//! The building blocks a greedy scheduler needs.

use apsis_resource::{ResourceId, ResourceSet};
use apsis_schedule::Schedule;
use apsis_task::{Task, TaskId};

/// Returns `true` when every resource of a task has the capacity to hold its
/// demand.
///
/// A task that demands more of a resource than the resource has can never be
/// scheduled on it, so this is checked before the search for a time. Returns
/// `false` when a required resource does not exist.
pub fn resources_have_capacity(task: &Task, resources: &ResourceSet) -> bool {
    task.requirements.iter().all(|requirement| {
        resources
            .get(requirement.resource)
            .is_some_and(|resource| requirement.demand <= resource.capacity() + 1e-9)
    })
}

/// Returns the total demand placed on each resource at an instant, given a
/// schedule and the tasks it contains.
///
/// The result is in the order the resources appear in the set. A resource
/// that is not in the set is omitted.
pub fn demand_at(
    schedule: &Schedule,
    tasks: &[Task],
    resources: &ResourceSet,
    time: f64,
) -> Vec<(ResourceId, f64)> {
    resources
        .resources()
        .iter()
        .map(|resource| {
            let total: f64 = schedule
                .placements()
                .iter()
                .filter(|placement| placement.covers(time))
                .filter_map(|placement| {
                    tasks
                        .iter()
                        .find(|task| task.id == placement.task)
                        .and_then(|task| task.demand_for(resource.id))
                })
                .sum();
            (resource.id, total)
        })
        .collect()
}

/// Returns the identifiers of the resources that are free at an instant.
///
/// A resource is free when its availability windows cover the instant and
/// the demand already placed on it leaves room for one more unit. This is
/// the query a watermark scheduler makes when it looks for a satellite with
/// spare capacity.
pub fn resources_free_at(
    schedule: &Schedule,
    tasks: &[Task],
    resources: &ResourceSet,
    time: f64,
) -> Vec<ResourceId> {
    resources
        .resources()
        .iter()
        .filter(|resource| resource.is_available_at(time))
        .filter(|resource| {
            let placed: f64 = schedule
                .placements()
                .iter()
                .filter(|placement| placement.covers(time))
                .filter_map(|placement| {
                    tasks
                        .iter()
                        .find(|task| task.id == placement.task)
                        .and_then(|task| task.demand_for(resource.id))
                })
                .sum();
            placed < resource.capacity() - 1e-9
        })
        .map(|resource| resource.id)
        .collect()
}

/// Returns the earliest time at which a task can be placed.
///
/// The earliest feasible start is always one of a small set of candidate
/// times: the start of the task's window, the start of an availability
/// window of one of its resources, or the end of a placement already on one
/// of those resources. Feasibility cannot change at any other instant,
/// because the demand on every resource is constant between those events and
/// the availability is constant between them too. Enumerating that set and
/// taking the smallest feasible member therefore finds the true earliest
/// start.
///
/// Returns `None` when the task cannot be placed anywhere in its window.
pub fn earliest_feasible_start(
    task: &Task,
    tasks: &[Task],
    schedule: &Schedule,
    resources: &ResourceSet,
) -> Option<f64> {
    if !task.is_valid() || !resources_have_capacity(task, resources) {
        return None;
    }
    if task.requirements.is_empty() {
        // A task with no requirements can start as soon as its window opens.
        return Some(task.earliest_start);
    }

    // The candidate times, collected from every event that can change
    // feasibility.
    let mut candidates: Vec<f64> = vec![task.earliest_start];
    for requirement in &task.requirements {
        let resource = resources.get(requirement.resource)?;
        for window in resource.windows() {
            candidates.push(window.start());
        }
        for placement in schedule.placements() {
            let overlaps = tasks
                .iter()
                .find(|other| other.id == placement.task)
                .is_some_and(|other| other.demand_for(resource.id).is_some());
            if overlaps {
                candidates.push(placement.end());
            }
        }
    }

    // Only times at or after the window opens, and no later than the last
    // start that could still fit, are worth testing.
    let latest_start = task.latest_end - task.duration;
    candidates
        .retain(|time| time.is_finite() && *time >= task.earliest_start && *time <= latest_start);
    candidates.sort_by(|first, second| {
        first
            .partial_cmp(second)
            .unwrap_or(core::cmp::Ordering::Equal)
    });
    candidates.dedup();

    candidates
        .into_iter()
        .find(|start| feasible_at(task, tasks, schedule, resources, *start))
}

/// Returns `true` when a task placed at a time is feasible.
///
/// The check is that every required resource is available for the whole
/// duration and that the demand already placed on it, plus the new demand,
/// stays within the capacity at every instant of the placement.
pub fn feasible_at(
    task: &Task,
    tasks: &[Task],
    schedule: &Schedule,
    resources: &ResourceSet,
    start: f64,
) -> bool {
    if !start.is_finite() {
        return false;
    }
    let end = start + task.duration;
    for requirement in &task.requirements {
        let Some(resource) = resources.get(requirement.resource) else {
            return false;
        };
        if !resource.is_available_over(start, task.duration) {
            return false;
        }
        // The demand of the new task must fit alongside the placements that
        // overlap it. Because the demand of each placement is constant over
        // its own interval, it is enough to test the boundaries of the
        // overlapping placements, where the total is greatest.
        let mut boundaries: Vec<f64> = vec![start];
        for placement in schedule.placements() {
            let Some(other) = tasks.iter().find(|other| other.id == placement.task) else {
                continue;
            };
            if other.demand_for(resource.id).is_none() {
                continue;
            }
            if placement.start < end && start < placement.end() {
                boundaries.push(placement.start);
            }
        }
        for boundary in boundaries {
            let placed: f64 = schedule
                .placements()
                .iter()
                .filter(|placement| placement.covers(boundary))
                .filter_map(|placement| {
                    tasks
                        .iter()
                        .find(|other| other.id == placement.task)
                        .and_then(|other| other.demand_for(resource.id))
                })
                .sum();
            if placed + requirement.demand > resource.capacity() + 1e-9 {
                return false;
            }
        }
    }
    true
}

/// Returns the identifiers of the tasks that a set of resources cannot
/// accommodate at all.
///
/// A task is infeasible when it demands more of a resource than the resource
/// has, when a resource it requires does not exist, or when its window is
/// shorter than its duration. These cannot be repaired by rescheduling and
/// are worth reporting separately from the tasks that were merely left out.
pub fn infeasible_tasks(tasks: &[Task], resources: &ResourceSet) -> Vec<TaskId> {
    tasks
        .iter()
        .filter(|task| !task.is_valid() || !resources_have_capacity(task, resources))
        .map(|task| task.id)
        .collect()
}