F}_p.$
This motivates us to construct another curve with scalar field $\mathbb{F}_p$, which has
an $\mathbb{F}_p$-arithmetic circuit that can efficiently verify proofs from the first
curve. As a bonus, if this second curve had base field $E_q/\mathbb{F}_q,$ it would
generate proofs that could be efficiently verified in the first curve's
$\mathbb{F}_q$-arithmetic circuit. In other words, we instantiate a second proof system
over $E_q/\mathbb{F}_q,$ forming a 2-cycle with the first:
![](https://i.imgur.com/bNMyMRu.png)
### TODO: Pallas-Vesta curves
Reference: https://github.com/zcash/pasta
## Hashing to curves
Sometimes it is useful to be able to produce a random point on an elliptic curve
$E_p/\mathbb{F}_p$ corresponding to some input, in such a way that no-one will know its
discrete logarithm (to any other base).
This is described in detail in the [Internet draft on Hashing to Elliptic Curves][cfrg-hash-to-curve].
Several algorithms can be used depending on efficiency and security requirements. The
framework used in the Internet Draft makes use of several functions:
* ``hash_to_field``: takes a byte sequence input and maps it to a element in the base
field $\mathbb{F}_p$
* ``map_to_curve``: takes an $\mathbb{F}_p$ element and maps it to $E_p$.
[cfrg-hash-to-curve]: https://datatracker.ietf.org/doc/draft-irtf-cfrg-hash-to-curve/?include_text=1
### TODO: Simplified SWU
Reference: https://eprint.iacr.org/2019/403.pdf
## References
[^complete-formulae]: Renes, J., Costello, C., & Batina, L. (2016, May). "Complete addition formulas for prime order elliptic curves." In Annual International Conference on the Theory and Applications of Cryptographic Techniques (pp. 403-428). Springer, Berlin, Heidelberg. https://eprint.iacr.org/2015/1060.pdf

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# Fields
A fundamental component of many cryptographic protocols is the algebraic structure known
as a [field]. Fields are sets of objects (usually numbers) with two associated binary
operators $+$ and $\times$ such that various [field axioms][field-axioms] hold. The real
numbers $\mathbb{R}$ are an example of a field with an uncountably infinite number of
elements.
[field]: https://en.wikipedia.org/wiki/Field_(mathematics)
[field-axioms]: https://en.wikipedia.org/wiki/Field_(mathematics)#Classic_definition
Halo makes use of _finite fields_ which have a finite number of elements. Finite fields
are fully classified as follows:
- if $\mathbb{F}$ is a finite field, it contains $|\mathbb{F}| = p^k$ elements for some
integer $k \geq 1$ and some prime $p$;
- any two finite fields with the same number of elements are isomorphic. In particular,
all of the arithmetic in a prime field $\mathbb{F}_p$ is isomorphic to addition and
multiplication of integers modulo $p$, i.e. in $\mathbb{Z}_p$. This is why we often
refer to $p$ as the _modulus_.
We'll write a field as $\mathbb{F}_q$ where $q = p^k$. The prime $p$ is called its
_characteristic_. In the cases where $k \gt 1$ the field $\mathbb{F}_q$ is a $k$-degree
extension of the field $\mathbb{F}_p$. (By analogy, the complex numbers
$\mathbb{C} = \mathbb{R}(i)$ are an extension of the real numbers.) However, in Halo we do
not care about extension fields. Whenever we write $\mathbb{F}_p$ we are referring to what
we call a _prime field_ which has a prime $p$ number of elements, i.e. $k = 1$.
Important notes:
* There are two special elements in any field: $\mathcal{O}$, the additive identity, and
$1$, the multiplicative identity.
* The least significant bit of a field element, when represented as an integer in binary
format, can be interpreted as its "sign" to help distinguish it from its additive
inverse (negation). This is because for some nonzero element $a$ which has a least
significant bit $0$ we have that $-a = p - a$ has a least significant bit $1$, and vice
versa. We could also use whether or not an element is larger than $(p - 1) / 2$ to give
it a "sign."
## Inverses and groups
Any non-zero element $a \in \mathbb{F}_p$ has a _multiplicative inverse_ $b = a^{-1}$,
which is the _unique_ element $b$ such that $ab = 1$.
A quick way of obtaining the inverse is $a^{-1} = a^{p - 2}$. The reason for this stems
from [Fermat's little theorem][fermat-little], which states that $a^p = a \pmod p$ for any
integer $a$. If $a$ is nonzero, we can divide by $a$ twice to get $a^{p-2} = a^{-1}.$
[fermat-little]: https://en.wikipedia.org/wiki/Fermat%27s_little_theorem
However, it may be more intuitive to understand the set of nonzero elements of
$\mathbb{F}_p$ as a [group], where the group operation is given by multiplication on the
field. We use the notation $\mathbb{F}_p^\times$ for the multiplicative group over the set
$\mathbb{F}_p - \{\mathcal{O}\}$. Groups are simpler and more limited than fields; they
have only _one_ operator $\cdot$ and fewer axioms.
[group]: https://en.wikipedia.org/wiki/Group_(mathematics)
> #### (aside) Additive vs multiplicative notation
> If $\cdot$ is written as $+$ and the identity as $0$ or $\mathcal{O}$, then we say the
> group is "written additively". If $\cdot$ is written as $\times$ or omitted (i.e.
> $a \cdot b$ written as $ab$) and the identity as $1$, then we say it is "written
> multiplicatively". It's conventional to use additive notation for
> [elliptic curve groups](curves.md), and multiplicative notation when (as in this case)
> the elements come from a finite field. When additive notation is used, we also write
>
> $$[k] A = \underbrace{A + A + \cdots + A}_{k \text{ times}}$$
>
> for nonnegative $k$ and call this "scalar multiplication"; we also often use uppercase
> letters for variables denoting group elements. When multiplicative notation is used, we
> also write
>
> $$a^k = \underbrace{a \times a \times \cdots \times a}_{k \text{ times}}$$
>
> and call this "exponentiation". In either case we call the scalar $k$ such that
> $[k] g = a$ or $g^k = a$ the "discrete logarithm" of $a$ to base $g$. We can extend
> scalars to negative integers by inversion, i.e. $[-k] A + [k] A = \mathcal{O}$ or
> $a^{-k} \times a^k = 1$.
