Lookup (Type 1) #

Recap of types #

Relation #

$ \mathcal{R}_{\mathtt{lookup1}} := \left\{ \begin{array}{l} (K_\mathsf{Arr}) \end{array} \middle | \begin{array}{l} \mathsf{Arr} = [a_0, a_1, a_2, \dots, a_{n-1}], \\ \mathsf{Poly}_\mathsf{Arr}=\mathsf{FFT.Interp}(\omega,\mathsf{Arr}), \\ K_\mathsf{Arr}=\mathsf{KZG.Commit}(\mathsf{Poly}_\mathsf{Arr}), \end{array} \right\} $

Intuition #

The prover ($\mathcal{P}$) holds an array $\mathsf{Arr}$ of $n$ integers from $\mathbb{Z}_q$: $[a_0, a_1, a_2, \dots, a_{n-1}]$. It will produce a succinct (independent of $n$) proof that each element in $\mathsf{Arr}$ is in $\{0,1\}$ (or another small set). The prover will encode $\mathsf{Arr}$ into the polynomial $\mathsf{Poly}_\mathsf{Arr}$ (using evaluation points on the domain $\mathcal{H}_\kappa$). It will commit to the polynomial: $K_\mathsf{Arr}$. The verifier ($\mathcal{V}$) cannot check any of the $\mathsf{Arr}$ values directly (they may contain secret information, and even if they do not, they are too long to check) so the verifier only sees $K_\mathsf{Arr}$.

In order to check that each element of $\mathsf{Arr}$ is either 0 or 1, consider the expression $\mathsf{Arr}[i] \cdot (\mathsf{Arr}[i] - 1)$. If $\mathsf{Arr}[i]$ is 0, then the first part of the product is zero, and if $\mathsf{Arr}[i]$ is 1, then the second part of the product is zero. If $\mathsf{Arr}[i]$ is not 0 or 1, then the expression will be non-zero. Thus, to check our condition, it suffices to check that $\mathsf{Arr}[i] \cdot (\mathsf{Arr}[i] - 1) = 0$ for $i$ from 0 to $n-1$.

Protocol Details #

Array Level #

• $\mathcal{P}$ holds an array $\mathsf{Arr} = [a_0, a_1, a_2, \dots, a_{n-1}]$ of $n$ integers ($a_i \in \mathbb{Z}_q$)

Polynomial Level #

We assume that $\mathsf{Arr}$ is encoded as the y-coordinates into a univariant polynomial where the x-coordinates (called the domain $\mathcal{H}_\kappa$) are chosen as the multiplicative group of order $\kappa$ with generator $\omega\in\mathbb{G}_\kappa$ (see Background for more). In short, $\omega^0$ is the first element and $\omega^{\kappa-1}$ is the last element of $\mathcal{H}_\kappa$. If $\kappa$ is larger than the length of the array, the array can be padded with elements of value 0 or 1.

Recall the constraint we want to prove:

1. $\mathsf{Arr}[i] \cdot (\mathsf{Arr}[i] - 1) = 0$ for $i$ from 0 to $n-1$

We write our constraint in polynomial form and label it:

1. For all $X$ from $\omega^0$ to $\omega^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Vanish}(X)=\mathsf{Poly}_\mathsf{Arr}(X) \cdot (\mathsf{Poly}_\mathsf{Arr}(X) - 1) =0$

This equation is true for every value of $X \in \mathcal{H}_\kappa$ (but not necessarily true outside of these values). To show this, we divide each polynomial by $X^\kappa - 1$, which is a minimal vanishing polynomial for $\mathcal{H}_\kappa$ that does not require interpolation to create. If the quotient is polynomial (and not a rational function), then $\mathsf{Poly}_\mathsf{Vanish}(X)$ must be vanishing on $\mathcal{H}_\kappa$ too. Specifically, the prover computes:

1. $Q(X) = \frac{\mathsf{Poly}_\mathsf{Vanish}(X)}{X^\kappa - 1}$

By rearranging, we can get $\mathsf{Poly}_\mathsf{Zero}(X)$ as a true zero polynomial (zero at every value both in $\mathcal{H}_\kappa$ and outside of it):

