Q7 — Waltz Line Labeling: Two Drawings, Both NIL

A house drawing and a podium drawing — the T-junction forces arrows, the boundary edges force W1, and both domains wipe out.

The Question

The paper's junction dictionary (same catalog as the 2025 papers): L1-L6, W1-W3, Y1-Y3, T1-T4, with in-plane rotations allowed. Two drawings to label; both official answers are NIL.

#DrawingOfficial answer
7.1"House": apex 5 over centre 7, walls to 3 and 1, inner Y at 2NIL
7.2"Podium": top bar 3-4-5, centre stem to 1, dangling edges at 2 and 6NIL

The Topic in 60 Seconds

Each junction must realize one legal dictionary pattern; shared edges agree at both ends (same label, same arrow direction). Any junction with an empty domain means NIL.

Key structural rules:

  • T-junction: the two collinear bar edges carry arrows pointing away from the T.
  • Edges running off the page must be occluding arrows (a +/- edge joins two visible faces and cannot dangle).
  • W1 is the only W pattern with arrows on its diagonals — and they point into the W.
  • Y patterns allow arrows pointing into the Y (Y1), or all minus (Y2), or all plus (Y3) — never arrows pointing away.

Deep-dive: Waltz Line Labeling.

7.1 — The house drawing → NIL

Junctions: apex 5 = W (fan of three: to 4, 7, 6), 7 = T (bar 5-7-2 collinear, stem 7-6), 2 = Y (edges to 7, 3, 1 at ~120 degrees), 4 = L, 6 = W, 3 = W, 1 = W.

Propagation:

  1. T(7): bar arrows point away from the T — edge 5-7 = arrow into 5, edge 7-2 = arrow into 2. Stem 7-6 still free (T1/T2/T3/T4).
  2. W(5): edge 7-5 arrives as an arrow into 5. W1 is the only W pattern with arrows on its diagonals, and they point into the junction. So W1 at 5: one of {4-5, 6-5} carries the other in-arrow.
  3. If 6-5 were that arrow: from 6's side it points away from 6 — no W pattern allows an away-arrow. Dead. So 4-5 = arrow into 5, and the stem 6-5 = +.
  4. L(4): 5-4 arrives as an arrow away from 4 — only L4 (both arrows away) permits that, so 4-3 = arrow into 3.
  5. Y(2): edge 7-2 arrives as an arrow into 2 (from the T bar). Y patterns need the remaining edges {3-2, 1-2} to be {arrow-into-2, minus}. An arrow along 3-2 would be an arrow away from 3 — no W pattern allows an away-arrow on any edge.
  6. So 3-2 = minus. At W(3) that forces the W3 shape (+, minus, +), which needs 4-3 = +. But step 4 fixed 4-3 as an arrow into 3. Contradiction — the domain at junction 3 empties.

Answer: NIL

1 / 6
Junctions: 5=W{W1,W2,W3} 7=T{T1-T4} 2=Y{Y1,Y2,Y3} 4=L{L1-L6} 6=W{W1-W3} 3=W{W1-W3} 1=W{W1-W3}
junction with open domainjunction being forcededge pinneddomain wiped out (NIL)

7.2 — The podium drawing → NIL

Same structure as the 2025 papers' podium: top bar 3-4-5, stem 4-1, diagonals 2-1 and 1-6, stems 2-3 and 5-6, dangling edges hanging off 2 and 6.

Junctions: 3 = L, 4 = T, 5 = L, 1 = Y, 2 = W, 6 = W.

  1. T(4): bar arrows forced outward — edge 3-4 = arrow into 3, edge 4-5 = arrow into 5.
  2. Boundary edges (dangling at 2 and 6) must be occluding arrows; the only W pattern with arrow diagonals is W1, pointing into the W. So at 2: diagonal 2-1 = arrow into 2 (and the dangle too), stem 2-3 = +; at 6: diagonal 1-6 = arrow into 6, stem 5-6 = +.
  3. The L junctions 3 and 5 survive (arrow-in + plus = legal L5/L6 shapes).
  4. Y(1): both diagonals 2-1 and 1-6 now carry arrows pointing away from junction 1. No Y pattern allows that (Y1 wants arrows in, Y2 all minus, Y3 all plus). Domain empties → NIL.
1 / 4
Junctions: 3=L 4=T{T1-T4} 5=L 1=Y{Y1-Y3} 2=W{W1-W3} 6=W{W1-W3}
open domainbeing forcedpinned edge / surviving junctionwiped out (NIL)

Tricks & Tips

Answers: 7.1 NIL · 7.2 NIL