Field operations

In this section we describe the AIR constraints for Miden VM field operations (i.e., arithmetic operations over field elements).

ADD

Assume $a$ and $b$ are the elements at the top of the stack. The ADD operation computes $c\leftarrow (a+b)$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^{\prime }-(s_0+s_1)=0\text{ | degree}=1$$

The effect on the rest of the stack is: * Left shift starting from position $2$.

NEG

Assume $a$ is the element at the top of the stack. The NEG operation computes $b\leftarrow (-a)$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^{\prime }+s_0=0\text{ | degree}=1$$

The effect on the rest of the stack is: * No change starting from position $1$.

MUL

Assume $a$ and $b$ are the elements at the top of the stack. The MUL operation computes $c\leftarrow (a\cdot b)$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^{\prime }-s_0\cdot s_1=0\text{ | degree}=2$$

The effect on the rest of the stack is: * Left shift starting from position $2$.

INV

Assume $a$ is the element at the top of the stack. The INV operation computes $b\leftarrow (a^{-1})$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$1-s_0^{\prime }\cdot s_0=0\text{ | degree}=2$$

Note that the above constraint can be satisfied only if the value in $s_0\ne 0$.

The effect on the rest of the stack is: * No change starting from position $1$.

INCR

Assume $a$ is the element at the top of the stack. The INCR operation computes $b\leftarrow (a+1)$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^{\prime }-(s_0+1)=0\text{ | degree}=1$$

The effect on the rest of the stack is: * No change starting from position $1$.

NOT

Assume $a$ is a binary value at the top of the stack. The NOT operation computes $b\leftarrow (\mathrm{\neg }a)$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^2-s_0=0\text{ | degree}=2$$

$$s_0^{\prime }-(1-s_0)=0\text{ | degree}=1$$

The first constraint ensures that the value in $s_0$ is binary, and the second constraint ensures the correctness of the boolean NOT operation.

The effect on the rest of the stack is: * No change starting from position $1$.

AND

Assume $a$ and $b$ are binary values at the top of the stack. The AND operation computes $c\leftarrow (a\wedge b)$. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_i^2-s_i=0\text{ for }i\in \{0,1\}\text{ | degree}=2$$

$$s_0^{\prime }-s_0\cdot s_1=0\text{ | degree}=2$$

The first two constraints ensure that the value in $s_0$ and $s_1$ are binary, and the third constraint ensures the correctness of the boolean AND operation.

The effect on the rest of the stack is: * Left shift starting from position $2$.

OR

Assume $a$ and $b$ are binary values at the top of the stack. The OR operation computes $c\leftarrow (a\vee b)$ The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_i^2-s_i=0\text{ for }i\in \{0,1\}\text{ | degree}=2$$

$$s_0^{\prime }-(s_1+s_0-s_1\cdot s_0)=0\text{ | degree}=2$$

The first two constraints ensure that the value in $s_0$ and $s_1$ are binary, and the third constraint ensures the correctness of the boolean OR operation.

The effect on the rest of the stack is: * Left shift starting from position $2$.

EQ

Assume $a$ and $b$ are the elements at the top of the stack. The EQ operation computes $c$ such that $c=1$ if $a=b$, and $0$ otherwise. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^{\prime }\cdot (s_0-s_1)=0\text{ | degree}=2$$

$$s_0^{\prime }-(1-(s_0-s_1)\cdot h_0)=0\text{ | degree}=2$$

To satisfy the above constraints, the prover must populate the value of helper register $h_0$ as follows:

• If $s_0\ne s_1$, set $h_0=\frac{1}{s_0-s_1}$.
• Otherwise, set $h_0$ to any value (e.g., $0$).

The effect on the rest of the stack is: * Left shift starting from position $2$.

EQZ

Assume $a$ is the element at the top of the stack. The EQZ operation computes $b$ such that $b=1$ if $a=0$, and $0$ otherwise. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

$$s_0^{\prime }\cdot s_0=0\text{ | degree}=2$$

$$s_0^{\prime }-(1-s_0\cdot h_0)=0\text{ | degree}=2$$

To satisfy the above constraints, the prover must populate the value of helper register $h_0$ as follows: * If $s_0\ne 0$, set $h_0=\frac{1}{s_0}$. * Otherwise, set $h_0$ to any value (e.g., $0$).

The effect on the rest of the stack is:

• No change starting from position $1$.

EXPACC

The EXPACC operation pops top $4$ elements from the top of the stack, performs a single round of exponent aggregation, and pushes the resulting $4$ values onto the stack. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

bit should be a binary.

$$s_0^{\prime 2}-s_0^{\prime }=0\text{ | degree}=2$$

The exp in the next frame should be the square of the exp in the current frame.

$$s_1^{\prime }-s_1^2=0\text{ | degree}=2$$

The value val in the helper register is computed correctly using the bit and exp in next and current frame respectively.

$$h_0-((s_1-1)\ast s_0^{\prime }+1)=0\text{ | degree}=2$$

The acc in the next frame is the product of val and acc in the current frame.

$$s_2^{\prime }-s_2\ast h_0=0\text{ | degree}=2$$

b in the next frame is the right shift of b in the current frame.

$$s_3^{\prime }-(s_3\ast 2+s_0^{\prime })=0\text{ | degree}=1$$

The effect on the rest of the stack is:

• No change starting from position $4$.

EXT2MUL

The EXT2MUL operation pops top $4$ values from the top of the stack, performs multiplication between the two extension field elements, and pushes the resulting $4$ values onto the stack. The diagram below illustrates this graphically.

Stack transition for this operation must satisfy the following constraints:

The first stack element should be unchanged in the next frame.

$$s_0^{\prime }-s_0=0\text{ | degree}=1$$

The second stack element should be unchanged in the next frame.

$$s_1^{\prime }-s_1=0\text{ | degree}=1$$

The third stack element should satisfy the following constraint.

$$s_2^{\prime }-(s_0+s_1)\cdot (s_2+s_3)+s_0\cdot s_2=0\text{ | degree}=2$$

The fourth stack element should satisfy the following constraint.

$$s_3^{\prime }-s_1\cdot s_3+2\cdot s_0\cdot s_2=0\text{ | degree}=2$$

The effect on the rest of the stack is:

• No change starting from position $4$.

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Last update: January 17, 2024

Authors: avenbreaks