# Algebraic Intermediate Representation (AIR)#

Arithmetization is the reduction of computational statements to a set of algebraic statements involving a set of bounded-degree polynomials.

In the STARK protocol, the output of arithmetization is an Algebraic Intermediate Representation of a computation.

Informally, AIR consists of the following three elements:

1. execution trace which is a two-dimensional matrix, in which each row represents the state of the computation at a single point in time and each column corresponds to an algebraic register tracked over all steps of the computation. Let $$T$$ denote the execution trace matrix.

2. transition constraints which define algebraic relationships between two (or more) rows of the execution trace.

3. boundary constraints which enforce equality between certain cells of the execution trace and a set of constant values. Boundary constraints can be thought of as defining a set of input and output values for the computation.

## Execution Trace#

Denote $$m$$ as the width of the execution trace and $$n$$ as the number of steps in the execution trace, i.e. $$T$$ is an $$m \times n$$ matrix.

We define the register trace of a register $$k$$ as the polynomial interpolation $$f_{k}$$ of the set $$\left\{\left(\omega^{i}, T[i][k]\right) \mid i \in[0, n)\right\}$$, where $$\omega$$ is a generator of a multiplicative subgroup of size $$n$$ in the base field specified for an instantiation of a STARK protocol.

The set $$\left\{f_{k} \mid k \in[0, m)\right\}$$ is called the set of trace polynomials.

Notice that if $$f_{k}(x)$$ is the value in the execution trace matrix in column $$k$$ and row $$i$$, then $$f_{k}(x \cdot \omega)$$ is the value in $$k$$ at step $$i+1$$.

For efficient execution of the STARK protocol, $$n$$ must be a power of two. This allows us to use FFT-based polynomial evaluation and interpolation, which have the complexity of $$O(n \log n)$$

Thus, the base field for the STARK protocol must be 2 -smooth , which indeed is the case for out selected field with modulus $$q=2^{128}-45 \cdot 2^{40}+1$$.

Note

A field is k-smooth if it contains a subgroup (multiplicative or additive) all of whose prime divisors are at most $$k$$. For example, a prime field of size $$q$$ such that $$q-1$$ is divisible by a large power of 2 , is 2 -smooth.

In addition to registers of the execution trace, we also use periodic registers (also called periodic columns), which are not included in the execution trace but can be referenced in transition constraints.

Periodic columns are typically used in STARKs to encode a small set of values which can be represented by succinct polynomials of size much smaller than $$n$$. One example is a register where values repeats in a cycle and the length of the cycle is a power of two.

## Constraints#

Both boundary and transition constraints are defined by rational functions of the form:

$\frac{p(x)}{z(x)}$

where, $$p(x)$$ defines the constraint relationship , and $$z(x)$$ defines the constraint domain (a set of steps at which the constraint should hold). This constraint is said to hold if the polynomial $$z$$ divides the polynomial $$p$$.

For boundary constraints, $$p(x)$$ has the following form:

$p(x)=c\left(f_{k}(x)\right)$

where, $$f_{k}(x)$$ is the trace polynomial for register $$k$$ against which the constraint is enforced.

Example 1

For example, to specify that the value in the first column of the first row in the execution trace must be 1 , we could use the following constraint:

$\frac{f_{0}(x)-1}{x-1}$

Example 2

Similarly, to specify that the value in the 7 th row of the second column must be 987 , we could use the following constraint:

$\frac{f_{1}(x)-987}{x-\omega^{7}}$

For transition constraints , $$p(x)$$ has the following form:

$p(x)=c\left(\left\{f_{0}(x), \ldots, f_{m-1}(x)\right\},\left\{f_{0}(x \cdot \omega), \ldots, f_{m-1}(x \cdot \omega)\right\}\right)$

that is, $$p(x)$$ is a function of all register values in two consecutive steps of a computation.

Example 3

For example, the following constraint enforces that a value in the first register of the execution trace must be incremented by 1 at every step:

$\frac{f_{0}(x \cdot \omega)-\left(f_{0}(x)+1\right)}{\prod_{i=0}^{n-1}\left(x-\omega^{i}\right)}$

Additionally, since trace polynomials are evaluated over a multiplicative subgroup of a field, the denominator of the constraint above can be expressed succinctly, and the constraint can be rewritten as:

$\frac{f_{0}(x \cdot \omega)-\left(f_{0}(x)+1\right)}{x^{n}-1}$