Multiplication (Type 2) #

Recap of types #

Relation #

$ \mathcal{R}_{\mathtt{mult2}} := \left\{ \begin{array}{l} (K_\mathsf{Arr},\mathsf{Prod}_\mathsf{Arr}) \end{array} \middle | \begin{array}{l} \mathsf{Arr} = [a_0, a_1, a_2, \dots, a_{n-1}], \\ \mathsf{Prod}_\mathsf{Arr} = \prod_{i = 0}^{n-1} a_i, \\ \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 $\mathsf{Arr} = [a_0, a_1, a_2, \dots, a_{n-1}]$ of $n$ integers (from $\mathbb{Z}_q$) and a disclosed integer $\mathsf{Prod}_\mathsf{Arr}$. It will produce a succinct (independent of $n$) proof that $\mathsf{Prod}_\mathsf{Arr}$ is the product of all the elements in the array. The prover will encode the array into a polynomial $\mathsf{Poly}_\mathsf{Arr}$ (using evaluation points on the domain $\mathcal{H}_\kappa$) and commit to the polynomial $K_\mathsf{Arr}$. The verifier ($\mathcal{V}$) cannot check $\textsf{Arr}$ or $\mathsf{Poly}_\mathsf{Arr}$ directly (they may contain secret information, and even if they do not, it is too long to check) so the verifier only sees $K_\mathsf{Arr}$ and the asserted value $\mathsf{Prod_\mathsf{Arr}}$.

In order to prove $K_\mathsf{Arr}$ and $\mathsf{Prod}_\mathsf{Arr}$ are consistent, the prover will build a helper array $\mathsf{Acc}_\mathsf{Arr}$ called an accumulator (or accumulating array or incremental array). This should not be confused with accumulators from cryptography, which are a concept related to succinct proofs but are distinct. As with $\mathsf{Arr}$, the prover will also encode $\mathsf{Acc}$ as a polynomial and provide a commitment of it to the verifier. The idea is that the prover will prove a relation between $\mathsf{Arr}$ and $\mathsf{Acc}$; and a relation between $\mathsf{Acc}$ and $\mathsf{Prod_\mathsf{Arr}}$. Put together, it will imply the correct relation between $\mathsf{Arr}$ and $\mathsf{Prod_\mathsf{Arr}}$.

Consider a small numeric example in $\mathbb{Z}_{97}$ where $\mathsf{Arr}= [84,67,11,92,36,67]$ and $\mathsf{Prod}_\mathsf{Arr}=72$. The first idea is to get $\mathsf{Prod}_\mathsf{Arr}$ into an array. Say, we just append it: $\mathsf{Arr}''= [84,67,11,92,36,67,72] $. How does the prover show $\mathsf{Arr}''$ is correct? The last value of the array depends on every single element that precedes it, which will be a complex constraint to prove.

An alternative idea is to create a new array that starts the same as $\mathsf{Arr}$ and ends up at $\mathsf{Prod}_\mathsf{Arr}$ by folding in the integers from $\mathsf{Arr}$ one-by-one with multiplication. Then each value in the new array will depend on only two values, as below.

The first value in $\mathsf{Acc}$ will be a copy of the first value from $\mathsf{Arr}$:

• $\mathsf{Arr}= [84,67,11,92,36,67]$

• $\mathsf{Acc}= [84, \bot,\bot,\bot,\bot,\bot] $

The next value will be the multiplication (mod 97) of: 67 (the value at the same index in $\mathsf{Arr}$) and 84 (the previous value in $\mathsf{Acc}$):

• $\mathsf{Arr}= [84,67,11,92,36,67]$

• $ \mathsf{Acc} = [84, (67\cdot84),\bot,\bot,\bot,\bot] = [84, 2,\bot,\bot,\bot,\bot]$

The next value will be the multiplication of: 11 (the value at the same index in $\mathsf{Arr}$) and 2 (the previous value in $\mathsf{Acc}$):

