Four Blocks, Four Officers: The Control Room Roster
A district police control room in Haryana must detail four officers - Kiran, Lalit, Meena and Naveen - to four consecutive two-hour PCR motorcycle patrol blocks: block 1 (06:00-08:00), block 2 (08:00-10:00), block 3 (10:00-12:00) and block 4 (12:00-14:00). Exactly one officer is placed in each block. The duty officer writes four constraints on the board:
- Meena's block must start later than Kiran's block.
- Naveen must take either block 1 or block 4, because he is the only one trained on the new e-challan device and the device is free only at the start and end of the day.
- Lalit must not take block 2, because he is attending a court deposition at 09:00.
- No officer may be assigned to more than one block.
The duty officer tells you the roster is 'obvious' and starts writing one arrangement. You are asked to check the reasoning before it goes to the inspector.
1. How many valid allotments exist? Enumerate them and show the case split that produces the count.
Read the worked answer
Fix Naveen first, because rule 2 restricts him to block 1 or block 4.
Case A - Naveen in block 1. Kiran, Lalit and Meena fill blocks 2, 3, 4. Meena must come after Kiran, so Kiran cannot be in block 3 (that would leave Meena in 4 and Lalit in 2, barred by rule 3). Kiran is therefore in block 2, and Meena takes 3 or 4:
- A1: N1, K2, M3, L4
- A2: N1, K2, M4, L3
Case B - Naveen in block 4. Kiran, Lalit and Meena fill blocks 1, 2, 3. Kiran can be in 1 or 2. Testing each Kiran-Meena pair: K1-M2 leaves L3 (valid); K1-M3 forces L2 (barred); K2-M3 leaves L1 (valid):
- B1: K1, L3, M2, N4
- B2: L1, K2, M3, N4
Total: 4 allotments. The method matters more than the number: fixing the most restricted variable (Naveen) and then testing the ordered pair (Kiran before Meena) exhausts the space without leaving a hidden branch. Writing one arrangement and stopping - as the duty officer was doing - ignores three equally lawful rosters.
2. Kiran is now detailed for a training at 07:00 and cannot take block 1. Which allotments survive, and why does the count fall by exactly one?
Read the worked answer
Filter the four rosters against the new condition:
- A1 (N1, K2, M3, L4): Kiran is in block 2 - survives.
- A2 (N1, K2, M4, L3): Kiran is in block 2 - survives.
- B1 (K1, L3, M2, N4): Kiran is in block 1 - eliminated.
- B2 (L1, K2, M3, N4): Kiran is in block 2 - survives.
Three allotments survive. The count falls by exactly one because Kiran occupies block 1 in only one of the four branches. That is a feature of the structure, not a coincidence: Naveen is pinned to the ends, so when Naveen is at the start Kiran is pushed to block 2 in every branch; Kiran reaches block 1 only in the branch where Naveen takes block 4. A useful check is to ask which variable is now most constrained - with Kiran barred from block 1 and Naveen taking only 1 or 4, block 1 can be filled only by Lalit or Meena, and Meena cannot lead because she must follow Kiran. That immediately points to the single surviving Naveen-at-4 roster and confirms the answer.
3. Naveen reports sick. Kiran, Lalit and Meena must now cover blocks 1, 2 and 3, with rules 1 and 3 still applying. Is the roster uniquely determined? What does this tell you about how the duty officer should proceed?
Read the worked answer
Test the pairs. Meena must follow Kiran, so possible Kiran-Meena placements among blocks 1-3 are K1-M2, K1-M3 and K2-M3. Lalit takes the remaining block and must not be in block 2.
- K1-M2 leaves Lalit in 3 - valid.
- K1-M3 leaves Lalit in 2 - barred.
- K2-M3 leaves Lalit in 1 - valid.
Two rosters remain, so the answer is no - it is not uniquely determined. The removal of Naveen did not simplify the problem; it merely removed the rule that had been doing the constraining work, leaving an under-determined puzzle with two lawful answers.
Practical consequence: the duty officer must not present a roster as 'the' roster when two satisfy the board. He should either add a genuine operational rule - for example, Lalit, who is deposing in court at 09:00, must be given the earliest block so he is free afterwards, which selects K2, L1, M3 - or state both options and let the inspector choose. Presenting an arbitrary branch as forced is exactly the reasoning error this roster was meant to catch.