Three-Layer Cancellation With a Triple Overlap
On a superposition item you have 75 seconds. Three transparent square layers are placed exactly on top of one another; the square has four outer edges (top, right, bottom, left) and two diagonals (top-left to bottom-right, and top-right to bottom-left). Layer 1 contains the top edge, the left edge, and the top-left-to-bottom-right diagonal. Layer 2 contains the top edge, the right edge, and the same top-left-to-bottom-right diagonal. Layer 3 contains the top edge, the bottom edge, and the left edge. The rule printed above the item is: a segment that appears in an even number of layers is removed; a segment that appears in an odd number of layers remains. The options are: A) top, right and bottom edges only; B) right and bottom edges only; C) top, left and right edges only; D) all four outer edges. You initially think 'any overlap cancels' and start to eliminate the top edge.
1. Apply the parity rule segment by segment. Which option is correct?
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Count appearances per segment:
| segment | layers | parity | result | | top | 1, 2, 3 | 3 odd | remains | | right | 2 | 1 odd | remains | | bottom | 3 | 1 odd | remains | | left | 1, 3 | 2 even | removed | | top-left to bottom-right diagonal | 1, 2 | 2 even | removed | | top-right to bottom-left diagonal | none | 0 | absent |
The composite is top, right and bottom only: option A. Option B is the trap produced by treating any overlap as cancellation: that would remove the top (three layers) and the left (two layers), leaving right and bottom. The printed rule is parity, so a triple overlap survives.
2. Why is the triple overlap the critical check in this item, and how would the answer change if layer 3 also contained the right edge?
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The critical check is the top edge: it appears in all three layers. A candidate who uses 'overlap cancels' removes it; the parity rule keeps it because 3 is odd. If layer 3 also contained the right edge, the right edge would appear in layers 2 and 3 (2, even) and would be removed. The remaining figure would then be top and bottom only, not option A. This shows why every segment must be counted separately rather than applying a blanket cancellation rule to the whole overlay.