The _order_ of a group element $a$ is defined (in multiplicative notation) as the smallest
positive integer $k$ such that $a^k = 1$.
Groups always have [generators] which are elements that, when the group operation is
applied repeatedly with the same element some number of times, produce every other element
of the group. That is, a generator has maximal order, which we also call the order of the
group.
[generators]: https://en.wikipedia.org/wiki/Generating_set_of_a_group
There can be many different generators. Let's assume that $\alpha$ is a generator of
$\mathbb{F}_p^\times$, so it has order $p-1$ (equal to the number of elements in
$\mathbb{F}_p^\times$). Therefore, for any element in $a \in \mathbb{F}_p^\times$ there is
a unique integer $i \in \{0..p-2\}$ such that $a = \alpha^i$.
Notice that $a \times b$ where $a, b \in \mathbb{F}_p^\times$ can really be interpreted as
$\alpha^i \times \alpha^j$ where $a = \alpha^i$ and $b = \alpha^j$. Indeed, it holds that
$\alpha^i \times \alpha^j = \alpha^{i + j}$ for all $0 \leq i, j \lt p - 1$. As a result
the multiplication of nonzero field elements can be interpreted as addition modulo $p - 1$
with respect to some fixed generator $\alpha$. The addition just happens "in the exponent."
This is another way to look at where $a^{p - 2}$ comes from for computing inverses in the
field:
$$p - 2 \equiv -1 \pmod{p - 1},$$
so $a^{p - 2} = a^{-1}$.
### Montgomery's Trick
Montgomery's trick, named after Peter Montgomery (RIP) is a way to compute many field
inversions at the same time. Imagine we need to compute the inverses of three nonzero
elements $a, b, c \in \mathbb{F}_p$. Instead, we'll compute the products $x = ab$ and
$y = xc = abc$, and compute the inversion
$$z = y^{p - 2} = \frac{1}{abc}.$$
We can now multiply $z$ by $x$ to obtain $\frac{1}{c}$ and multiply $z$ by $c$ to obtain
$\frac{1}{ab}$, which we can then multiply by $a, b$ to obtain their respective inverses.
This technique generalizes to arbitrary numbers of field elements with just a single
inversion necessary.
## Multiplicative subgroups
A _subgroup_ of a group $G$ with operation $\cdot$, is a subset of elements of $G$ that
also form a group under $\cdot$.
In the previous section we said that $\alpha$ is a generator of the $p - 1$ order
multiplicative group $\mathbb{F}_p^\times$. This is a _composite_ order group, and so by
the Chinese remainder theorem[^chinese-remainder] it has strict subgroups. As an example
let's imagine that $p = 11$, and so $p - 1$ factors into $5 \cdot 2$. Thus, there is a
generator $\beta$ of the $5$-order subgroup and a generator $\gamma$ of the $2$-order
subgroup. All elements in $\mathbb{F}_p^\times$, therefore, can be written uniquely as
$\beta^i \cdot \gamma^j$ for some $i$ (modulo $5$) and some $j$ (modulo $2$).
If we have $a = \beta^i \cdot \gamma^j$ notice what happens when we compute
$$
a^5 = (\beta^i \cdot \gamma^j)^5
= \beta^{i \cdot 5} \cdot \gamma^{j \cdot 5}
= \beta^0 \cdot \gamma^{j \cdot 5}
= \gamma^{j \cdot 5};
$$
we have effectively "killed" the $5$-order subgroup component, producing a value in the
$2$-order subgroup.
[Lagrange's theorem (group theory)][lagrange-group] states that the order of any subgroup
$H$ of a finite group $G$ divides the order of $G$. Therefore, the order of any subgroup
of $\mathbb{F}_p^\times$ must divide $p-1.$
[lagrange-group]: https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)
## Square roots
In a field $\mathbb{F}_p$ exactly half of all nonzero elements are squares; the remainder
are non-squares or "quadratic non-residues". In order to see why, consider an $\alpha$
that generates the $2$-order multiplicative subgroup of $\mathbb{F}_p^\times$ (this always
exists because $p - 1$ is divisible by $2$ since $p$ is prime) and $\beta$ that generates
the $t$-order multiplicative subgroup of $\mathbb{F}_p^\times$ where $p - 1 = 2t$. Then
every element $a \in \mathbb{F}_p^\times$ can be written uniquely as
$\alpha^i \cdot \beta^j$ with $i \in \mathbb{Z}_2$ and $j \in \mathbb{Z}_t$. Half of all
elements will have $i = 0$ and the other half will have $i = 1$.
Let's consider the simple case where $p \equiv 3 \pmod{4}$ and so $t$ is odd (if $t$ is
even, then $p - 1$ would be divisible by $4$, which contradicts $p$ being $3 \pmod{4}$).
If $a \in \mathbb{F}_p^\times$ is a square, then there must exist
$b = \alpha^i \cdot \beta^j$ such that $b^2 = a$. But this means that
$$a = (\alpha^i \cdot \beta^j)^2 = \alpha^{2i} \cdot \beta^{2j} = \beta^{2j}.$$
In other words, all squares in this particular field do not generate the $2$-order
multiplicative subgroup, and so since half of the elements generate the $2$-order subgroup
then at most half of the elements are square. In fact exactly half of the elements are
square (since squaring each nonsquare element gives a unique square). This means we can
assume all squares can be written as $\beta^m$ for some $m$, and therefore finding the
square root is a matter of exponentiating by $2^{-1} \pmod{t}$.
