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| 1 | +/* SPDX-License-Identifier: MIT OR Apache-2.0 */ |
| 2 | + |
| 3 | +//! To keep the equations somewhat concise, the following conventions are used: |
| 4 | +//! - all integer operations are in the mathematical sense, without overflow |
| 5 | +//! - concatenation means multiplication: `2xq = 2 * x * q` |
| 6 | +//! - `R = (1 << U::BITS)` is the modulus of wrapping arithmetic in `U` |
| 7 | +
|
| 8 | +use crate::support::int_traits::NarrowingDiv; |
| 9 | +use crate::support::{DInt, HInt, Int}; |
| 10 | + |
| 11 | +/// Compute the remainder `(x << e) % y` with unbounded integers. |
| 12 | +/// Requires `x < 2y` and `y.leading_zeros() >= 2` |
| 13 | +#[allow(dead_code)] |
| 14 | +pub fn linear_mul_reduction<U>(x: U, mut e: u32, mut y: U) -> U |
| 15 | +where |
| 16 | + U: HInt + Int<Unsigned = U>, |
| 17 | + U::D: NarrowingDiv, |
| 18 | +{ |
| 19 | + assert!(y <= U::MAX >> 2); |
| 20 | + assert!(x < (y << 1)); |
| 21 | + let _0 = U::ZERO; |
| 22 | + let _1 = U::ONE; |
| 23 | + |
| 24 | + // power of two divisors |
| 25 | + if (y & (y - _1)).is_zero() { |
| 26 | + if e < U::BITS { |
| 27 | + // shift and only keep low bits |
| 28 | + return (x << e) & (y - _1); |
| 29 | + } else { |
| 30 | + // would shift out all the bits |
| 31 | + return _0; |
| 32 | + } |
| 33 | + } |
| 34 | + |
| 35 | + // Use the identity `(x << e) % y == ((x << (e + s)) % (y << s)) >> s` |
| 36 | + // to shift the divisor so it has exactly two leading zeros to satisfy |
| 37 | + // the precondition of `Reducer::new` |
| 38 | + let s = y.leading_zeros() - 2; |
| 39 | + e += s; |
| 40 | + y <<= s; |
| 41 | + |
| 42 | + // `m: Reducer` keeps track of the remainder `x` in a form that makes it |
| 43 | + // very efficient to do `x <<= k` modulo `y` for integers `k < U::BITS` |
| 44 | + let mut m = Reducer::new(x, y); |
| 45 | + |
| 46 | + // Use the faster special case with constant `k == U::BITS - 1` while we can |
| 47 | + while e >= U::BITS - 1 { |
| 48 | + m.word_reduce(); |
| 49 | + e -= U::BITS - 1; |
| 50 | + } |
| 51 | + // Finish with the variable shift operation |
| 52 | + m.shift_reduce(e); |
| 53 | + |
| 54 | + // The partial remainder is in `[0, 2y)` ... |
| 55 | + let r = m.partial_remainder(); |
| 56 | + // ... so check and correct, and compensate for the earlier shift. |
| 57 | + r.checked_sub(y).unwrap_or(r) >> s |
| 58 | +} |
| 59 | + |
| 60 | +/// Helper type for computing the reductions. The implementation has a number |
| 61 | +/// of seemingly weird choices, but everything is aimed at streamlining |
| 62 | +/// `Reducer::word_reduce` into its current form. |
| 63 | +/// |
| 64 | +/// Implicitly contains: |
| 65 | +/// n in (R/8, R/4) |
| 66 | +/// x in [0, 2n) |
| 67 | +/// The value of `n` is fixed for a given `Reducer`, |
| 68 | +/// but the value of `x` is modified by the methods. |
| 69 | +#[derive(Debug, Clone, PartialEq, Eq)] |
