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The second CSP flavor asks 2D line drawings of trihedral objects: junctions and edges get labels from a fixed catalog, and the question is which junction labels 1 to 4 make every shared edge agree? One chain answers it: type each junction by arm count, rotate a catalog pattern onto it, and check each shared edge from both ends. No consistent labeling means NIL.
Edges take \(+\) (convex), \(-\) (concave), or an arrow (occluding edge; the head points along the edge). IN means the head points toward the junction, OUT away from it. In-plane rotations of every pattern are allowed. Read each pattern left arm first, then middle (if any), then right.
| Junction | Arms, left to right (middle = stem) |
|---|---|
| L1 | arrow OUT, arrow OUT |
| L2 | arrow OUT, minus |
| L3 | minus, arrow OUT |
| L4 | arrow IN, arrow OUT |
| L5 | arrow OUT, plus |
| L6 | plus, arrow OUT |
| W1 | arrow IN, plus, arrow IN |
| W2 | minus, plus, minus |
| W3 | plus, minus, plus |
| Y1 | arrow IN, arrow OUT, minus |
| Y2 | minus, minus, minus |
| Y3 | plus, plus, plus |
| T1 | arrow OUT, arrow OUT, arrow IN (stem) |
| T2 | arrow OUT, arrow OUT, arrow OUT (stem) |
| T3 | arrow OUT, arrow OUT, plus (stem) |
| T4 | arrow OUT, arrow OUT, minus (stem) |
3a. The drawing, redrawn (solve from this, then check the walkthrough)
Q75. Junctions 1, 2: L. Junction 3: W. Junction 4: T. Shared edges: 1-2, 2-3, 3-4.
| Shared edge | Junction on each side |
|---|---|
| 1-2 | Junction 1 (L) and junction 2 (L) |
| 2-3 | Junction 2 (L) and junction 3 (W) |
| 3-4 | Junction 3 (W) and junction 4 (T, stem up to 3) |
Cover the walkthrough. Agree each shared edge from both ends; arms to junction 5 and free ends take what is left.
3b. Walkthrough: one shared edge per step, both official answers
Why this step
Tracker
3c. The question
Assign consistent labels to all the edges and junctions in the 2D line drawing above. Enter the labels of the junctions 1, 2, 3, 4 in order, or NIL if no consistent labeling exists.
3d. Answers first, then the working
Accepted answers: L4,L2,W2,T3 and L6,L6,W1,T3.
Solve each with a pen before opening the answer. Each drill is self contained.
Junction 1: L (arms to 2, to 6). Junction 2: Y (arms to 5, to 1, to 3). Junction 3: L (arms to 2, to 4). Junction 4: W (arms to 5, to 3, to 6). Enter 1 to 4.
Why this step
Tracker
L2,Y1,L3,W3. Edge 1-2: L2 minus meets Y1 stem minus. Edge 2-3: Y1 left IN-arrow meets L3 OUT-arrow as one physical arrowhead. Edge 3-4: L3 minus meets W3 middle minus. Remaining arms (to 5, to 6, free ends) take what fits: junction 4 puts plus on both.
Back on the Q75 drawing: junction 1 = L4 with IN on the 1-2 arm forces junction 2's 1-2 arm to carry what? And junction 3 = W2 with plus on the 3-4 arm forces junction 4's stem to carry what?
Why this step
Tracker
OUT-arrow (one IN plus one OUT makes a single physical arrowhead; minus or IN would clash), so junction 2 is L2 with OUT on that arm. Plus (only plus meets plus; an arrow or minus on the stem kills it), so junction 4 is T3. Each shared edge is a two-way check: read the forced label, then keep only catalog patterns carrying it on the matching arm.