406 lines
15 KiB
Rust
406 lines
15 KiB
Rust
//! The earliest start search, done in one pass over the schedule.
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//!
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//! The sequential form asks the same question in a loop: it lists every time
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//! at which feasibility could change, sorts them, and tests each in turn until
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//! one passes. That is correct but it recomputes the busy profile of every
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//! resource once per candidate, so a plan with many placements on one resource
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//! spends quadratic work deciding each task and cubic work across the plan.
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//!
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//! The profile does not depend on the candidate. A placement is busy between
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//! its start and its end whatever time the new task is proposed for, so the
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//! intervals during which the resource has no room are the same for every
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//! candidate. This module computes them once, subtracts them from the
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//! availability windows, and reads the earliest surviving start off the front.
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//! That is one sweep per resource and a linear merge, rather than a sweep per
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//! candidate.
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//!
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//! The result is confirmed with [`feasible_at`] before it is returned, and the
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//! candidate enumeration is kept as the fallback for the rare case where the
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//! sweep and the direct test disagree because of floating point rounding at a
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//! capacity boundary. The fast path is therefore an accelerator, never a
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//! second definition of what is feasible.
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use crate::greedy::{feasible_at_indexed, resources_have_capacity};
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use apsis_resource::ResourceId;
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use apsis_resource::{Resource, ResourceSet};
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use apsis_schedule::Schedule;
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use apsis_task::Task;
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use apsis_task::TaskId;
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/// The slack allowed when comparing a load against a capacity.
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///
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/// It matches the tolerance in [`feasible_at`], so the sweep and the direct
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/// test draw the same line at a boundary.
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///
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/// 比较负载与容量时允许的余量。
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///
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/// 它与 [`feasible_at`] 中的容差一致，故扫掠与直接测试在边界处划同一条
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/// 线。
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const CAPACITY_SLACK: f64 = 1e-9;
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/// Returns the earliest time at which a task can be placed.
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///
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/// The earliest feasible start is always one of a small set of candidate
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/// times: the start of the task's window, the start of an availability
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/// window of one of its resources, or the end of a placement already on one
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/// of those resources. Feasibility cannot change at any other instant,
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/// because the demand on every resource is constant between those events and
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/// the availability is constant between them too.
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///
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/// Returns `None` when the task cannot be placed anywhere in its window.
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///
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/// 返回某任务可被放置的最早时刻。
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///
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/// 最早可行起点总是某个小候选集合之一：任务窗口的起点、其某资源可用窗口
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/// 的起点，或其某资源上已有放置的结束时刻。可行性在其他任何瞬时都不会
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/// 改变，因为各资源的需求在这些事件之间恒定，可用性亦然。
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///
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/// 该任务在其窗口内无处可放时返回 `None`。
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pub fn earliest_feasible_start(
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task: &Task,
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tasks: &[Task],
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schedule: &Schedule,
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resources: &ResourceSet,
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) -> Option<f64> {
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if !task.is_valid() || !resources_have_capacity(task, resources) {
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return None;
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}
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if task.requirements.is_empty() {
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// A task with no requirements can start as soon as its window opens.
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// 无需求的任务可在其窗口开启时即开始。
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return Some(task.earliest_start);
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}
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let latest_start = task.latest_end - task.duration;
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// The demand of every task, looked up by identifier once. The sweep
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// visits every placement of every required resource, and reading the
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// demand by scanning the task list at each of those visits would make
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// deciding one task quadratic in the plan. The map is built once per
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// task, which is linear, and shared by every visit.
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// 各任务的需求，按标识符一次性建好索引。扫掠会访问每个所需资源上
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// 的每处放置，若每次访问都扫描任务列表来读需求，判定一个任务就会
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// 对整个计划呈二次。该索引每个任务构建一次，是线性的，并由所有
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// 访问共享。
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let demands = demand_index(tasks);
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// The sweep, confirmed by the direct test.
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// 扫掠，其结果由直接测试确认。
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if let Some(candidate) = swept_earliest_start(task, schedule, resources, latest_start, &demands)
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{
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if feasible_at_indexed(task, schedule, resources, candidate, &demands) {
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return Some(candidate);
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}
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}
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// The enumeration, which is exact. It runs when the sweep could not
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// produce a confirmed answer, which is when the task does not fit at all
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// or when rounding made the two disagree.
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// 枚举，它是精确的。当扫掠无法给出经确认的答案时运行，即该任务
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// 完全放不下，或舍入使二者不一致时。
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enumerate_earliest_start(task, schedule, resources, latest_start, &demands)
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}
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/// Returns the earliest start by testing every candidate in turn.
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///
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/// This is the original formulation, kept as the fallback for the sweep.
