The Analogy Card With One Example
A tutor is writing a single analogy card for a Friday reasoning session: 'A is to B as C is to ?'. Card A shows a large white circle containing two small black squares. Card B shows a large black circle containing four small white squares. Card C shows a large black square containing three small white triangles.
The tutor's intended rule is one uniform transformation applied to the whole figure: every filled region becomes white and every white region becomes filled, and the number of small inner shapes doubles. A student, working from the single A-to-B pair, proposes a different reading: 'Nothing swaps colour. The only change that matters is that two shapes are added, so C should gain two as well.'
The tutor has no second example pair on the card and one minute to decide whether to keep the item, rewrite it, or drop it. Four candidate answers are already printed, including the intended one.
1. Give the intended answer for the fourth figure and justify each attribute.
Read the worked answer
The answer is a large white square containing six small black triangles.
- Container fill: A's container is white and B's is black, so fill inverts. C's container is black, so the answer's container is white.
- Container shape: unchanged across the pair - circle stays circle - so a square stays a square.
- Inner count: two black squares become four white squares, a doubling of three to six.
- Inner fill: black inner shapes become white, so white triangles become black triangles.
- Inner shape: unchanged, as the container shape was.
The attributes are independent, which is why the answer can be assembled rather than guessed: change only what the exemplar pair changes, and leave alone what it leaves alone. A wrong option usually alters two attributes at once, or alters an attribute the pair never touched, such as the container's shape or the inner shape's outline.
2. Show why the student's 'add two' reading cannot be refuted from the card as written, and say precisely what the tutor should change.
Read the worked answer
From one pair the card cannot decide the count. Two is the only number for which doubling and adding two agree, since 2 × 2 = 4 and 2 + 2 = 4. So the single transition 2 → 4 is equally consistent with 'double' and with 'add two', and nothing else on the card eliminates the additive reading.
The tiebreak the tutor is using is convention, not deduction: an analogy item normally applies one uniform operation to the whole figure, and doubling is the operation that also explains the colour inversion as a uniform 'everything flips'. That is a reason to prefer six, not a proof.
The fix is structural. Print a second pair whose starting count is anything other than two, for example three shapes becoming six. The additive reading then predicts five and doubling predicts six, so the card becomes decidable. The second pair must keep the same colour behaviour, or the tutor merely moves the dispute from the count to the shading.
3. The tutor adds a second pair: a large white triangle holding one small black circle becomes a large black triangle holding two small white circles. Does this rescue the card? What could still go wrong?
Read the worked answer
It does settle the count, and the check is arithmetic. The new pair runs one shape to two. Doubling predicts 2; 'add two' predicts 3. The observed 2 kills the additive reading and leaves doubling as the only rule fitting both pairs.
Two cautions before the tutor relaxes. First, the new pair must not create a rival rule of its own: 'add one' fits 1 → 2 but fails 2 → 4, so the two pairs together are still uniquely explained by doubling. Second, the colours on the new pair must invert the same way - white container to black, black inner to white. If the tutor prints a pair where the fill does not flip, the card now carries two conflicting colour rules and the dispute simply relocates.
The general principle: a second example fixes an item only when it varies the disputed attribute while holding every other attribute's rule constant.