In the event that $p \equiv 1 \pmod{4}$ then things get more complicated because
$2^{-1} \pmod{t}$ does not exist. Let's write $p - 1$ as $2^k \cdot t$ with $t$ odd. The
case $k = 0$ is impossible, and the case $k = 1$ is what we already described, so consider
$k \geq 2$. $\alpha$ generates a $2^k$-order multiplicative subgroup and $\beta$ generates
the odd $t$-order multiplicative subgroup. Then every element $a \in \mathbb{F}_p^\times$
can be written as $\alpha^i \cdot \beta^j$ for $i \in \mathbb{Z}_{2^k}$ and
$j \in \mathbb{Z}_t$. If the element is a square, then there exists some $b = \sqrt{a}$
which can be written $b = \alpha^{i'} \cdot \beta^{j'}$ for $i' \in \mathbb{Z}_{2^k}$ and
$j' \in \mathbb{Z}_t$. This means that $a = b^2 = \alpha^{2i'} \cdot \beta^{2j'}$,
therefore we have $i \equiv 2i' \pmod{2^k}$, and $j \equiv 2j' \pmod{t}$. $i$ would have
to be even in this case because otherwise it would be impossible to have
$i \equiv 2i' \pmod{2^k}$ for any $i'$. In the case that $a$ is not a square, then $i$ is
odd, and so half of all elements are squares.
In order to compute the square root, we can first raise the element
$a = \alpha^i \cdot \beta^j$ to the power $t$ to "kill" the $t$-order component, giving
$$a^t = \alpha^{it \pmod 2^k} \cdot \beta^{jt \pmod t} = \alpha^{it \pmod 2^k}$$
and then raise this result to the power $t^{-1} \pmod{2^k}$ to undo the effect of the
original exponentiation on the $2^k$-order component:
$$(\alpha^{it \bmod 2^k})^{t^{-1} \pmod{2^k}} = \alpha^i$$
(since $t$ is relatively prime to $2^k$). This leaves bare the $\alpha^i$ value which we
can trivially handle. We can similarly kill the $2^k$-order component to obtain
$\beta^{j \cdot 2^{-1} \pmod{t}}$, and put the values together to obtain the square root.
It turns out that in the cases $k = 2, 3$ there are simpler algorithms that merge several
of these exponentiations together for efficiency. For other values of $k$, the only known
way is to manually extract $i$ by squaring until you obtain the identity for every single
bit of $i$. This is the essence of the [Tonelli-Shanks square root algorithm][ts-sqrt] and
describes the general strategy. (There is another square root algorithm that uses
quadratic extension fields, but it doesn't pay off in efficiency until the prime becomes
quite large.)
> TODO: describe more recent algorithms, e.g. Bernstein's table-based method and
> [eprint 2020/1407](https://eprint.iacr.org/2020/1407).
[ts-sqrt]: https://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm
## Roots of unity
In the previous sections we wrote $p - 1 = 2^k \cdot t$ with $t$ odd, and stated that an
element $\alpha \in \mathbb{F}_p^\times$ generated the $2^k$-order subgroup. For
convenience, let's denote $n := 2^k.$ The elements $\{1, \alpha, \alpha^2, \alpha^{n-1}\}$
are known as the $n$th [roots of unity](https://en.wikipedia.org/wiki/Root_of_unity).
The **primitive root of unity**, $\omega,$ is an $n$th root of unity such that
$\omega^i \neq 1$ except when $i \equiv 0 \pmod{n}$.
Important notes:
- if $\alpha$ is an $n$th root of unity, $\alpha$ satisfies $\alpha^n - 1 = 0.$ If
$\alpha \neq 1,$ then
$$1 + \alpha + \alpha^2 + \cdots + \alpha^{n-1} = 0.$$
- equivalently, the roots of unity are solutions to the equation
$$X^n - 1 = (X - 1)(X - \alpha)(X - \alpha^2) \cdots (X - \alpha^{n-1}).$$
- **$\boxed{\omega^{\frac{n}{2}+i} = -\omega^i}$ ("Negation lemma")**. Proof:
$$
\begin{aligned}
\omega^n = 1 &\implies \omega^n - 1 = 0 \\
&\implies (\omega^{n/2} + 1)(\omega^{n/2} - 1) = 0.
\end{aligned}
$$
Since the order of $\omega$ is $n$, $\omega^{n/2} \neq 1.$ Therefore, $\omega^{n/2} = -1.$
- **$\boxed{(\omega^{\frac{n}{2}+i})^2 = (\omega^i)^2}$ ("Halving lemma")**. Proof:
$$
(\omega^{\frac{n}{2}+i})^2 = \omega^{n + 2i} = \omega^{n} \cdot \omega^{2i} = \omega^{2i} = (\omega^i)^2.
$$
In other words, if we square each element in the $n$th roots of unity, we would get back
only half the elements, $\{(\omega_n^i)^2\} = \{\omega_{n/2}\}$ (i.e. the $n/2$th roots
of unity). There is a two-to-one mapping between the elements and their squares.
## References
[^chinese-remainder]: Friedman, R. (n.d.) "Cyclic Groups and Elementary Number Theory II" (p. 5). http://www.math.columbia.edu/~rf/numbertheory2.pdf

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# Cryptographic groups
In the section [Inverses and groups](fields.md#inverses-and-groups) we introduced the
concept of *groups*. A group has an identity and a group operation. In this section we
will write groups additively, i.e. the identity is $\mathcal{O}$ and the group operation
is $+$.
Some groups can be used as *cryptographic groups*. At the risk of oversimplifying, this
means that the problem of finding a discrete logarithm of a group element $P$ to a given
base $G$, i.e. finding $x$ such that $P = [x] G$, is hard in general.
## Pedersen commitment
The Pedersen commitment [[P99]] is a way to commit to a secret message in a verifiable
way. It uses two random public generators $G, H \in \mathbb{G},$ where $\mathbb{G}$ is a
cryptographic group of order $p$. A random secret $r$ is chosen in $\mathbb{Z}_q$, and the
message to commit to $m$ is from any subset of $\mathbb{Z}_q$. The commitment is
$$c = \text{Commit}(m,r)=[m]G + [r]H.$$
To open the commitment, the committer reveals $m$ and $r,$ thus allowing anyone to verify
that $c$ is indeed a commitment to $m.$
[P99]: https://link.springer.com/content/pdf/10.1007%2F3-540-46766-1_9.pdf#page=3
Notice that the Pedersen commitment scheme is homomorphic:
$$
\begin{aligned}
\text{Commit}(m,r) + \text{Commit}(m',r') &= [m]G + [r]H + [m']G + [r']H \\
&= [m + m']G + [r + r']H \\
&= \text{Commit}(m + m',r + r').
