# Elliptic nets Let $E$ be an elliptic curve $y^2 = x^3 + ax +b$ over some finite field $\mathbb{F}_p$, where $p$ is a prime $\ne 2, 3$. For simplicity, let's focus on $j$-invariant $0$ *i.e.* $y^2=x^3+b$. With [division polynomials](https://en.wikipedia.org/wiki/Division_polynomials), one can come up with the following formulas due to [Katherine E. Stange](https://arxiv.org/pdf/0710.1316.pdf). - *input*: $P = (x_1,y_1)$ - *output*: $[n]P = (x_n, y_n)$ where $x_n=x_1−\frac{W(n-1, 0)\cdot W(n+ 1, 0)}{W(n, 0)^2}$ $y_n=\frac{W(n− 1, 0)^2 \cdot W(n+ 2, 0) −W(n+1, 0)^2 \cdot W(n− 2, 0)}{4y_1 \cdot W(n, 0)^3}$ With: \begin{align} W(-2, 0) &= − 2y_1 \\ W( − 1, 0) &= − 1 \\ W(0, 0) &= 0 \\ W(1, 0)&= 1 \\ W(2, 0) &= 2y_1 \\ W(3, 0) &= 3x_1^3+12b \cdot x_1 \\ W(4, 0) &= 4y_1(x_1^6 + 20b \cdot x_1^3- 8b^2)\\ W(0, 1) &= 1\\ W(1, 1) &= 1\\ W(2, 1) &= 2x_1+x_2− (\frac{y_2−y_1}{x_2−x_1})^2\\ W( − 1, 1) &=x_1−x_2\\ W(2, − 1) &= (y_1+y_2)^2− (2x_1+x_2)(x_1−x_2)^2 \end{align} When $i \geq 5$, $W(i, 0)$ and $W(i, 1)$ can be computed using some recursive formulas (cf. [[Stange07](https://arxiv.org/pdf/0710.1316.pdf)]). For example: \begin{align} W(2i− 1, 0) &=W(i+ 1, 0)W(i− 1, 0)^3−W(i− 2, 0)W(i, 0)^3 \\ W(2i, 0) &=\frac{W(i, 0)W(i+ 2, 0)W(i− 1, 0)^2}{W(2, 0)}−\frac{W(i, 0)W(i− 2, 0)W(i+ 1, 0)^2}{W(2, 0)} \end{align} --- Doubling ($n=2$): \begin{align} x_2&=x_1−\frac{W(1, 0)\cdot W(3, 0)}{W(2, 0)^2} \\ x_1-x_2&=\frac{W(3, 0)}{W(2, 0)^2}\\ y_2&=\frac{W(1, 0)^2 \cdot W(4, 0) −W(3, 0)^2 \cdot W(0, 0)}{4y_1 \cdot W(2, 0)^3} \\ 4y_1\cdot y_2&=\frac{W(4, 0)}{W(2, 0)^3} \end{align} ## More Stange depicted two algorithms Double and DoubleAdd to compute Tate pairing. [Kanayama et al.](https://www.jstage.jst.go.jp/article/transfun/E97.A/1/E97.A_300/_article) adapted Stange's algorithm to compute scalar multiplication. [Chen et al.](https://eprint.iacr.org/2015/284) later improved on this. If we denote the multiplication and squaring in the finite field under consideration by M and S, respectively, it can be seen that the number of operations for both the Double and DoubleAdd steps is 26M+ 6S for Kanyama et al. and 18M+ 10S for Chen et al. It can be seen that these formulas are more expensive than the traditional double-and-add algorithms. For SNARKs (where an inverse costs almost as much as a multiplication), these formulas are even more expensive in comparison with double-and-add in affine coordinates. However, my point is to see if there is a verification-friendly shortcut of the division sequences which would make them SNARK-friendly.