| 70 | +struct Reducer<U: HInt> { |
| 71 | + // m = 2n |
| 72 | + m: U, |
| 73 | + // q = (RR/2) / m |
| 74 | + // r = (RR/2) % m |
| 75 | + // Then RR/2 = qm + r, where `0 <= r < m` |
| 76 | + // The value `q` is only needed during construction, so isn't saved. |
| 77 | + r: U, |
| 78 | + // The value `x` is implicitly stored as `2 * q * x`: |
| 79 | + _2xq: U::D, |
| 80 | +} |
| 81 | + |
| 82 | +impl<U> Reducer<U> |
| 83 | +where |
| 84 | + U: HInt, |
| 85 | + U: Int<Unsigned = U>, |
| 86 | +{ |
| 87 | + /// Construct a reducer for `(x << _) mod n`. |
| 88 | + /// |
| 89 | + /// Requires `R/8 < n < R/4` and `x < 2n`. |
| 90 | + fn new(x: U, n: U) -> Self |
| 91 | + where |
| 92 | + U::D: NarrowingDiv, |
| 93 | + { |
| 94 | + let _1 = U::ONE; |
| 95 | + assert!(n > (_1 << (U::BITS - 3))); |
| 96 | + assert!(n < (_1 << (U::BITS - 2))); |
| 97 | + let m = n << 1; |
| 98 | + assert!(x < m); |
| 99 | + |
| 100 | + // We need to compute the parameters |
| 101 | + // `q = (RR/2) / m` |
| 102 | + // `r = (RR/2) % m` |
| 103 | + |
| 104 | + // Since `m` is in `(R/4, R/2)`, the quotient `q` is in `[R, 2R)`, and |
| 105 | + // it would overflow in `U` if computed directly. Instead, we compute |
| 106 | + // `f = q - R`, which is in `[0, R)`. To do so, we simply subtract `Rm` |
| 107 | + // from the dividend, which doesn't change the remainder: |
| 108 | + // `f = R(R/2 - m) / m` |
| 109 | + // `r = R(R/2 - m) % m` |
| 110 | + let dividend = ((_1 << (U::BITS - 1)) - m).widen_hi(); |
| 111 | + let (f, r) = dividend.checked_narrowing_div_rem(m).unwrap(); |
| 112 | + |
| 113 | + // As `x < m`, `xq < qm <= RR/2` |
| 114 | + // Thus `2xq = 2xR + 2xf` does not overflow in `U::D`. |
| 115 | + let _2x = x + x; |
| 116 | + let _2xq = _2x.widen_hi() + _2x.widen_mul(f); |
| 117 | + Self { m, r, _2xq } |
| 118 | + } |
| 119 | + |
| 120 | + /// Extract the current remainder `x` in the range `[0, 2n)` |
| 121 | + fn partial_remainder(&self) -> U { |
| 122 | + // `RR/2 = qm + r`, where `0 <= r < m` |
| 123 | + // `2xq = uR + v`, where `0 <= v < R` |
| 124 | + |
| 125 | + // The goal is to extract the current value of `x` from the value `2xq` |
| 126 | + // that we actually have. A bit simplified, we could multiply it by `m` |
| 127 | + // to obtain `2xqm == 2x(RR/2 - r) == xRR - 2xr`, where `2xr < RR`. |
| 128 | + // We could just round that up to the next multiple of `RR` to get `x`, |
| 129 | + // but we can avoid having to multiply the full double-wide `2xq` by |
| 130 | + // making a couple of adjustments: |
| 131 | + |
| 132 | + // First, let's only use the high half `u` for the product, and |
| 133 | + // include an additional error term due to the truncation: |
| 134 | + // `mu = xR - (2xr + mv)/R` |
| 135 | + |
| 136 | + // Next, show bounds for the error term |
| 137 | + // `0 <= mv < mR` follows from `0 <= v < R` |
| 138 | + // `0 <= 2xr < mR` follows from `0 <= x < m < R/2` and `0 <= r < m` |