\end{aligned}
$$
Assuming the discrete log assumption holds, Pedersen commitments are also perfectly hiding
and computationally binding:
* **hiding**: the adversary chooses messages $m_0, m_1.$ The committer commits to one of
these messages $c = \text{Commit}(m_b;r), b \in \{0,1\}.$ Given $c,$ the probability of
the adversary guessing the correct $b$ is no more than $\frac{1}{2}$.
* **binding**: the adversary cannot pick two different messages $m_0 \neq m_1,$ and
randomness $r_0, r_1,$ such that $\text{Commit}(m_0,r_0) = \text{Commit}(m_1,r_1).$
### Vector Pedersen commitment
We can use a variant of the Pedersen commitment scheme to commit to multiple messages at
once, $\mathbf{m} = (m_1, \cdots, m_n)$. This time, we'll have to sample a corresponding
number of random public generators $\mathbf{G} = (G_0, \cdots, G_{n-1}),$ along with a
single random generator $H$ as before (for use in hiding). Then, our commitment scheme is:
$$
\begin{aligned}
\text{Commit}(\mathbf{m}; r) &= \text{Commit}((m_0, \cdots, m_{n-1}); r) \\
&= [r]H + [m_0]G_0 + \cdots + [m_{n-1}]G_{n-1} \\
&= [r]H + \sum_{i= 0}^{n-1} [m_i]G_i.
\end{aligned}
$$
> TODO: is this positionally binding?
## Diffie--Hellman
An example of a protocol that uses cryptographic groups is Diffie--Hellman key agreement
[[DH1976]]. The Diffie--Hellman protocol is a method for two users, Alice and Bob, to
generate a shared private key. It proceeds as follows:
1. Alice and Bob publicly agree on two prime numbers, $p$ and $G,$ where $p$ is large and
$G$ is a primitive root $\pmod p.$ (Note that $g$ is a generator of the group
$\mathbb{F}_p^\times.$)
2. Alice chooses a large random number $a$ as her private key. She computes her public key
$A = [a]G \pmod p,$ and sends $A$ to Bob.
3. Similarly, Bob chooses a large random number $b$ as his private key. He computes his
public key $B = [b]G \pmod p,$ and sends $B$ to Alice.
4. Now both Alice and Bob compute their shared key $K = [ab]G \pmod p,$ which Alice
computes as
$$K = [a]B \pmod p = [a]([b]G) \pmod p,$$
and Bob computes as
$$K = [b]A \pmod p = [b]([a]G) \pmod p.$$
[DH1976]: https://ee.stanford.edu/~hellman/publications/24.pdf
A potential eavesdropper would need to derive $K = [ab]g \pmod p$ knowing only
$g, p, A = [a]G,$ and $B = [b]G$: in other words, they would need to either get the
discrete logarithm $a$ from $A = [a]G$ or $b$ from $B = [b]G,$ which we assume to be
computationally infeasible in $\mathbb{F}_p^\times.$
More generally, protocols that use similar ideas to Diffie--Hellman are used throughout
cryptography. One way of instantiating a cryptographic group is as an
[elliptic curve](curves.md). Before we go into detail on elliptic curves, we'll describe
some algorithms that can be used for any group.
## Multiscalar multiplication
### TODO: Pippenger's algorithm
Reference: https://jbootle.github.io/Misc/pippenger.pdf

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# Polynomial commitment using inner product argument
We want to commit to some polynomial $p(X) \in \mathbb{F}_p[X]$, and be able to provably
evaluate the committed polynomial at arbitrary points. The naive solution would be for the
prover to simply send the polynomial's coefficients to the verifier: however, this
requires $O(n)$ communication. Our polynomial commitment scheme gets the job done using
$O(\log n)$ communication.
### `Setup`
Given a parameter $d = 2^k,$ we generate the common reference string
$\sigma = (\mathbb{G}, \mathbf{G}, H, \mathbb{F}_p)$ defining certain constants for this
scheme:
* $\mathbb{G}$ is a group of prime order $p;$
* $\mathbf{G} \in \mathbb{G}^d$ is a vector of $d$ random group elements;
* $H \in \mathbb{G}$ is a random group element; and
* $\mathbb{F}_p$ is the finite field of order $p.$
### `Commit`
The Pedersen vector commitment $\text{Commit}$ is defined as
$$\text{Commit}(\sigma, p(X); r) = \langle\mathbf{a}, \mathbf{G}\rangle + [r]H,$$
for some polynomial $p(X) \in \mathbb{F}_p[X]$ and some blinding factor
$r \in \mathbb{F}_p.$ Here, each element of the vector $\mathbf{a}_i \in \mathbb{F}_p$ is
the coefficient for the $i$th degree term of $p(X),$ and $p(X)$ is of maximal degree
$d - 1.$
### `Open` (prover) and `OpenVerify` (verifier)
The modified inner product argument is an argument of knowledge for the relation
$$\boxed{\{((P, x, v); (\mathbf{a}, r)): P = \langle\mathbf{a}, \mathbf{G}\rangle + [r]H, v = \langle\mathbf{a}, \mathbf{b}\rangle\}},$$
where $\mathbf{b} = (1, x, x^2, \cdots, x^{d-1})$ is composed of increasing powers of the
evaluation point $x.$ This allows a prover to demonstrate to a verifier that the
polynomial contained “inside” the commitment $P$ evaluates to $v$ at $x,$ and moreover,
that the committed polynomial has maximum degree $d − 1.$
The inner product argument proceeds in $k = \log_2 d$ rounds. For our purposes, it is
sufficient to know about its final outputs, while merely providing intuition about the
intermediate rounds. (Refer to Section 3 in the [Halo] paper for a full explanation.)
