Rebuilding a Smudged Sentry Duty Roster
Six constables, A, B, C, D, E and F, must be detailed for night sentry duty at a small installation over three consecutive nights. Exactly two men stand duty each night and each man stands duty on exactly one night. The standing restrictions are: C must be on the second night; A and B must not be on the same night; D can be posted only on the first or the third night; E and F must always be posted together; F cannot be posted on the third night; and A must be posted on an earlier night than B. The duty NCO's chart has been smudged and the adjutant wants the roster rebuilt from the rules alone before 1800 hours. While this is being done, the company commander adds a further demand: every night's pair must include at least one senior man, and among those available for this duty only A, C and E hold that rank.
1. Reconstruct the roster from the six rules and set out the deduction step by step.
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Step 1. E and F are posted together, so they occupy one whole night. They cannot be on night 2, because C already holds one of that night's two places. F cannot be on night 3 by rule (v). Therefore night 1 = E and F.
Step 2. The remaining men A, B, C and D fill nights 2 and 3. C is on night 2. C's partner must come from A, B or D, but D is barred from night 2 by rule (iii), so the partner is A or B.
Step 3. Test C with B: then night 3 would be A and D. But A would be on night 3 while B is on night 2, which breaks rule (vi). Rejected.
Step 4. Therefore night 2 = A and C, and night 3 = B and D.
Check every rule. C on night 2 correct; A and B on different nights correct; D on night 3 correct; E and F together on night 1 correct; F not on night 3 correct; A (night 2) earlier than B (night 3) correct. The roster is E,F / A,C / B,D and it is the only arrangement consistent with all six rules.
2. Can the commander's senior-man condition be met without breaking any existing rule? Prove your answer.
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Test the ready roster. Night 1 has E (senior) and night 2 has A and C (senior), but night 3 has B and D, and neither is senior. Since every senior is already posted, night 3 cannot acquire one by juggling within this roster.
Test whether any roster can satisfy it. C is fixed on night 2, so night 2 already holds one senior. E is tied to F and F is barred from night 3, so E cannot be moved to night 3 without breaking rules (iv) and (v). That leaves only A available to strengthen night 3, and A can reach night 3 only if B takes the night 2 slot, which contradicts rule (vi). Alternatively, night 3 keeps a non-senior and the condition fails.
Conclusion. The senior condition is incompatible with the rules as they stand; no roster satisfies all of them simultaneously. The commander must be told this, with the reasoning, and asked to relax one rule. The cheapest relaxation is rule (vi): post B with C on night 2 and A with D on night 3, giving seniors on all three nights. Record the relaxation and the authority for it.
3. Suppose D reports sick and G, who carries no restrictions, is detailed in his place. Does a valid roster now exist, and is it unique?
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Existence. Yes. Using the same deduction: night 1 = E and F; night 2 = C plus one of A, B or G; night 3 takes the remaining two. Placing A with C gives night 1 E,F; night 2 A,C; night 3 B,G, which satisfies every rule (G carries no restriction, and rule (iii) about D no longer applies).
Uniqueness. Test the alternatives. If B partners C on night 2, night 3 becomes A and G, but A would then be posted later than B, breaking rule (vi). If G partners C on night 2, night 3 becomes A and B, breaking rule (ii). So both alternatives fail and the roster above is the only valid one.
Practical point. Uniqueness here is a consequence of the rules, not of D's absence; if G had restrictions of his own the answer could change. Publish the reasoning with the roster so that the next duty NCO can re-derive it rather than guess.