mirror of https://github.com/zcash/halo2.git
Merge pull request #155 from zcash/book-selector-optimisations
[Book] Undo selector optimisations
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ying tong
GitHub
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a2ed3f1b52
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@ -165,79 +165,77 @@ We need six advice columns to witness $(x_T, y_T, \lambda_1, \lambda_2, x_{A,i},
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- the second, covering the $lo$ half for the remaining $i$ from $129$ down to $4$, with a special case at $i = 4$.
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$$
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\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
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\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
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\hline
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x_T & y_T & z^{hi} & x_A^{hi} & \lambda_1^{hi} & \lambda_2^{hi} &q_{mul}^{hi}& z^{lo} & x_A^{lo} & \lambda_1^{lo} & \lambda_2^{lo} &q_{mul}^{lo}\\\hline
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& & \mathbf{z}_{255} = 0 & & y_{A,254}=2[T]_y & & 1 & \mathbf{z}_{130} & & y_{A,129} & & 1 \\\hline
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x_T & y_T & \mathbf{z}_{254} & x_{A,254} = 2[T]_x & \lambda_{1,254} & \lambda_{2,254} & 2 & \mathbf{z}_{129} & x_{A,129} & \lambda_{1,129} & \lambda_{2,129} & 2 \\\hline
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x_T & y_T & \mathbf{z}_{253} & x_{A,253} & \lambda_{1,253} & \lambda_{2,253} & 2 & \mathbf{z}_{128} & x_{A,128} & \lambda_{1,128} & \lambda_{2,128} & 2 \\\hline
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\\hline
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x_T & y_T & \mathbf{z}_{130} & x_{A,130} & \lambda_{1,130} & \lambda_{2,130} & 3 & \mathbf{z}_5 & x_{A,5} & \lambda_{1,5} & \lambda_{2,5} & 2 \\\hline
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& & & x_{A,129} & y_{A,129} & & & \mathbf{z}_4 & x_{A,4} & \lambda_{1,4} & \lambda_{2,4} & 3 \\\hline
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& & & & & & & & x_{A,3} & y_{A,3} & & \\\hline
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x_T & y_T & z^{hi} & x_A^{hi} & \lambda_1^{hi} & \lambda_2^{hi} & q_1^{hi} & q_2^{hi} & q_3^{hi} & z^{lo} & x_A^{lo} & \lambda_1^{lo} & \lambda_2^{lo} & q_1^{lo} & q_2^{lo} & q_3^{lo} \\\hline
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& & \mathbf{z}_{255} = 0 & & y_{A,254}=2[T]_y & & 1 & 0 & 0 & \mathbf{z}_{130} & & y_{A,129} & & 1 & 0 & 0 \\\hline
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x_T & y_T & \mathbf{z}_{254} & x_{A,254} = 2[T]_x & \lambda_{1,254} & \lambda_{2,254} & 0 & 1 & 0 & \mathbf{z}_{129} & x_{A,129} & \lambda_{1,129} & \lambda_{2,129} & 0 & 1 & 0 \\\hline
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x_T & y_T & \mathbf{z}_{253} & x_{A,253} & \lambda_{1,253} & \lambda_{2,253} & 0 & 1 & 0 & \mathbf{z}_{128} & x_{A,128} & \lambda_{1,128} & \lambda_{2,128} & 0 & 1 & 0 \\\hline
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\\hline
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x_T & y_T & \mathbf{z}_{130} & x_{A,130} & \lambda_{1,130} & \lambda_{2,130} & 0 & 0 & 1 & \mathbf{z}_5 & x_{A,5} & \lambda_{1,5} & \lambda_{2,5} & 0 & 1 & 0 \\\hline
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& & & x_{A,129} & y_{A,129} & & & & & \mathbf{z}_4 & x_{A,4} & \lambda_{1,4} & \lambda_{2,4} & 0 & 0 & 1 \\\hline
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& & & & & & & & & & x_{A,3} & y_{A,3} & & & & \\\hline
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\end{array}
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$$
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For each $hi$ and $lo$ half, we have three sets of gates. Note that $i$ is going from $255..=3$; $i$ is NOT indexing the rows.