[Halo]: https://eprint.iacr.org/2019/1021.pdf
Before beginning the argment, the verifier selects a random group element $U$ and sends it
to the prover. We initialise the argument at round $k,$ with the vectors
$\mathbf{a}^{(k)} := \mathbf{a},$ $\mathbf{G}^{(k)} := \mathbf{G}$ and
$\mathbf{b}^{(k)} := \mathbf{b}.$ In each round $j = k, k-1, \cdots, 1$:
* the prover computes two values $L_j$ and $R_j$ by taking some inner product of
$\mathbf{a}^{(j)}$ with $\mathbf{G}^{(j)}$ and $\mathbf{b}^{(j)}$. Note that are in some
sense "cross-terms": the lower half of $\mathbf{a}$ is used with the higher half of
$\mathbf{G}$ and $\mathbf{b}$, and vice versa:
$$
\begin{aligned}
L_j &= \langle\mathbf{a_{lo}^{(j)}}, \mathbf{G_{hi}^{(j)}}\rangle + [l_j]H + [\langle\mathbf{a_{lo}^{(j)}}, \mathbf{b_{hi}^{(j)}}\rangle] U\\
R_j &= \langle\mathbf{a_{hi}^{(j)}}, \mathbf{G_{lo}^{(j)}}\rangle + [l_j]H + [\langle\mathbf{a_{hi}^{(j)}}, \mathbf{b_{lo}^{(j)}}\rangle] U\\
\end{aligned}
$$
* the verifier issues a random challenge $u_j$;
* the prover uses $u_j$ to compress the lower and higher halves of $\mathbf{a}^{(j)}$,
thus producing a new vector of half the original length
$$\mathbf{a}^{(j-1)} = \mathbf{a_{hi}^{(j)}}\cdot u_j^{-1} + \mathbf{a_{lo}^{(j)}}\cdot u_j.$$
The vectors $\mathbf{G}^{(j)}$ and $\mathbf{b}^{(j)}$ are similarly compressed to give
$\mathbf{G}^{(j-1)}$ and $\mathbf{b}^{(j-1)}$.
* $\mathbf{a}^{(j-1)}$, $\mathbf{G}^{(j-1)}$ and $\mathbf{b}^{(j-1)}$ are input to the
next round $j - 1.$
Note that at the end of the last round $j = 1,$ we are left with $a := \mathbf{a}^{(0)}$,
$G := \mathbf{G}^{(0)}$, $b := \mathbf{b}^{(0)},$ each of length 1. The intuition is that
these final scalars, together with the challenges $\{u_j\}$ and "cross-terms"
$\{L_j, R_j\}$ from each round, encode the compression in each round. Since the prover did
not know the challenges $U, \{u_j\}$ in advance, they would have been unable to manipulate
the round compressions. Thus, checking a constraint on these final terms should enforce
that the compression had been performed correctly, and that the original $\mathbf{a}$
satisfied the relation before undergoing compression.
Note that $G, b$ are simply rearrangements of the publicly known $\mathbf{G}, \mathbf{b},$
with the round challenges $\{u_j\}$ mixed in: this means the verifier can compute $G, b$
independently and verify that the prover had provided those same values.

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# Polynomials
Let $A(X)$ be a polynomial over $\mathbb{F}_p$ with formal indeterminate $X$. As an example,
$$
A(X) = a_0 + a_1 X + a_2 X^2 + a_3 X^3
$$
defines a degree-$3$ polynomial. $a_0$ is referred to as the constant term. Polynomials of
degree $n-1$ have $n$ coefficients.
Important notes:
* Multiplication of polynomials produces a product polynomial that is the sum of the
degrees of its factors. Polynomial division subtracts from the degree.
$$\deg(A(X)B(X)) = \deg(A(X)) + \deg(B(X)),$$
$$\deg(A(X)/B(X)) = \deg(A(X)) -\deg(B(X)).$$
* Given a polynomial $A(X)$ of degree $n-1$, if we obtain $n$ evaluations of the
polynomial at distinct points then these evaluations perfectly define the polynomial. In
other words, given these evaluations we can obtain a unique polynomial $A(X)$ of degree
$n-1$ via polynomial interpolation.
* $[a_0, a_1, \cdots, a_{n-1}]$ is the **coefficient representation** of the polynomial
$A(X)$. Equivalently, we could use its **evaluation representation**
$$[(x_0, A(x_0)), (x_1, A(x_1)), \cdots, (x_{n-1}, A(x_{n-1}))]$$
at $n$ distinct points. Either representation uniquely specifies the same polynomial.
> #### (aside) Horner's rule
> Horner's rule allows for efficient evaluation of a polynomial of degree $n-1$, using
> only $n-1$ multiplications and $n-1$ additions. It is the following identity:
> $$\begin{aligned}a_0 &+ a_1X + a_2X^2 + \cdots + a_{n-1}X^{n-1} \\ &= a_0 + X\bigg( a_1 + X \Big( a_2 + \cdots + X(a_{n-2} + X a_{n-1}) \Big)\!\bigg),\end{aligned}$$
## Fast Fourier Transform (FFT)
The FFT is an efficient way of converting between the coefficient and evaluation
representations of a polynomial. It evaluates the polynomial at the $n$th roots of unity
$\{\omega^0, \omega^1, \cdots, \omega^{n-1}\},$ where $\omega$ is a primitive $n$th root
of unity. By exploiting symmetries in the roots of unity, each round of the FFT reduces
the evaluation into a problem only half the size. Most commonly we use polynomials of
length some power of two, $n = 2^k$, and apply the halving reduction recursively.
### Motivation: Fast polynomial multiplication
In the coefficient representation, it takes $O(n^2)$ operations to multiply two
polynomials $A(X)\cdot B(X) = C(X)$:
$$
\begin{aligned}
A(X) &= a_0 + a_1X + a_2X^2 + \cdots + a_{n-1}X^{n-1}, \\
B(X) &= b_0 + b_1X + b_2X^2 + \cdots + b_{n-1}X^{n-1}, \\
C(X) &= a_0\cdot (b_0 + b_1X + b_2X^2 + \cdots + b_{n-1}X^{n-1}) \\
&+ a_1X\cdot (b_0 + b_1X + b_2X^2 + \cdots + b_{n-1}X^{n-1})\\
&+ \cdots \\
&+ a_{n-1}X^{n-1} \cdot (b_0 + b_1X + b_2X^2 + \cdots + b_{n-1}X^{n-1}),
\end{aligned}
$$
where each of the $n$ terms in the first polynomial has to be multiplied by the $n$ terms
of the second polynomial.
