bitcoin-bitcoin-core/doc/ellswift.md

# ElligatorSwift for secp256k1 explained

In this document we explain how the `ellswift` module implementation is related to the
construction in the
["SwiftEC: Shallue–van de Woestijne Indifferentiable Function To Elliptic Curves"](https://eprint.iacr.org/2022/759)
paper by Jorge Chávez-Saab, Francisco Rodríguez-Henríquez, and Mehdi Tibouchi.

* [1. Introduction](#1-introduction)
* [2. The decoding function](#2-the-decoding-function)
  + [2.1 Decoding for `secp256k1`](#21-decoding-for-secp256k1)
* [3. The encoding function](#3-the-encoding-function)
  + [3.1 Switching to *v, w* coordinates](#31-switching-to-v-w-coordinates)
  + [3.2 Avoiding computing all inverses](#32-avoiding-computing-all-inverses)
  + [3.3 Finding the inverse](#33-finding-the-inverse)
  + [3.4 Dealing with special cases](#34-dealing-with-special-cases)
  + [3.5 Encoding for `secp256k1`](#35-encoding-for-secp256k1)
* [4. Encoding and decoding full *(x, y)* coordinates](#4-encoding-and-decoding-full-x-y-coordinates)
  + [4.1 Full *(x, y)* coordinates for `secp256k1`](#41-full-x-y-coordinates-for-secp256k1)

## 1. Introduction

The `ellswift` module effectively introduces a new 64-byte public key format, with the property
that (uniformly random) public keys can be encoded as 64-byte arrays which are computationally
indistinguishable from uniform byte arrays. The module provides functions to convert public keys
from and to this format, as well as convenience functions for key generation and ECDH that operate
directly on ellswift-encoded keys.

The encoding consists of the concatenation of two (32-byte big endian) encoded field elements $u$
and $t.$ Together they encode an x-coordinate on the curve $x$, or (see further) a full point $(x, y)$ on
the curve.

**Decoding** consists of decoding the field elements $u$ and $t$ (values above the field size $p$
are taken modulo $p$), and then evaluating $F_u(t)$, which for every $u$ and $t$ results in a valid
x-coordinate on the curve. The functions $F_u$ will be defined in [Section 2](#2-the-decoding-function).

**Encoding** a given $x$ coordinate is conceptually done as follows:
* Loop:
  * Pick a uniformly random field element $u.$
  * Compute the set $L = F_u^{-1}(x)$ of $t$ values for which $F_u(t) = x$, which may have up to *8* elements.
  * With probability $1 - \dfrac{\\#L}{8}$, restart the loop.
  * Select a uniformly random $t \in L$ and return $(u, t).$

This is the *ElligatorSwift* algorithm, here given for just x-coordinates. An extension to full
$(x, y)$ points will be given in [Section 4](#4-encoding-and-decoding-full-x-y-coordinates).
The algorithm finds a uniformly random $(u, t)$ among (almost all) those
for which $F_u(t) = x.$ Section 3.2 in the paper proves that the number of such encodings for
almost all x-coordinates on the curve (all but at most 39) is close to two times the field size
(specifically, it lies in the range $2q \pm (22\sqrt{q} + O(1))$, where $q$ is the size of the field).

## 2. The decoding function

First some definitions:
* $\mathbb{F}$ is the finite field of size $q$, of characteristic 5 or more, and $q \equiv 1 \mod 3.$
  * For `secp256k1`, $q = 2^{256} - 2^{32} - 977$, which satisfies that requirement.
* Let $E$ be the elliptic curve of points $(x, y) \in \mathbb{F}^2$ for which $y^2 = x^3 + ax + b$, with $a$ and $b$
  public constants, for which $\Delta_E = -16(4a^3 + 27b^2)$ is a square, and at least one of $(-b \pm \sqrt{-3 \Delta_E} / 36)/2$ is a square.
  This implies that the order of $E$ is either odd, or a multiple of *4*.
  If $a=0$, this condition is always fulfilled.
  * For `secp256k1`, $a=0$ and $b=7.$
* Let the function $g(x) = x^3 + ax + b$, so the $E$ curve equation is also $y^2 = g(x).$
* Let the function $h(x) = 3x^3 + 4a.$
* Define $V$ as the set of solutions $(x_1, x_2, x_3, z)$ to $z^2 = g(x_1)g(x_2)g(x_3).$
* Define $S_u$ as the set of solutions $(X, Y)$ to $X^2 + h(u)Y^2 = -g(u)$ and $Y \neq 0.$
* $P_u$ is a function from $\mathbb{F}$ to $S_u$ that will be defined below.
* $\psi_u$ is a function from $S_u$ to $V$ that will be defined below.

