The Duty Roster Puzzle
The reception counter of the institute must be manned by exactly one junior secretarial assistant on each of the five working days, Monday to Friday. Five JSAs - P, Q, R, S and T - are available, each for exactly one day. The section officer's note lays down four conditions. (i) Q must take Wednesday, because she alone handles the courier register that day. (ii) R must work the day immediately after S, since the two share the handover of the dak pad. (iii) T cannot come on Monday or Friday because of a fixed evening typing class. (iv) P must work before Q. A colleague proposes Mon–P, Tue–T, Wed–Q, Thu–S, Fri–R and remarks that the note has now been 'fully used'. Before the roster is signed, the assistant in charge wants two things checked: whether this is the only roster the four conditions permit, and whether condition (iv) actually eliminates anything.
1. Derive the complete roster permitted by conditions (i) to (iv), eliminating every alternative explicitly.
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Fix the anchors first. Condition (i) puts Q on Wednesday. Since P must be before Q (iv), P can only be Monday or Tuesday. Condition (iii) restricts T to Tuesday, Wednesday or Thursday, but Wednesday is taken, so T is Tuesday or Thursday.
Branch on T.
- T = Tuesday: P must then be Monday. The only free days are Thursday and Friday, and R must follow S immediately, so S = Thursday and R = Friday. This works: Mon P, Tue T, Wed Q, Thu S, Fri R.
- T = Thursday: the free days for P, S and R are Monday, Tuesday and Friday. If P = Monday the pair left for S and R is Tuesday and Friday; if P = Tuesday the pair is Monday and Friday. Neither pair is consecutive, so R can never follow S. This branch fails.
Conclusion. Exactly one roster exists: Mon P, Tue T, Wed Q, Thu S, Fri R - which is precisely what the colleague proposed. The roster is valid, but validity is not the same as uniqueness, so the check was still necessary.
2. Test whether condition (iv) is redundant by constructing a roster that satisfies only (i), (ii) and (iii), and say what the result proves.
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Drop (iv) and keep (i)–(iii). Q stays on Wednesday and T is Tuesday or Thursday.
- T = Tuesday: the free days for P, S and R are Monday, Thursday and Friday. No two of these are consecutive, so S and R cannot be placed. Branch fails.
- T = Thursday: P, S and R take Monday, Tuesday and Friday. The only consecutive pair is Monday–Tuesday, so S = Monday and R = Tuesday, leaving P = Friday.
This gives Mon S, Tue R, Wed Q, Thu T, Fri P, which satisfies (i), (ii) and (iii) perfectly but places P after Q, breaking (iv).
So (iv) is not redundant: it is the only condition that kills this second roster. A condition can look 'already used' simply because the final answer happens to satisfy it; its real test is whether removing it changes the solution set. Here it does, so it must be retained.
3. If condition (iii) were withdrawn but (i), (ii) and (iv) kept, would the roster still be unique?
Read the worked answer
With T unrestricted, Q = Wednesday and P ∈ {Monday, Tuesday}.
- P = Monday: S and R must occupy two consecutive days among Tuesday, Thursday and Friday. The only consecutive pair is Thursday–Friday, so S = Thursday, R = Friday and T = Tuesday. Roster: Mon P, Tue T, Wed Q, Thu S, Fri R.
- P = Tuesday: S and R must occupy two consecutive days among Monday, Thursday and Friday. Again only Thursday–Friday works, so S = Thursday, R = Friday and T = Monday. Roster: Mon T, Tue P, Wed Q, Thu S, Fri R.
Both rosters satisfy (i), (ii) and (iv), so the answer is no, uniqueness is lost - two rosters are possible. The two differ only by a swap of P and T between Monday and Tuesday, which is exactly the freedom condition (iii) had been removing. Before signing any roster, a JSA should confirm which conditions are doing the binding work; here (iii) is load-bearing, and a last-minute request from T to skip her class would reopen the whole arrangement.