Set A and the Feature That Was Never Tested
Set A is a collection of tiles from a reasoning test. Each tile is described completely by three visible features: the number of sides of the outer shape, the shading of that shape (dark or white), and the number of small dots printed inside it. The four tiles shown so far are: (1) a dark square with 2 dots; (2) a dark hexagon with 2 dots; (3) a dark octagon with 2 dots; (4) a dark decagon with 2 dots. Candidates to join Set A are then offered: P, a dark triangle with 2 dots; Q, a dark pentagon with 2 dots; R, a white square with 2 dots; S, a dark hexagon with 5 dots. Test-takers are told only to 'use the features shown'. Nothing is said about whether a feature must be shared by every member of the set.
1. Is there a single rule supported by the four example tiles? Set out competing rules and what each implies for P, Q, R and S.
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At least three rules fit all four examples equally well:
- R1 - dark shading and exactly 2 dots (sides irrelevant). Admits P and Q; rejects R (white) and S (5 dots).
- R2 - dark shading and an even number of sides (dots irrelevant). Admits S; rejects P (3 sides) and Q (5 sides); rejects R on shading.
- R3 - dark shading and at least 4 sides. Admits Q and S; rejects P and R.
The four examples share three features at once (dark, 2 dots, even sides ≥ 4), which is precisely why they cannot separate these rules. The verdicts diverge for P, Q and S, so none of those three can be certified either way. All three rules reject R, but that verdict rests on shading mattering - and since every example happens to be dark, a rule such as '2 dots and even sides' would readmit R. The honest answer is that the data underdetermine membership.
2. Suppose the examiner declares one of the four candidates to be a genuine member of Set A. Which declaration would be most informative, and which least? Justify by counting what survives.
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Rank by how many live rules the declaration kills.
- R declared a member - most informative. R is white with an even side count and 2 dots. Admitting it destroys every rule that requires dark shading (R1, R2, R3 as stated) and leaves only rules built on dots and side counts, e.g. '2 dots and even sides'. A whole dimension is eliminated at a stroke.
- S declared a member kills R1 (5 dots) but leaves R2 and R3 alive - moderate information.
- P declared a member kills R2 (3 sides, an odd count) and R3 (fewer than 4 sides), leaving R1 - also strong, killing two rules.
- Q declared a member - least informative. Q (dark, 5 sides, 2 dots) is already admitted by R1 and R3; only the parity rule R2 dies. One rule out of three is poor value, because most of the uncertainty survives.
So the same one-tile budget buys very different amounts of information; the efficient test targets the feature that has never varied.
3. Write one sentence of instruction that would have made the item well posed, and say what changes for the solver.
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A fair instruction would be something like: 'Every tile in Set A is dark and has exactly 2 dots; the outer shape may have any number of sides' - that is, the item must state which features are essential and which are free to vary, or supply examples that vary the incidental ones.
What changes for the solver: with the free and fixed features separated, membership becomes decidable and the four candidates can be judged rather than ranked by plausibility. Without that separation, the solver is silently asked to guess the examiner's intent from a sample in which three features are perfectly correlated - a task no amount of careful looking can settle. In the version as given, the best available answer is a ranked assessment of informative tests plus an explicit statement that no candidate is certain.