3.3 Visual Classification, Topological Invariants & Odd-One-Out (Farklı Olanı Bulma)

3.3 Visual Classification, Topological Invariants & Odd-One-Out (Farklı Olanı Bulma)

Quick Method: In a TR-YÖS Odd-One-Out (Farklı Olanı Bulma) question, you are given 5 figures (A, B, C, D, E) with no sequence arrows and no analogy colon. Four of the five figures share a strict geometric, topological, or combinatorial invariant, while one figure breaks that rule. Never guess based on vague visual aesthetics. Run a fast, prioritized diagnostic checklist across the six classification checks used in this guide: (1) Rigid Rotation vs. Mirror Chirality, (2) Closed Regions & Intersection Counts, (3) Single-Stroke Graph Traversability, (4) Reflection vs. Rotational Symmetry, (5) Convexity vs. Concavity & Element Ratios, and (6) Shaded Area Fractions.

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How Farklı Olanı Bulma Works on TR-YÖS

Unlike Şekil Dizileri (where order matters) or Analojiler (where a reference pair defines an operator), Farklı Olanı Bulma tests equivalence class recognition. Two figures belong to the same equivalence class if they share a preserved property $P(F)$ under a specific set of transformations.

Because a candidate only needs to examine 3 figures to identify the majority invariant whenever the first two disagree ($P(F_A) \neq P(F_B) \implies P(F_C)$ breaks the tie immediately!), a mathematical diagnostic method lets you solve these items in 10 to 15 seconds.

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Six Useful Classification Checks

Invariant Class 1: Rigid Rotation vs. Reflection (Chirality / Yönelim ve Simetri)

When all 5 options (A–E) appear to be the exact same asymmetric drawing tilted at different angles, the question is testing 2D chirality (planar handedness). Four of the figures can be obtained from one another by pure 2D rotation in the plane of the page without lifting the figure into 3D space. The fifth figure is a mirror reflection (enantiomorph) that can only be produced by flipping the shape over.

The 10-Second Clockwise Vertex-Labeling Method ($A \to B \to C$)

Do not crane your neck or rotate the exam booklet wildly. Instead, use the Cyclic Landmark Test:

1. Choose three unmistakable asymmetric landmarks on the figure and label them $A$, $B$, and $C$ (for example: $A =$ the black dot, $B =$ the sharp acute tip, $C =$ the square hook).
2. Stand inside the centroid of each figure and trace the shortest perimeter path from $A \to B \to C$.
3. Record whether the loop $A \to B \to C$ runs Clockwise (Saat Yönü / CW) or Counterclockwise (Saat Yönünün Tersi / CCW).
4. Because pure 2D rotations preserve cyclic orientation and any single reflection reverses it, 4 options will share the same cyclic direction (e.g., CW) and 1 option will have the opposite direction (CCW). That single opposite option is guaranteed to be the odd one out!

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Invariant Class 2: Euler Characteristic, Closed Regions & Intersection Counts

When the 5 figures look completely different in shape (e.g., curved loops, overlapping polygons, or abstract line networks), the invariant is topological or combinatorial:

• Closed Bounded Regions (Kapalı Bölge Sayısı): Count the number of completely enclosed interior white cells (minimal non-overlapping regions bounded by edges). Four figures might each divide the plane into exactly $5$ closed regions, while the fifth divides it into $6$ closed regions.
• Euler's Planar Graph Formula: For any connected planar drawing with $V$ vertices (endpoints and intersection nodes) and $E$ non-crossing edge segments between vertices, Euler's formula states:

$$V - E + F = 2 \quad \Longleftrightarrow \quad F_{\text{bounded}} = E - V + 1$$

where $F$ includes the unbounded exterior face and $F_{\text{bounded}} = F - 1$ is the exact count of enclosed interior regions.

• Line-Segment Intersections (Kesişim Noktası): Four figures contain lines that cross each other at exactly $k$ interior points, whereas the fifth has $k-1$ or $k+1$ crossings.
• Open Endpoints (Açık Uç Sayısı): Count the number of degree-1 dangling line tips.