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#### $q_{mul} = 1$
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#### $q_1 = 1$
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This gate is only used on the first row (before the for loop). We check that $\lambda_1, \lambda_2$ are initialized to values consistent with the initial $y_A.$
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$$
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\begin{array}{|c|l|}
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\hline
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\text{Degree} & \text{Constraint} \\\hline
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5 & q_{mul}^{one} \cdot \left(y_{A,n}^\text{witnessed} - y_{A,n}\right) = 0 \\\hline
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3 & q_1 \cdot \left(y_{A,n}^\text{witnessed} - y_{A,n}\right) = 0 \\\hline
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\end{array}
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$$
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where
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$$
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\begin{aligned}
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q_{mul}^{one} &= q_{mul} \cdot (2 - q_{mul}) \cdot (3 - q_{mul}),\\
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y_{A,n} &= \frac{(\lambda_{1,n} + \lambda_{2,n}) \cdot (x_{A,n} - (\lambda_{1,n}^2 - x_{A,n} - x_T))}{2},\\
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y_{A,n}^\text{witnessed} &\text{ is witnessed.}
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\end{aligned}
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$$
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#### $q_{mul} = 2$
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#### $q_2 = 1$
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This gate is used on all rows corresponding to the for loop except the last.
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$$
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\begin{array}{|c|l|}
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\hline
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\text{Degree} & \text{Constraint} \\\hline
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4 & q_{mul}^{two} \cdot \left(x_{T,cur} - x_{T,next}\right) = 0 \\\hline
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4 & q_{mul}^{two} \cdot \left(y_{T,cur} - y_{T,next}\right) = 0 \\\hline
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5 & q_{mul}^{two} \cdot \mathbf{k}_i \cdot (\mathbf{k}_i - 1) = 0, \text{ where } \mathbf{k}_i = \mathbf{z}_{i} - 2\mathbf{z}_{i+1} \\\hline
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6 & q_{mul}^{two} \cdot \left(\lambda_{1,i} \cdot (x_{A,i} - x_{T,i}) - y_{A,i} + (2\mathbf{k}_i - 1) \cdot y_{T,i}\right) = 0 \\\hline
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5 & q_{mul}^{two} \cdot \left(\lambda_{2,i}^2 - x_{A,i-1} - \lambda_{1,i}^2 + x_{T,i}\right) = 0 \\\hline
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5 & q_{mul}^{two} \cdot \left(\lambda_{2,i} \cdot (x_{A,i} - x_{A,i-1}) - y_{A,i} - y_{A,i-1}\right) = 0 \\\hline
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2 & q_2 \cdot \left(x_{T,cur} - x_{T,next}\right) = 0 \\\hline
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2 & q_2 \cdot \left(y_{T,cur} - y_{T,next}\right) = 0 \\\hline
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3 & q_2 \cdot \mathbf{k}_i \cdot (\mathbf{k}_i - 1) = 0, \text{ where } \mathbf{k}_i = \mathbf{z}_{i} - 2\mathbf{z}_{i+1} \\\hline
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4 & q_2 \cdot \left(\lambda_{1,i} \cdot (x_{A,i} - x_{T,i}) - y_{A,i} + (2\mathbf{k}_i - 1) \cdot y_{T,i}\right) = 0 \\\hline
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3 & q_2 \cdot \left(\lambda_{2,i}^2 - x_{A,i-1} - \lambda_{1,i}^2 + x_{T,i}\right) = 0 \\\hline
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3 & q_2 \cdot \left(\lambda_{2,i} \cdot (x_{A,i} - x_{A,i-1}) - y_{A,i} - y_{A,i-1}\right) = 0 \\\hline
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\end{array}
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$$
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where
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$$
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\begin{aligned}
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q_{mul}^{two} &= q_{mul} \cdot (1 - q_{mul}) \cdot (3 - q_{mul}),\\
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y_{A,i} &= \frac{(\lambda_{1,i} + \lambda_{2,i}) \cdot (x_{A,i} - (\lambda_{1,i}^2 - x_{A,i} - x_T))}{2}, \\
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y_{A,i-1} &= \frac{(\lambda_{1,i-1} + \lambda_{2,i-1}) \cdot (x_{A,i-1} - (\lambda_{1,i-1}^2 - x_{A,i-1} - x_T))}{2}, \\
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\end{aligned}
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$$
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#### $q_{mul} = 3$
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#### $q_3 = 1$
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This gate is used on the final iteration of the for loop, handling the special case where we check that the output $y_A$ has been witnessed correctly.