**Note**: In the paper:
* $F_u$ corresponds to $F_{0,u}$ there.
* $P_u(t)$ is called $P$ there.
* All $S_u$ sets together correspond to $S$ there.
* All $\psi_u$ functions together (operating on elements of $S$) correspond to $\psi$ there.

Note that for $V$, the left hand side of the equation $z^2$ is square, and thus the right
hand must also be square. As multiplying non-squares results in a square in $\mathbb{F}$,
out of the three right-hand side factors an even number must be non-squares.
This implies that exactly *1* or exactly *3* out of
$\\{g(x_1), g(x_2), g(x_3)\\}$ must be square, and thus that for any $(x_1,x_2,x_3,z) \in V$,
at least one of $\\{x_1, x_2, x_3\\}$ must be a valid x-coordinate on $E.$ There is one exception
to this, namely when $z=0$, but even then one of the three values is a valid x-coordinate.

**Define** the decoding function $F_u(t)$ as:
* Let $(x_1, x_2, x_3, z) = \psi_u(P_u(t)).$
* Return the first element $x$ of $(x_3, x_2, x_1)$ which is a valid x-coordinate on $E$ (i.e., $g(x)$ is square).

$P_u(t) = (X(u, t), Y(u, t))$, where:

$$
\begin{array}{lcl}
X(u, t) & = & \left\\{\begin{array}{ll}
  \dfrac{g(u) - t^2}{2t} & a = 0 \\
  \dfrac{g(u) + h(u)(Y_0(u) - X_0(u)t)^2}{X_0(u)(1 + h(u)t^2)} & a \neq 0
\end{array}\right. \\
Y(u, t) & = & \left\\{\begin{array}{ll}
  \dfrac{X(u, t) + t}{u \sqrt{-3}} = \dfrac{g(u) + t^2}{2tu\sqrt{-3}} & a = 0 \\
  Y_0(u) + t(X(u, t) - X_0(u)) & a \neq 0
\end{array}\right.
\end{array}
$$

$P_u(t)$ 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
random element in it, so we do not need to have all $\bot$ values at the end.
As we have 8 distinct formulas for finding $(v, w)$ (taking the variants due to $\pm$ into account),
we can associate every index in $T$ with exactly one of those formulas, making sure that:
* Formulas that yield no solutions (due to division by zero or non-existing square roots) or invalid solutions are made to return $\bot.$
* For the $x_1$ and $x_2$ cases, if $g(-u-x)$ is a square, $\bot$ is returned instead (the round-trip check).
* In case multiple formulas would return the same non- $\bot$ result, all but one of those must be turned into $\bot$ to avoid biasing those.

The last condition above only occurs with negligible probability for cryptographically-sized curves, but is interesting
to take into account as it allows exhaustive testing in small groups. See [Section 3.4](#34-dealing-with-special-cases)
for an analysis of all the negligible cases.

If we define $T = (G_{0,u}(x), G_{1,u}(x), \ldots, G_{7,u}(x))$, with each $G_{i,u}$ matching one of the formulas,
the loop can be simplified to only compute one of the inverses instead of all of them:

**Define** *ElligatorSwift(x)* as:
* Loop:
  * Pick a uniformly random field element $u.$
  * Pick a uniformly random integer $c$ in $[0,8).$
  * Let $t = G_{c,u}(x).$
  * If $t \neq \bot$, return $(u, t)$; restart loop otherwise.

This is implemented in `secp256k1_ellswift_xelligatorswift_var`.

