# RE8 | moonshotai/kimi-k3 | run 1

## Prompt

Minimize total time to finish these tasks, one at a time, respecting dependencies: A (2h, no deps); B (3h, needs A); C (1h, no deps); D (2h, needs B and C). Give a valid order and the total time.

## Response

**Total time: 8 hours.** Because tasks run one at a time and dependencies never force idle time, the minimum equals the sum of all durations (2 + 3 + 1 + 2 = 8h). Any topological order achieves this.

## Valid schedule: A → B → C → D

| Order | Task | Duration | Start | Finish | Dependency check |
|-------|------|----------|-------|--------|------------------|
| 1 | A | 2h | 0:00 | 2:00 | None needed |
| 2 | B | 3h | 2:00 | 5:00 | A done at 2:00 ✓ |
| 3 | C | 1h | 5:00 | 6:00 | None needed |
| 4 | D | 2h | 6:00 | 8:00 | B (5:00) and C (6:00) done ✓ |

Other equally valid orders include **C → A → B → D** and **A → C → B → D** — all finish at 8 hours with zero idle time. Note that while the critical path (A → B → D) is only 7 hours, that bound only applies if tasks could run in parallel; with a single worker, 8 hours is optimal.

Would you like to see how the schedule changes if two tasks could run simultaneously?