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$$
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\begin{array}{|c|l|}
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\hline
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\text{Degree} & \text{Constraint} \\\hline
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5 & q_{mul}^{three} \cdot \mathbf{k}_i \cdot (\mathbf{k}_i - 1) = 0, \text{ where } \mathbf{k}_i = \mathbf{z}_{i} - 2\mathbf{z}_{i+1} \\\hline
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6 & q_{mul}^{three} \cdot \left(\lambda_{1,i} \cdot (x_{A,i} - x_{T,i}) - y_{A,i} + (2\mathbf{k}_i - 1) \cdot y_{T,i}\right) = 0 \\\hline
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5 & q_{mul}^{three} \cdot \left(\lambda_{2,i}^2 - x_{A,i-1} - \lambda_{1,i}^2 + x_{T,i}\right) = 0 \\\hline
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5 & q_{mul}^{three} \cdot \left(\lambda_{2,i} \cdot (x_{A,i} - x_{A,i-1}) - y_{A,i} - y_{A,i-1}^\text{witnessed}\right) = 0 \\\hline
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3 & q_3 \cdot \mathbf{k}_i \cdot (\mathbf{k}_i - 1) = 0, \text{ where } \mathbf{k}_i = \mathbf{z}_{i} - 2\mathbf{z}_{i+1} \\\hline
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4 & q_3 \cdot \left(\lambda_{1,i} \cdot (x_{A,i} - x_{T,i}) - y_{A,i} + (2\mathbf{k}_i - 1) \cdot y_{T,i}\right) = 0 \\\hline
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3 & q_3 \cdot \left(\lambda_{2,i}^2 - x_{A,i-1} - \lambda_{1,i}^2 + x_{T,i}\right) = 0 \\\hline
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3 & q_3 \cdot \left(\lambda_{2,i} \cdot (x_{A,i} - x_{A,i-1}) - y_{A,i} - y_{A,i-1}^\text{witnessed}\right) = 0 \\\hline
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\end{array}
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$$
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where
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$$
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\begin{aligned}
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q_{mul}^{three} &= q_{mul} \cdot (1 - q_{mul}) \cdot (2 - q_{mul}),\\
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y_{A,i} &= \frac{(\lambda_{1,i} + \lambda_{2,i}) \cdot (x_{A,i} - (\lambda_{1,i}^2 - x_{A,i} - x_T))}{2},\\
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y_{A,i-1}^\text{witnessed} &\text{ is witnessed.}
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\end{aligned}
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@ -324,11 +322,11 @@ $$
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\begin{array}{|c|l|}
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\hline
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\text{Degree} & \text{Constraint} \\\hline
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2 & q_\text{mul}^\text{overflow} \cdot \left(s - (\alpha + \mathbf{k}_{254} \cdot 2^{130})\right) = 0 \\\hline
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2 & q_\text{mul}^\text{overflow} \cdot \left(\mathbf{z}_0 - \alpha - t_q\right) = 0 \\\hline
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3 & q_\text{mul}^\text{overflow} \cdot \left(\mathbf{k}_{254} \cdot (\mathbf{z}_{130} - 2^{124})\right) = 0 \\\hline
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3 & q_\text{mul}^\text{overflow} \cdot \left(\mathbf{k}_{254} \cdot (s - \sum\limits_{i=0}^{129} 2^i \cdot \mathbf{s}_i)/2^{130}\right) = 0 \\\hline
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5 & q_\text{mul}^\text{overflow} \cdot \left((1 - \mathbf{k}_{254}) \cdot (1 - \mathbf{z}_{130} \cdot \eta) \cdot (s - \sum\limits_{i=0}^{129} 2^i \cdot \mathbf{s}_i)/2^{130}\right) = 0 \\\hline