In the evaluation representation, however, polynomial multiplication only requires $O(n)$
operations:
$$
\begin{aligned}
A&: \{(x_0, A(x_0)), (x_1, A(x_1)), \cdots, (x_{n-1}, A(x_{n-1}))\}, \\
B&: \{(x_0, B(x_0)), (x_1, B(x_1)), \cdots, (x_{n-1}, B(x_{n-1}))\}, \\
C&: \{(x_0, A(x_0)B(x_0)), (x_1, A(x_1)B(x_1)), \cdots, (x_{n-1}, A(x_{n-1})B(x_{n-1}))\},
\end{aligned}
$$
where each evaluation is multiplied pointwise.
This suggests the following strategy for fast polynomial multiplication:
1. Evaluate polynomials at all $n$ points;
2. Perform fast pointwise multiplication in the evaluation representation ($O(n)$);
3. Convert back to the coefficient representation.
The challenge now is how to **evaluate** and **interpolate** the polynomials efficiently.
Naively, evaluating a polynomial at $n$ points would require $O(n^2)$ operations (we use
the $O(n)$ Horner's rule at each point):
$$
\begin{bmatrix}
A(1) \\
A(\omega) \\
A(\omega^2) \\
\vdots \\
A(\omega^{n-1})
\end{bmatrix} =
\begin{bmatrix}
1&1&1&\dots&1 \\
1&\omega&\omega^2&\dots&\omega^{n-1} \\
1&\omega^2&\omega^{2\cdot2}&\dots&\omega^{2\cdot(n-1)} \\
\vdots&\vdots&\vdots& &\vdots \\
1&\omega^{n-1}&\omega^{2(n-1)}&\cdots&\omega^{(n-1)^2}\\
\end{bmatrix} \cdot
\begin{bmatrix}
a_0 \\
a_1 \\
a_2 \\
\vdots \\
a_{n-1}
\end{bmatrix}.
$$
For convenience, we will denote the matrices above as:
$$\hat{\mathbf{A}} = \mathbf{V}_\omega \cdot \mathbf{A}. $$
($\hat{\mathbf{A}}$ is known as the *Discrete Fourier Transform* of $\mathbf{A}$;
$\mathbf{V}_\omega$ is also called the *Vandermonde matrix*.)
### The (radix-2) Cooley-Tukey algorithm
Our strategy is to divide a DFT of size $n$ into two interleaved DFTs of size $n/2$. Given
the polynomial $A(X) = a_0 + a_1X + a_2X^2 + \cdots + a_{n-1}X^{n-1},$ we split it up into
even and odd terms:
$$
\begin{aligned}
A_{\text{even}} &= a_0 + a_2X + \cdots + a_{n-2}X^{\frac{n}{2} - 1}, \\
A_{\text{odd}} &= a_1 + a_3X + \cdots + a_{n-1}X^{\frac{n}{2} - 1}. \\
\end{aligned}
$$
To recover the original polynomial, we do
$A(X) = A_{\text{even}} (X^2) + X A_{\text{odd}}(X^2).$
Trying this out on points $\omega_n^i$ and $\omega_n^{\frac{n}{2} + i}$,
$i \in [0..\frac{n}{2}-1],$ we start to notice some symmetries:
$$
\begin{aligned}
A(\omega_n^i) &= A_{\text{even}} ((\omega_n^i)^2) + \omega_n^i A_{\text{odd}}((\omega_n^i)^2), \\
A(\omega_n^{\frac{n}{2} + i}) &= A_{\text{even}} ((\omega_n^{\frac{n}{2} + i})^2) + \omega_n^{\frac{n}{2} + i} A_{\text{odd}}((\omega_n^{\frac{n}{2} + i})^2) \\
&= A_{\text{even}} ((-\omega_n^i)^2) - \omega_n^i A_{\text{odd}}((-\omega_n^i)^2) \leftarrow\text{(negation lemma)} \\
&= A_{\text{even}} ((\omega_n^i)^2) - \omega_n^i A_{\text{odd}}((\omega_n^i)^2).
\end{aligned}
$$
Notice that we are only evaluating $A_{\text{even}}(X)$ and $A_{\text{odd}}(X)$ over half
the domain $\{(\omega_n^0)^2, (\omega_n)^2, \cdots, (\omega_n^{\frac{n}{2} -1})^2\} = \{\omega_{n/2}^i\}, i = [0..\frac{n}{2}-1]$ (halving lemma).
This gives us all the terms we need to reconstruct $A(X)$ over the full domain
$\{\omega^0, \omega, \cdots, \omega^{n -1}\}$: which means we have transformed a
length-$n$ DFT into two length-$\frac{n}{2}$ DFTs.
We choose $n = 2^k$ to be a power of two (by zero-padding if needed), and apply this
divide-and-conquer strategy recursively. By the Master Theorem[^master-thm], this gives us
an evaluation algorithm with $O(n\log_2n)$ operations, also known as the Fast Fourier
Transform (FFT).
### Inverse FFT
So we've evaluated our polynomials and multiplied them pointwise. What remains is to
convert the product from the evaluation representation back to coefficient representation.
To do this, we simply call the FFT on the evaluation representation. However, this time we
also:
- replace $\omega^i$ by $\omega^{-i}$ in the Vandermonde matrix, and
- multiply our final result by a factor of $1/n$.
In other words:
$$\mathbf{A} = \frac{1}{n} \mathbf{V}_{\omega^{-1}} \cdot \hat{\mathbf{A}}. $$
(To understand why the inverse FFT has a similar form to the FFT, refer to Slide 13-1 of
[^ifft]. The below image was also taken from [^ifft].)
![](https://i.imgur.com/lSw30zo.png)
## The Schwartz-Zippel lemma
The Schwartz-Zippel lemma informally states that "different polynomials are different at
most points." Formally, it can be written as follows:
> Let $p(x_1, x_2, \cdots, x_n)$ be a nonzero polynomial of $n$ variables with degree $d$.
> Let $S$ be a finite set of numbers with at least $d$ elements in it. If we choose random
> $\alpha_1, \alpha_1, \cdots, \alpha_n$ from $S$,
> $$\text{Pr}[p(\alpha_1, \alpha_2, \cdots, \alpha_n) = 0] \leq \frac{d}{|S|}.$$
In the familiar univariate case $p(X)$, this reduces to saying that a nonzero polynomial
of degree $d$ has at most $d$ roots.