• $\mathsf{Arr}= [84,67,11,92,36,67]$
• $ \mathsf{Acc} = [84, 2,(11\cdot2),\bot,\bot,\bot] = [84,2,22,\bot,\bot,\bot]$
• $ \mathsf{Acc} = [84, 2,22,(92\cdot22),\bot,\bot] = [84,2,22,84,\bot,\bot]$
• $ \mathsf{Acc} = [84, 2,22,84,(36\cdot84),\bot] = [84,2,22,84,17,\bot]$
• $ \mathsf{Acc} = [84,2,22,84,17,(67\cdot17)] = [84, 2, 22, 84, 17, 72]$
• $\mathsf{Prod}_\mathsf{Arr}=72$

Notice the last value in $\mathsf{Acc}$ is $\mathsf{Prod_\mathsf{Arr}}$. The prover wants to show three constraints:

1. The first value in $\mathsf{Acc}$ matches the first value in $\mathsf{Arr}$,
2. The rest of the values in $\mathsf{Acc}$ are of the form $\mathsf{Acc}[i]=\mathsf{Arr}[i]\cdot\mathsf{Acc}[i-1]$,
3. The last value in $\mathsf{Acc}$ matches $\mathsf{Prod}_\mathsf{Arr}$.

If all three constraints are true, then $\mathsf{Prod}_\mathsf{Arr}$ is the product of the elements of $\mathsf{Arr}$.

Last, while it is not necessary to do, it is often convenient to hold the the value $\mathsf{Prod}_\mathsf{Arr}$ at the start of the array $\mathsf{Acc}$ instead of the end. For this reason, the mathematical explanation below will construct $\mathsf{Acc}$ “backwards” (or right-to-left) from the above example, where the last value of $\mathsf{Acc}$ matches the last value of $\mathsf{Arr}$, the values are folded in from right to left, and the first (leftmost) value of $\mathsf{Acc}$ is $\mathsf{Prod}_\mathsf{Arr}$:

• $\mathsf{Arr}= [84,67,11,92,36,67]$
• $ \mathsf{Acc} = [\bot, \bot, \bot, \bot, \bot, 67]$
• $ \mathsf{Acc} = [\bot, \bot, \bot, \bot, 84, 67]$
• $\ldots$
• $ \mathsf{Acc} = [72, 84, 36, 65, 84, 67]$
• $\mathsf{Prod}_\mathsf{Arr}=72$

Protocol Details #

Array Level #

• $\mathcal{P}$ holds an array $\mathsf{Arr} = [a_0, a_1, a_2, \dots, a_{n-1}]$ of $n$ integers ($a_i \in \mathbb{Z}_q$)
• $\mathcal{P}$ computes array $\mathsf{Acc}$ as follows:
  • $\mathsf{Acc}[n-1]\leftarrow\mathsf{Arr}[n-1]$
  • $\mathsf{Acc}[i]\leftarrow\mathsf{Arr}[i]\cdot\mathsf{Acc}[i+1]$ for $i$ from $n-2$ to 0
• $\mathcal{P}$ computes $\mathsf{Prod}_\mathsf{Arr}\leftarrow\mathsf{Acc}[0]$

Polynomial Level #

We assume arrays $\mathsf{Arr}$ and $\mathsf{Acc}$ are 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 1 (which will not change the product).

Recall the three constraints we want to prove (now adjusted to fit with an $\mathsf{Acc}$ that is constructed “backwards,” as noted above):

1. The last value in $\mathsf{Acc}$ matches the last value in $\mathsf{Arr}$,
2. The rest of the values in $\mathsf{Acc}$ are of the form $\mathsf{Acc}[i]=\mathsf{Arr}[i]\cdot\mathsf{Acc}[i-1]$,
3. The first value in $\mathsf{Acc}$ matches $\mathsf{Prod}_\mathsf{Arr}$.