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2 & \text{q\_mul}^\text{overflow} \cdot \left(s - (\alpha + \mathbf{k}_{254} \cdot 2^{130})\right) = 0 \\\hline
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2 & \text{q\_mul}^\text{overflow} \cdot \left(\mathbf{z}_0 - \alpha - t_q\right) = 0 \\\hline
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3 & \text{q\_mul}^\text{overflow} \cdot \left(\mathbf{k}_{254} \cdot (\mathbf{z}_{130} - 2^{124})\right) = 0 \\\hline
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3 & \text{q\_mul}^\text{overflow} \cdot \left(\mathbf{k}_{254} \cdot (s - \sum\limits_{i=0}^{129} 2^i \cdot \mathbf{s}_i)/2^{130}\right) = 0 \\\hline
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5 & \text{q\_mul}^\text{overflow} \cdot \left((1 - \mathbf{k}_{254}) \cdot (1 - \mathbf{z}_{130} \cdot \eta) \cdot (s - \sum\limits_{i=0}^{129} 2^i \cdot \mathbf{s}_i)/2^{130}\right) = 0 \\\hline
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\end{array}
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$$
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where $(s - \sum\limits_{i=0}^{129} 2^i \cdot \mathbf{s}_i)/2^{130}$ can be computed by another running sum. Note that the factor of $1/2^{130}$ has no effect on the constraint, since the RHS is zero.
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@ -80,59 +80,70 @@ In other words, $z_{n-i} = \sum\limits_{h=0}^{i-1} 2^{kh} \cdot m_{h+1}$.
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>
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> In order to support chaining multiple field elements without a gap, we will use a slightly more complicated expression for $m_{i+1}$ that effectively forces $\mathbf{z}_n$ to zero for the last step of each element, as indicated by $q_{S2}$. This allows the cell that would have been $\mathbf{z}_n$ to be used to reinitialize the running sum for the next element.
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### Layout
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Note: $q_{S3}$ is synthesized from $q_{S1}$ and $q_{S2}$; it is shown here only for clarity.
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### Generator lookup table
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The Sinsemilla circuit makes use of $2^{10}$ pre-computed random generators. These are loaded into a lookup table:
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$$
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\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
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\begin{array}{|c|c|c|}
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\hline
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\text{Step} & x_A & x_P & bits & \lambda_1 & \lambda_2 & q_{S1} & q_{S2} & q_{S3} & fixed\_y_Q & table_{idx} & table_x & table_y \\\hline
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0 & x_Q & x_{P[m_1]} & z_0 & \lambda_{1,0} & \lambda_{2,0} & 1 & 1 & 0 & y_Q & 0 & x_{P[0]} & y_{P[0]} \\\hline
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1 & x_{A,1} & x_{P[m_2]} & z_1 & \lambda_{1,1} & \lambda_{2,1} & 1 & 1 & 0 & 0 & 1 & x_{P[1]} & y_{P[1]} \\\hline
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2 & x_{A,2} & x_{P[m_3]} & z_2 & \lambda_{1,2} & \lambda_{2,2} & 1 & 1 & 0 & 0 & 2 & x_{P[2]} & y_{P[2]} \\\hline
