//! Is our index list one element late? A capture-free check of the `pad` choice.
//!
//! `anchor_pool_mesh` tries `ib = vb − idx_count*2 − pad` for `pad` in `0..=3`,
//! **pad 0 first and with the looser winding gate** (`XBG7_PAD0_CONSISTENCY`,
//! 0.70) than the pad ≥ 1 path (0.85). The 2026-08-13 index-byte comparison
//! against the runtime showed 17 draw batches whose captured indices equal our
//! decoded list shifted by exactly one element — all of them on buffers whose
//! real index data sits at pad 2. A one-element shift re-wires every triangle,
//! and it is invisible to every count-based metric (the count, the coverage and
//! the positions are all still right).
//!
//! This reproduces that finding from the container alone: for each resource it
//! reads the index run at pad 0 and at pad 2 and scores both on two properties a
//! correct triangle list has — **no degenerate triangles** (two equal indices)
//! and consistent winding against the stored normals.
//!
//! Usage: index_pad_check <container.xpr> [name-substring]
use sylpheed_formats::mesh::Xbg7Model;

fn be16(b: &[u8], at: usize) -> u32 {
    ((b[at] as u32) << 8) | b[at + 1] as u32
}

/// Degenerate triangles (a repeated index) and the fraction of triangles whose
/// face normal agrees with their vertices' stored normals.
fn score(idx: &[u32], pos: &[[f32; 3]], nrm: &[[f32; 3]]) -> (usize, f32, usize) {
    let mut degen = 0usize;
    let mut agree = 0usize;
    let mut counted = 0usize;
    for t in idx.chunks_exact(3) {
        let (a, b, c) = (t[0] as usize, t[1] as usize, t[2] as usize);
        if a == b || b == c || a == c {
            degen += 1;
            continue;
        }
        if a >= pos.len() || b >= pos.len() || c >= pos.len() {
            continue;
        }
        let e1 = [pos[b][0] - pos[a][0], pos[b][1] - pos[a][1], pos[b][2] - pos[a][2]];
        let e2 = [pos[c][0] - pos[a][0], pos[c][1] - pos[a][1], pos[c][2] - pos[a][2]];
        let f = [
            e1[1] * e2[2] - e1[2] * e2[1],
            e1[2] * e2[0] - e1[0] * e2[2],
            e1[0] * e2[1] - e1[1] * e2[0],
        ];
        if nrm.len() <= a {
            continue;
        }
        let n = nrm[a];
        let dot = f[0] * n[0] + f[1] * n[1] + f[2] * n[2];
        counted += 1;
        if dot > 0.0 {
            agree += 1;
        }
    }
    let frac = if counted == 0 { 0.0 } else { agree as f32 / counted as f32 };
    (degen, frac.max(1.0 - frac), counted)
}

fn main() {
    let a: Vec<String> = std::env::args().collect();
    let bytes = std::fs::read(&a[1]).expect("container");
    let filter = a.get(2).cloned().unwrap_or_default();

    let models = Xbg7Model::stage_models(&bytes);
    println!(
        "{:<22} {:>6} {:>6}  {:<26} {:<26} verdict",
        "resource", "verts", "idx", "pad 0 (what we decode)", "pad 2 (2 bytes earlier)"
    );
    let (mut better_p2, mut better_p0, mut tie) = (0usize, 0usize, 0usize);
    let mut hist_ok: Vec<(usize, String, usize)> = Vec::new();
    for m in &models {
        if !filter.is_empty() && !m.name.contains(&filter) {
            continue;
        }
        if m.meshes.len() != 1 {
            continue; // single-block path only; grouped pools use another formula
        }
        let sm = &m.meshes[0];
        let Some(vb) = sm.vbuf_offset else { continue };
        let n = sm.indices.len();
        if n < 6 || sm.normals.is_empty() || vb < n * 2 + 2 {
            continue;
        }
        let read_at = |start: usize| -> Vec<u32> { (0..n).map(|k| be16(&bytes, start + k * 2)).collect() };
        let l0 = read_at(vb - n * 2);
        let l2 = read_at(vb - n * 2 - 2);
        // Sanity: l0 must be what the decoder emitted, or the assumption that we
        // took pad 0 is wrong for this resource and the comparison is meaningless.
        let took_pad0 = l0 == sm.indices;
        if !took_pad0 {
            continue;
        }
        let (d0, c0, _) = score(&l0, &sm.positions, &sm.normals);
        let (d2, c2, _) = score(&l2, &sm.positions, &sm.normals);
        let max_i0 = l0.iter().max().copied().unwrap_or(0);
        let max_i2 = l2.iter().max().copied().unwrap_or(0);
        let v = sm.positions.len() as u32;
        // A shifted list typically also overruns the pool by one vertex or leaves
        // the last vertex unreferenced, so carry the coverage as evidence too.
        let verdict = if d2 < d0 && c2 >= c0 {
            better_p2 += 1;
            "pad 2 is the real one"
        } else if d0 < d2 || c0 > c2 {
            better_p0 += 1;
            "pad 0 fine"
        } else {
            tie += 1;
            "indistinguishable"
        };
        if verdict == "pad 0 fine" {
            hist_ok.push((d0, m.name.clone(), n));
        }
        if verdict != "pad 0 fine" || !filter.is_empty() {
            println!(
                "{:<22} {:>6} {:>6}  degen {:>5} wind {:.3} max {:<5}  degen {:>5} wind {:.3} max {:<5}  {}",
                m.name, v, n, d0, c0, max_i0, d2, c2, max_i2, verdict
            );
        }
    }
    println!("\npad 2 wins: {better_p2} · pad 0 wins: {better_p0} · indistinguishable: {tie}");
    // Is "no degenerate triangle" a safe invariant for a correctly anchored
    // block? Distribution over the resources where pad 0 is the better choice.
    let zero = hist_ok.iter().filter(|(d, _, _)| *d == 0).count();
    println!(
        "of the {} pad-0-correct resources, {zero} have ZERO degenerate triangles",
        hist_ok.len()
    );
    let mut worst: Vec<_> = hist_ok.iter().filter(|(d, _, _)| *d > 0).collect();
    worst.sort_by_key(|(d, _, _)| std::cmp::Reverse(*d));
    for (d, n, idx) in worst.iter().take(10) {
        println!("  {n}: {d} degenerate of {} triangles", idx / 3);
    }
}