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| 1 | +// Copyright 2025 Developers of the Rand project. |
| 2 | +// |
| 3 | +// Licensed under the Apache License, Version 2.0 <LICENSE-APACHE or |
| 4 | +// https://www.apache.org/licenses/LICENSE-2.0> or the MIT license |
| 5 | +// <LICENSE-MIT or https://opensource.org/licenses/MIT>, at your |
| 6 | +// option. This file may not be copied, modified, or distributed |
| 7 | +// except according to those terms. |
| 8 | + |
| 9 | +use rand::SeedableRng; |
| 10 | +use rand::rngs::SmallRng; |
| 11 | +use rand_distr::{Bernoulli, Binomial, Distribution, Geometric}; |
| 12 | + |
| 13 | +/// Say `n` samples are drawn independently from a Bernoulli distribution |
| 14 | +/// with probability `q` in `[q_lb, q_ub]` of outputting 1. Let X be their sum. |
| 15 | +/// |
| 16 | +/// If `k > q_ub n`, this function returns an estimate (some inaccuracy possible |
| 17 | +/// due to floating point error) of an upper bound on the log-probability that X >= k; |
| 18 | +/// if `k < q_lb n`, the function returns an estimate of an upper bound for the |
| 19 | +/// log-probability that X <= k. Otherwise, `k` might be equal to `q n` and the |
| 20 | +/// function returns log-probability 0.0 (probability 1.0). |
| 21 | +/// |
| 22 | +/// Note: the value returned is the logarithm of the probability bound estimate. |
| 23 | +fn bernouilli_ln_tail_prob_bound(q_lb: f64, q_ub: f64, k: u64, n: u64) -> f64 { |
| 24 | + fn kl_div_lb(r: f64, p: f64) -> f64 { |
| 25 | + // Note: this calculation can be inaccurate when r, p are tiny |
| 26 | + if p <= 0.0 { |
| 27 | + if r > p { f64::INFINITY } else { 0.0 } |
| 28 | + } else if 1.0 - p <= 0.0 { |
| 29 | + if r < p { f64::INFINITY } else { 0.0 } |
| 30 | + } else { |
| 31 | + if r == 0.0 { |
| 32 | + (1.0 - r) * f64::ln((1.0 - r) / (1.0 - p)) |
| 33 | + } else if 1.0 - r == 0.0 { |
| 34 | + r * f64::ln(r / p) |
| 35 | + } else { |
| 36 | + r * f64::ln(r / p) + (1.0 - r) * f64::ln((1.0 - r) / (1.0 - p)) |
| 37 | + } |
| 38 | + } |
| 39 | + } |
| 40 | + |
| 41 | + assert!(k <= n); |
| 42 | + assert!(0.0 <= q_lb && q_ub <= 1.0); |
| 43 | + |
| 44 | + let r = (k as f64) / (n as f64); |
| 45 | + if r < q_lb { |
| 46 | + -(n as f64) * kl_div_lb(r, q_lb) |
| 47 | + } else if r > q_ub { |
| 48 | + -(n as f64) * kl_div_lb(1.0 - r, 1.0 - q_ub) |
| 49 | + } else { |
| 50 | + 0.0 |
| 51 | + } |
| 52 | +} |
| 53 | + |
| 54 | +/// For y=e^x, z = min(2 y, 1), return ln(z) |
| 55 | +fn min_2x_under_ln(x: f64) -> f64 { |
| 56 | + if x.is_nan() { |
| 57 | + x |
| 58 | + } else { |
| 59 | + 0.0f64.min(x + f64::ln(2.0)) |
| 60 | + } |
| 61 | +} |
| 62 | + |
| 63 | +// Threshold probability for the output of a test possibly indicating |
| 64 | +// a discrepancy between the actual and ideal distribution. |
| 65 | +// (The event could happen by chance on a 1e-3 fraction of seeds even |
