//! Test the Decoder's UNTESTED reading of a residual they recorded as odd.
//!
//! `GP_DIALOG` holds 140 entries against a 70-record dialog table — a 2:1 ratio
//! that would make the id→entry join an ordering question. It does not hold:
//! adjacent pairing gives identical element-name sets on **2 of 65** pairs,
//! halves pairing on **0**. In `GP_TITLE` a language pair shares its element set
//! exactly, so identical sets are the signature there and almost nothing matches
//! here.
//!
//! The residual: the only two adjacent pairs that DO match are entries `0/1` and
//! `2/3` — and `2/3` is the DIFFICULTY build. Their plausible reading is that
//! dialog text is baked into language-specific sprites, so EN/JP entries differ
//! by construction. ⚠️ **They flagged it as untested and did not assert it**, and
//! it has a hole they named themselves: it would explain the 63 that differ and
//! leave the 2 that match needing their own explanation.
//!
//! This prints what the differences actually look like, so the reading is judged
//! against the names rather than accepted as plausible.
use std::collections::BTreeSet;
use sylpheed_formats::{pak, ratc, ui_layout};

fn main() {
    let root =
        std::env::var("SYLPHEED_DISC").expect("set SYLPHEED_DISC to the extracted disc root");
    let ar = pak::PakArchive::open(format!("{root}/dat/GP_DIALOG.pak")).expect("GP_DIALOG.pak");
    let sets: Vec<Option<BTreeSet<String>>> = ar
        .entries()
        .iter()
        .map(|e| {
            let by = ar.read(e).ok()?;
            if !ratc::is_ratc(&by) {
                return None;
            }
            let b = ui_layout::parse_build(&by)?;
            Some(b.elements.iter().map(|el| el.name.clone()).collect())
        })
        .collect();

    let (mut same, mut diff, mut pairs) = (0usize, 0usize, 0usize);
    let mut shown = 0;
    for i in (0..sets.len().saturating_sub(1)).step_by(2) {
        let (Some(a), Some(b)) = (&sets[i], &sets[i + 1]) else {
            continue;
        };
        pairs += 1;
        if a == b {
            same += 1;
            println!(
                "  entries {i:>3}/{:<3} IDENTICAL sets, {} element(s)",
                i + 1,
                a.len()
            );
            continue;
        }
        diff += 1;
        // The stage-dialog pairs, checked by name and by SPRITE COUNT. A
        // translation of one dialog carries the same amount of text; a
        // different stage does not. This is the Decoder's closing evidence for
        // the 37 pairs that differ WITHOUT a button-count mismatch, re-derived
        // here because it settles a bound I had recorded as unlikely to be
        // tested -- and saying so is what got it tested.
        if (10..=15).contains(&i) {
            let sp = |x: &BTreeSet<String>| x.iter().filter(|n| n.ends_with(".t32")).count();
            let stage = |x: &BTreeSet<String>| -> Vec<String> {
                let mut v: Vec<String> = x
                    .iter()
                    .filter_map(|n| {
                        n.strip_prefix("pzstg")
                            .and_then(|r| r.get(..2))
                            .map(|s| s.to_string())
                    })
                    .collect();
                v.sort();
                v.dedup();
                v
            };
            println!(
                "  entries {i:>3}/{:<3} stages {:?} vs {:?}   sprites {} vs {}",
                i + 1,
                stage(a),
                stage(b),
                sp(a),
                sp(b)
            );
        }
        if shown < 3 {
            shown += 1;
            let only_a: Vec<_> = a.difference(b).cloned().collect();
            let only_b: Vec<_> = b.difference(a).cloned().collect();
            println!(
                "  entries {i:>3}/{:<3} differ: {} only-in-first, {} only-in-second",
                i + 1,
                only_a.len(),
                only_b.len()
            );
            println!("      first  : {:?}", &only_a[..only_a.len().min(4)]);
            println!("      second : {:?}", &only_b[..only_b.len().min(4)]);
        }
    }
    // 🔴 THE DECISIVE DETAIL, not the impressionistic one. Two languages of one
    // dialog cannot differ in BUTTON COUNT. If adjacent entries do, they are
    // different dialogs and the whole adjacent-pairing premise is wrong -- which
    // is a stronger statement than "the language reading is untested".
    let btns = |s: &Option<BTreeSet<String>>| -> usize {
        s.as_ref()
            .map_or(0, |x| x.iter().filter(|n| n.contains("btn")).count())
    };
    let mut mismatched = 0;
    for i in (0..sets.len().saturating_sub(1)).step_by(2) {
        if sets[i].is_none() || sets[i + 1].is_none() {
            continue;
        }
        if btns(&sets[i]) != btns(&sets[i + 1]) {
            mismatched += 1
        }
    }
    println!("\n  adjacent pairs whose BUTTON COUNTS differ: {mismatched}");
    println!("  A language pair cannot. Every one of these is two different dialogs.");
    println!("\n  {pairs} adjacent pair(s): {same} identical, {diff} differing");
    println!("  Their reading -- text baked into language-specific sprites -- predicts");
    println!("  the differing names look SYSTEMATIC (a locale suffix, a parallel set).");
    println!("  Judge it against the names above rather than against its plausibility.");
}