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Merge #200

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200: Fix plane sweep algorithm r=Kerollmops a=LegendreM

Fix plain sweep algorithm after creating some tests on proximity.

Co-authored-by: many <maxime@meilisearch.com>
This commit is contained in:
bors[bot] 2021-05-26 11:36:24 +00:00 committed by GitHub
commit 76b9178b16
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1 changed files with 20 additions and 13 deletions
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33
milli/src/search/criteria/proximity.rs
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@ -324,18 +324,30 @@ fn resolve_plane_sweep_candidates(
// take the inner proximity of the first group as initial
let (_, (_, mut proximity, _)) = groups.first()?;
let (_, (left_most_pos, _, _)) = groups.first()?;
let (_, (_, _, right_most_pos)) = groups.last()?;
let (_, (_, _, right_most_pos)) = groups.iter().max_by_key(|(_, (_, _, right_most_pos))| right_most_pos)?;
for pair in groups.windows(2) {
if let [(i1, (_, _, rpos1)), (i2, (lpos2, prox2, _))] = pair {
// if a pair overlap, meaning that they share at least a word, we return None
if rpos1 >= lpos2 { return None }
if let [(i1, (lpos1, _, rpos1)), (i2, (lpos2, prox2, rpos2))] = pair {
// if two positions are equal, meaning that they share at least a word, we return None
if rpos1 == rpos2 || lpos1 == lpos2 || rpos1 == lpos2 || lpos1 == rpos2 {
return None
}
let pair_proximity = {
// if intervals are disjoint [..].(..)
if lpos2 > rpos1 { lpos2 - rpos1 }
// if the second interval is a subset of the first [.(..).]
else if rpos2 < rpos1 { (lpos2 - lpos1).min(rpos1 - rpos2) }
// if intervals overlaps [.(..].)
else { (lpos2 - lpos1).min(rpos2 - rpos1) }
};
// if groups are in the good order (query order) we remove 1 to the proximity
// the proximity is clamped to 7
let pair_proximity = if i1 < i2 {
(*lpos2 - *rpos1 - 1).min(7)
(pair_proximity - 1).min(7)
} else {
(*lpos2 - *rpos1).min(7)
pair_proximity.min(7)
};
proximity += pair_proximity as u8 + prox2;
@ -385,26 +397,21 @@ fn resolve_plane_sweep_candidates(
// let q be the position q of second group of the interval.
let q = current[1];
let mut leftmost_index = 0;
// If p > r, then the interval [l, r] is minimal and
// we insert it into the heap according to its size.
if p.map_or(true, |p| p.1 > rightmost.1) {
leftmost_index = current[0].0;
if let Some(group) = compute_groups_proximity(&current, consecutive) {
output.push(group);
}
}
// TODO not sure about breaking here or when the p list is found empty.
let p = match p {
Some(p) => p,
None => break,
};
// Remove the leftmost group P in the interval,
// and pop the same group from a list.
current[leftmost_index] = p;
// Replace the leftmost group P in the interval.
current[0] = p;
if p.1 > rightmost.1 {
// if [l, r] is minimal, let r = p and l = q.