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 1 & 1 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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n-1 & x_{A,n-1} & x_{P[m_n]} & z_{n-1} & \lambda_{1,n-1} & \lambda_{2,n-1} & 1 & 0 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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0' & x'_{A,0} & x_{P[m'_1]} & z'_0 & \lambda'_{1,0} & \lambda'_{2,0} & 1 & 1 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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1' & x'_{A,1} & x_{P[m'_2]} & z'_1 & \lambda'_{1,1} & \lambda'_{2,1} & 1 & 1 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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2' & x'_{A,2} & x_{P[m'_3]} & z'_2 & \lambda'_{1,2} & \lambda'_{2,2} & 1 & 1 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 1 & 1 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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n-1' & x'_{A,n-1} & x_{P[m'_n]} & z'_{n-1} & \lambda'_{1,n-1} & \lambda'_{2,n-1} & 1 & 2 & 2 & 0 & \vdots & \vdots & \vdots \\\hline
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n' & x'_{A,n} & & & y_{A,n} & & 0 & 0 & 0 & 0 & \vdots & \vdots & \vdots \\\hline
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table_{idx} & table_x & table_y \\\hline
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0 & x_{P[0]} & y_{P[0]} \\\hline
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1 & x_{P[1]} & y_{P[1]} \\\hline
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2 & x_{P[2]} & y_{P[2]} \\\hline
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\vdots & \vdots & \vdots \\\hline
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2^{10} - 1 & x_{P[2^{10}-1]} & y_{P[2^{10}-1]} \\\hline
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\end{array}
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$$
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### Layout
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$$
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\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
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\hline
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\text{Step} & x_A & x_P & bits & \lambda_1 & \lambda_2 & q_{S1} & q_{S2} & q_{S3} & q_{S4} & \textsf{fixed\_y\_Q}\\\hline
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0 & x_Q & x_{P[m_1]} & z_0 & \lambda_{1,0} & \lambda_{2,0} & 1 & 1 & 0 & 1 & y_Q \\\hline
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1 & x_{A,1} & x_{P[m_2]} & z_1 & \lambda_{1,1} & \lambda_{2,1} & 1 & 1 & 0 & 0 & 0 \\\hline
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2 & x_{A,2} & x_{P[m_3]} & z_2 & \lambda_{1,2} & \lambda_{2,2} & 1 & 1 & 0 & 0 & 0 \\\hline
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 1 & 1 & 0 & 0 & 0 \\\hline
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n-1 & x_{A,n-1} & x_{P[m_n]} & z_{n-1} & \lambda_{1,n-1} & \lambda_{2,n-1} & 1 & 0 & 0 & 0 & 0 \\\hline
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0' & x'_{A,0} & x_{P[m'_1]} & z'_0 & \lambda'_{1,0} & \lambda'_{2,0} & 1 & 1 & 0 & 0 & 0 \\\hline
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1' & x'_{A,1} & x_{P[m'_2]} & z'_1 & \lambda'_{1,1} & \lambda'_{2,1} & 1 & 1 & 0 & 0 & 0 \\\hline
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2' & x'_{A,2} & x_{P[m'_3]} & z'_2 & \lambda'_{1,2} & \lambda'_{2,2} & 1 & 1 & 0 & 0 & 0 \\\hline
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 1 & 1 & 0 & 0 & 0 \\\hline
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n-1' & x'_{A,n-1} & x_{P[m'_n]} & z'_{n-1} & \lambda'_{1,n-1} & \lambda'_{2,n-1} & 1 & 0 & 1 & 0 & 0 \\\hline
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n' & x'_{A,n} & & & y_{A,n} & & 0 & 0 & 0 & 0 & 0 \\\hline
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\end{array}
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$$
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$x_Q$, $z_0$, $z'_0$, etc. would be copied in using equality constraints.