| 66 | +// if the distributions match.) |
| 67 | +const POSSIBLE_DISCREPANCY_THRESHOLD: f64 = -3.0 * std::f64::consts::LN_10; |
| 68 | + |
| 69 | +// Threshold probability for the output of a test certainly indicating |
| 70 | +// a discrepancy between the actual and ideal distribution |
| 71 | +// (Hardware failures are many orders of magnitude more likely |
| 72 | +// than the entire system being correct.) |
| 73 | +const CERTAIN_DISCREPANCY_THRESHOLD: f64 = -40.0 * std::f64::consts::LN_10; |
| 74 | + |
| 75 | +#[derive(Debug)] |
| 76 | +enum TestFailure { |
| 77 | + #[allow(unused)] |
| 78 | + Possible(f64), |
| 79 | + Certain, |
| 80 | +} |
| 81 | + |
| 82 | +fn test_binary( |
| 83 | + seed: u64, |
| 84 | + ideal_prob_lb: f64, |
| 85 | + ideal_prob_ub: f64, |
| 86 | + sample_size: u64, |
| 87 | + sample_fn: &dyn Fn(&mut SmallRng) -> bool, |
| 88 | +) -> Result<(), TestFailure> { |
| 89 | + let mut rng = rand::rngs::SmallRng::seed_from_u64(seed); |
| 90 | + let mut ones: u64 = 0; |
| 91 | + for _ in 0..sample_size { |
| 92 | + ones += if sample_fn(&mut rng) { 1 } else { 0 }; |
| 93 | + } |
| 94 | + |
| 95 | + let ln_single_tail_p = |
| 96 | + bernouilli_ln_tail_prob_bound(ideal_prob_lb, ideal_prob_ub, ones, sample_size); |
| 97 | + // Double the probability to correct for the fact that there are two tails |
| 98 | + let ln_p = min_2x_under_ln(ln_single_tail_p); |
| 99 | + |
| 100 | + if ln_p < CERTAIN_DISCREPANCY_THRESHOLD { |
| 101 | + Err(TestFailure::Certain) |
| 102 | + } else if ln_p < POSSIBLE_DISCREPANCY_THRESHOLD { |
| 103 | + Err(TestFailure::Possible(f64::exp(ln_p))) |
| 104 | + } else { |
| 105 | + Ok(()) |
| 106 | + } |
| 107 | +} |
| 108 | + |
| 109 | +/// Verify that the re-exported Bernoulli sampler is |
| 110 | +/// not clearly far from the correct distribution |
| 111 | +#[test] |
| 112 | +fn test_bernouilli() { |
| 113 | + let sample_size = 1000000; |
| 114 | + let seed = 0x1; |
| 115 | + |
| 116 | + // Check that the Bernouilli sampler is not far from correct |
| 117 | + for p_base in [0.0, 1e-9, 1e-3, 1.0 / 3.0, 0.5] { |
| 118 | + for p in [p_base, 1.0 - p_base] { |
| 119 | + test_binary(seed, p, p, sample_size, &|rng| { |
| 120 | + let b = Bernoulli::new(p).unwrap(); |
| 121 | + b.sample(rng) |
| 122 | + }) |
| 123 | + .unwrap(); |
| 124 | + } |
| 125 | + } |
| 126 | + |
| 127 | + // Check that the test will actually catch clear discrepancies. |
| 128 | + assert!(matches!( |
| 129 | + test_binary(seed, 0.4, 0.4, sample_size, &|rng| { |
| 130 | + let b = Bernoulli::new(0.6).unwrap(); |
| 131 | + b.sample(rng) |
| 132 | + }), |
| 133 | + Err(TestFailure::Certain) |
| 134 | + )); |
| 135 | +} |
| 136 | + |
| 137 | +/// For X ~ Binomial(n; p), returns Pr[X mod 2 = 1] |
| 138 | +fn binomial_last_bit_probability(n: u64, p: f64) -> f64 { |
| 139 | + /* Since |
| 140 | + * |
| 141 | + * 1 = (p + (1-p))^n = ∑_k \binom{n}{k} p^k (1-p)^{n-k} , |
