Commit d51ff03e authored by Niels Möller's avatar Niels Möller

Notes on the Montgomery ladder.

parent ac1e6e5a
......@@ -63,6 +63,75 @@ y_2)$:
Again, very similar to the Weierstraß formulas, with only an
additional $b$ term in the formula for $x_3$.
\subsection{Montgomery ladder}
It's possible to do operations on a Montgomery curve in terms of the
$x$ coordinate only. Or, with homogeneous coordinates, use $X$ and $Z$
with $x = X/Z$.
For doubling,
x' &= (x^2 - z^2)^2 = (x-z)^2 (x+z)^2 \\
t &= (x+z)^2 - (x-z)^2 \\
z' &= 4 xz (x^2 + bzx + z^2) = t \left((x+z)^2 + b't\right)
with $b' = (b-2)/4$.
Addition is a bit trickier. If we have $x$ and $z$ for points $Q_1$,
$Q_2$ and $Q_3$, with $Q_3 = Q_1 + Q_3$, and $x_1, z_1 \neq 0$, we
get the coordinates for $Q_2 + Q_3$ as
x' &= 4 (x_2 x_3 - z_2 z_3)^2 z_1 = \left((x_2 - z_2)(x_3 + z_3) +
(x_2 + z_2)(x_3 - z_3)\right)^2 z_1 \\
z' &= 4 (x_2 z_3 - z_2 x_3)^2 x_1 = \left((x_2 - z_2)(x_3 + z_3) -
(x_2 + z_2)(x_3 - z_3)\right)^2 x_1
Note that the doubling formula is symmetric in $Q_2$ and $Q_3$. Which
is consistent with negating of $Q_1$, which really is the negatiion of
the $y$-coordinate, which doesn't appear in the formula.
This can be used for a binary ``Montgomery ladder'' to compute $n Q$
for any $n$. If we have the points $Q$, $n Q$, and $(n+1) Q$, we can
compute the three points
(2n) Q &= 2 (nQ) && \text{doubling} \\
(2n+1) Q &= (nQ) + (n+1)Q && \text{addition} \\
(2n+2) Q &= 2((n+1) Q) && \text{doubling}
The following algorithm is suggested by dj (see
x2,z2,x3,z3 = 1,0,x1,1
for i in reversed(range(255)):
bit = 1 & (n >> i)
x2,x3 = cswap(x2,x3,bit)
z2,z3 = cswap(z2,z3,bit)
x3,z3 = ((x2*x3-z2*z3)^2,x1*(x2*z3-z2*x3)^2)
x2,z2 = ((x2^2-z2^2)^2,4*x2*z2*(x2^2+A*x2*z2+z2^2))
x2,x3 = cswap(x2,x3,bit)
z2,z3 = cswap(z2,z3,bit)
return x2*z2^(p-2)
It's not too hard to decipher this. The update for $x_2, z_2$ is the
doubling. The update for $x_3, z_3$ is an addition.
If the bit is zero, we get $x_2', z_2'$ representing $Q_2' = 2 Q_2$,
and $x_3', z_3'$ representing $Q_3' = Q_2 + Q_3 = 2 Q_2 + Q_1$.
What if the bit is set? For the doubling, we get it applied to $Q_3$
instead, so we get $x_3', z_3'$ representing $Q_3' = 2 Q_3 = 2 Q_2 + 2
Q_1$. For the add, the initial swap flips the sign of one of the
intermediate values, but the end result is the same, so we get $x_2',
z_2'$ representing $Q_2' = Q_2 + Q_3 = 2 Q_2 + Q_1$, as desired.
Note that the initial conditional swap doesn't have to be a full swap;
if that's convenient in the implementation, a conditional assignment
should be sufficient to get the duplication formula appplied to the
right point. It looks like, in all cases, one will start by computing
$x_2 \pm z_2$ and $x_3 \pm z_3$, so maybe one can apply conditional
assignment to these values instead.
\section{Edwards curve}
For an Edwards curve, we consider the special case
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