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  Y_0(u) + t(X(u, t) - X_0(u)) & a \neq 0
\end{array}\right.
\end{array}
$$

$P_u(t)$ is defined:
* For $a=0$, unless:
  * $u = 0$ or $t = 0$ (division by zero)
  * $g(u) = -t^2$ (would give $Y=0$).
* For $a \neq 0$, unless:
  * $X_0(u) = 0$ or $h(u)t^2 = -1$ (division by zero)
  * $Y_0(u) (1 - h(u)t^2) = 2X_0(u)t$ (would give $Y=0$).

The functions $X_0(u)$ and $Y_0(u)$ are defined in Appendix A of the paper, and depend on various properties of $E.$

The function $\psi_u$ is the same for all curves: $\psi_u(X, Y) = (x_1, x_2, x_3, z)$, where:

$$
\begin{array}{lcl}
  x_1 & = & \dfrac{X}{2Y} - \dfrac{u}{2} && \\
  x_2 & = & -\dfrac{X}{2Y} - \dfrac{u}{2} && \\
  x_3 & = & u + 4Y^2 && \\
  z   & = & \dfrac{g(x_3)}{2Y}(u^2 + ux_1 + x_1^2 + a) = \dfrac{-g(u)g(x_3)}{8Y^3}
\end{array}
$$

### 2.1 Decoding for `secp256k1`

Put together and specialized for $a=0$ curves, decoding $(u, t)$ to an x-coordinate is:

**Define** $F_u(t)$ as:
* Let $X = \dfrac{u^3 + b - t^2}{2t}.$
* Let $Y = \dfrac{X + t}{u\sqrt{-3}}.$
* Return the first $x$ in $(u + 4Y^2, \dfrac{-X}{2Y} - \dfrac{u}{2}, \dfrac{X}{2Y} - \dfrac{u}{2})$ for which $g(x)$ is square.

To make sure that every input decodes to a valid x-coordinate, we remap the inputs in case
$P_u$ is not defined (when $u=0$, $t=0$, or $g(u) = -t^2$):

**Define** $F_u(t)$ as:
* Let $u'=u$ if $u \neq 0$; $1$ otherwise (guaranteeing $u' \neq 0$).
* Let $t'=t$ if $t \neq 0$; $1$ otherwise (guaranteeing $t' \neq 0$).
* Let $t''=t'$ if $g(u') \neq -t'^2$; $2t'$ otherwise (guaranteeing $t'' \neq 0$ and $g(u') \neq -t''^2$).
* Let $X = \dfrac{u'^3 + b - t''^2}{2t''}.$
* Let $Y = \dfrac{X + t''}{u'\sqrt{-3}}.$
* Return the first $x$ in $(u' + 4Y^2, \dfrac{-X}{2Y} - \dfrac{u'}{2}, \dfrac{X}{2Y} - \dfrac{u'}{2})$ for which $x^3 + b$ is square.

The choices here are not strictly necessary. Just returning a fixed constant in any of the undefined cases would suffice,
but the approach here is simple enough and gives fairly uniform output even in these cases.

**Note**: in the paper these conditions result in $\infty$ as output, due to the use of projective coordinates there.
We wish to avoid the need for callers to deal with this special case.

This is implemented in `secp256k1_ellswift_xswiftec_frac_var` (which decodes to an x-coordinate represented as a fraction), and
in `secp256k1_ellswift_xswiftec_var` (which outputs the actual x-coordinate).

## 3. The encoding function

To implement $F_u^{-1}(x)$, the function to find the set of inverses $t$ for which $F_u(t) = x$, we have to reverse the process:
* Find all the $(X, Y) \in S_u$ that could have given rise to $x$, through the $x_1$, $x_2$, or $x_3$ formulas in $\psi_u.$
* Map those $(X, Y)$ solutions to $t$ values using $P_u^{-1}(X, Y).$
* For each of the found $t$ values, verify that $F_u(t) = x.$
* Return the remaining $t$ values.

The function $P_u^{-1}$, which finds $t$ given $(X, Y) \in S_u$, is significantly simpler than $P_u:$

$$
P_u^{-1}(X, Y) = \left\\{\begin{array}{ll}
Yu\sqrt{-3} - X & a = 0 \\
\dfrac{Y-Y_0(u)}{X-X_0(u)} & a \neq 0 \land X \neq X_0(u) \\
\dfrac{-X_0(u)}{h(u)Y_0(u)} & a \neq 0 \land X = X_0(u) \land Y = Y_0(u)
\end{array}\right.
$$

The third step above, verifying that $F_u(t) = x$, is necessary because for the $(X, Y)$ values found through the $x_1$ and $x_2$ expressions,
it is possible that decoding through $\psi_u(X, Y)$ yields a valid $x_3$ on the curve, which would take precedence over the
$x_1$ or $x_2$ decoding. These $(X, Y)$ solutions must be rejected.