| 139 | + // Adding those together, we have: |
| 140 | + // `0 <= (mv + 2xr)/R < 2m` |
| 141 | + // Which also implies: |
| 142 | + // `0 < 2m - (mv + 2xr)/R <= 2m < R` |
| 143 | + |
| 144 | + // For that reason, we can use `u + 2` as the factor to obtain |
| 145 | + // `m(u + 2) = xR + (2m - (mv + 2xr)/R)` |
| 146 | + // By the previous inequality, the second term fits neatly in the lower |
| 147 | + // half, so we get exactly `x` as the high half. |
| 148 | + let u = self._2xq.hi(); |
| 149 | + let _2 = U::ONE + U::ONE; |
| 150 | + self.m.widen_mul(u + _2).hi() |
| 151 | + |
| 152 | + // Additionally, we should ensure that `u + 2` cannot overflow: |
| 153 | + // Since `x < m` and `2qm <= RR`, |
| 154 | + // `2xq <= 2q(m-1) <= RR - 2q` |
| 155 | + // As we also have `q > R`, |
| 156 | + // `2xq < RR - 2R` |
| 157 | + // which is sufficient. |
| 158 | + } |
| 159 | + |
| 160 | + /// Replace the remainder `x` with `(x << k) - un`, |
| 161 | + /// for a suitable quotient `u`, which is returned. |
| 162 | + /// |
| 163 | + /// Requires that `k < U::BITS`. |
| 164 | + fn shift_reduce(&mut self, k: u32) -> U { |
| 165 | + assert!(k < U::BITS); |
| 166 | + |
| 167 | + // First, split the shifted value: |
| 168 | + // `2xq << k = aRR/2 + b`, where `0 <= b < RR/2` |
| 169 | + let a = self._2xq.hi() >> (U::BITS - 1 - k); |
| 170 | + let (low, high) = (self._2xq << k).lo_hi(); |
| 171 | + let b = U::D::from_lo_hi(low, high & (U::MAX >> 1)); |
| 172 | + |
| 173 | + // Then, subtract `2anq = aqm`: |
| 174 | + // ``` |
| 175 | + // (2xq << k) - aqm |
| 176 | + // = aRR/2 + b - aqm |
| 177 | + // = a(RR/2 - qm) + b |
| 178 | + // = ar + b |
| 179 | + // ``` |
| 180 | + self._2xq = a.widen_mul(self.r) + b; |
| 181 | + a |
| 182 | + |
| 183 | + // Since `a` is at most the high half of `2xq`, we have |
| 184 | + // `a + 2 < R` (shown above, in `partial_remainder`) |
| 185 | + // Using that together with `b < RR/2` and `r < m < R/2`, |
| 186 | + // we get `(a + 2)r + b < RR`, so |
| 187 | + // `ar + b < RR - 2r = 2mq` |
| 188 | + // which shows that the new remainder still satisfies `x < m`. |
| 189 | + } |
| 190 | + |
| 191 | + // NB: `word_reduce()` is just the special case `shift_reduce(U::BITS - 1)` |
| 192 | + // that optimizes especially well. The correspondence is that `a == u` and |
| 193 | + // `b == (v >> 1).widen_hi()` |
| 194 | + // |
| 195 | + /// Replace the remainder `x` with `x(R/2) - un`, |
| 196 | + /// for a suitable quotient `u`, which is returned. |
| 197 | + fn word_reduce(&mut self) -> U { |
| 198 | + // To do so, we replace `2xq = uR + v` with |
| 199 | + // ``` |
| 200 | + // 2 * (x(R/2) - un) * q |
| 201 | + // = xqR - 2unq |
| 202 | + // = xqR - uqm |
| 203 | + // = uRR/2 + vR/2 - uRR/2 + ur |
| 204 | + // = ur + (v/2)R |
| 205 | + // ``` |
| 206 | + let (v, u) = self._2xq.lo_hi(); |
| 207 | + self._2xq = u.widen_mul(self.r) + U::widen_hi(v >> 1); |
| 208 | + u |
| 209 | + |
