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Polish test

trunk
Sarah Jamie Lewis 1 year ago
parent
commit
e6be185d3d
  1. 23
      src/lib.rs

23
src/lib.rs

@ -234,11 +234,12 @@ mod tests {
use std::fmt::{Display, Formatter};
// Generate a Prime Field to Test with...
// This is way larger than the 64 bit prime the paper specifies, but wth
#[derive(PrimeField)]
#[PrimeFieldModulus = "65537"]
#[PrimeFieldGenerator = "3"]
#[PrimeFieldModulus = "52435875175126190479447740508185965837690552500527637822603658699938581184513"]
#[PrimeFieldGenerator = "7"]
#[PrimeFieldReprEndianness = "little"]
struct Fp([u64; 1]);
struct Fp([u64; 4]);
impl Display for Fp {
fn fmt(&self, f: &mut Formatter<'_>) -> std::fmt::Result {
@ -249,18 +250,18 @@ mod tests {
#[test]
fn it_works() {
let rng = OsRng;
let s = 3u64;
let t = 6u64;
let s = 10u64;
let t = 30u64;
let dhf = HashKey::<Fp>::new(rng, s, t);
let mut solver = Solver::new(s, t);
for i in 0..10 {
for i in 0..40 {
// These are the indexes which are to be random...you can try swapping them around..
// REMEMBER: the number of true hash needs to be AT LEAST `t+1` AND the number of fake hashes
// must be AT-MOST `s`.
// If |random hashes| > s this this procedure will *not* work
// There is an extended algorithm for extract assuming |true hashes| > t though - we do not implement it.
if i != 2 && i != 5 && i != 8 {
if i != 2 && i != 5 && i != 8 && i < 34 {
let x0 = Fp::random(rng);
let hash = dhf.generate_hash(x0);
solver.add_hash(hash);
@ -269,6 +270,12 @@ mod tests {
}
}
assert_eq!(solver.attempt_solve().unwrap(), vec![0, 1, 3, 4, 6, 7, 9]);
assert_eq!(
solver.attempt_solve().unwrap(),
vec![
0, 1, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30, 31, 32, 33
]
);
}
}

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