1. $\mathsf{Poly}_\mathsf{Zero}(X)=\mathsf{Poly}_\mathsf{Vanish}(X) - Q(X)\cdot (X^\kappa - 1)=0$

Ultimately the lookup1 argument will satisfy the following constraints at the Commitment Level:

1. Show $Q(X)$ exists (as a polynomial that evenly divides the divisor)
2. Show $\mathsf{Poly}_\mathsf{Zero}(X)$ is correctly constructed from $\mathsf{Poly}_\mathsf{Arr}(X)$
3. Show $\mathsf{Poly}_\mathsf{Zero}(X)$ is the zero polynomial

Commitment Level #

The verifier will never see the arrays or polynomials themselves. They are undisclosed because they either (i) contain private data or (ii) they are too large to examine and maintain a succinct proof system. Instead the prover will use commitments.

The prover will write the following commitments to the transcript:

• $K_\mathsf{Arr}=\mathsf{KZG.Commit}(\mathsf{Poly}_\mathsf{Arr}(X))$
• $K_Q=\mathsf{KZG.Commit}(Q(X))$

The prover will generate a random challenge evaluation point (using strong Fiat-Shamir) on the polynomial that is outside of $\mathcal{H}_\kappa$. Call this point $\zeta$. The prover will write the point and opening proofs to the transcript:

• $\zeta$
• $\mathsf{Poly}_\mathsf{Arr}(\zeta)=\mathsf{KZG.Open}(K_\mathsf{Arr},\zeta)$
• $Q(\zeta)=\mathsf{KZG.Open}(K_Q,\zeta)$

To check the proof, the verifier uses the transcript to construct the value $Y_\mathsf{Zero}$ as follows:

• $Y_\mathsf{Vanish}=\mathsf{Poly}_\mathsf{Arr}(\zeta) \cdot (\mathsf{Poly}_\mathsf{Arr}(\zeta) - 1)$
• $Y_\mathsf{Zero}=Y_\mathsf{Vanish} - Q(\zeta)\cdot (\zeta^\kappa - 1)$

Finally, if the constraint system is true, the following constraint will be true (and will be false otherwise with overwhelming probability, due to the Schwartz-Zippel lemma on $\zeta$) :

• $Y_\mathsf{Zero}\overset{?}{=}0$

Implementations #

Security Proof #

Completeness #

If $Y_\mathsf{Zero}$ is zero, then $\mathcal{V}$ will accept. Therefore, to show completeness, we show that any prover who holds $\mathsf{Arr}$ such that $\mathsf{Arr}[i] \in \{0, 1\} \forall 0 \leq i \leq n$, can follow the steps outlined in the above protocol and the resulting $Y_\mathsf{Zero}$ will be equal to zero. To see this, observed that $Y_\mathsf{Zero}$

$= Y_\mathsf{Vanish} - Q(\zeta)\cdot (\zeta^\kappa - 1)$

$ = \mathsf{Poly}_\mathsf{Arr}(\zeta) \cdot (\mathsf{Poly}_\mathsf{Arr}(\zeta) - 1) - Q(\zeta)\cdot (\zeta^\kappa - 1)$

$ = \mathsf{Poly}_\mathsf{Arr}(\zeta) \cdot (\mathsf{Poly}_\mathsf{Arr}(\zeta) - 1) - \frac{\mathsf{Poly}_\mathsf{Vanish}(\zeta)}{\zeta^\kappa - 1}\cdot (\zeta^\kappa - 1)$

$ = \mathsf{Poly}_\mathsf{Arr}(\zeta) \cdot (\mathsf{Poly}_\mathsf{Arr}(\zeta) - 1) - [\mathsf{Poly}_\mathsf{Arr}(\zeta) \cdot (\mathsf{Poly}_\mathsf{Arr}(\zeta) - 1)]$