The Schwartz-Zippel lemma is used in polynomial equality testing. Given two multi-variate
polynomials $p_1(x_1,\cdots,x_n)$ and $p_2(x_1,\cdots,x_n)$ of degrees $d_1, d_2$
respectively, we can test if
$p_1(\alpha_1, \cdots, \alpha_n) - p_2(\alpha_1, \cdots, \alpha_n) = 0$ for random
$\alpha_1, \cdots, \alpha_n \leftarrow S,$ where the size of $S$ is at least
$|S| \geq (d_1 + d_2).$ If the two polynomials are identical, this will always be true,
whereas if the two polynomials are different then the equality holds with probability at
most $\frac{\max(d_1,d_2)}{|S|}$.
## Vanishing polynomial
Consider the order-$n$ multiplicative subgroup $\mathcal{H}$ with primitive root of unity
$\omega$. For all $\omega^i \in \mathcal{H}, i \in [n-1],$ we have
$(\omega^i)^n = (\omega^n)^i = (\omega^0)^i = 1.$ In other words, every element of
$\mathcal{H}$ fulfils the equation
$$
\begin{aligned}
Z_H(X) &= X^n - 1 \\
&= (X-\omega^0)(X-\omega^1)(X-\omega^2)\cdots(X-\omega^{n-1}),
\end{aligned}
$$
meaning every element is a root of $Z_H(X).$ We call $Z_H(X)$ the **vanishing polynomial**
over $\mathcal{H}$ because it evaluates to zero on all elements of $\mathcal{H}.$
This comes in particularly handy when checking polynomial constraints. For instance, to
check that $A(X) + B(X) = C(X)$ over $\mathcal{H},$ we simply have to check that
$A(X) + B(X) - C(X)$ is some multiple of $Z_H(X)$. In other words, if dividing our
constraint by the vanishing polynomial still yields some polynomial
$\frac{A(X) + B(X) - C(X)}{Z_H(X)} = H(X),$ we are satisfied that $A(X) + B(X) - C(X) = 0$
over $\mathcal{H}.$
## Lagrange basis functions
Polynomials are commonly written in the monomial basis (e.g. $X, X^2, ... X^n$). However,
when working over a multiplicative subgroup, we find a more natural expression in the
Lagrange basis.
Consider the order-$n$ multiplicative subgroup $\mathcal{H}$ with primitive root of unity
$\omega$. The Lagrange basis corresponding to this subgroup is a set of functions
$\{\mathcal{L}_i\}_{i = 0}^{n-1}$, where
$$
\mathcal{L_i}(\omega^j) = \begin{cases}
1 & \text{if } i = j, \\
0 & \text{otherwise.}
\end{cases}
$$
We can write this more compactly as $\mathcal{L_i}(\omega^j) = \delta_{ij},$ where
$\delta$ is the Kronecker delta function.
Now, we can write our polynomial as a linear combination of Lagrange basis functions,
$$A(X) = \sum_{i = 0}^{n-1} a_i\mathcal{L_i}(X), X \in \mathcal{H},$$
which is equivalent to saying that $p(X)$ evaluates to $a_0$ at $\omega^0$,
$p(\omega^1) = a_1, p(\omega^2) = a_2, \cdots,$ and so on.
When working over a multiplicative subgroup, the Lagrange basis function has a convenient
sparse representation of the form
$$
\mathcal{L}_i(X) = \frac{c_i\cdot(X^{n} - 1)}{X - \omega^i},
$$
where $c_i$ is the barycentric weight. (To understand how this form was derived, refer to
[^barycentric].) For $i = 0,$ we have
$c = 1/n \implies \mathcal{L}_0(X) = \frac{1}{n} \frac{(X^{n} - 1)}{X - 1}$.
Since we cannot assume that the $x_i$'s form a multiplicative subgroup, we consider also
the Lagrange polynomials $\mathcal{L}_i$'s in the general case. Given a set of evaluation
points $\{x_0, x_1, \cdots, x_{n-1}\},$ we can construct
$$
\mathcal{L}_i(X) = \prod_{j\neq i}\frac{X - x_j}{x_i - x_j}, i \in [0..n-1].
$$
Here, every $X = x_j \neq x_i$ will produce a zero numerator term $(x_j - x_j),$ causing
the whole product to evaluate to zero. On the other hand, $X= x_i$ will evaluate to
$\frac{x_i - x_j}{x_i - x_j}$ at every term, resulting in an overall product of one. This
gives the desired Kronecker delta behaviour $\mathcal{L_i}(x_j) = \delta_{ij}.$
### Lagrange interpolation
Given a polynomial in its evaluation representation
$$A: \{(x_0, A(x_0)), (x_1, A(x_1)), \cdots, (x_{n-1}, A(x_{n-1}))\},$$
we can reconstruct its coefficient form in the Lagrange basis:
$$A(X) = \sum_{i = 0}^{n-1} A(x_i)\mathcal{L_i}(X), $$
where $X \in \{x_0, x_1,\cdots, x_{1-n}\}.$
## References
[^master-thm]: Dasgupta, S., Papadimitriou, C. H., & Vazirani, U. V. (2008). "Algorithms" (ch. 2). New York: McGraw-Hill Higher Education. https://people.eecs.berkeley.edu/~vazirani/algorithms/chap2.pdf
[^ifft]: http://www.cs.ust.hk/mjg_lib/Classes/COMP3711H_Fall16/lectures/FFT_Slides.pdf
[^barycentric]: Berrut, J. and Trefethen, L. (2004). "Barycentric Lagrange Interpolation." https://people.maths.ox.ac.uk/trefethen/barycentric.pdf

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## Recursion
> Alternative terms: Induction; Accumulation scheme; Proof-carrying data
However, the computation of $G$ requires a length-$2^k$ multiexponentiation
$\langle \mathbf{G}, \mathbf{s}\rangle,$ where $\mathbf{s}$ is composed of the round
challenges $u_1, \cdots, u_k$ arranged in a binary counting structure. This is the
linear-time computation that we want to amortise across a batch of proof instances.
