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Support 5 mod 8 case in SqrtPrecomputation #968

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Support 5 mod 8 case in SqrtPrecomputation #968

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8918f1d
aditional logic for special case of sqrt
serhii023 Apr 2, 2025
bebc3d8
bug fix for zero case
serhii023 Apr 2, 2025
20f03bd
Merge branch 'arkworks-rs:master' into master
serhii023 Apr 2, 2025
087aad3
cargo fmt
serhii023 Apr 2, 2025
d0fa65c
warning reduce
serhii023 Apr 2, 2025
b665879
cargo fmt
serhii023 Apr 2, 2025
0137ed3
Merge branch 'master' into master
serhii023 Apr 10, 2025
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7 changes: 7 additions & 0 deletions ff/src/biginteger/mod.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -157,6 +157,13 @@ impl<const N: usize> BigInt<N> {
(((self.0[0] << 62) >> 62) % 4) as u8
}

#[doc(hidden)]
pub const fn mod_8(&self) -> u8 {
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So, this and the above mod_4 generalize to any modulo power two and could be rewritten in a general way correct?

https://www.geeksforgeeks.org/compute-modulus-division-by-a-power-of-2-number

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Exactly, the value a % (2^d) is equal to a & (2^d - 1) and this evaluation is more efficient, however I wrote implementation similar to the existing code for the sake of simplicity.

// To compute n % 8, we need to simply look at the
// 3 least significant bits of n, and check their value mod 8.
(((self.0[0] << 61) >> 61) % 8) as u8
}

/// Compute a right shift of `self`
/// This is equivalent to a (saturating) division by 2.
#[doc(hidden)]
Expand Down
58 changes: 52 additions & 6 deletions ff/src/fields/models/fp/montgomery_backend.rs
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Expand Up @@ -93,6 +93,37 @@ pub trait MontConfig<const N: usize>: 'static + Sync + Send + Sized {
}
};

/// (MODULUS + 3) / 8 when MODULUS % 8 == 5. Used for square root precomputations.
#[doc(hidden)]
const MODULUS_PLUS_THREE_DIV_EIGHT: Option<BigInt<N>> = {
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Very interesting! I too think it would be helpful to reference a source—perhaps from Hacker's Delight, a university lecture, or an ePrint paper—for further context.

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It is just a partial case of Tonneli-Shanks algorithm https://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm
For the case p=3 mod 4 the solution is straightforward and simple, for the case p=5 mod 8 we need one probing Legendre symbol evaluation and after that multiply by non-residue in appropriate power if needed. The most complicated case is p=1 mod 8 where the number of Legendre symbol evaluations can be different for different values.

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Hi @serhii023, thanks for the link—this adds some context. And BTW in general, optimizations are great and appreciated.

That being said, a reasonable concern here is that proof by example is not generally sufficient for a cryptographic library (which has a higher standard than general purpose software).

What are your thoughts? Would you be interested in writing a proof for the cases you've referenced?

TBH, maybe a cool side-project in Lean etc.

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Here is a proof:
Let $x^2 \equiv a \pmod p$ is our quadratic equation where we need to find $x$ if one exists. Firstly, note that solutions modullo p exists iff $a$ is quadratic residue modullo p. Recall that $a \in \mathbb{F}_p$ is quadratic residue iff $a^{(p-1)/2} \equiv 1 \pmod p$ by Euler criterion or equivalently Legendre symbol $(\frac{a}{p}) = 1$ so that solutions to the equation $x^2 \equiv a \pmod p$ exist.

Consider than partial case where $p$ takes form $p=4k+3$ (case that arkworks already implements):
$x^2 \equiv a \pmod p$, so that $a$ is quadratic residue iff $a^{2k+1} \equiv 1 \pmod p$, multiplying by $a$ both sides of conjugation we obtain $a^{2k+2} \equiv a \pmod p$ so that $a^{2k+2} \equiv x^2 \pmod p$ so we can easily take square root of both sides as every exponent is even: $x = \pm a^{k+1} \pmod p \quad \square$

Consider our case $p=8k+5$ (@serhii023 proposal):
$x^2 \equiv a \pmod p$, so that $a$ is quadratic residue iff $a^{4k+2} \equiv 1 \pmod p$, here we could not simply multiply each side by $a$ as in previous case since the exponent will be odd. But instead since $a^{4k+2} \equiv 1^2 \equiv 1 \pmod p$ taking square root of $1$ gives us either $a^{2k+1} \equiv 1 \pmod p$ or $a^{2k+1} \equiv -1 \pmod p$. Consider each case separately:

