Crate stacking in the dry provision store
Leading Seaman Verma must stack five sealed crates - P, Q, R, S and T - on a five-tier rack in the dry provision store of a shore establishment. Shelf 1 is the topmost tier and shelf 5 is deck level, with one crate per shelf. The standing instructions and the physical nature of the crates impose these conditions: crate Q, the heaviest, must be reachable without a ladder and therefore has to sit on shelf 5; crate R is fragile and must be placed immediately above crate P so that P takes part of its weight; crate S holds salted provisions and must be kept above crate T so that seepage never reaches T; and crate S must not occupy the topmost shelf because the sprinkler head is mounted there. A newly posted sailor who has never handled this rack proposes putting R on shelf 2 and P on shelf 3, and asks whether his proposal can be accepted. The storekeeper-in-charge is on two days' leave, and Verma must finalise the layout before the inventory inspection the next morning.
1. Work out the only stacking order that satisfies every condition, and show how each alternative is eliminated.
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Step 1 - fix what is forced. Q is on shelf 5. R must be immediately above P, so the pair R-P can occupy only (1,2), (2,3) or (3,4); it can never be (4,5) because shelf 5 already holds Q.
Step 2 - test each pair position.
- Pair on (1,2): shelves 3 and 4 remain for S and T. S above T forces S = 3, T = 4. S is not on the top shelf, so the sprinkler rule holds. Valid.
- Pair on (2,3): shelves 1 and 4 remain. S above T forces S = 1, T = 4 - but S may not be on the top shelf. Rejected.
- Pair on (3,4): shelves 1 and 2 remain. S above T forces S = 1, T = 2 - S on top again. Rejected.
Step 3 - result. Shelf 1 R, shelf 2 P, shelf 3 S, shelf 4 T, shelf 5 Q. The junior's proposal (R on 2, P on 3) is one of the rejected cases, so it must be refused and the reason explained to him, not simply overruled.
2. The junior then argues that if the sprinkler restriction were removed, exactly two additional layouts would become valid. Is he right, and does that justify ignoring the restriction?
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Checking the arithmetic. Removing the 'S not on top' rule re-admits the two cases rejected earlier:
- R2, P3, S1, T4, Q5 - S above T is satisfied.
- R3, P4, S1, T2, Q5 - S above T is satisfied. With R1, P2, S3, T4, Q5 that makes three valid layouts, i.e. two more than before, so the junior's count is correct.
Judging the conclusion. Being right about the count does not make the layouts acceptable. The shelf-1 restriction is a fire-safety clearance for the sprinkler head, not a preference: blocking it defeats the sprinkler in exactly the store where dry provisions burn well. The correct response is to accept the arithmetic, reject the proposal, and record the reason in the store diary so the point is not reopened.