| 142 | + * |
| 143 | + * and |
| 144 | + * |
| 145 | + * (-p + (1-p))^n = ∑_k (-1)^k \binom{n}{k} p^k (1-p)^{n-k} , |
| 146 | + * |
| 147 | + * adding them together gives: |
| 148 | + * |
| 149 | + * 1 + (1 - 2p)^n = ∑_k (1 + (-1)^k) \binom{n}{k} p^k (1-p)^{n-k} |
| 150 | + * = ∑_k 2 ⋅ 1_{k mod 2 = 0} \binom{n}{k} p^k (1-p)^{n-k} |
| 151 | + * = 2 Pr[k mod 2 = 0] . |
| 152 | + * |
| 153 | + * So: |
| 154 | + * |
| 155 | + * Pr[k mod 2 = 1] = 1 - ½ (1 + (1 - 2p)^n) = ½ (1 - (1 - 2p)^n) |
| 156 | + */ |
| 157 | + |
| 158 | + 0.5 * (1.0 - (1.0 - 2.0 * p).powi(n.try_into().unwrap_or(i32::MAX))) |
| 159 | +} |
| 160 | + |
| 161 | +/// Do samples from a binomial distribution, taken mod 2, match the expected distribution? |
| 162 | +/// (This is a likely failure mode of samplers using floating point.) |
| 163 | +#[test] |
| 164 | +fn test_binomial_last_bit() { |
| 165 | + let sample_size = 100000; |
| 166 | + let seed = 0x1; |
| 167 | + |
| 168 | + let mut sizes = Vec::<u64>::new(); |
| 169 | + // Binomial::new(n, p) currently panics for when a quantity derived from n is >= i64::MAX |
| 170 | + for i in 1..=62 { |
| 171 | + sizes.push((1 << i) - 1); |
| 172 | + sizes.push(1 << i); |
| 173 | + sizes.push((1 << i) + 1); |
| 174 | + sizes.push(3 * (1 << (i - 1))); |
| 175 | + } |
| 176 | + |
| 177 | + for s in sizes { |
| 178 | + for p in [0.1, 0.5, 0.9] { |
| 179 | + let t = binomial_last_bit_probability(s, p); |
| 180 | + |
| 181 | + let Ok(dist) = Binomial::new(s, p) else { |
| 182 | + println!("Failed to create Binomial with n={}, p={}", s, p); |
| 183 | + continue; |
| 184 | + }; |
| 185 | + |
| 186 | + let res = test_binary(seed, t, t, sample_size, &|rng| dist.sample(rng) % 2 == 1); |
| 187 | + |
| 188 | + // Binomial::new()'s documentation only promises accuracy up to n=~2^53 |
| 189 | + // Using `p` closer to 0 or 1 produces a narrower peak which is easier to sample correctly |
| 190 | + if s <= 1 << 53 { |
| 191 | + assert!(res.is_ok()); |
| 192 | + } else { |
| 193 | + println!( |
| 194 | + "Binomial distribution with n={}, p={:.4}, last bit prob {:.4}, log2(n)={:.3}: last bit result {:?}", |
| 195 | + s, |
| 196 | + p, |
| 197 | + t, |
| 198 | + (s as f64).log2(), |
| 199 | + res |
| 200 | + ); |
| 201 | + } |
| 202 | + } |
| 203 | + } |
| 204 | +} |
| 205 | + |
| 206 | +/// For X ~ Geometric(p), returns Pr[X mod 2 = 1] |
| 207 | +fn geometric_last_bit_probability(p: f64) -> f64 { |
| 208 | + /* The geometric probabilities are |
| 209 | + * 0 1 2 3 |
| 210 | + * p, (1-p)p, (1-p)^2 p, (1-p)^3 p, ... |
| 211 | + * |
| 212 | + * As Pr[X mod 2 = 1] = (1 - p) Pr[X mod 2 = 0], |
| 213 | + * and Pr[X mod 2 = 1] = 1 - Pr[X mod 2 = 0], |
| 214 | + * it follows: |
| 215 | + * |
| 216 | + * Pr[X mod 2 = 1] = 1 - 1/(2 - p) |
| 217 | + */ |
| 218 | + (1.0 - p) / (2.0 - p) |
| 219 | +} |
| 220 | + |
| 221 | +#[test] |
