Merge pull request #91 from zcash/book-multipoint

[book] Add multipoint opening + small set interpolation

book/src/SUMMARY.md

@@ -2,6 +2,7 @@
 
 [halo2](README.md)
 - [Concepts](concepts.md)
+  - [Multipoint opening argument](concepts/multipoint-opening.md)
 - [User Documentation](user.md)
   - [A simple example](user/simple-example.md)
   - [Lookup tables](user/lookup-tables.md)

book/src/concepts/multipoint-opening.md

@@ -0,0 +1,80 @@
+# Multipoint opening argument
+
+Consider the commitments $A, B, C, D$ to polynomials $a(X), b(X), c(X), d(X)$.
+Let's say that $a$ and $b$ were queried at the point $x$, while $c$ and $d$
+were queried at both points $x$ and $\omega x$. (Here, $\omega$ is the primitive
+root of unity in the multiplicative subgroup over which we constructed the
+polynomials).
+
+We can group the commitments in terms of the sets of points at which they were
+queried:
+$$
+\begin{array}{cccc}
+&\{x\}&     &\{x, \omega x\}& \\
+ &A&            &C& \\
+ &B&            &D&
+\end{array}
+$$
+
+The multipoint opening optimisation proceeds as such:
+
+1. Sample random $x_3$, at which we evaluate $a(X), b(X), c(X), d(X)$.
+2. The prover provides evaluations of each polynomial at each point of interest:
+   $a(x_3), b(x_3), c(x_3), d(x_3), c(\omega x_3), d(\omega x_3)$
+3. Sample random $x_4$, to keep $a, b, c, d$ linearly independent.
+4. Accumulate polynomials and their corresponding evaluations according
+   to the point set at which they were queried:
+    `q_polys`:
+    $$
+    \begin{array}{rccl}
+    q_1(X) &=& a(X) &+& x_4 b(X) \\
+    q_2(X) &=& c(X) &+& x_4 d(X)
+    \end{array}
+    $$
+    `q_eval_sets`:
+    ```math
+            [
+                [a(x_3) + x_4 b(x_3)],
+                [
+                    c(x_3) + x_4 d(x_3),
+                    c(\omega x_3) + x_4 d(\omega x_3)
+                ]
+            ]
+    ```
+    NB: `q_eval_sets` is a vector of sets of evaluations, where the outer vector
+    goes over the point sets, and the inner vector goes over the points in each set.
+5. Interpolate each set of values in `q_eval_sets`:
+    `r_polys`:
+    $$
+    \begin{array}{cccc}
+    r_1(X) s.t.&&& \\
+        &r_1(x_3) &=& a(x_3) + x_4 b(x_3) \\
+    r_2(X) s.t.&&& \\
+        &r_2(x_3) &=& c(x_3) + x_4 d(x_3) \\
+        &r_2(\omega x_3) &=& c(\omega x_3) + x_4 d(\omega x_3) \\
+    \end{array}
+    $$
+6. Construct `f_polys` which check the correctness of `q_polys`:
+    `f_polys`
+    $$
+    \begin{array}{rcl}
+    f_1(X) &=& \frac{ q_1(X) - r_1(X)}{X - x_3} \\
+    f_2(X) &=& \frac{ q_2(X) - r_2(X)}{(X - x_3)(X - \omega x_3)} \\
+    \end{array}
+    $$
+
+    If $q_1(x_3) = r_1(x_3)$, then $f_1(X)$ should be a polynomial.
+    If $q_2(x_3) = r_2(x_3)$ and $q_2(\omega x_3) = r_2(\omega x_3)$
+    then $f_2(X)$ should be a polynomial.
+7. Sample random $x_5$ to keep the `f_polys` linearly independent.
+8. Construct $f(X) = f_1(X) + x_5 f_2(X)$.
+9. Sample random $x_6$, at which we evaluate $f(X)$:
+    $$
+    \begin{array}{rcccl}
+    f(x_6) &=& f_1(x_6) &+& x_5 f_2(x_6) \\
+           &=& \frac{q_1(x_6) - r_1(x_6)}{x_6 - x_3} &+& x_5\frac{q_2(x_6) - r_2(x_6)}{(x_6 - x_3)(x_6 - \omega x_3)}
+    \end{array}
+    $$
+10. Sample random $x_7$ to keep $f(X)$ and `q_polys` linearly independent.
+11. Construct `final_poly`, $$final\_poly(X) = f(X) + x_7 q_1(X) + x_7^2 q_2(X),$$
+    which is the polynomial we commit to in the inner product argument.

book/src/user/tips-and-tricks.md

@@ -25,3 +25,47 @@ $$(7 - c) \cdot (13 - c) = 0$$
 > higher-degree constraints are more expensive to use.
 
 Note that the roots don't have to be constants; for example $(a - x) \cdot (a - y) \cdot (a - z) = 0$ will constrain $a$ to be equal to one of $\{ x, y, z \}$ where the latter can be arbitrary polynomials, as long as the whole expression stays within the maximum degree bound.
+
+## Small set interpolation
+We can use Lagrange interpolation to create a polynomial constraint that maps
+$f(X) = Y$ for small sets of $X \in \{x_i\}, Y \in \{y_i\}$. 
+
+For instance, say we want to map a 2-bit value to a "spread" version interleaved
+with zeros. We first precompute the evaluations at each point:
+
+$$
+\begin{array}{cc}
+00 &\rightarrow 0000 \implies 0 \rightarrow 0 \\
+01 &\rightarrow 0001 \implies 1 \rightarrow 1 \\
+10 &\rightarrow 0100 \implies 2 \rightarrow 4 \\
+11 &\rightarrow 0101 \implies 3 \rightarrow 5 
+\end{array}
+$$
+
+Then, we construct the Lagrange basis polynomial for each point using the
+identity:
+$$\mathcal{l}_j(X) = \prod_{0 \leq m \leq k, m \neq j} \frac{x - x_m}{x_j - x_m},$$
+where $k + 1$ is the number of data points. ($k = 3$ in our example above.)
+
+Recall that the Lagrange basis polynomial $\mathcal{l}_j(X)$ evaluates to $1$ at
+$X = x_j$ and $0$ at all other $x_i, j \neq i.$
+
+Continuing our example, we get four Lagrange basis polynomials:
+
+$$
+\begin{array}{ccc}
+l_0(X) &=& \frac{(X - 3)(X - 2)(X - 1)}{(-3)(-2)(-1)} \\
+l_1(X) &=& \frac{(X - 3)(X - 2)(X)}{(-2)(-1)(1)} \\
+l_2(X) &=& \frac{(X - 3)(X - 1)(X)}{(-1)(1)(2)} \\
+l_3(X) &=& \frac{(X - 2)(X - 1)(X)}{(1)(2)(3)}
+\end{array}
+$$
+
+Our polynomial constraint is then
+
+$$
+\begin{array}{ccccccccc}
+&&f(0)l_0(X) &+& f(1)l_1(X) &+& f(2)l_2(X) &+& f(3)l_3(X) - f(X) &=& 0 \\
+&\implies& 0 \cdot l_0(X) &+& 1 \cdot l_1(X) &+& 4 \cdot l_2(X) &+& 5 \cdot l_3(X) - f(X) &=& 0. \\
+\end{array}
+$$