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Invariant Class 3: Graph Traversability (Eulerian Path / Tek Hamlede Çizilebilirlik)

A classic topological invariant asks which of five complex wireframes can (or cannot) be drawn in a single continuous stroke without lifting the pencil or retracing any line segment (Tek Hamlede Çizilebilirlik).

Instead of trying to trace the figure by hand—which wastes time and fails on complex grids—apply Euler's Degree Theorem by counting the degree $d(v)$ of each vertex $v$ (the exact number of line ends meeting at node $v$). Specifically, count how many vertices have an odd degree ($d(v) \in \{1, 3, 5, 7, \dots\}$, called tek dereceli düğümler):

$$N_{\text{odd}} = \text{number of vertices where an odd number of line ends meet}$$

By the Handshaking Lemma ($\sum d(v) = 2E$), $N_{\text{odd}}$ is always an even integer ($0, 2, 4, 6, \dots$):

1. $N_{\text{odd}} = 0$ (All vertices have even degree $2, 4, 6$): The figure has an Eulerian circuit. It can be drawn in one continuous stroke starting at any vertex and ending back at that same starting vertex.
2. $N_{\text{odd}} = 2$ (Exactly two odd-degree vertices): The figure has an Eulerian path. It can be drawn in one continuous stroke, provided you start at one odd-degree vertex and finish at the other odd-degree vertex.
3. $N_{\text{odd}} \ge 4$ ($4, 6, 8, \dots$ odd-degree vertices): The figure cannot be drawn in a single stroke without lifting the pen! In fact, a connected graph with $2k$ odd vertices ($k \ge 1$) requires a minimum of exactly $k = \frac{N_{\text{odd}}}{2}$ separate pen strokes.

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Invariant Class 4: Symmetry Axis Count & Rotational vs. Reflectional Symmetry

When the 5 figures are geometric medallions, stars, or polygons, compare their symmetry groups:

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Invariant Class 5: Convexity vs. Concavity & Quantitative Element Ratios

• Convex (Dışbükey) vs. Concave (İçbükey) Polygons: A polygon is convex (dışbükey) if every interior angle is strictly less than $180^\circ$ (so any line segment connecting two interior points stays completely inside the polygon). A polygon is concave (içbükey) if at least one interior angle is a reflex angle ($> 180^\circ$), creating an inward-pointing dent or cave.
• Sides-to-Dots Algebraic Invariant: Each figure consists of a polygon with $S$ sides accompanied by $D_{\text{in}}$ interior dots and $D_{\text{out}}$ exterior dots. Check the algebraic relation across all 5 choices, such as $S = D_{\text{in}} + D_{\text{out}}$, or $S - D_{\text{in}} = 2$, or even/odd parity of $S + D$.

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Invariant Class 6: Shading Area Fractions

When each of the 5 figures is a square, circle, or regular hexagon partitioned by grid lines or diagonals into shaded and unshaded pieces, calculate the exact fraction of total area shaded:

$$\text{Area Fraction} = \frac{\text{Area}_{\text{shaded}}}{\text{Area}_{\text{total}}}$$

Subdivide each figure into congruent unit triangles (for example, drawing both diagonals and both midlines of a square creates $8$ congruent right triangles, each equal to $\frac{1}{8}$ of the square's area). Four figures will have shaded triangles totaling $\frac{4}{8} = \frac{1}{2}$ (or $\frac{2}{8} = \frac{1}{4}$), while the fifth figure will have $\frac{3}{8}$ shaded.

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Worked Step-by-Step Examples

Worked Example 1: The 10-Second $A \to B \to C$ Chirality Test

Problem Setup: Five options (A–E) each show a circle containing three peripheral symbols attached to its inner rim: a Black Square ($A$), a White Triangle ($B$), and a Cross ($C$), spaced at $90^\circ$ intervals with one empty quadrant.

• Option A: $A$ at North, $B$ at East, $C$ at South (West is empty).
• Option B: $A$ at East, $B$ at South, $C$ at West (North is empty).
• Option C: $A$ at South, $B$ at East, $C$ at North (West is empty).
• Option D: $A$ at West, $B$ at North, $C$ at East (South is empty).
• Option E: $A$ at South, $B$ at West, $C$ at North (East is empty).