| 210 | + // Additional notes: |
| 211 | + // 1. As `v` is the low bits of `2xq`, it is even and can be halved. |
| 212 | + // 2. The new remainder is `(xr + mv/2) / R` (see below) |
| 213 | + // and since `v < R`, `r < m`, `x < m < R/2`, |
| 214 | + // that is also strictly less than `m`. |
| 215 | + // ``` |
| 216 | + // (x(R/2) - un)R |
| 217 | + // = xRR/2 - (m/2)uR |
| 218 | + // = x(qm + r) - (m/2)(2xq - v) |
| 219 | + // = xqm + xr - xqm + mv/2 |
| 220 | + // = xr + mv/2 |
| 221 | + // ``` |
| 222 | + } |
| 223 | +} |
| 224 | + |
| 225 | +#[cfg(test)] |
| 226 | +mod test { |
| 227 | + use crate::support::linear_mul_reduction; |
| 228 | + use crate::support::modular::Reducer; |
| 229 | + |
| 230 | + #[test] |
| 231 | + fn reducer_ops() { |
| 232 | + for n in 33..=63_u8 { |
| 233 | + for x in 0..2 * n { |
| 234 | + let temp = Reducer::new(x, n); |
| 235 | + let n = n as u32; |
| 236 | + let x0 = temp.partial_remainder() as u32; |
| 237 | + assert_eq!(x as u32, x0); |
| 238 | + for k in 0..=7 { |
| 239 | + let mut red = temp.clone(); |
| 240 | + let u = red.shift_reduce(k) as u32; |
| 241 | + let x1 = red.partial_remainder() as u32; |
| 242 | + assert_eq!(x1, (x0 << k) - u * n); |
| 243 | + assert!(x1 < 2 * n); |
| 244 | + assert!((red._2xq as u32).is_multiple_of(2 * x1)); |
| 245 | + |
| 246 | + // `word_reduce` is equivalent to |
| 247 | + // `shift_reduce(U::BITS - 1)` |
| 248 | + if k == 7 { |
| 249 | + let mut alt = temp.clone(); |
| 250 | + let w = alt.word_reduce(); |
| 251 | + assert_eq!(u, w as u32); |
| 252 | + assert_eq!(alt, red); |
| 253 | + } |
| 254 | + } |
| 255 | + } |
| 256 | + } |
| 257 | + } |
| 258 | + #[test] |
| 259 | + fn reduction_u8() { |
| 260 | + for y in 1..64u8 { |
| 261 | + for x in 0..2 * y { |
| 262 | + let mut r = x % y; |
| 263 | + for e in 0..100 { |
| 264 | + assert_eq!(r, linear_mul_reduction(x, e, y)); |
| 265 | + // maintain the correct expected remainder |
| 266 | + r <<= 1; |
| 267 | + if r >= y { |
| 268 | + r -= y; |
| 269 | + } |
| 270 | + } |
| 271 | + } |
| 272 | + } |
| 273 | + } |
| 274 | + #[test] |
| 275 | + fn reduction_u128() { |
| 276 | + assert_eq!( |
| 277 | + linear_mul_reduction::<u128>(17, 100, 123456789), |
| 278 | + (17 << 100) % 123456789 |
| 279 | + ); |
| 280 | + |
| 281 | + // power-of-two divisor |
| 282 | + assert_eq!( |
| 283 | + linear_mul_reduction(0xdead_beef, 100, 1_u128 << 116), |
| 284 | + 0xbeef << 100 |
| 285 | + ); |
| 286 | + |
| 287 | + let x = 10_u128.pow(37); |
| 288 | + let y = 11_u128.pow(36); |
| 289 | + assert!(x < y); |
| 290 | + let mut r = x; |
| 291 | + for e in 0..1000 { |
| 292 | + assert_eq!(r, linear_mul_reduction(x, e, y)); |
| 293 | + // maintain the correct expected remainder |
| 294 | + r <<= 1; |
| 295 | + if r >= y { |
| 296 | + r -= y; |
| 297 | + } |
| 298 | + assert!(r != 0); |
| 299 | + } |
| 300 | + } |
| 301 | +} |
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