Building a six-night roster, then testing it with an exchange request
You are preparing the night-duty roster for the six staff nurses of a 30-bed surgical ward - Anita, Bhavna, Chetna, Divya, Esha and Farah - for Monday to Saturday. Exactly one nurse takes the night shift each day, and no nurse takes more than one night in the week. The ward sister has left five conditions: (i) Chetna must take Tuesday; (ii) Anita's night must fall on the day immediately before Bhavna's; (iii) Esha must take Saturday; (iv) Divya must take the day immediately before Esha; (v) Farah must not be on Wednesday or Friday. While you are drafting, two of the nurses ask whether they can simply exchange their allotted nights with each other, because of a family function. The nursing superintendent will approve an exchange only if all five conditions still hold after the exchange, since the conditions reflect supervision levels and leave already sanctioned. You must settle the roster first, and only then answer the exchange request with reasons.
1. Work out the night roster for Monday to Saturday.
Read the worked answer
Roster: Monday – Farah; Tuesday – Chetna; Wednesday – Anita; Thursday – Bhavna; Friday – Divya; Saturday – Esha.
Reasoning:
- From (iii) Esha = Saturday. From (iv) Divya is the day immediately before Esha, so Divya = Friday. Two days fixed.
- From (i) Chetna = Tuesday. Three days fixed; Monday, Wednesday and Thursday remain for Anita, Bhavna and Farah.
- From (ii) Anita is immediately before Bhavna. The only adjacent free pair among Monday, Wednesday, Thursday is Wednesday–Thursday (Monday and Wednesday are not adjacent, Tuesday sits between them). So Anita = Wednesday, Bhavna = Thursday.
- Monday is the only slot left, so Farah = Monday.
- Verify (v): Farah is on Monday, not Wednesday or Friday - satisfied. Every nurse appears exactly once. Trap to avoid: placing Anita on Monday and Bhavna on Tuesday, which is impossible because Chetna already holds Tuesday.
2. Is there any pair of nurses whose exchange of nights leaves all five conditions satisfied? Justify your answer.
Read the worked answer
No - every possible exchange breaks at least one condition.
Test each partner against Farah (Monday) as an illustration, then close the argument:
- Farah ↔ Chetna: Chetna moves off Tuesday → breaks (i).
- Farah ↔ Anita: Anita moves to Monday, no longer immediately before Bhavna → breaks (ii).
- Farah ↔ Bhavna: Bhavna moves to Monday, Anita (Wednesday) is no longer immediately before her → breaks (ii).
- Farah ↔ Divya: Divya moves to Monday, no longer immediately before Esha → breaks (iv); and Farah lands on Friday → also breaks (v).
- Farah ↔ Esha: Esha moves off Saturday → breaks (iii); Divya no longer precedes her → breaks (iv).
Now the pairs not involving Farah:
- Anita ↔ Bhavna: reverses the required order → breaks (ii).
- Anita ↔ Divya: Divya to Wednesday, no longer before Esha → breaks (iv); Anita to Friday leaves Bhavna without her predecessor → breaks (ii).
- Anita ↔ Chetna, Bhavna ↔ Chetna, Divya ↔ Chetna, Esha ↔ Chetna: Chetna leaves Tuesday → breaks (i) every time.
- Anita ↔ Esha, Bhavna ↔ Esha, Divya ↔ Esha: Esha leaves Saturday → breaks (iii) in all three, and each also disturbs (ii) or (iv).
- Bhavna ↔ Divya: Divya to Thursday, not immediately before Saturday → breaks (iv); Bhavna to Friday → breaks (ii). Conclusion: the roster is rigid; the correct managerial answer is to refuse the exchange and offer a compensatory off-duty or a day-shift adjustment instead.
3. Which one of the five conditions is redundant, and why?
Read the worked answer
Condition (v) - “Farah must not be on Wednesday or Friday” - is redundant.
Reasoning: conditions (i) to (iv) alone already force a unique roster.
- Esha = Saturday (iii), therefore Divya = Friday (iv).
- Chetna = Tuesday (i).
- Anita and Bhavna occupy the only adjacent free pair, Wednesday and Thursday (ii).
- Only Monday survives for Farah. Because Farah necessarily lands on Monday, the prohibition on Wednesday and Friday can never be violated; it adds no information and cannot exclude any candidate roster that survives (i)–(iv). Exam point: a condition is redundant when it is entailed by the others - test it by solving without it. Note the contrast: removing (v) changes nothing, whereas removing (ii) or (iv) collapses the puzzle into several valid rosters.