Reasoning · Grade 6-1 Deductive Logic (2)

Problem

Seat people from adjacency clues

Five people sit evenly spaced around a round table. A already sits in the top seat. A and D are not neighbours, E sits immediately to D's left, B and C are not neighbours, and A is not immediately to C's right. Work out who sits in the other four seats.
table A
Your answer
How to solve
Strategy Draw a Diagram — The figure is already the diagram, so the real work is deciding what the clues mean on it. First I settle the left-and-right question by imagining myself in one of the chairs facing the table — that one act of visualising converts clues 2 and 4 into 'clockwise' and 'anticlockwise', which are things I can point at in the picture. Then I do not use the clues in the order given: I start with the clue that has the fewest possibilities. Clue 1 leaves D only two seats, clue 2 then forces E, and one of the two branches dies against clue 3, so a short elimination beats writing out all 24 seatings. I keep the systematic list in reserve as the check at the end.
1STEP 1

Number the seats and say what 'left' means in a view from above

Number the seats and fix which way is left.

seat 1 → 2 → 3 → 4 → 5 → 1 is clockwise = the direction each person's left hand points
2STEP 2

Clue 1 gives D only two possible seats

The first clue leaves D two seats.

D ∈ {3, 4}
3STEP 3

Clue 2 attaches E to D, so only two branches remain

The second clue attaches E to D.

D = 3 → E = 4, D = 4 → E = 5
4STEP 4

Clue 3 kills the branch D in seat 4

The third clue kills one branch.

D = 4 → {B, C} = {2, 3} adjacent → contradiction
5STEP 5

Clue 4 decides between the last two seats

The fourth clue settles the last two seats.

C = 2 → right of C = seat 1 = A → contradiction, so C = 5, B = 2
6STEP 6

Write the seating into the boxes and read all four clues back

The order is A, B, D, E, C.

clockwise from the top: A, B, D, E, C
7STEP 7

Confirm no other seating works

Of 24 seatings only one survives.

4 × 3 × 2 × 1 = 24 seatings, 1 survives
Answer
A, B, D, E, C
4 × 3 × 2 × 1 = 24
Five different letters fill five different boxes, so nobody is seated twice and no box is empty. Each clue was checked against the finished picture and each holds: A's neighbours are B and C (not D), the seat to D's left holds E, B's neighbours are A and D (not C), and the seat to C's right holds E rather than A. The size of the answer is sensible too — the puzzle promises one seating and the elimination closed 23 of the 24 possibilities, leaving exactly one. It is worth noting where the mirror image A, C, E, D, B would go wrong: it satisfies clues 1 and 3, which say nothing about direction, but it puts E to D's right instead of D's left and puts A to C's right, so it fails clues 2 and 4 together. That is the check that confirms 'left' was read as clockwise, from the point of view of someone facing the table.
Takeaway

Sit yourself in one of the chairs facing the table first — then 'to my left' becomes a direction you can point at in the picture.

  • Number the seats and say what 'left' means in a view from above
  • Clue 1 gives D only two possible seats
  • Clue 2 attaches E to D, so only two branches remain
  • Clue 3 kills the branch D in seat 4
  • Clue 4 decides between the last two seats
  • Write the seating into the boxes and read all four clues back
  • Confirm no other seating works