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### Optimized Sinsemilla gate
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$\begin{array}{lrcl}
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$$
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\begin{array}{lrcl}
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\text{For } i \in [0, n), \text{ let} &x_{R,i} &=& \lambda_{1,i}^2 - x_{A,i} - x_{P,i} \\
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&Y_{A,i} &=& (\lambda_{1,i} + \lambda_{2,i}) \cdot (x_{A,i} - x_{R,i}) \\
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&y_{P,i} &=& Y_{A,i}/2 - \lambda_{1,i} \cdot (x_{A,i} - x_{P,i}) \\
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&m_{i+1} &=& z_{i} - 2^k \cdot (q_{S2,i} - q_{S3,i}) \cdot z_{i+1} \\
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&q_{S3} &=& q_{S2} \cdot (q_{S2} - 1)
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\end{array}$
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&m_{i+1} &=& z_{i} - 2^k \cdot q_{S2,i} \cdot z_{i+1}
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\end{array}
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$$
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The Halo 2 circuit API can automatically substitute $y_{P,i}$, $x_{R,i}$, $y_{A,i}$, and $y_{A,i+1}$, so we don't need to do that manually.
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- $x_{A,0} = x_Q$
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- $2 \cdot y_Q = Y_{A,0}$
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- for $i$ from $0$ up to $n-1$:
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- $(m_{i+1},\, x_{P,i},\, y_{P,i}) \in \mathcal{P}$
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- $\lambda_{2,i}^2 = x_{A,i+1} + x_{R,i} + x_{A,i}$
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- $4 \cdot \lambda_{2,i} \cdot (x_{A,i} - x_{A,i+1}) = 2 \cdot Y_{A,i} + (2 - q_{S3}) \cdot Y_{A,i+1} + 2 q_{S3} \cdot y_{A,n}$
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$x_{A,0} = x_Q$
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$2 \cdot y_Q = Y_{A,0}$
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for $i$ from $0$ up to $n-1$:
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$(m_{i+1},\, x_{P,i},\, y_{P,i}) \in \mathcal{P}$
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$\lambda_{2,i}^2 = x_{A,i+1} + x_{R,i} + x_{A,i}$
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$2 \cdot \lambda_{2,i} \cdot (x_{A,i} - x_{A,i+1}) = Y_{A,i} + (1 - q_{S3}) \cdot Y_{A,i+1} + 2 \cdot q_{S3} \cdot y_{A,n}$
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Note that each term of the last constraint is multiplied by $4$ relative to the constraint program given earlier. This is a small optimization that avoids divisions by $2$.
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Note that each term of the last constraint is multiplied by $2$ relative to the constraint program given earlier. This is a small optimization that avoids divisions by $2$.
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$$
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\begin{array}{|c|l|}
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\hline
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\text{Degree} & \text{Constraint} \\\hline
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4 & fixed\rule{0.4em}{0.02ex}y_Q \cdot (2 \cdot fixed\rule{0.4em}{0.02ex}y_Q - Y_{A,0}) = 0 \\\hline
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2 & q_{S4} \cdot (2 \cdot y_Q - Y_{A,0}) = 0 \\\hline
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5 & q_{S1,i} \Rightarrow (m_{i+1},\, x_{P,i},\, y_{P,i}) \in \mathcal{P} \\\hline
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3 & q_{S1,i} \cdot \big(\lambda_{2,i}^2 - (x_{A,i+1} + x_{R,i} + x_{A,i})\big) \\\hline
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6 & q_{S1,i} \cdot \left(4 \cdot \lambda_{2,i} \cdot (x_{A,i} - x_{A,i+1}) - (2 \cdot Y_{A,i} + (2 - q_{S3,i}) \cdot Y_{A,i+1} + 2 \cdot q_{S3,i} \cdot y_{A,n})\right) = 0 \\\hline
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4 & q_{S1,i} \cdot \left(2 \cdot \lambda_{2,i} \cdot (x_{A,i} - x_{A,i+1}) - (Y_{A,i} + (1 - q_{S3,i}) \cdot Y_{A,i+1} + 2 \cdot q_{S3,i} \cdot y_{A,n})\right) = 0 \\\hline
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\end{array}
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$$
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Loading…
Reference in New Issue