| 222 | +fn test_geometric_last_bit() { |
| 223 | + let sample_size = 100000; |
| 224 | + let seed = 0x1; |
| 225 | + |
| 226 | + // Test on distributions with values of p of various closeness to 0.0 |
| 227 | + for i in 0..=128 { |
| 228 | + let p = 0.5_f64.powf(i as f64 / 2.0); |
| 229 | + |
| 230 | + if 1.0 - p == 1.0 { |
| 231 | + // The current implementation of Geometric always returns u64::MAX when 1.0 - p = 1.0 |
| 232 | + continue; |
| 233 | + } |
| 234 | + |
| 235 | + let t = geometric_last_bit_probability(p); |
| 236 | + let clipped_prob = (special::Primitive::ln_1p(-p) * 2.0f64.powi(64)).exp(); |
| 237 | + |
| 238 | + let Ok(dist) = Geometric::new(p) else { |
| 239 | + println!("Failed to create Geometric with p={}", p); |
| 240 | + continue; |
| 241 | + }; |
| 242 | + |
| 243 | + let res = test_binary( |
| 244 | + seed, |
| 245 | + t - clipped_prob, |
| 246 | + t + clipped_prob, |
| 247 | + sample_size, |
| 248 | + &|rng| dist.sample(rng) % 2 == 1, |
| 249 | + ); |
| 250 | + |
| 251 | + println!( |
| 252 | + "Geometric distribution with p={:e}, last bit prob {}-{}: last bit result {:?}", |
| 253 | + p, |
| 254 | + t - clipped_prob, |
| 255 | + t + clipped_prob, |
| 256 | + res |
| 257 | + ); |
| 258 | + |
| 259 | + assert!(res.is_ok(), "{:?}", res); |
| 260 | + } |
| 261 | +} |
| 262 | + |
| 263 | +/// How accurate are the probabilities for outputting 0 and n for Binomial(n, p)? |
| 264 | +/// |
| 265 | +/// Uncareful boundary handling can make the endpoint probabilities wrong by a |
| 266 | +/// constant factor, although the tails have low enough probability that this |
| 267 | +/// can be hard to detect. |
| 268 | +#[test] |
| 269 | +#[ignore] |
| 270 | +fn test_binomial_endpoints() { |
| 271 | + let p = 0.5; |
| 272 | + // Note: conclusively indicating an error tends to require more than 1 / event probability |
| 273 | + // samples, so this test needs ~100M samples to find any issues in the below at n >= 20 |
| 274 | + let sample_size = 400_000_000; |
| 275 | + let seed = 0x1; |
| 276 | + |
| 277 | + // With the current implementation, n <= 20 always uses BINV or Poisson sampling, so to check |
| 278 | + // the BTPE implementation larger `s` is needed. |
| 279 | + |
| 280 | + for s in 1..25u64 { |
| 281 | + let Ok(dist) = Binomial::new(s, p) else { |
| 282 | + println!("Failed to create Binomial with n={}, p={}", s, p); |
| 283 | + continue; |
| 284 | + }; |
| 285 | + |
| 286 | + let t = p.powi(s as i32) + (1.0 - p).powi(s as i32); |
| 287 | + let res = test_binary(seed, t, t, sample_size, &|rng| { |
| 288 | + let v = dist.sample(rng); |
| 289 | + v == 0 || v == s |
| 290 | + }); |
| 291 | + println!( |
| 292 | + "Binomial({}, {}) endpoint test, evt prob {:e}, {:?}", |
| 293 | + s, p, t, res |
| 294 | + ); |
| 295 | + assert!(res.is_ok()); |
| 296 | + } |
| 297 | +} |
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