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///
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/// 依次测试每个候选，返回最早起点。
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///
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/// 这是最初的表述，保留为扫掠的回退。
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fn enumerate_earliest_start(
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task: &Task,
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schedule: &Schedule,
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resources: &ResourceSet,
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latest_start: f64,
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index: &DemandIndex,
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) -> Option<f64> {
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let mut candidates: Vec<f64> = vec![task.earliest_start];
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for requirement in &task.requirements {
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let resource = resources.get(requirement.resource)?;
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for window in resource.windows() {
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candidates.push(window.start());
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}
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for placement in schedule.placements() {
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let overlaps = demand_on(index, placement.task, resource.id).is_some();
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if overlaps {
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candidates.push(placement.end());
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}
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}
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}
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candidates
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.retain(|time| time.is_finite() && *time >= task.earliest_start && *time <= latest_start);
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candidates.sort_by(|first, second| {
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first
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.partial_cmp(second)
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.unwrap_or(core::cmp::Ordering::Equal)
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});
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candidates.dedup();
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candidates
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.into_iter()
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.find(|start| feasible_at_indexed(task, schedule, resources, *start, index))
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}
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/// Returns the earliest start derived from the swept profile.
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///
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/// The intervals returned by [`starts_fitting_on`] for each resource are
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/// intersected, and the first interval that lies inside the task's window
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/// gives the answer.
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///
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/// 由扫掠所得的负载曲线导出最早起点。
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///
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/// [`starts_fitting_on`] 为各资源返回的区间被求交，落在该任务窗口内的
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/// 第一个区间即给出答案。
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fn swept_earliest_start(
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task: &Task,
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schedule: &Schedule,
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resources: &ResourceSet,
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latest_start: f64,
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demands: &DemandIndex,
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) -> Option<f64> {
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let mut allowed = vec![(f64::NEG_INFINITY, f64::INFINITY)];
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for requirement in &task.requirements {
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let resource = resources.get(requirement.resource)?;
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let fitting = starts_fitting_on(
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task.duration,
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requirement.demand,
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requirement.resource,
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schedule,
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resource,
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demands,
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);
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allowed = intersect(&allowed, &fitting);
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if allowed.is_empty() {
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return None;
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}
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}
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let &(low, high) = allowed.first()?;
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let start = low.max(task.earliest_start);
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let end = high.min(latest_start);
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(start <= end).then_some(start)
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}
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/// Returns the intervals of start times at which a placement of the given
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/// duration and demand fits on one resource.
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///
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/// A start is in the result when the whole placement lies inside one
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/// availability window and the demand already placed on the resource never
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/// exceeds the capacity less the new demand. The result is sorted and the
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/// intervals are disjoint.
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///
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/// 返回可使给定时长与需求的放置落在某资源上的起点区间。
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///
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/// 当整个放置位于某个可用窗口内、且资源上已放置的需求始终不超过容量减去
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/// 新需求时，该起点被计入结果。结果有序且各区间互不相交。
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fn starts_fitting_on(
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duration: f64,
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demand: f64,
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resource_id: ResourceId,
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schedule: &Schedule,
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resource: &Resource,
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demands: &DemandIndex,
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) -> Vec<(f64, f64)> {
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let blocked = blocked_intervals(demand, resource_id, schedule, resource, demands);
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let mut result = Vec::new();
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for window in resource.windows() {
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for (from, to) in subtract(&blocked, window.start(), window.end()) {
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// The placement has to fit inside, so the last start is the end
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// of the free interval less the duration.
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// 放置须容纳于其中，故最晚起点为该空闲区间的终点减去时长。
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let last_start = to - duration;
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if last_start >= from {
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result.push((from, last_start));
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}
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}
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}
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result
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}
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/// Returns the intervals during which a resource has no room for a demand.
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///
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/// The intervals are where the demand already placed on the resource exceeds
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/// the capacity less `demand`. They are computed with a single sweep over the
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/// placement boundaries, and are sorted and disjoint.
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///
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/// 返回某资源无余量容纳某需求的区间。
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///
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/// 这些区间即资源上已放置的需求超过容量减去 `demand` 的时段。它们由
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/// 对各放置边界的一次扫掠算出，有序且互不相交。
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fn blocked_intervals(
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demand: f64,
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resource_id: ResourceId,
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schedule: &Schedule,
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resource: &Resource,
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demands: &DemandIndex,
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) -> Vec<(f64, f64)> {
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let spare = resource.capacity() - demand;
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// The load profile changes only where a placement starts or ends, so the
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// sweep needs only those two instants per placement.
