Six-Pump Servicing Roster: Solving It and Testing Whether Every Clue Is Needed
A candidate is revising linear arrangement through a plant-maintenance roster. Six pumps, labelled A to F, must each be serviced on exactly one day, Monday through Saturday, with every day used. The maintenance log records five conditions: (i) Pump A is serviced on Monday; (ii) Pump F is serviced on Saturday; (iii) Pump E is serviced on Tuesday; (iv) Pump D is serviced on the day immediately after Pump B; (v) Pump C is serviced on a day after Pump D. The institute's answer key states that the puzzle has a unique solution and that all five conditions are essential - an assertion the candidate doubts. In the test she is asked to fix the roster and to comment on the independence of the conditions, without listing all possible arrangements.
1. Fix the complete Monday-to-Saturday roster and state, with reasons, whether any of the five conditions can be deleted without changing it.
Read the worked answer
Work from the fixed days first. (i) and (iii) place A on day 1 and E on day 2, leaving days 3, 4, 5 and 6 for B, C, D and F. Condition (iv) makes B and D a consecutive block, and (v) needs a free day after D for C.
- B on 3, D on 4 → C can take day 5 and F takes day 6 ✔
- B on 4, D on 5 → C would need day 6, which is the only day left for F ✗
Roster: Mon A, Tue E, Wed B, Thu D, Fri C, Sat F.
Condition (ii) is not essential. Once A, E, B, D and C occupy days 1–5, day 6 has only F left, so "F on Saturday" follows from the other clues and can be deleted without disturbing the solution. The key's claim that all five clues are essential is wrong. A clue is essential only if removing it admits a second valid arrangement.
2. The supervisor deletes condition (iii). How many valid rosters now exist, and what does that show about the status of the clue? Also check what happens if (iii) is instead replaced by "E is serviced before D".
Read the worked answer
With A on day 1 and F on day 6 fixed, test each consecutive B–D block, remembering that C must come after D and E must go somewhere:
- B 2, D 3 → days 4 and 5 hold C and E; C after D allows C = 4, E = 5 or C = 5, E = 4 → 2 rosters
- B 3, D 4 → C = 5 (day 6 is F), leaving E = 2 → 1 roster
- B 4, D 5 → C needs a day after 5, none free → 0 rosters
Three valid rosters, so uniqueness collapses. Condition (iii) is load-bearing.
Now replace (iii) with "E before D". The block B 2, D 3 fails, because E would have to sit on day 4 or 5, after D. Only B 3, D 4, C 5, E 2 survives, so the same unique roster appears - but for a different reason. That is the trap: two clues can give the same final answer while doing completely different work, so "the answer did not change" is not proof that a clue was redundant.