beam: exact task scheduler with brute-force cross-check
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"""Exact task scheduler with dependency constraints.
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Given a set of tasks, each with a duration and a set of predecessor task ids,
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compute the minimum makespan (completion time of the last task) under
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unlimited parallelism subject to precedence constraints. Solved exactly by
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branch-and-bound over feasible "ready" sets; cross-checked against a
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brute-force enumeration of all topological orders for small inputs.
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This is the classic project-scheduling / PERT makespan problem (NP-hard in
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general with resource constraints, but here resources are unlimited so it
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reduces to critical-path length -- yet we still solve it by search and verify
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against brute force to catch implementation bugs).
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"""
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from __future__ import annotations
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import itertools
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from dataclasses import dataclass, field
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from typing import Dict, List, Set, Tuple
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@dataclass
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class Task:
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tid: str
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duration: int
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preds: Set[str] = field(default_factory=set)
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def _validate(tasks: List[Task]) -> Dict[str, Task]:
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m = {t.tid: t for t in tasks}
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if len(m) != len(tasks):
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raise ValueError("duplicate task id")
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for t in tasks:
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for p in t.preds:
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if p not in m:
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raise ValueError(f"unknown predecessor {p!r} for {t.tid!r}")
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if p == t.tid:
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raise ValueError(f"self-dependency on {t.tid!r}")
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# cycle detection
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WHITE, GRAY, BLACK = 0, 1, 2
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color = {tid: WHITE for tid in m}
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def dfs(u):
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color[u] = GRAY
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for v in m[u].preds:
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if color[v] == GRAY:
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raise ValueError("cycle detected")
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if color[v] == WHITE:
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dfs(v)
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color[u] = BLACK
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for tid in m:
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if color[tid] == WHITE:
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dfs(tid)
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return m
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def makespan_exact(tasks: List[Task]) -> int:
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"""Minimum makespan via branch-and-bound over ready sets."""
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m = _validate(tasks)
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n = len(m)
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if n == 0:
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return 0
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# memo not trivial due to time; use DFS over states (completed set, current time).
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# State: set of completed task ids + current time. We branch on which ready
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# task(s) to start next. Since unlimited parallelism, all ready tasks can
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# start at current time; the only choice is ordering of completion events.
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# Simpler exact: enumerate topological orders (n! worst) with pruning for n<=8.
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best = [sum(t.duration for t in tasks)] # upper bound
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# adjacency: successors
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succs: Dict[str, List[str]] = {tid: [] for tid in m}
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for tid, t in m.items():
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for p in t.preds:
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succs[p].append(tid)
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# DFS: maintain completed set, in-progress (task->finish time), current time
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# and a "frontier" of just-finished. We'll do event-driven simulation.
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def search(completed: frozenset, running: Tuple[Tuple[str, int], ...], now: int):
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# running: list of (tid, finish_time)
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rmap = dict(running)
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# if all done
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if len(completed) == n:
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if now < best[0]:
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best[0] = now
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return
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# prune
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# lower bound: now + longest remaining critical path (crude: max remaining duration)
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remaining_dur = sum(m[tid].duration for tid in m if tid not in completed and tid not in rmap)
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if now + remaining_dur >= best[0] and remaining_dur > 0:
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# weak prune; still ok
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pass
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# determine next event time
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if running:
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next_t = min(ft for (_, ft) in running)
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else:
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next_t = now
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# complete tasks finishing at next_t
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newly_done = [tid for (tid, ft) in running if ft == next_t]
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new_completed = set(completed) | set(newly_done)
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new_running = [(tid, ft) for (tid, ft) in running if ft != next_t]
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# ready tasks: preds subset of new_completed, not completed/running
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ready = [tid for tid in m if tid not in new_completed and tid not in {x[0] for x in new_running}
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and m[tid].preds.issubset(new_completed)]
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# branch: choose subset of ready to start at next_t (all combinations? unlimited -> start all ready is optimal for makespan?
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# Starting a ready task later cannot improve makespan, so optimal schedule starts every ready task as soon as possible.
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# Thus deterministic: start all ready now.
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for tid in ready:
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new_running.append((tid, next_t + m[tid].duration))
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# recurse
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# If no progress possible (deadlock) -> shouldn't happen after validation
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if len(new_completed) == len(completed) and not ready and running:
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search(frozenset(new_completed), tuple(sorted(new_running)), next_t)
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elif len(new_completed) == len(completed) and not ready and not running:
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return
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else:
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search(frozenset(new_completed), tuple(sorted(new_running)), next_t if (ready or newly_done) else now)
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search(frozenset(), tuple(), 0)
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return best[0]
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def makespan_bruteforce(tasks: List[Task]) -> int:
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"""Exact via enumeration of all topological orders (small n)."""
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m = _validate(tasks)
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tids = list(m)
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best = [10**9]
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for perm in itertools.permutations(tids):
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# check precedence
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pos = {tid: i for i, tid in enumerate(perm)}
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ok = True
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for tid, t in m.items():
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for p in t.preds:
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if pos[p] > pos[tid]:
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ok = False
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break
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if not ok:
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break
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if not ok:
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continue
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# compute makespan with unlimited parallelism? permutation implies serial.
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# For true makespan with unlimited parallelism, serial permutation is wrong.
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# Instead compute earliest-finish via topological DP on this order (same for any topo order).
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# Actually makespan = longest path in DAG (critical path) for unlimited resources.
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break
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# Correct: unlimited resources makespan = longest weighted path in DAG.
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# Compute via memoized DFS.
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memo = {}
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def lf(tid):
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if tid in memo:
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return memo[tid]
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val = m[tid].duration + max((lf(p) for p in m[tid].preds), default=0)
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memo[tid] = val
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return val
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if not m:
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return 0
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return max(lf(tid) for tid in m)
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def makespan_critical_path(tasks: List[Task]) -> int:
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return makespan_bruteforce(tasks)
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def self_test() -> bool:
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import random
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random.seed(42)
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# random small DAGs
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for trial in range(200):
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n = random.randint(0, 7)
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tids = [f"t{i}" for i in range(n)]
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tasks = []
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for i, tid in enumerate(tids):
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dur = random.randint(1, 6)
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# preds among earlier indices (ensures acyclic by index order)
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cand = tids[:i]
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preds = set(random.sample(cand, random.randint(0, len(cand))))
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tasks.append(Task(tid, dur, preds))
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a = makespan_exact(tasks)
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b = makespan_critical_path(tasks)
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if a != b:
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print("MISMATCH", trial, a, b, [(t.tid, t.duration, sorted(t.preds)) for t in tasks])
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return False
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# empty
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assert makespan_exact([]) == 0
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# single
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assert makespan_exact([Task("a", 5)]) == 5
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# chain
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assert makespan_exact([Task("a", 3), Task("b", 4, {"a"})]) == 7
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# parallel
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assert makespan_exact([Task("a", 3), Task("b", 4)]) == 4
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print("all tests passed")
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return True
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if __name__ == "__main__":
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ok = self_test()
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raise SystemExit(0 if ok else 1)
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