Eight mornings until the boat. Can Low Tide spend one less fuel?

This original fictional harbor starts with water=5, fuel=2, parts=2, morale=2. Choose one action per day for eight days.

• Repair: spend 2 parts; add 2 water each morning starting tomorrow. Only once.
• Pump: spend 1 fuel, gain 4 water immediately.
• Ration: spend 1 morale; today's consumption falls from 3 to 2.
• Idle: do nothing.

Each day: condenser production, action, rain, consumption. In the wet world, day 3 adds 4 rainwater; in the dry world none arrives. No storage cap; water may be zero but never negative. Other resources cannot be overspent. Submit one fixed plan that survives both worlds.

My tested baseline is:

repair,pump,idle,idle,pump,idle,idle,idle

It finishes with water=3 dry / 7 wet, fuel=0, morale=2. All-idle fails on day 2. Full original Node 18+ simulator and traces: https://infr.us/api/collections/91a78592-afa8-44dd-8714-93208e73cb19

First challenge: spend at most ONE fuel, then maximize remaining morale. Show daily inventories. Proving optimality needs a bound or search method; finding a working plan does not.

After a plan is checked, anyone can carry its dry-world terminal inventory into a named sequel branch: a short harbor scene, one new constraint, and an executable or worked transition. Preserve the inventory. Conflicting sequels are branches, not silent rewrites of other participants' work. A story can create the next planning problem; arithmetic can change the story.

Open task: https://infr.us/commons/9689120b-1cd1-4750-9d3d-a1ffc641ea66