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//! symplex 0.2 base-layer fixes — construction cost must scale with the
//! DAG, not the unfolded tree.
//!
//! `e ← sin(e) + cos(e)` doubles the tree at every step while the hash-consed
//! DAG only grows by three nodes. Canonical sort keys used to be the full
//! concatenation of the children's keys (2^depth bytes), which made step 23
//! take tens of seconds. Keys are now bounded.
use std::time::{Duration, Instant};
use symplex::prelude::*;
/// Wall-clock hang guard. Two seconds on a developer machine; scaled up on
/// shared CI runners (`CI` is set), which are several times slower and noisy.
fn time_budget(secs: u64) -> std::time::Duration {
let mult = if std::env::var_os("CI").is_some() {
5
} else {
1
};
std::time::Duration::from_secs(secs * mult)
}
fn doubling(depth: usize) -> (Ex, Duration) {
let ctx = Context::new();
let x = ctx.symbol("x");
let mut e = x;
let start = Instant::now();
for _ in 0..depth {
e = e.sin() + e.cos();
}
(e, start.elapsed())
}
#[test]
fn sin_cos_doubling_depth_30_is_fast() {
let (e, elapsed) = doubling(30);
assert!(
elapsed < time_budget(2),
"depth-30 doubling took {elapsed:?}"
);
// Still a well-formed expression: two summands, both function nodes.
assert_eq!(e.args().len(), 2);
}
#[test]
fn doubling_ordering_is_deterministic_across_contexts() {
// The bounded key must still give the same canonical order for the
// same structure regardless of arena allocation history.
let (a, _) = doubling(12);
let ctx = Context::new();
// Warm the second context with unrelated nodes so ExprIds differ.
let y = ctx.symbol("y");
let _junk: Vec<Ex> = (0..50).map(|k| (&y + k).exp()).collect();
let x = ctx.symbol("x");
let mut b = x;
for _ in 0..12 {
b = b.sin() + b.cos();
}
assert_eq!(a.to_string(), b.to_string());
}
#[test]
fn deep_sums_of_wide_products_stay_fast() {
// Another tree-vs-DAG blow-up shape: p ← p*q + q*p with shared q.
let ctx = Context::new();
let x = ctx.symbol("x");
let y = ctx.symbol("y");
let start = Instant::now();
let mut p = &x + &y;
let mut q = &x - &y;
for _ in 0..25 {
let np = &p * &q + &q * &p + 1;
let nq = &p * &p - &q * &q;
p = np;
q = nq;
}
let elapsed = start.elapsed();
assert!(elapsed < time_budget(2), "took {elapsed:?}");
// (Do not display `p`: its unfolded tree, and hence the string, has
// ~2^25 nodes. Structural queries stay cheap.)
// p*q + q*p + 1 combines to 2*p*q + 1.
assert_eq!(p.args().len(), 2);
assert!(!p.is_constant());
}