Picket Puzzle: Testing a Trainee's Two Claims
Six constables - A, B, C, D, E and F - must be posted at two pickets, North and South, exactly three at each, for a festival. The sub-inspector's written instructions are: (i) C must be at the North picket; (ii) A and D must be posted at the same picket; (iii) B and E must not be posted together; (iv) F must be posted with B, that is, at the same picket. A trainee reads the file and makes two claims: that instruction (iv) is unnecessary because the posting can be settled without it, and that instruction (iii) alone is enough to rule out any arrangement in which A and D are at the North picket. Before signing the duty roster, the sub-inspector also learns that E has been taken ill and will be replaced by G, and that henceforth G must not be posted at the South picket.
1. Work out the posting of all six constables.
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Test A and D at North first. North would then hold C, A, D - already three, so North is full. South must hold B, E, F. But that places B with E, breaking (iii). So A and D cannot be at North.
Therefore A and D are at South. South needs one more man; North holds C plus the remaining two.
Now apply (iv): F must be with B. If B and F were both at South, South would hold A, D, B, F - four men, not allowed. So B and F go to North, and the last place at South goes to E.
Final posting:
- North: C, B, F
- South: A, D, E
Checks: C at North ✓; A and D together ✓; B and E separated ✓; F with B ✓; three at each picket ✓. The solution is unique.
2. Evaluate the trainee's two claims.
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Claim 1 - that (iv) is unnecessary - is wrong. Drop (iv) and work only with the other three rules. A and D are at South (as shown by the same argument), and South's third man can be B, E or F:
- third man B → South {A, D, B}, North {C, E, F} - valid;
- third man E → South {A, D, E}, North {C, B, F} - valid;
- third man F → South {A, D, F}, North {C, B, E} - invalid, B with E.
Two valid arrangements survive, so without (iv) the roster is not settled. The claim is false.
Claim 2 - that (iii) alone rules out A and D at North - is right. If A, D, C occupy North, South is forced to be {B, E, F}, and (iii) is breached. No other rule is needed for this step.
So the trainee is half right: correct on the exclusion, wrong on the sufficiency of the instructions. A good answer separates the two claims rather than accepting or rejecting both together.
3. E is replaced by G and G must not be posted at South. Is a valid posting possible? If not, what is the least change to the rules?
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The six men are now A, B, C, D, F, G, with rule (iii) automatically void (E is gone), rule (iv) still requiring F with B, and the new restriction that G cannot be at South. A and D must still be together.
- A, D at North: with C, North is full as {A, C, D}; South becomes {B, F, G} - but G at South is barred. Invalid.
- A, D at South: South is {A, D, X}.
- X = B → North {C, F, G}; F is not with B, breaking (iv). Invalid.
- X = F → (iv) drags B to South as well, giving four men. Invalid.
- X = G → barred by the new restriction. Invalid.
No valid posting exists. The roster is over-constrained, and the sub-inspector must relax something. The cheapest relaxation is to drop the new "G not at South" restriction: then A, D, G sit at South and C, B, F at North, which satisfies every other rule and is unique. The alternative - dropping (iv) - yields two valid rosters and so is a worse fix. This is the practical lesson: when a roster is impossible, identify the single binding rule and get it amended in writing rather than quietly ignoring it.