Files
serai/coins/monero/ringct/bulletproofs/src/core.rs
Luke Parker 77a2496ade Tidy and document monero-bulletproofs
I still don't like the impl of the original Bulletproofs...
2024-07-04 02:18:36 -04:00

75 lines
2.4 KiB
Rust

use std_shims::{vec, vec::Vec};
use curve25519_dalek::{
traits::{MultiscalarMul, VartimeMultiscalarMul},
scalar::Scalar,
edwards::EdwardsPoint,
};
pub(crate) use monero_generators::{MAX_COMMITMENTS, COMMITMENT_BITS, LOG_COMMITMENT_BITS};
pub(crate) fn multiexp(pairs: &[(Scalar, EdwardsPoint)]) -> EdwardsPoint {
let mut buf_scalars = Vec::with_capacity(pairs.len());
let mut buf_points = Vec::with_capacity(pairs.len());
for (scalar, point) in pairs {
buf_scalars.push(scalar);
buf_points.push(point);
}
EdwardsPoint::multiscalar_mul(buf_scalars, buf_points)
}
pub(crate) fn multiexp_vartime(pairs: &[(Scalar, EdwardsPoint)]) -> EdwardsPoint {
let mut buf_scalars = Vec::with_capacity(pairs.len());
let mut buf_points = Vec::with_capacity(pairs.len());
for (scalar, point) in pairs {
buf_scalars.push(scalar);
buf_points.push(point);
}
EdwardsPoint::vartime_multiscalar_mul(buf_scalars, buf_points)
}
/*
This has room for optimization worth investigating further. It currently takes
an iterative approach. It can be optimized further via divide and conquer.
Assume there are 4 challenges.
Iterative approach (current):
1. Do the optimal multiplications across challenge column 0 and 1.
2. Do the optimal multiplications across that result and column 2.
3. Do the optimal multiplications across that result and column 3.
Divide and conquer (worth investigating further):
1. Do the optimal multiplications across challenge column 0 and 1.
2. Do the optimal multiplications across challenge column 2 and 3.
3. Multiply both results together.
When there are 4 challenges (n=16), the iterative approach does 28 multiplications
versus divide and conquer's 24.
*/
pub(crate) fn challenge_products(challenges: &[(Scalar, Scalar)]) -> Vec<Scalar> {
let mut products = vec![Scalar::ONE; 1 << challenges.len()];
if !challenges.is_empty() {
products[0] = challenges[0].1;
products[1] = challenges[0].0;
for (j, challenge) in challenges.iter().enumerate().skip(1) {
let mut slots = (1 << (j + 1)) - 1;
while slots > 0 {
products[slots] = products[slots / 2] * challenge.0;
products[slots - 1] = products[slots / 2] * challenge.1;
slots = slots.saturating_sub(2);
}
}
// Sanity check since if the above failed to populate, it'd be critical
for product in &products {
debug_assert!(*product != Scalar::ZERO);
}
}
products
}