### 3.3 Finding the inverse

To implement $G_{c,u}$, we map $c=0$ to the $x_1$ formula, $c=1$ to the $x_2$ formula, and $c=2$ and $c=3$ to the $x_3$ formula.
Those are then repeated as $c=4$ through $c=7$ for the other sign of $w$ (noting that in each formula, $w$ is a square root of some expression).
Ignoring the negligible cases, we get:

**Define** $G_{c,u}(x)$ as:
* If $c \in \\{0, 1, 4, 5\\}$ (for $x_1$ and $x_2$ formulas):
  * If $g(-u-x)$ is square, return $\bot$ (as $x_3$ would be valid and take precedence).
  * If $c \in \\{0, 4\\}$ (the $x_1$ formula) let $v = x$, otherwise let $v = -u-x$ (the $x_2$ formula)
  * Let $s = -g(u)/(u^2 + uv + v^2 + a)$ (using $s = w^2$ in what follows).
* Otherwise, when $c \in \\{2, 3, 6, 7\\}$ (for $x_3$ formulas):
  * Let $s = x-u.$
  * Let $r = \sqrt{-s(4g(u) + sh(u))}.$
  * Let $v = (r/s - u)/2$ if $c \in \\{3, 7\\}$; $(-r/s - u)/2$ otherwise.
* Let $w = \sqrt{s}.$
* Depending on $c:$
  * If $c \in \\{0, 1, 2, 3\\}:$ return $P_u^{'-1}(v, w).$
  * If $c \in \\{4, 5, 6, 7\\}:$ return $P_u^{'-1}(v, -w).$

Whenever a square root of a non-square is taken, $\bot$ is returned; for both square roots this happens with roughly
50% on random inputs. Similarly, when a division by 0 would occur, $\bot$ is returned as well; this will only happen
with negligible probability. A division by 0 in the first branch in fact cannot occur at all, because $u^2 + uv + v^2 + a = 0$
implies $g(-u-x) = g(x)$ which would mean the $g(-u-x)$ is square condition has triggered
and $\bot$ would have been returned already.

**Note**: In the paper, the $case$ variable corresponds roughly to the $c$ above, but only takes on 4 possible values (1 to 4).
The conditional negation of $w$ at the end is done randomly, which is equivalent, but makes testing harder. We choose to
have the $G_{c,u}$ be deterministic, and capture all choices in $c.$

Now observe that the $c \in \\{1, 5\\}$ and $c \in \\{3, 7\\}$ conditions effectively perform the same $v \rightarrow -u-v$
transformation. Furthermore, that transformation has no effect on $s$ in the first branch
as $u^2 + ux + x^2 + a = u^2 + u(-u-x) + (-u-x)^2 + a.$ Thus we can extract it out and move it down:

**Define** $G_{c,u}(x)$ as:
* If $c \in \\{0, 1, 4, 5\\}:$
  * If $g(-u-x)$ is square, return $\bot.$
  * Let $s = -g(u)/(u^2 + ux + x^2 + a).$
  * Let $v = x.$
* Otherwise, when $c \in \\{2, 3, 6, 7\\}:$
  * Let $s = x-u.$
  * Let $r = \sqrt{-s(4g(u) + sh(u))}.$
  * Let $v = (r/s - u)/2.$
* Let $w = \sqrt{s}.$
* Depending on $c:$
  * If $c \in \\{0, 2\\}:$ return $P_u^{'-1}(v, w).$
  * If $c \in \\{1, 3\\}:$ return $P_u^{'-1}(-u-v, w).$
  * If $c \in \\{4, 6\\}:$ return $P_u^{'-1}(v, -w).$
  * If $c \in \\{5, 7\\}:$ return $P_u^{'-1}(-u-v, -w).$

This shows there will always be exactly 0, 4, or 8 $t$ values for a given $(u, x)$ input.
There can be 0, 1, or 2 $(v, w)$ pairs before invoking $P_u^{'-1}$, and each results in 4 distinct $t$ values.

