Q8 — Waltz Line Labeling

The Question

Same junction dictionary (L1–L6, W1–W3, Y1–Y3, T1–T4, rotations allowed). A six-junction drawing to label.

     3 --------- 4 --------- 5
     | \       / |           |
     |  \     /  |           |
     |   \   /   |           |
     |     1     |           |
     |   /   \   |           |
     |  /     \  |           |
     | /       \ |           |
     2 --------- 6 -----------

Official answer: NIL

The Topic in 60 Seconds

Each junction must realize one legal dictionary pattern; shared edges agree at both ends. Empty domain ⇒ NIL.

Deep-dive: Waltz Line Labeling.

Step 0 — Count edges per junction

The dictionary only contains L (2 edges), T/W/Y (3 edges each). Junction 1 has a fourth spoke: the drawing connects it to 3 and 4 (upper diagonals) and to 2 and 6 (lower diagonals).

Why the answer is NIL

1. A CSP assignment needs a dictionary pattern for every junction. Junction 1's domain starts empty: no L, W, Y or T pattern has four incident edges, no rotation fixes that, and no neighbor can remove an edge for it.
2. Propagation never even gets a chance — the wipe-out is at the initial domain, not mid-search. Every other junction could label fine (T4 would fire its two outward bar arrows on 3-4 and 4-5, the L at 5 takes a matching arrow), but one unassignable junction makes the whole labeling inconsistent.
3. So the drawing cannot come from a trihedral scene → NIL ✓

Exam takeaway: before propagating anything, count edges at every junction. A 4-edge fan is an instant NIL.

Interactive check

Tricks & Tips

Answer: 8.1 NIL