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// 负载曲线只在某处放置开始或结束时改变，故扫掠每个放置只需那两个
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// 瞬时。
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let mut events: Vec<(f64, f64)> = Vec::new();
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for placement in schedule.placements() {
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let Some(placed) = demand_on(demands, placement.task, resource_id) else {
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continue;
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};
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events.push((placement.start, placed));
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events.push((placement.end(), -placed));
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}
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if events.is_empty() {
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return Vec::new();
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}
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events.sort_by(|first, second| {
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first
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.0
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.partial_cmp(&second.0)
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.unwrap_or(core::cmp::Ordering::Equal)
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.then_with(|| {
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first
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.1
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.partial_cmp(&second.1)
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.unwrap_or(core::cmp::Ordering::Equal)
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})
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});
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let limit = spare + CAPACITY_SLACK;
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let mut blocked: Vec<(f64, f64)> = Vec::new();
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let mut load = 0.0;
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let mut opened: Option<f64> = None;
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let mut index = 0;
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while index < events.len() {
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let time = events[index].0;
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// The load is constant on the interval that ends at this instant.
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// 负载在以该瞬时结束的区间上恒定。
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if load > limit {
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opened.get_or_insert(time);
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}
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// Apply every change at this instant at once, so the load compared
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// above belongs to the interval before it and the one after belongs
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// to the next comparison.
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// 一次性施加该瞬时的全部变化，故上面比较的负载属于其前的区间，
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// 其后的负载则用于下一次比较。
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let mut next = index;
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while next < events.len() && events[next].0 == time {
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load += events[next].1;
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next += 1;
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}
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if load <= limit {
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if let Some(start) = opened.take() {
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blocked.push((start, time));
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}
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}
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index = next;
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}
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// A run that never closes extends past every placement, so it is
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// recorded as open ended and swallows the rest of the horizon.
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// 始终不闭合的区段延伸越过所有放置，故记为开区间，吞没时域的
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// 其余部分。
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if let Some(start) = opened {
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blocked.push((start, f64::INFINITY));
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}
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blocked
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}
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/// The demand each task puts on each resource, by task identifier.
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///
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/// A task identifier is a small dense integer, so the demands are held in a
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/// vector indexed by it rather than in a map. Reading one is then an array
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/// index rather than a hash, and a hash of a small integer costs more than
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/// the scan it replaces.
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///
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/// 各任务在各资源上的需求，按任务标识符索引。
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///
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/// 任务标识符是小的稠密整数，故需求存于以其为下标的向量而非映射中。读取
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/// 时即数组下标而非哈希，而哈希一个小整数的代价高于它所取代的扫描。
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pub(crate) type DemandIndex = Vec<Option<std::collections::HashMap<ResourceId, f64>>>;
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/// Builds a [`DemandIndex`] in one pass over the task list.
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///
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/// 由任务列表一趟构建 [`DemandIndex`]。
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pub(crate) fn demand_index(tasks: &[Task]) -> DemandIndex {
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let highest = tasks.iter().map(|task| task.id.index()).max().unwrap_or(0);
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let mut index: DemandIndex = (0..=highest).map(|_| None).collect();
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for task in tasks {
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if task.requirements.is_empty() {
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continue;
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}
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let demands: std::collections::HashMap<ResourceId, f64> = task
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.requirements
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.iter()
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.map(|requirement| (requirement.resource, requirement.demand))
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.collect();
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index[task.id.index()] = Some(demands);
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}
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index
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}
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/// Returns the demand a placed task puts on a resource, if any.
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///
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/// 返回某已放置任务在某资源上的需求，若有。
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pub(crate) fn demand_on(index: &DemandIndex, task: TaskId, resource: ResourceId) -> Option<f64> {
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index
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.get(task.index())
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.and_then(Option::as_ref)
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.and_then(|required| required.get(&resource))
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.copied()
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}
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/// Returns the pieces of an interval that no blocked interval covers.
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///
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/// Both inputs are sorted and disjoint. The result is sorted and disjoint.
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///
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/// 返回某区间中未被任何阻塞区间覆盖的部分。
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///
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/// 两个输入均有序且互不相交，结果亦然。
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fn subtract(blocked: &[(f64, f64)], from: f64, to: f64) -> Vec<(f64, f64)> {
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let mut pieces = Vec::new();
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let mut cursor = from;
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for &(start, end) in blocked {
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if end <= cursor {
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continue;
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}
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if start >= to {
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break;
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}
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if start > cursor {
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pieces.push((cursor, start.min(to)));
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}
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cursor = cursor.max(end);
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if cursor >= to {
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break;
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}
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}
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if cursor < to {
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pieces.push((cursor, to));
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}
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pieces
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}
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/// Returns the intersection of two sorted, disjoint lists of intervals.
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///
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/// The result is sorted and disjoint.
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///
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/// 返回两份有序且互不相交的区间列表之交。
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///
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/// 结果有序且互不相交。
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fn intersect(left: &[(f64, f64)], right: &[(f64, f64)]) -> Vec<(f64, f64)> {
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let mut result = Vec::new();
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let mut i = 0;
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let mut j = 0;
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while i < left.len() && j < right.len() {
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let start = left[i].0.max(right[j].0);
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let end = left[i].1.min(right[j].1);
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if start < end {
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result.push((start, end));
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}
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if left[i].1 < right[j].1 {
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i += 1;
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} else {
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j += 1;
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}
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}
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result
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}
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