In polynomial form, the constraints are:

1. For $X=w^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Acc}(X)=\mathsf{Poly}_\mathsf{Arr}(X)$,
2. For all $X$ except $X=\omega^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Acc}(X)=\mathsf{Poly}_\mathsf{Arr}(X)\cdot\mathsf{Poly}_\mathsf{Acc}(\omega\cdot X)$
3. For $X=w^0$: $\mathsf{Poly}_\mathsf{Acc}(X)=\mathsf{Prod}_\mathsf{Arr}$

In constraint 2, $\mathsf{Poly}_\mathsf{Acc}(\omega\cdot X)$ can also be conceptualized as rotate applied to $\mathsf{Poly}_\mathsf{Acc}(X)$ by one element (rightward in the array view). Also note that constraint 2 does not hold at $X=\omega^{\kappa-1}$ because this value is defined by constraint 1 (for the last value of $X$, the “next” value, $\omega X$, wraps back to the first element of the array which is a boundary condition).

We adjust each of these constraints to show an equality with 0:

1. For $X=w^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Poly}_\mathsf{Arr}(X)=0$,
2. For all $X$ except $X=\omega^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Poly}_\mathsf{Arr}(X)\cdot\mathsf{Poly}_\mathsf{Acc}(\omega\cdot X)=0$
3. For $X=w^0$: $\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Prod}_\mathsf{Arr}=0$

Next we take care of the “for $X$” conditions by zeroing out the rest of the polynomial that is not zero. See the gadget zero1 for more on why this works.

1. $\mathsf{Poly}_\mathsf{Vanish1}(X)=(\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Poly}_\mathsf{Arr}(X))\cdot\frac{(X^\kappa-1)}{(X-\omega^{\kappa-1})}=0$,
2. $\mathsf{Poly}_\mathsf{Vanish2}(X)=(\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Poly}_\mathsf{Arr}(X)\cdot\mathsf{Poly}_\mathsf{Acc}(\omega\cdot X))\cdot(X-\omega^{\kappa-1})=0$
3. $\mathsf{Poly}_\mathsf{Vanish3}(X)=(\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Prod}_\mathsf{Arr})\cdot\frac{(X^\kappa-1)}{(X-\omega^0)}=0$

These equations are 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 quotients are polynomials (and not rational functions), then $\mathsf{Poly}_\mathsf{Vanish1}(X)$, $\mathsf{Poly}_\mathsf{Vanish2}(X)$, and $\mathsf{Poly}_\mathsf{Vanish3}(X)$ must be vanishing on $\mathcal{H}_\kappa$ too. Specifically, the prover computes,

1. $Q_1(X) = \frac{\mathsf{Poly}_\mathsf{Vanish1}(X)}{X^\kappa - 1}$
2. $Q_2(X) = \frac{\mathsf{Poly}_\mathsf{Vanish2}(X)}{X^\kappa - 1}$
3. $Q_3(X) = \frac{\mathsf{Poly}_\mathsf{Vanish3}(X)}{X^\kappa - 1}$

We can replace polynomials $Q_1(X)$, $Q_2(X)$, and $Q_3(X)$ with a single polynomial $Q(X)$. We can do this because all three constraints have the same format: $\mathsf{Poly}_\mathsf{Vanish_i}(X)=0$. The batching technique is to create a new polynomial with all three $\mathsf{Poly}_\mathsf{Vanish_i}(X)$ values as coefficients. If and (overwhelmingly) only if all three are vanishing, then so will the new polynomial. This polynomial will be evaluated at a random challenge point $\rho$ selected after the commitments to the earlier polynomials are fixed.

$Q(X) = \frac{\mathsf{Poly}_\mathsf{Vanish1}(X) + \mathsf{Poly}_\mathsf{Vanish2}(X) \rho + \mathsf{Poly}_\mathsf{Vanish3}(X)\rho^2}{X^\kappa - 1}$

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):

$\mathsf{Poly}_\mathsf{Zero}(X)=\mathsf{Poly}_\mathsf{Vanish1}(X) + \rho \mathsf{Poly}_\mathsf{Vanish2}(X) + \rho^2 \mathsf{Poly}_\mathsf{Vanish3}(X) - Q(X)\cdot (X^\kappa - 1)=0$

Ultimately the mult2 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 $\mathsf{Poly}_\mathsf{Acc}(X)$, $\mathsf{Poly}_\mathsf{Acc}(\omega X)$, $\mathsf{Poly}_\mathsf{Arr}(X)$, and $\mathsf{Prod}_\mathsf{Arr}$
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:

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

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 $\rho$. It will be used by the prover to create polynomial $Q(X)$ (see above) and the prover will write to the transcript:

• $\rho$
• $K_Q=\mathsf{KZG.Commit}(Q(X))$

The prover will generate a second 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:

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

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

• $Y_\mathsf{Vanish1}=(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta))\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^{\kappa-1})}$
• $Y_\mathsf{Vanish2}=(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta)\cdot\mathsf{Poly}_\mathsf{Acc}(\omega\cdot \zeta))\cdot(\zeta-\omega^{\kappa-1})$
• $Y_\mathsf{Vanish3}=(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Prod}_\mathsf{Arr})\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^0)}$
• $Y_\mathsf{Zero}=Y_\mathsf{Vanish1} + \rho Y_\mathsf{Vanish2} + \rho^2 Y_\mathsf{Vanish3} - Q(\zeta)\cdot (\zeta^\kappa - 1)$

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 $\rho$ and $\zeta$) :

• $Y_\mathsf{Zero}\overset{?}{=}0$

Implementations #

• Rust
• Mathematica (Toy Example)

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 $\mathsf{Prod}_\mathsf{Arr}=\prod_{i = 0}^{n-1} \mathsf{Arr}[i] \space \forall i \in [0, n - 1]$ 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 $Y_\mathsf{Zero}$

$= Y_\mathsf{Vanish1} + \rho Y_\mathsf{Vanish2} + \rho^2 Y_\mathsf{Vanish3} - Q(\zeta)\cdot (\zeta^\kappa - 1)$

$= (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta))\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^{\kappa-1})}) + \rho(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta)+\mathsf{Poly}_\mathsf{Acc}(\omega\cdot \zeta))\cdot(\zeta-\omega^{\kappa-1}) + \rho^2 (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Sum}_\mathsf{Arr})\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^0)} - Q(\zeta)\cdot (\zeta^\kappa - 1)$

$= (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta))\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^{\kappa-1})}) + \rho(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta)+\mathsf{Poly}_\mathsf{Acc}(\omega\cdot \zeta))\cdot(\zeta-\omega^{\kappa-1}) + \rho^2 (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Sum}_\mathsf{Arr})\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^0)} \newline - \frac{\mathsf{Poly}_\mathsf{Vanish1}(\zeta) + \mathsf{Poly}_\mathsf{Vanish2}(\zeta) \rho + \mathsf{Poly}_\mathsf{Vanish3}(\zeta)\rho^2}{\zeta^\kappa - 1} \cdot(\zeta^\kappa - 1)$

$= (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta))\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^{\kappa-1})}) + \rho(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta)+\mathsf{Poly}_\mathsf{Acc}(\omega\cdot \zeta))\cdot(\zeta-\omega^{\kappa-1}) + \rho^2 (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Sum}_\mathsf{Arr})\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^0)} \newline - ((\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta))\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^{\kappa-1})}) + \rho(\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Poly}_\mathsf{Arr}(\zeta)+\mathsf{Poly}_\mathsf{Acc}(\omega\cdot \zeta))\cdot(\zeta-\omega^{\kappa-1}) \newline + \rho^2 (\mathsf{Poly}_\mathsf{Acc}(\zeta)-\mathsf{Sum}_\mathsf{Arr})\cdot\frac{(\zeta^\kappa-1)}{(\zeta-\omega^0)})$

$= 0$

Where the third equality relies on the fact that $\mathsf{Poly}_\mathsf{Vanish1}(\zeta) + \mathsf{Poly}_\mathsf{Vanish2}(\zeta) \rho + \mathsf{Poly}_\mathsf{Vanish3}(\zeta)\rho^2$ is divisible by $X^\kappa - 1$. This is true if $\mathsf{Poly_{Vanish_1}}(X), \mathsf{Poly_{Vanish_2}}(X)$ and $\mathsf{Poly_{Vanish_3}}(X),$ are all vanishing on $\mathcal{H_\kappa}$, i.e. if all three of the following conditions hold for all $X \in \mathcal{H}_\kappa$:

1. $(\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Poly}_\mathsf{Arr}(X))\cdot\frac{(X^\kappa-1)}{(X-\omega^{\kappa-1})}=0$
2. $ (\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Poly}_\mathsf{Arr}(X) \cdot \mathsf{Poly}_\mathsf{Acc}(\omega\cdot X))\cdot(X-\omega^{\kappa-1})=0$
3. $ (\mathsf{Poly}_\mathsf{Acc}(X)-\mathsf{Prod}_\mathsf{Arr})\cdot\frac{(X^\kappa-1)}{(X-\omega^0)}=0$

These conditions, in turn, hold if:

1. For $X=w^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Acc}(X)=\mathsf{Poly}_\mathsf{Arr}(X)$
2. For all $X$ except $X=\omega^{\kappa-1}$: $\mathsf{Poly}_\mathsf{Acc}(X)=\mathsf{Poly}_\mathsf{Arr}(X) \cdot \mathsf{Poly}_\mathsf{Acc}(\omega\cdot X)$
3. For $X=w^0$: $\mathsf{Poly}_\mathsf{Acc}(X)=\mathsf{Prod}_\mathsf{Arr}$

Where we get the “For $X$” due to zeroing parts of the polynomials (see zero1). Since $\mathsf{Poly_j}(\omega^i) = \mathsf{Arr_j}[i] \space \forall i \in [0, \kappa - 1]$, the above conditions are true if:

1. The last value in $\mathsf{Acc}$ matches the last value in $\mathsf{Arr}$
2. The rest of the values in $\mathsf{Acc}$ are of the form $\mathsf{Acc}[i]=\mathsf{Arr}[i] \cdot \mathsf{Acc}[i-1]$
3. The first value in $\mathsf{Acc}$ matches $\mathsf{Prod}_\mathsf{Arr}$

Which are precisely the conditions the Intuitions sections explains will hold if the prover constructs their Accumulator by following the protocol and using $\mathsf{Arr}$ such that $\mathsf{Prod}_\mathsf{Arr}=\prod_{i = 0}^{n-1} \mathsf{Arr}[i] \space \forall i \in [0, n - 1]$. This is what we assumed true about the prover, thus the $Y_\mathsf{Zero}$ it creates by following the protocol is zero, and its 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 $\mathsf{Poly}_\mathsf{Acc}(X)$, $\mathsf{Poly}_\mathsf{Arr}(X)$, $Q(X)$

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

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) $\mathsf{Prod}_\mathsf{Arr}\neq\prod_{i = 0}^{n-1} \mathsf{Arr}[i]$

Our proof is as follows:

For the second win condition to be fulfilled, one of the three constraints must be false. But then the $\mathsf{Poly}_\mathsf{Vanish}$ corresponding to that constraint 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 commitments to $\mathsf{Poly}_\mathsf{Acc}(X)$, and $\mathsf{Poly}_\mathsf{Arr}(X)$. Each commitment $\mathcal{A}$ has written is a linear combination 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, $\mathsf{Poly}_\mathsf{Acc}(\zeta)$, $\mathsf{Poly}_\mathsf{Arr}(\zeta)$, and $\mathsf{Poly}_\mathsf{Acc}(\zeta \cdot \omega)$ can only feasibly be opened to one value each. For $\mathcal{A}$ to have the verifier accept, it must produce a proof that $Q(\zeta)$ opens to $Q(\zeta) = \frac{Y_\mathsf{Vanish1} + \rho Y_\mathsf{Vanish2} + \rho^2Y_\mathsf{Vanish3}}{\zeta^\kappa - 1}$. This means being able to produce $g^{q(\tau)}$ where $q(\tau) = \frac{Q(\tau) - Q(\zeta)}{\tau - \zeta}$ and $Q(\zeta) = \frac{Y_\mathsf{Vanish1} + \rho Y_\mathsf{Vanish2} + \rho^2Y_\mathsf{Vanish3}}{\zeta^\kappa - 1}$. 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 #