Instead of computing $G,$ notice that we can express $G$ as a commitment to a polynomial
$$G = \text{Commit}(\sigma, g(X, u_1, \cdots, u_k)),$$
where $g(X, u_1, \cdots, u_k) := \prod_{i=1}^k (u_i + u_i^{-1}X^{2^{i-1}})$ is a
polynomial with degree $2^k - 1.$
| | |
| -------- | -------- |
| <img src="https://i.imgur.com/vMXKFDV.png" width=1900> | Since $G$ is a commitment, it can be checked in an inner product argument. The verifier circuit witnesses $G$ and brings $G, u_1, \cdots, u_k$ out as public inputs to the proof $\pi.$ The next verifier instance checks $\pi$ using the inner product argument; this includes checking that $G = \text{Commit}(g(X, u_1, \cdots, u_k))$ evaluates at some random point to the expected value for the given challenges $u_1, \cdots, u_k.$ Recall from the [previous section](#Polynomial-commitment-using-inner-product-argument) that this check only requires $\log d$ work. <br><br> At the end of checking $\pi$ and $G,$ the circuit is left with a new $G',$ along with the $u_1', \cdots, u_k'$ challenges sampled for the check. To fully accept $\pi$ as valid, we should perform a linear-time computation of $G' = \langle\mathbf{G}, \mathbf{s}'\rangle$. Once again, we delay this computation by witnessing $G'$ and bringing $G, u_1, \cdots, u_k$ out as public inputs to the proof $\pi.$ <br><br> This goes on from one proof instance to the next, until we are satisfied with the size of our batch of proofs. We finally perform a single linear-time computation, thus deciding the validity of the whole batch. |
We recall from the section [Cycles of curves](curves.md#cycles-of-curves) that we can
instantiate this protocol over a two-cycle, where a proof produced by one curve is
efficiently verified in the circuit of the other curve. However, some of these verifier
checks can actually be efficiently performed in the native circuit; these are "deferred"
to the next native circuit (see diagram below) instead of being immediately passed over to
the other curve.
![](https://i.imgur.com/l4HrYgE.png)

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# [WIP] UltraPLONK arithmetisation
We work over a multiplicative subgroup
$\mathcal{H} =\{1, \omega, \omega^2, \cdots, \omega^{n-1}\},$ where $\omega$ is primitive
root of unity, in the Lagrange basis corresponding to these points.
## Polynomial rules
A polynomial rule defines a constraint that must hold between its specified columns at
every row (i.e. at every point in the multiplicative subgroup).
e.g.
```text
a * sa + b * sb + a * b * sm + c * sc + PI = 0
```
## Columns
- **fixed (i.e. "selector") columns**: fixed for all instances of a particular circuit.
These columns toggle parts of a polynomial rule "on" or "off" to form a "custom gate".
- **advice columns**: variable values assigned in each instance of the circuit.
Corresponds to the prover's secret witness.
- **public input**: like advice columns, but publicly known values.
Each column is a vector of $n$ values, e.g. $\mathbf{a} = [a_0, a_1, \cdots, a_{n-1}]$. We
can think of the vector as the evaluation form of the column polynomial
$a(X), X \in \mathcal{H}.$ To recover the coefficient form, we can use
[Lagrange interpolation](polynomials.md#lagrange-interpolation), such that
$a(\omega^i) = a_i.$
## Copy constraints
- Define permutation between a set of columns, e.g. $\sigma(a, b, c)$
- Copy specific cells between these columns, e.g. $b_1 = c_0$
- Construct permuted columns which should evaluate to same value as original columns
## Permutation grand product
$$Z(\omega^i) := \prod_{0 \leq j \leq i} \frac{C_k(\omega^j) + \beta\delta^k \omega^j + \gamma}{C_k(\omega^j) + \beta S_k(\omega^j) + \gamma},$$
where $i = 0, \cdots, n-1$ indexes over the size of the multiplicative subgroup, and
$k = 0, \cdots, m-1$ indexes over the advice columns involved in the permutation. This is
a running product, where each term includes the cumulative product of the terms before it.
> TODO: what is $\delta$? keep columns linearly independent
Check the constraints:
1. First term is equal to one
$$\mathcal{L}_0(X) \cdot (1 - Z(X)) = 0$$
2. Running product is well-constructed. For each row, we check that this holds:
$$Z(\omega^i) \cdot{(C(\omega^i) + \beta S_k(\omega^i) + \gamma)} - Z(\omega^{i-1}) \cdot{(C(\omega^i) + \delta^k \beta \omega^i + \gamma)} = 0$$
Rearranging gives
$$Z(\omega^i) = Z(\omega^{i-1}) \frac{C(\omega^i) + \beta\delta^k \omega^i + \gamma}{C(\omega^i) + \beta S_k(\omega^i) + \gamma},$$
which is how we defined the grand product polynomial in the first place.
### Lookup
Reference: [Generic Lookups with PLONK (DRAFT)](/LTPc5f-3S0qNF6MtwD-Tdg?view)
### Vanishing argument
We want to check that the expressions defined by the gate constraints, permutation
constraints and loookup constraints evaluate to zero at all points in the multiplicative
subgroup. To do this, the prover collapses all the expressions into one polynomial
$$H(X) = \sum_{i=0}^e y^i E_i(X),$$
where $e$ is the number of expressions and $y$ is a random challenge used to keep the
constraints linearly independent. The prover then divides this by the vanishing polynomial
(see section: [Vanishing polynomial](polynomials.md#vanishing-polynomial)) and commits to
the resulting quotient
$$\text{Commit}(Q(X)), \text{where } Q(X) = \frac{H(X)}{Z_H(X)}.$$
The verifier responds with a random evaluation point $x,$ to which the prover replies with
the claimed evaluations $q = Q(x), \{e_i\}_{i=0}^e = \{E_i(x)\}_{i=0}^e.$ Now, all that
remains for the verifier to check is that the evaluations satisfy
$$q \stackrel{?}{=} \frac{\sum_{i=0}^e y^i e_i}{Z_H(x)}.$$
Notice that we have yet to check that the committed polynomials indeed evaluate to the
claimed values at
$x, q \stackrel{?}{=} Q(x), \{e_i\}_{i=0}^e \stackrel{?}{=} \{E_i(x)\}_{i=0}^e.$
This check is handled by the polynomial commitment scheme (described in the next section).