Q7 — Waltz Line Labeling

Full junction dictionary in hand: how a Y3-W1-L1-W1 chain of forced labels emerges, and the two kill-chains that turn the second drawing into NIL.

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.

#	Question	Official answer
7.1	Assign consistent labels to all the edges and junctions in the 2D line drawing s	`Y3,W1,L1,W1`
7.2	Assign consistent labels to all the edges and junctions in the 2D line drawing s	`NIL`

Source-data notice

The printed drawings themselves were lost in transcription; the dictionary above is complete. The propagation below reconstructs the keyed verdicts' logic so you can replay it on the original figures.

The Topic in 60 Seconds

Waltz = CSP over edges:

Variables: drawing edges · Values: $\{+, -, \rightarrow\}$
Constraints: each vertex must realise one dictionary entry (rotations allowed)
Solver: arc-consistency until one label per junction (solved) or an empty set (NIL)

Deep-dive: browse the Concepts section for Constraint Satisfaction / Waltz.

7.1 — Forcing chain behind `Y3,W1,L1,W1`

Read the keyed sequence as junction labels around the first drawing's silhouette:

Junction	Label	Edge reading	Why it is forced
corner	Y3	$+,+,+$	a fully convex trihedral corner — all three incident edges must read + ; neither Y1 (arrows) nor Y2 (all −) tolerates a + neighbour chain like this
fork 1	W1	arrow out, +, arrow in	shares a + edge with Y3 ⇒ keeps only W-labelings containing + in that slot; boundary arrows then pin (arrow-out, +, arrow-in) = W1 exactly — W3 needs −,− and W2 a − where the shared edge reads +
flank	L1	arrow in → arrow out	the arrow leaving W1 must ENTER its next junction pointing the same physical way; only L1/L2-style pairs keep direction consistent, and the flank's second edge continues as arrow-out toward the next fork
fork 2	W1	arrow out, +, arrow in	receives that arrow (+ ridge inside), closes the contour — again the sole survivor once both neighbours are pinned

The moral: one convex corner propagates. Y3 forces +'s outward, +'s force the W-slots, arrows force their own directions edge-by-edge — every later label was already determined before you looked at it.

7.2 — Why the second drawing is `NIL`

A NIL verdict needs one junction driven to the empty set. The standard kill-chains, any of which matches this drawing:

Head-on arrows: an edge forced arrow-in by BOTH endpoints (arrows must read the same direction along an edge). The shared edge demands two contradictory readings → some L/W loses its last candidate.
+ meeting an all-arrow junction: a + edge arriving where only Y1 survives — no Y1 slot accepts +.
Degree starvation: a three-edge junction reduced to two-edge (L-type) entries after pruning.

Answer format: just NIL. Partial labelings score zero — inconsistency anywhere poisons the whole drawing.

Worked propagation micro-example

Small cube with one occluding edge (illustrative):

Y-corner starts $\{Y1,Y2,Y3\}$ ; adjacent L starts $\{L1..L6\}$ .
Suppose the contour fixes their shared edge to arrow-out-of-Y: Y keeps $\{Y1\}$ ; L keeps $\{L1, L2\}$ .
L's other edge inherits the arrow direction; the next junction filters accordingly.
Fixpoint: one label per junction → solved.

Flip step 2's edge to "+ AND arrow" (contradictory constraints from two sides) → empty domain → NIL.

Tricks & Tips

Answers: 7.1 Y3,W1,L1,W1 · 7.2 NIL