Since we know that exactly one or exactly three out of $\\{x_1, x_2, x_3\\}$ are valid x-coordinates for any $t$,
the case where either $x_1$ or $x_2$ is valid and in addition also $x_3$ is valid must mean that all three are valid.
This means that instead of checking whether $x_3$ is on the curve, it is also possible to check whether the other one out of
$x_1$ and $x_2$ is on the curve. This is significantly simpler, as it turns out.

Observe that $\psi_u$ guarantees that $x_1 + x_2 = -u.$ So given either $x = x_1$ or $x = x_2$, the other one of the two can be computed as
$-u - x.$ Thus, when encoding $x$ through the $x_1$ or $x_2$ expressions, one can simply check whether $g(-u-x)$ is a square,
and if so, not include the corresponding $t$ values in the returned set. As this does not need $X$, $Y$, or $t$, this condition can be determined
before those values are computed.

It is not possible that an encoding found through the $x_1$ expression decodes to a different valid x-coordinate using $x_2$ (which would
take precedence), for the same reason: if both $x_1$ and $x_2$ decodings were valid, $x_3$ would be valid as well, and thus take
precedence over both. Because of this, the $g(-u-x)$ being square test for $x_1$ and $x_2$ is the only test necessary to guarantee the found $t$
values round-trip back to the input $x$ correctly. This is the reason for choosing the $(x_3, x_2, x_1)$ precedence order in the decoder;
any order which does not place $x_3$ first requires more complicated round-trip checks in the encoder.

### 3.1 Switching to *v, w* coordinates

Before working out the formulas for all this, we switch to different variables for $S_u.$ Let $v = (X/Y - u)/2$, and
$w = 2Y.$ Or in the other direction, $X = w(u/2 + v)$ and $Y = w/2:$
* $S_u'$ becomes the set of $(v, w)$ for which $w^2 (u^2 + uv + v^2 + a) = -g(u)$ and $w \neq 0.$
* For $a=0$ curves, $P_u^{-1}$ can be stated for $(v,w)$ as $P_u^{'-1}(v, w) = w\left(\frac{\sqrt{-3}-1}{2}u - v\right).$
* $\psi_u$ can be stated for $(v, w)$ as $\psi_u'(v, w) = (x_1, x_2, x_3, z)$, where

$$
\begin{array}{lcl}
  x_1 & = & v \\
  x_2 & = & -u - v \\
  x_3 & = & u + w^2 \\
  z   & = & \dfrac{g(x_3)}{w}(u^2 + uv + v^2 + a) = \dfrac{-g(u)g(x_3)}{w^3}
\end{array}
$$

We can now write the expressions for finding $(v, w)$ given $x$ explicitly, by solving each of the $\\{x_1, x_2, x_3\\}$
expressions for $v$ or $w$, and using the $S_u'$ equation to find the other variable:
* Assuming $x = x_1$, we find $v = x$ and $w = \pm\sqrt{-g(u)/(u^2 + uv + v^2 + a)}$ (two solutions).
* Assuming $x = x_2$, we find $v = -u-x$ and $w = \pm\sqrt{-g(u)/(u^2 + uv + v^2 + a)}$ (two solutions).
* Assuming $x = x_3$, we find $w = \pm\sqrt{x-u}$ and $v = -u/2 \pm \sqrt{-w^2(4g(u) + w^2h(u))}/(2w^2)$ (four solutions).

### 3.2 Avoiding computing all inverses

The *ElligatorSwift* algorithm as stated in Section 1 requires the computation of $L = F_u^{-1}(x)$ (the
set of all $t$ such that $(u, t)$ decode to $x$) in full. This is unnecessary.

Observe that the procedure of restarting with probability $(1 - \frac{\\#L}{8})$ and otherwise returning a
uniformly random element from $L$ is actually equivalent to always padding $L$ with $\bot$ values up to length 8,
picking a uniformly random element from that, restarting whenever $\bot$ is picked:

**Define** *ElligatorSwift(x)* as:
* Loop:
  * Pick a uniformly random field element $u.$
  * Compute the set $L = F_u^{-1}(x).$
  * Let $T$ be the 8-element vector consisting of the elements of $L$, plus $8 - \\#L$ times $\\{\bot\\}.$
  * Select a uniformly random $t \in T.$
  * If $t \neq \bot$, return $(u, t)$; restart loop otherwise.

Now notice that the order of elements in $T$ does not matter, as all we do is pick a uniformly