Add small set interpolation to tips-and-tricks

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
+
+$$
+\begin{array}{ccccccccc}
+&&f(0)l_0(X) &+& f(1)l_1(X) &+& f(2)l_2(X) &+& f(3)l_3(X) - f(X) &=& 0 \\
+&\implies& 0 \cdot l_0(X) &+& 1 \cdot l_1(X) &+& 4 \cdot l_2(X) &+& 5 \cdot l_3(X) - f(X) &=& 0. \\
+\end{array}
+$$