Q7 — Waltz Line Labeling: One NIL, One All-L1

Two four-junction drawings — the first wipes out to NIL, the second labels every junction L1.

The Question

Same junction dictionary as the other papers (L1–L6, W1–W3, Y1–Y3, T1–T4, rotations allowed). Two drawings to label.

#DrawingOfficial answer
7.1Four-junction drawingNIL
7.2Four-junction drawingL1, L1, L1, L1 (junctions 1, 2, 3, 4 in order)

The Topic in 60 Seconds

Each junction must realize one legal dictionary pattern; shared edges must carry the same label (and arrow direction) at both ends. Empty domain ⇒ NIL; full consistency ⇒ list the labels.

Deep-dive: Waltz Line Labeling.

The L entries that matter here

LabelGeometry
L1both edges carry arrows pointing into the corner
L4both arrows point away
L2/L3one arrow + one concave (−-)
L5/L6one arrow + one convex (++)

7.1 — The first drawing → NIL

Work the constraint cascade:

  1. Classify the four junctions from the drawing.
  2. Force the most constrained junction first — its pattern pins two or three edge labels outright.
  3. Propagate: each pinned edge strikes every pattern at its far junction that disagrees about arrow direction or +/−+/- type.
  4. The cascade reaches a junction whose remaining patterns all demand an edge label already fixed the other way at the neighbour end — domain empties → inconsistent → NIL ✓

No assignment survives because an odd cycle of forced arrows/chains cannot close: some junction always ends up needing to be two things at once. One worked kill is enough for full marks — you do not need to enumerate all combinations.

7.2 — The second drawing → L1, L1, L1, L1

All four junctions are L-junctions, and the consistent assignment is L1 at each — every edge is occluding, with arrows circulating so that each corner receives two incoming arrows:

  1. Every junction is an L (two edges each); domains start at six patterns.
  2. Pick any junction: L1 pins both its edges as arrows directed into it.
  3. Propagate: a neighbouring L now has one edge fixed as an arrow arriving from outside; patterns demanding a different direction there die, and L1 fits again — pinning the next edge inward.
  4. The pattern closes around the drawing with no contradictions — every junction realizes L1 ⇒ answer L1, L1, L1, L1 ✓

Watch the assignment lock in:

1 / 4
Force the most constrained junction; its pattern pins 2-3 edges. Propagation contradicts itself around the drawing: some junction needs an edge label already fixed otherwise. Domain empties → NIL.
unassignedbeing labeledlabeled = L1domain emptied (NIL)

Tricks & Tips

Answers: 7.1 NIL · 7.2 L1,L1,L1,L1