### 3.4 Dealing with special cases

As mentioned before there are a few cases to deal with which only happen in a negligibly small subset of inputs.
For cryptographically sized fields, if only random inputs are going to be considered, it is unnecessary to deal with these. Still, for completeness
we analyse them here. They generally fall into two categories: cases in which the encoder would produce $t$ values that
do not decode back to $x$ (or at least cannot guarantee that they do), and cases in which the encoder might produce the same
$t$ value for multiple $c$ inputs (thereby biasing that encoding):

* In the branch for $x_1$ and $x_2$ (where $c \in \\{0, 1, 4, 5\\}$):
  * When $g(u) = 0$, we would have $s=w=Y=0$, which is not on $S_u.$ This is only possible on even-ordered curves.
    Excluding this also removes the one condition under which the simplified check for $x_3$ on the curve
    fails (namely when $g(x_1)=g(x_2)=0$ but $g(x_3)$ is not square).
    This does exclude some valid encodings: when both $g(u)=0$ and $u^2+ux+x^2+a=0$ (also implying $g(x)=0$),
    the $S_u'$ equation degenerates to $0 = 0$, and many valid $t$ values may exist. Yet, these cannot be targeted uniformly by the
    encoder anyway as there will generally be more than 8.
  * When $g(x) = 0$, the same $t$ would be produced as in the $x_3$ branch (where $c \in \\{2, 3, 6, 7\\}$) which we give precedence
    as it can deal with $g(u)=0$.
    This is again only possible on even-ordered curves.
* In the branch for $x_3$ (where $c \in \\{2, 3, 6, 7\\}$):
  * When $s=0$, a division by zero would occur.
  * When $v = -u-v$ and $c \in \\{3, 7\\}$, the same $t$ would be returned as in the $c \in \\{2, 6\\}$ cases.
    It is equivalent to checking whether $r=0$.
    This cannot occur in the $x_1$ or $x_2$ branches, as it would trigger the $g(-u-x)$ is square condition.
    A similar concern for $w = -w$ does not exist, as $w=0$ is already impossible in both branches: in the first
    it requires $g(u)=0$ which is already outlawed on even-ordered curves and impossible on others; in the second it would trigger division by zero.
* Curve-specific special cases also exist that need to be rejected, because they result in $(u,t)$ which is invalid to the decoder, or because of division by zero in the encoder:
  * For $a=0$ curves, when $u=0$ or when $t=0$. The latter can only be reached by the encoder when $g(u)=0$, which requires an even-ordered curve.
  * For $a \neq 0$ curves, when $X_0(u)=0$, when $h(u)t^2 = -1$, or when $w(u + 2v) = 2X_0(u)$ while also either $w \neq 2Y_0(u)$ or $h(u)=0$.

**Define** a version of $G_{c,u}(x)$ which deals with all these cases:
* If $a=0$ and $u=0$, return $\bot.$
* If $a \neq 0$ and $X_0(u)=0$, return $\bot.$
* If $c \in \\{0, 1, 4, 5\\}:$
  * If $g(u) = 0$ or $g(x) = 0$, return $\bot$ (even curves only).
  * If $g(-u-x)$ is square, return $\bot.$
  * Let $s = -g(u)/(u^2 + ux + x^2 + a)$ (cannot cause division by zero).
  * Let $v = x.$
* Otherwise, when $c \in \\{2, 3, 6, 7\\}:$
  * Let $s = x-u.$
  * Let $r = \sqrt{-s(4g(u) + sh(u))}$; return $\bot$ if not square.
  * If $c \in \\{3, 7\\}$ and $r=0$, return $\bot.$
  * If $s = 0$, return $\bot.$
  * Let $v = (r/s - u)/2.$
* Let $w = \sqrt{s}$; return $\bot$ if not square.
* If $a \neq 0$ and $w(u+2v) = 2X_0(u)$ and either $w \neq 2Y_0(u)$ or $h(u) = 0$, return $\bot.$
* Depending on $c:$
  * If $c \in \\{0, 2\\}$, let $t = P_u^{'-1}(v, w).$
  * If $c \in \\{1, 3\\}$, let $t = P_u^{'-1}(-u-v, w).$
  * If $c \in \\{4, 6\\}$, let $t = P_u^{'-1}(v, -w).$
  * If $c \in \\{5, 7\\}$, let $t = P_u^{'-1}(-u-v, -w).$
* If $a=0$ and $t=0$, return $\bot$ (even curves only).
* If $a \neq 0$ and $h(u)t^2 = -1$, return $\bot.$
* Return $t.$