  • If $a^{2k+1} \equiv 1 \pmod p$ than using the same technique as in $p=4k+3$ case multiplying both sides by $a$ we get $a^{2k+2} \equiv a \equiv x^2 \pmod p$ so that $x = \pm a^{k+1} \pmod p \quad \square$
  • If $a^{2k+1} \equiv -1 \pmod p$ than we could not obtain square root using the same technique, but recalling Tonneli-Shanks trick of multiplying each side by some non-residue we could obtain the square root. Firstly, $2 \in \mathbb{F}_p$ is always quadratic non-residue modullo $p$ as by Legendre symbol properties $(\frac{2}{p})=(-1)^{\frac{p^2 - 1}{8}} = (-1)^{8k^2 +10k + 3} = -1$. So that by Euler criterion: $2^{(p-1)/2} \equiv 2^{4k+2} \equiv -1 \pmod p$. So multiplying both sides of $a^{2k+1} \equiv -1 \pmod p$ by $2^{4k+2}$ we get the following conjugation: $a^{2k+1} 2^{4k+2} \equiv 1 \pmod p$. Multiplying both sides by $a$ we get $a^{2k+2} 2^{4k+2} \equiv a \equiv x^2 \pmod p$ so that $x = \pm a^{k+1} 2^{2k+1} \pmod p \quad \square$

Above proof is not a proof by example, but a general proof for all primes of form $p=8k+5$ and exactly this construction is implemented in @serhii023 PR.

match Self::MODULUS.mod_8() == 5 {
true => {
let (modulus_plus_three, carry) = Self::MODULUS.const_add_with_carry(&BigInt!("3"));
let mut result = modulus_plus_three.divide_by_2_round_down();
// Since modulus_plus_one is even, dividing by 2 results in a MSB of 0.
// Thus we can set MSB to `carry` to get the correct result of (MODULUS + 1) // 2:
result.0[N - 1] |= (carry as u64) << 63;
result = result.divide_by_2_round_down();

Some(result.divide_by_2_round_down())
},
false => None,
}
};

/// (MODULUS - 1) / 4 when MODULUS % 8 == 5. Used for square root precomputations.
#[doc(hidden)]
const MODULUS_MINUS_ONE_DIV_FOUR: Option<BigInt<N>> = {
match Self::MODULUS.mod_8() == 5 {
true => {
let (modulus_plus_three, _) = Self::MODULUS.const_sub_with_borrow(&BigInt::one());
let result = modulus_plus_three.divide_by_2_round_down();
Some(result.divide_by_2_round_down())
},
false => None,
}
};

/// Sets `a = a + b`.
#[inline(always)]
fn add_assign(a: &mut Fp<MontBackend<Self, N>, N>, b: &Fp<MontBackend<Self, N>, N>) {
Expand Down Expand Up @@ -548,12 +579,27 @@ pub const fn sqrt_precomputation<const N: usize, T: MontConfig<N>>(
}),
None => None,
},
_ => Some(SqrtPrecomputation::TonelliShanks {
two_adicity: <MontBackend<T, N>>::TWO_ADICITY,
quadratic_nonresidue_to_trace: T::TWO_ADIC_ROOT_OF_UNITY,
trace_of_modulus_minus_one_div_two:
&<Fp<MontBackend<T, N>, N>>::TRACE_MINUS_ONE_DIV_TWO.0,
}),
_ => match T::MODULUS.mod_8() {
5 => match (
T::MODULUS_PLUS_THREE_DIV_EIGHT.as_ref(),
T::MODULUS_MINUS_ONE_DIV_FOUR.as_ref(),
) {
(
Some(BigInt(modulus_plus_three_div_eight)),
Some(BigInt(modulus_minus_one_div_four)),
) => Some(SqrtPrecomputation::Case5Mod8 {
modulus_plus_three_div_eight,
modulus_minus_one_div_four,
}),
_ => None,
},
_ => Some(SqrtPrecomputation::TonelliShanks {
two_adicity: <MontBackend<T, N>>::TWO_ADICITY,
quadratic_nonresidue_to_trace: T::TWO_ADIC_ROOT_OF_UNITY,
trace_of_modulus_minus_one_div_two:
&<Fp<MontBackend<T, N>, N>>::TRACE_MINUS_ONE_DIV_TWO.0,
}),
},
}
}

Expand Down
30 changes: 30 additions & 0 deletions ff/src/fields/sqrt.rs
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Expand Up @@ -75,6 +75,11 @@ pub enum SqrtPrecomputation<F: crate::Field> {
Case3Mod4 {
modulus_plus_one_div_four: &'static [u64],
},
/// To be used when the modulus is 5 mod 8.
Case5Mod8 {
modulus_plus_three_div_eight: &'static [u64],
modulus_minus_one_div_four: &'static [u64],
},
}

impl<F: crate::Field> SqrtPrecomputation<F> {
Expand Down Expand Up @@ -144,6 +149,31 @@ impl<F: crate::Field> SqrtPrecomputation<F> {
let result = elem.pow(modulus_plus_one_div_four.as_ref());
(result.square() == *elem).then_some(result)
},
Self::Case5Mod8 {
modulus_plus_three_div_eight,
modulus_minus_one_div_four,
} => {
if elem.is_zero() {
return Some(F::zero());
}

let result;

// We have different solutions, if check_value is 1 or -1.
let check_value = elem.pow(modulus_minus_one_div_four.as_ref());
if check_value.is_one() {
result = elem.pow(modulus_plus_three_div_eight.as_ref());
} else if check_value.neg().is_one() {
let two: F = 2.into();
result = elem
.pow(modulus_plus_three_div_eight.as_ref())
.mul(two.pow(modulus_minus_one_div_four.as_ref()))
} else {
return None;
}

(result.square() == *elem).then_some(result)
},
}
}
}
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