Q7 — Waltz Line Labeling

Dictionary in the arrow notation: how a Y-W-Y-W chain of forced labels emerges around a silhouette, and what makes each next label inevitable.

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	`Y1,W1,Y3,W2`

Notation note

This FN sheet's dictionary uses the arrow/pipe notation (e.g., T1: (->, <-, |), Y1: (-,-,-)); the AN sibling's catalogue is the same object in +/- spelling. The propagation below works identically — map +↔convex-out, -↔concave-in, arrows↔occlusion.

The Topic in 60 Seconds

Waltz = CSP over edges:

Variables: drawing edges · Values: convex / concave / occlusion-arrows
Constraints: each vertex must realise one dictionary entry (rotations allowed)
Solver: arc consistency until unique labels everywhere (solved) or an empty domain (NIL)

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

Forcing chain behind `Y1,W1,Y3,W2`

Read the keyed answer as four junction labels around the printed silhouette:

Order	Label	Reading	Why forced
1	Y1	all-concave corner (−,−,−)	the start vertex sits in a concave notch: no arrow can terminate there and no + pair fits its three slots
2	neighbouring fork	shares a − edge with Y1 ⇒ middle/side slot pinned to −; the boundary contour supplies the arrow pair	W1
3	next Y corner	the two − edges leaving W1 arrive together; the third edge closes a fully convex tip	Y3 (+,+,+) — the only Y-entry tolerating + on all slots once neither − nor arrows survive
4	closing fork	receives the convex edge from Y3 in its −-free slot; boundary arrow fixes orientation	W2

The chain is deterministic: each dictionary entry leaves exactly one survivor given its already-labelled neighbours. That uniqueness — not luck — is what the key encodes.

The general procedure (what to quote in an answer)

Initialise every junction domain from the dictionary (rotations included).
Fix the most constrained junction first (a Y or T).
Propagate across shared edges: signs must MATCH along an edge; arrows must keep ONE direction.
Repeat to fixpoint → one label per junction = the answer; any empty domain = reset all labels to NIL.

Worked micro-example

Concave notch meeting a ridge (illustrative):

Y starts $\{Y1, Y2, Y3\}$ ; adjacent W starts $\{W1, W2, W3\}$ .
Shared edge fixed concave (−): Y keeps $\{Y1\}$ ; W keeps entries with a − slot: $\{W2\}$ (and W3-style if rotated).
W's other two edges inherit arrow/+ readings; propagate outward.
Fixpoint with singleton domains → solved, e.g., exactly a Y1 … W2 tail like the key's.

Flip step 2 so the edge must be BOTH − and + → empty domain → NIL.

Tricks & Tips