Given any $u$, using this algorithm over all $x$ and $c$ values, every $t$ value will be reached exactly once,
for an $x$ for which $F_u(t) = x$ holds, except for these cases that will not be reached:
* All cases where $P_u(t)$ is not defined:
  * For $a=0$ curves, when $u=0$, $t=0$, or $g(u) = -t^2.$
  * For $a \neq 0$ curves, when $h(u)t^2 = -1$, $X_0(u) = 0$, or $Y_0(u) (1 - h(u) t^2) = 2X_0(u)t.$
* When $g(u)=0$, the potentially many $t$ values that decode to an $x$ satisfying $g(x)=0$ using the $x_2$ formula. These were excluded by the $g(u)=0$ condition in the $c \in \\{0, 1, 4, 5\\}$ branch.

These cases form a negligible subset of all $(u, t)$ for cryptographically sized curves.

### 3.5 Encoding for `secp256k1`

Specialized for odd-ordered $a=0$ curves:

**Define** $G_{c,u}(x)$ as:
* If $u=0$, return $\bot.$
* If $c \in \\{0, 1, 4, 5\\}:$
  * If $(-u-x)^3 + b$ is square, return $\bot$
  * Let $s = -(u^3 + b)/(u^2 + ux + x^2)$ (cannot cause division by 0).
  * Let $v = x.$
* Otherwise, when $c \in \\{2, 3, 6, 7\\}:$
  * Let $s = x-u.$
  * Let $r = \sqrt{-s(4(u^3 + b) + 3su^2)}$; return $\bot$ if not square.
  * If $c \in \\{3, 7\\}$ and $r=0$, return $\bot.$
  * If $s = 0$, return $\bot.$
  * Let $v = (r/s - u)/2.$
* Let $w = \sqrt{s}$; return $\bot$ if not square.
* Depending on $c:$
  * If $c \in \\{0, 2\\}:$ return $w(\frac{\sqrt{-3}-1}{2}u - v).$
  * If $c \in \\{1, 3\\}:$ return $w(\frac{\sqrt{-3}+1}{2}u + v).$
  * If $c \in \\{4, 6\\}:$ return $w(\frac{-\sqrt{-3}+1}{2}u + v).$
  * If $c \in \\{5, 7\\}:$ return $w(\frac{-\sqrt{-3}-1}{2}u - v).$

This is implemented in `secp256k1_ellswift_xswiftec_inv_var`.

And the x-only ElligatorSwift encoding algorithm is still:

**Define** *ElligatorSwift(x)* as:
* Loop:
  * Pick a uniformly random field element $u.$
  * Pick a uniformly random integer $c$ in $[0,8).$
  * Let $t = G_{c,u}(x).$
  * If $t \neq \bot$, return $(u, t)$; restart loop otherwise.

Note that this logic does not take the remapped $u=0$, $t=0$, and $g(u) = -t^2$ cases into account; it just avoids them.
While it is not impossible to make the encoder target them, this would increase the maximum number of $t$ values for a given $(u, x)$
combination beyond 8, and thereby slow down the ElligatorSwift loop proportionally, for a negligible gain in uniformity.

## 4. Encoding and decoding full *(x, y)* coordinates

So far we have only addressed encoding and decoding x-coordinates, but in some cases an encoding
for full points with $(x, y)$ coordinates is desirable. It is possible to encode this information
in $t$ as well.

Note that for any $(X, Y) \in S_u$, $(\pm X, \pm Y)$ are all on $S_u.$ Moreover, all of these are
mapped to the same x-coordinate. Negating $X$ or negating $Y$ just results in $x_1$ and $x_2$
being swapped, and does not affect $x_3.$ This will not change the outcome x-coordinate as the order
of $x_1$ and $x_2$ only matters if both were to be valid, and in that case $x_3$ would be used instead.

