Q7 — Waltz Line Labeling: Two Drawings
Two line drawings, both official answers NIL — classify the junctions, force the T, propagate, and watch a domain wipe out.
The Question
Same junction dictionary as the other papers (L1–L6, W1–W3, Y1–Y3, T1–T4, rotations allowed). Two drawings; both official answers are NIL.
| # | Drawing | Official answer |
|---|---|---|
| 7.1 | L-shaped block: 5 over 6 over 3, wing to 2, base 4–1 | NIL |
| 7.2 | Second line drawing | NIL |
The Topic in 60 Seconds
Each junction must realize one legal dictionary pattern; shared edges agree at both ends. Any empty domain ⇒ NIL.
Structural rules that do the heavy lifting:
- T-junction: bar arrows point away from the T.
- Dangling/boundary edges must be occluding arrows.
- W1 is the only W with arrow diagonals (pointing into the W); W2/W3 carry only +/−.
- Y patterns: arrows-into (Y1), all − (Y2), all + (Y3) — never arrows pointing away.
- An L corner couples its two edges: once one edge's (label, direction) is fixed, the L pattern pins the second edge completely.
Deep-dive: Waltz Line Labeling.
7.1 — The L-block → NIL
Junctions (from the figure): 5 = L (edges to 6, 4), 6 = L (5, 3), 2 = L (3, 1), 1 = L (2, 4), 4 = W (edges to 5, 1, 3 — fan of three), 3 = W or Y (edges to 6, 2, 4).
Edges: 5–6, 5–4, 6–3, 3–2, 2–1, 1–4 form the boundary loop; 3–4 is the inner chord. Seven edges, every one shared by exactly two junctions — there is no T (no collinear triple) and no dangling edge to force anything for free. All the pressure comes from the two triple-junctions 4 and 3.
Why no labeling survives:
- L(5), L(6), L(2), L(1) each couple their two edges: fixing the label/direction on one side of the loop pins the other side. Propagating around the loop 5→6→3 and 4→1→2 means the labels on 6–3 and 2–3 are fully determined by whatever W(4) puts on 4–5 and 4–1.
- W(4) must be W1, W2 or W3. Each shape fixes the (label, direction) on all three of its edges — in particular on the chord 4–3 and, through the L-chain, on 6–3 and 2–3.
- Junction 3 must realize one of {W1, W2, W3, Y1, Y2, Y3} over edges {6–3, 2–3, 4–3}. Take the three cases:
- W1 at 4: an arrow leaves 4 along the chord — it arrives at 3 as an arrow pointing into 3, while the L-coupled neighbours force a +/− pairing on 6–3 and 2–3 that no single W/Y multiset accepts together with that arrow.
- W2 at 4 (+,−,+): the minus on 4–3 arrives at 3, but the same coupling pins a minus on 6–3 or 2–3 as well; a W/Y node admits at most one − outside the all-minus Y2, and Y2 is barred because the third edge is already pinned non-minus.
- W3 at 4 (+,−,+ mirrored): symmetrically strands a second +/arrow conflict at 3.
- Every case empties the domain at junction 3 → wipeout → NIL ✓
7.2 — The second drawing → NIL
Same method, same outcome: classify the junctions, find the most constrained one (a T whose bar arrows are forced, or a boundary-constrained W), commit it, then propagate. Here too the forced arrows and +/− pairings reach a junction whose every remaining dictionary pattern demands an edge label already ruled out elsewhere — its domain empties → NIL ✓
The general lesson of this paper: when a drawing contains neither a T nor a dangling edge, the forcing must come from the W/L interplay around closed loops — enumerate the W-shapes (only three), let each one speak through the chord, and watch the opposite junction die.
Tricks & Tips
Answers: 7.1 NIL · 7.2 NIL