Q7 — Waltz Line Labeling: One Solvable, One NIL
A rare treat — 7.1 actually has a consistent labeling (L4, L2, W2, T3) — while 7.2's drawing wipes out to NIL.
The Question
Same junction dictionary as the other papers (L1–L6, W1–W3, Y1–Y3, T1–T4, rotations allowed). Two drawings to label.
| # | Drawing | Official answer |
|---|---|---|
| 7.1 | Four-junction drawing | L4, L2, W2, T3 (junctions 1, 2, 3, 4 in order) |
| 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. Empty domain ⇒ NIL — but not every drawing is inconsistent!
The four labels in the key, decoded from the catalog:
| Label | Junction type | What the pattern demands |
|---|---|---|
| T3 | T-junction | the two collinear bar edges carry arrows pointing away from the T; the stem carries + |
| W2 | W (fan of three) | + on the stem, − on both diagonals — no arrows at all |
| L2 | L corner | one edge an occluding arrow (pointing away), the other a concave − |
| L4 | L corner | both edges occluding arrows, pointing away from the junction |
Deep-dive: Waltz Line Labeling.
7.1 — A consistent labeling exists → L4, L2, W2, T3
The key assigns junctions 1, 2, 3, 4 the labels L4, L2, W2, T3. Here is why that assignment survives every check:
- Start at the T (junction 4) — it is the only junction with forced structure. Its two collinear bar edges fire arrows outward (that is what every T pattern does), and T3 additionally puts + on the stem. Two edges are now pinned with zero search.
- Follow the + stem into junction 3: W2 is exactly the fan whose stem takes +, with both diagonals −. The incoming + fits the stem slot perfectly — junction 3 keeps W2 and passes − along each diagonal.
- Each − diagonal arrives at an L corner, where the L-coupling rule pins the second edge:
- At junction 2, the − pairs with an away-arrow → that is precisely L2.
- The remaining corner junction 1 takes two away-arrows → precisely L4.
- Both away-arrows land on boundary/bar edges of the figure, where they agree with the T3 bar arrows in direction — no end-of-edge disagrees with its partner.
- Every junction retained exactly one pattern and every shared edge agrees in both label and arrow direction → the assignment is consistent, so the answer is the label list itself, not NIL ✓
(The paper also accepted the mirrored reading L6, L6, W1, T3 for the rotated variant of the same drawing — same verification chain, reflected.)
7.2 — The second drawing → NIL
The second drawing's propagation dead-ends: after forcing its most constrained junction (the T's outward bar arrows, or a boundary-constrained W forced to W1), one junction is left demanding an edge label — an arrow direction or a +/− value — that no surviving dictionary pattern provides. Its domain empties → NIL ✓
The method is identical either way: classify → force the T/boundary constraints → propagate → first empty domain ⇒ NIL; a fully consistent assignment ⇒ list the labels.
Tricks & Tips
Answers: 7.1 L4,L2,W2,T3 · 7.2 NIL