Circular Seating Puzzle: When Is a Clue Actually Doing Work?
Six probationary officers - P, Q, R, S, T and U - are seated around a circular table in the branch training room, all facing the centre, for a reasoning drill. The seating conditions are: (i) Q is second to the right of P; (ii) S is immediately to the left of Q; (iii) R is not adjacent to P; (iv) T is second to the left of U. A trainee submits a seating plan and argues that the puzzle has more than one solution, so the answer key should list every possibility. The trainer disagrees, but before signing off she wants two things checked properly: whether the four conditions together fix a single arrangement, and whether any one condition is doing no work at all. She also wants to know what happens to the answer if the same six officers are asked to face outward instead of the centre, because a previous batch had challenged exactly this point.
1. Fix the seating arrangement from the four conditions and show how each seat is forced.
Read the worked answer
Number the seats 1 to 6 clockwise. For people facing the centre, a person's right is anticlockwise (seat number minus 1) and left is clockwise (seat number plus 1).
Put P at seat 1.
- Q is second to the right of P: right is anticlockwise, so Q is at seat 5.
- S is immediately left of Q: left is clockwise, so S is at seat 6.
- Remaining seats 2, 3, 4 are for R, T and U. T is second to the left of U, i.e. T = U + 2 (clockwise). Testing: U = 2 gives T = 4 (free, valid); U = 3 gives T = 5 (occupied by Q); U = 4 gives T = 6 (occupied by S). So U = 2, T = 4, and R takes seat 3.
- Check condition (iii): R at seat 3 is not adjacent to P at seat 1 (neighbours of seat 1 are 2 and 6). Satisfied.
Clockwise order: P – U – R – T – Q – S. The solution is unique, so the trainee's claim of multiple solutions is wrong.
2. Can the trainer delete exactly one condition and still keep a unique arrangement? Test the candidates and state the general principle she should apply.
Read the worked answer
Test by deleting one condition at a time and counting solutions.
Delete (iii): (i) and (ii) still fix P1, Q5, S6; (iv) alone forces U2, T4, leaving R at 3, which happens not to be adjacent to P. Exactly one arrangement - (iii) is redundant.
Delete (iv): with only (i), (ii), (iii), P1, Q5, S6 are fixed and R must avoid seats 2 and 6, so R is at 3 or 4; in each case T and U can swap over the two free seats. That is 4 arrangements - uniqueness is lost.
Delete (i) or (ii): Q or S becomes free, and several arrangements survive.
Conclusion: only condition (iii) can be removed safely. The principle is not 'which clue looks weakest' but enumeration - a clue is redundant only if the remaining clues still admit exactly one arrangement. Redundancy is a property of the whole set, not of a single line.
3. If the same six officers were asked to face outward instead of the centre, how would the answer change, and what error would produce a 'second' solution?
Read the worked answer
When people face outward, each person's right becomes clockwise (seat number plus 1) and left becomes anticlockwise (seat number minus 1); every directional clue reverses.
This is equivalent to reflecting the whole arrangement in a mirror. Put P at seat 1, then Q, second to the right (now clockwise), sits at seat 3, S immediately left (anticlockwise) at seat 2, and T second to the left of U gives U at 6 and T at 4, leaving R at seat 5 - again not adjacent to P. The clockwise order is P – S – Q – R – T – U, the mirror image of the centre-facing solution.
Uniqueness is preserved, because reflection is a one-to-one mapping: one arrangement in, one arrangement out. The 'second solution' the earlier batch found came from mixing conventions - applying centre-facing left/right rules to an outward-facing table. The correct handling is to fix one convention at the start and apply it consistently to all four clues.