Q8 — Waltz Line Labeling
Waltz labeling on a six-junction drawing — answer Y3,L5,L4,T3
The Question
The set of junctions (L, W, Y and T type junctions) that occur in a 2D line drawing of trihedral objects is provided below. The in-plane clockwise/counterclockwise rotations of these junctions are valid as well. These junctions provide constraints on the possible edge assignments (convex, concave, arrow) for the edges/lines in 2D line drawings of trihedral objects.
The junctions carry unique labels: L1, L2, L3, L4, L5, L6, T1, T2, T3, T4, W1, W2, W3, Y1, Y2, Y3. When required, use the labels in short answers.
Apply a suitable algorithm to assign labels to edges/junctions in the 2D line drawings given in the sub-questions, process the edges and junctions in any order you see fit.
| # | Question | Official answer |
|---|---|---|
| 8.1 | Assign consistent labels; enter the labels of junctions 1, 2, 3, 4 in that order (NIL if inconsistent) | Y3,L5,L4,T3 |
The Topic in 60 Seconds
Waltz labeling = constraint satisfaction: every junction must realize one legal dictionary pattern (rotations allowed), and a shared edge must agree in label and arrow direction at both ends. Fully consistent ⇒ the answer is the label list; any emptied domain ⇒ NIL.
Deep-dive: Waltz Line Labeling.
Step 0 — Classify every junction from the figure
Edges: 3–4, 4–5, 3–2, 4–1, 5–6, 2–1, 6–1.
| Junction | Incident edges | Geometry | Type |
|---|---|---|---|
| 1 | 4, 2, 6 | three edges fanning from the centre | Y |
| 2 | 3, 1 | corner | L |
| 3 | 4, 2 | corner | L |
| 4 | 3, 5, 1 | collinear bar 3–4–5 plus stem 4–1 | T |
| 5 | 4, 6 | corner | L |
| 6 | 5, 1 | corner | L |
Step 1 — Force the T first → T3 at junction 4
A T's two collinear bar edges always fire occluding arrows away from the T (heads toward 3 and toward 5); the catalog entry whose stem carries + is T3, so stem 4–1 takes a convex +. Two edges are pinned with zero search.
Step 2 — Propagate into the fork → Y3 at junction 1
The + arrives on the vertical spoke 1–4. Among the three legal fork patterns only one places + on a leg with − companions on the other two — that is the entry the key calls Y3: spoke +, legs 1–2 and 1–6 concave (−). Both legs are now pinned.
Step 3 — The L corners lock → L5, L4
Each corner receives one pinned edge and must complete it with a catalog-consistent partner:
| Corner | Receives | Catalog completion | Key |
|---|---|---|---|
| junction 2 | − on edge 2–1 | boundary edge 2–3 pairs as the arrow-side of the mixed corner pattern | L5 |
| junction 3 | bar arrow into 3 on edge 3–4 | boundary edge 3–2 completes the two-occluding-arrow corner | L4 |
| junctions 5, 6 | mirrored copies of the same pins | mirror images of the above | consistent |
Every shared edge now agrees at both ends (the T-bar arrows run 4→3 and 4→5 matching what corners 3 and 5 expect; the + stem feeds Y3's + spoke; the − legs land cleanly in the L patterns), and no junction ran out of patterns → the drawing is consistent → answer is the list itself.
Answer: Y3, L5, L4, T3 ✓
Interactive propagation
Tricks & Tips
Answer: 8.1 Y3,L5,L4,T3