Beat Allocation: What Is Valid Versus What Is Provable
The station house officer of a highway police station must post five constables - A, B, C, D and E - at five beat points numbered 1 to 5 in order along the highway, one constable per beat. An informer's input, cross-checked against the duty roster, yields four conditions: (i) B is not posted at beat 1 or beat 5. (ii) E's beat number is exactly two more than B's beat number. (iii) C is posted at an even-numbered beat. (iv) A and D are posted at adjacent beats, and together they form the night patrol pair. The officer writes down the arrangement A-1, D-2, B-3, C-4, E-5 and forwards it for approval. The inspector returns it with the remark that the paper does not actually prove this arrangement and asks for the full set of possible arrangements plus the one conclusion that can be relied on without further information.
1. Test the officer's proposed arrangement against all four conditions and determine the complete set of valid arrangements.
Read the worked answer
Step 1 - squeeze condition (ii), the strongest clue. E = B + 2, and from (i) B ∈ {2, 3}. So only two branches exist.
Branch 1: B = 2, E = 4. Remaining beats {1, 3, 5} for A, C, D. Condition (iv) needs A and D on adjacent beats, but 1, 3, 5 are all two apart. Branch fails.
Branch 2: B = 3, E = 5. Remaining beats {1, 2, 4} for A, C, D.
- (iii) C is even → C ∈ {2, 4}.
- (iv) A and D adjacent → only the pair (1, 2) exists in this set.
- Therefore A, D occupy beats 1 and 2 (either order) and C = 4.
Complete valid set - exactly two arrangements:
| Arrangement | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| I | A | D | B | C | E |
| II | D | A | B | C | E |
Checking the officer's paper: A-1, D-2, B-3, C-4, E-5 satisfies all four conditions - B is at 3 (not 1 or 5), E = 5 = 3 + 2, C = 4 is even, and A(1)/D(2) are adjacent.
So the arrangement is not wrong - it is under-determined. The officer wrote down one of two valid models and presented it as if the clues forced it. That is precisely the inspector's objection.
2. Identify the conclusion that holds in every valid arrangement, and explain why a statement such as 'A is at beat 1' cannot be relied on.
Read the worked answer
Conclusions that hold in every valid arrangement (entailed):
- B is at beat 3.
- E is at beat 5.
- C is at beat 4.
- A and D occupy beats 1 and 2 (as a pair, order unknown).
Conclusions that do NOT hold (not entailed):
- 'A is at beat 1' - true in model I, false in model II.
- 'D is at beat 2' - true in model I, false in model II.
The reasoning rule being tested. A conclusion is entailed only if it is true in every model that satisfies the premises. Once two models satisfy all four conditions and disagree about who stands at beats 1 and 2, any statement naming A or D specifically has a counter-example and must be withdrawn. Statements about B, C and E survive because both models agree on them.
Why this matters operationally. The beat at position 1 is the first point a vehicle from the highway reaches; if the officer tells the control room 'A is on beat 1' and the roster is later built on model II, the night patrol pair is deployed in the wrong order and the informer's input is wrongly blamed. The safe order is to state what the evidence fixes, and to request one more input to resolve A versus D.
3. A fifth input arrives: A's beat number is less than D's. State the final arrangement and explain which earlier conclusion, if any, becomes false.
Read the worked answer
Applying the new input. A < D, and A, D are known to occupy beats 1 and 2. Therefore A = 1 and D = 2. All other findings are untouched.
Final arrangement:
| Beat | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Constable | A | D | B | C | E |
Re-check all conditions: (i) B at 3, not 1 or 5 ✓ (ii) E = 5 = B + 2 ✓ (iii) C = 4, even ✓ (iv) A(1) and D(2) adjacent ✓
Which earlier conclusions change?
- None becomes false. B-3, E-5, C-4 and 'A and D occupy beats 1 and 2' were true in both models and remain true.
- The new input does not contradict anything; it resolves the ambiguity by eliminating model II. It upgrades 'A and D are on beats 1 and 2 in some order' to 'A is on beat 1 and D on beat 2'.
General point: a genuinely new, non-redundant constraint can only shrink the solution set. If adding a constraint ever makes a previously entailed conclusion false, the earlier reasoning was faulty, not merely incomplete. Here the constraint happens to shrink the set from two models to one, which is the ideal outcome - a deterministic roster that can be defended if questioned.