Constable Deployment Puzzle with a Late Cap
The district control room must deploy 14 constables for a two-day Ganesh immersion bandobast across four police stations: A, B, C, and D. Each station must send at least 2 constables. Station A must send fewer than B. Station B must send exactly twice the number sent by C. Station D must send one more than A. No station may send more than 6. The total must be exactly 14. The SP later adds a temporary cap of 5 per station because of leave vacancies. The control room also receives a request to spare two constables from A for a court production duty on the first day. A clerk's draft shows A=3, B=6, C=3, D=2, total 14. The deployment order must be issued by 18:00 today. No station can send a fraction of a constable, and no constable can be counted at two stations.
1. Using the original constraints, find the only valid allocation and show why the clerk's draft A=3, B=6, C=3, D=2 is wrong.
Read the worked answer
Let C = x. Then B = 2x. Since B <= 6, x can be 1, 2, or 3. If x = 1, B = 2; then A must be at least 2 and fewer than B, which is impossible. If x = 2, B = 4; the total gives A + 4 + 2 + (A + 1) = 14, so 2A + 7 = 14 and A = 3.5, not an integer. If x = 3, B = 6; the total gives A + 6 + 3 + (A + 1) = 14, so 2A + 10 = 14 and A = 2, D = 3. This satisfies A < B (2 < 6), D = A + 1, at least 2 each, and no station above 6. The clerk's draft A=3, B=6, C=3, D=2 has total 14 and B=2C, but D is not A+1 (2 is not 4). So the draft fails a mandatory condition and cannot be issued.
2. The SP adds a temporary cap of 5 per station. Is the original allocation still feasible? If not, identify the conflict and propose the smallest change that preserves the total of 14.
Read the worked answer
No. The original allocation needs B=6, which violates the cap of 5. With B=2C and B<=5, C can only be 1 or 2. If C=1, B=2, and A must be at least 2 but fewer than 2, impossible. If C=2, B=4; A can be 2 or 3. A=2 gives D=3 and total 11; A=3 gives D=4 and total 13. Neither reaches 14. The constraints are therefore inconsistent. The smallest change is to relax the cap for B from 5 to 6, which restores the original solution. Relaxing the B exactly twice C rule would also work but requires reworking two linked conditions. If the cap must stay at 5, the control room must either reduce the total to 13 or borrow at least one constable from the district reserve and relax another condition.