Step-by-Step Solution: Trace the cyclic direction from $A \to B \to C$ through the occupied quadrant path:

• Option A: $\text{North} \to \text{East} \to \text{South}$ = Clockwise (CW).
• Option B: $\text{East} \to \text{South} \to \text{West}$ = Clockwise (CW).
• Option C: $\text{South} \to \text{East} \to \text{North}$ = Counterclockwise (CCW).
• Option D: $\text{West} \to \text{North} \to \text{East}$ = Clockwise (CW).
• Option E: $\text{South} \to \text{West} \to \text{North}$ = Clockwise (CW).
• Conclusion: Option C reverses the cyclic order to CCW and is the mirror-reflected outlier.

Worked Example 2: Eulerian Traversability (Tek Hamlede Çizilebilirlik)

Problem Setup: You are given 5 connected wireframe graphs and asked to find the odd one out:

• Figure A: A square with both diagonals drawn, plus a rooftop triangle attached to the top two corners.
• Figure B: Two squares sharing one common vertical edge (forming a $1 \times 2$ grid of squares).
• Figure C: A regular pentagon with an inscribed 5-pointed star (every vertex connects to 4 edges).
• Figure D: Two triangles sharing a single central vertex (an hourglass/bowtie figure).
• Figure E: A circle divided into 4 quadrants by one vertical diameter and one horizontal diameter.

Step-by-Step Solution: Count the odd-degree vertices ($N_{\text{odd}}$) in each graph:

• Figure A (House with X): Top peak has degree $2$ (even). Top-left and top-right corners each have $4$ edges meeting (even). Bottom-left and bottom-right corners each have $3$ edges meeting (odd). Center crossing has degree $4$ (even). Total $N_{\text{odd}} = 2 \implies$ Traversable in 1 stroke.
• Figure B (Two adjacent squares): Outer 4 corners have degree $2$ (even). The 2 shared middle vertices each have $3$ edges meeting (odd). Total $N_{\text{odd}} = 2 \implies$ Traversable in 1 stroke.
• Figure C (Pentagon + Star): Every vertex has degree $4$ (and any interior crossings have degree $4$). Total $N_{\text{odd}} = 0 \implies$ Traversable in 1 stroke.
• Figure D (Hourglass/Bowtie): The 4 outer corners have degree $2$ (even) and the shared center vertex has degree $4$ (even). Total $N_{\text{odd}} = 0 \implies$ Traversable in 1 stroke.
• Figure E (Circle with + cross): Center has degree $4$ (even), but the 4 rim junctions where diameters meet the circle each have $3$ branches meeting (2 arc segments + 1 radius). Thus $N_{\text{odd}} = 4 \implies$ Requires $\frac{4}{2} = 2$ strokes (NOT traversable in 1 stroke).
• Conclusion: Figure E is the unique non-traversable outlier.

Connectivity and Relabeling: Preserve Edges, Not Positions

A graph picture represents objects by vertices and relationships by edges. Moving a vertex on the page does not change the graph if exactly the same vertex pairs remain connected. The released 2026-/1 booklet includes this type of relationship-preservation task; use the original example below to learn the method without copying the official item.

Suppose the first drawing has edges AB, AC, AD, BC and DE. Its neighbour sets are:

A rearranged drawing labels its positions U, V, W, X, Y and has edges UV, UW, UX, VW and XY. U is the only degree-3 vertex, so it maps to A. Y is the only degree-1 vertex, so it maps to E. The neighbour of E must be D, hence X maps to D. V and W map to B and C in either order: both connect to A and to each other. Without another distinguishing condition, their individual labels cannot be uniquely determined. This is a useful stopping point, not a reason to guess a label.

Check all five edges after any proposed mapping. Equal degree counts are necessary but not sufficient: two graphs can have identical degree lists yet different adjacency. Crossings of edges are not automatically vertices in a relationship diagram; follow the question's convention about marked junctions. This differs from a single-stroke drawing where an actual junction is a vertex. Preserve both required edges and required non-edges, and state any remaining ambiguity.

Review the Method

Check symmetry axis counts (reflectional vs. pure rotational pinwheel symmetry), polygon convexity vs. concavity (dışbükey vs. içbükey), and exact shaded area fractions when chirality and topology do not isolate a unique outlier