$= 0$

Where the third equality relies on the fact that $\mathsf{Poly_{Vanish}}(X)$ is divisible by $X^\kappa -1$. This is true if $\mathsf{Poly_{Vanish}}(\zeta)$ is vanishing on $\mathcal{H}_\kappa$, i.e. if $\mathsf{Poly}_\mathsf{Arr}(X) \cdot (\mathsf{Poly}_\mathsf{Arr}(X) - 1) = 0 \space \forall X \in \mathcal{H}_\kappa$. This is true if $\mathsf{Arr}[i] \cdot (\mathsf{Arr}[i] - 1) = 0 \space \forall i \in [0, \kappa -1]$, since $\mathsf{Poly}(\omega^i) = \mathsf{Arr}[i] \space \forall i \in [0, \kappa - 1]$. But this is precisely the condition we assumed held for the prover (since the array gets padded with $1$’s or $0$’s if $n \lt \kappa$), so the $Y_\mathsf{Zero}$ it creates by following the protocol is zero, and the transcript will be accepted.

Soundness #

We prove knowledge soundness in the Algebraic Group Model (AGM). To do so, we must prove that there exists an efficient extractor $\mathcal{E}$ such that for any algebraic adversary $\mathcal{A}$ the probability of $\mathcal{A}$ winning the following game is $\mathsf{negl}(\lambda)$.

1. Given $[g, g^\tau, g^{\tau^2}, \dots,g^{\tau^{n-1}}]$ $\mathcal{A}$ outputs commitments to $\mathsf{Poly}_\mathsf{Arr}(X)$, $Q(X)$

2. $\mathcal{E}$, given access to $\mathcal{A}$’s outputs from the previous step, outputs $\mathsf{Poly}_\mathsf{Arr}(X)$, $Q(X)$.

3. $\mathcal{A}$ plays the part of the prover in showing that $Y_{\mathsf{Zero}}$ is zero at a random challenge $\zeta$

4. $\mathcal{A}$ wins if:

   i) $\mathcal{V}$ accepts at the end of the protocol

   ii) $\mathsf{Arr}[i]\neq 0$ and $\mathsf{Arr}[i]\neq 1$ for some $i \in [0, n-1]$

Our proof is as follows:

For the second win condition to be fulfilled, there must be at least one entry is $\mathsf{Arr}$ that is not 0 or 1. But then $\mathsf{Poly}_\mathsf{Vanish}(X)$ is not vanishing on $\mathcal{H}_\kappa$, so $Q(X)$ is not a polynomial (it is a rational function). This means that $\mathcal{A}$ cannot calculated the correct commitment value $g^{Q(\tau)}$ without solving the t-SDH. Thus, $\mathcal{A}$ chooses an arbitrary value for $Q(\tau)$ and writes $K_Q = g^{Q(\tau)}$ to the transcript. Before this, it also writes a commitment to $\mathsf{Poly}_\mathsf{Arr}(X)$. Both commitments $\mathcal{A}$ has written are linear combinations of the elements in $[g, g^\tau, g^{\tau^2}, \dots,g^{\tau^{n-1}}]$. $\mathcal{E}$ is given these coefficients (since $\mathcal{A}$ is an algebraic adversary) so $\mathcal{E}$ can output the original polynomials.

$\mathcal{A}$ then obtains the random challenge $\zeta$ (using strong Fiat-Shamir). By the binding property of KZG commitments, $\mathsf{Poly}_\mathsf{Arr}(\zeta)$, can only feasibly be opened to one value. For $\mathcal{A}$ to have the verifier accept, it must produce a proof that $Q(\zeta)$ opens to $Q(\zeta) = \frac{Y_\mathsf{Vanish1}}{(\zeta^\kappa - 1)}$. This means being able to produce $g^{q(\tau)}$ where $q(\tau) = \frac{Q(\tau) - Q(\zeta)}{\tau - \zeta}$ and $Q(\zeta) = \frac{Y_\mathsf{Vanish1}}{(\zeta^\kappa - 1)}$. Since $Q(\tau)$ and $Q(\zeta)$ are known, this implies knowing $g^{\frac{1}{\tau - \zeta}}$. Thus $\mathcal{A}$ would have found $\langle\zeta,g^{\frac{1}{\tau - \zeta}}\rangle$, which is the t-SDH problem. We have shown that creating an accepting proof reduces to the t-SDH, so $\mathcal{A}$’s probability of success is negligible.

Zero-Knowledge #