Still, these four $(X, Y)$ combinations all correspond to distinct $t$ values, so we can encode
the sign of the y-coordinate in the sign of $X$ or the sign of $Y.$ They correspond to the
four distinct $P_u^{'-1}$ calls in the definition of $G_{u,c}.$

**Note**: In the paper, the sign of the y coordinate is encoded in a separately-coded bit.

To encode the sign of $y$ in the sign of $Y:$

**Define** *Decode(u, t)* for full $(x, y)$ as:
* Let $(X, Y) = P_u(t).$
* Let $x$ be the first value in $(u + 4Y^2, \frac{-X}{2Y} - \frac{u}{2}, \frac{X}{2Y} - \frac{u}{2})$ for which $g(x)$ is square.
* Let $y = \sqrt{g(x)}.$
* If $sign(y) = sign(Y)$, return $(x, y)$; otherwise return $(x, -y).$

And encoding would be done using a $G_{c,u}(x, y)$ function defined as:

**Define** $G_{c,u}(x, y)$ as:
* If $c \in \\{0, 1\\}:$
  * If $g(u) = 0$ or $g(x) = 0$, return $\bot$ (even curves only).
  * If $g(-u-x)$ is square, return $\bot.$
  * Let $s = -g(u)/(u^2 + ux + x^2 + a)$ (cannot cause division by zero).
  * Let $v = x.$
* Otherwise, when $c \in \\{2, 3\\}:$
  * Let $s = x-u.$
  * Let $r = \sqrt{-s(4g(u) + sh(u))}$; return $\bot$ if not square.
  * If $c = 3$ and $r = 0$, return $\bot.$
  * Let $v = (r/s - u)/2.$
* Let $w = \sqrt{s}$; return $\bot$ if not square.
* Let $w' = w$ if $sign(w/2) = sign(y)$; $-w$ otherwise.
* Depending on $c:$
  * If $c \in \\{0, 2\\}:$ return $P_u^{'-1}(v, w').$
  * If $c \in \\{1, 3\\}:$ return $P_u^{'-1}(-u-v, w').$

Note that $c$ now only ranges $[0,4)$, as the sign of $w'$ is decided based on that of $y$, rather than on $c.$
This change makes some valid encodings unreachable: when $y = 0$ and $sign(Y) \neq sign(0)$.

In the above logic, $sign$ can be implemented in several ways, such as parity of the integer representation
of the input field element (for prime-sized fields) or the quadratic residuosity (for fields where
$-1$ is not square). The choice does not matter, as long as it only takes on two possible values, and for $x \neq 0$ it holds that $sign(x) \neq sign(-x)$.

### 4.1 Full *(x, y)* coordinates for `secp256k1`

For $a=0$ curves, there is another option. Note that for those,
the $P_u(t)$ function translates negations of $t$ to negations of (both) $X$ and $Y.$ Thus, we can use $sign(t)$ to
encode the y-coordinate directly. Combined with the earlier remapping to guarantee all inputs land on the curve, we get
as decoder:

**Define** *Decode(u, t)* as:
* Let $u'=u$ if $u \neq 0$; $1$ otherwise.
* Let $t'=t$ if $t \neq 0$; $1$ otherwise.
* Let $t''=t'$ if $u'^3 + b + t'^2 \neq 0$; $2t'$ otherwise.
* Let $X = \dfrac{u'^3 + b - t''^2}{2t''}.$
* Let $Y = \dfrac{X + t''}{u'\sqrt{-3}}.$
* Let $x$ be the first element of $(u' + 4Y^2, \frac{-X}{2Y} - \frac{u'}{2}, \frac{X}{2Y} - \frac{u'}{2})$ for which $g(x)$ is square.
* Let $y = \sqrt{g(x)}.$
* Return $(x, y)$ if $sign(y) = sign(t)$; $(x, -y)$ otherwise.

This is implemented in `secp256k1_ellswift_swiftec_var`. The used $sign(x)$ function is the parity of $x$ when represented as in integer in $[0,q).$

The corresponding encoder would invoke the x-only one, but negating the output $t$ if $sign(t) \neq sign(y).$

This is implemented in `secp256k1_ellswift_elligatorswift_var`.

Note that this is only intended for encoding points where both the x-coordinate and y-coordinate are unpredictable. When encoding x-only points
where the y-coordinate is implicitly even (or implicitly square, or implicitly in $[0,q/2]$), the encoder in
[Section 3.5](#35-encoding-for-secp256k1) must be used, or a bias is reintroduced that undoes all the benefit of using ElligatorSwift
in the first place.