3.1 Figural Series, Angular Rotations, Translations & Element-Counting Rules

3.1 Figural Series, Angular Rotations, Translations & Element-Counting Rules

Quick Method: On the TR-YÖS Sayısal Yetenek Testi, an original Figural Series (Şekil Dizileri) exercise can present several sequential frames ($F_1, F_2, F_3, F_4$, and optionally $F_5$) and asks you to identify either the next frame in the sequence or a missing interior frame. Never stare at the entire drawing at once. Instead, apply Independent Element Tracking (Bağımsız Eleman Takibi): isolate a single micro-feature (such as a corner dot, an internal arrow, or a polygon boundary), determine its numerical rule across the frames, and immediately cross out any answer choices that violate that single rule.

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How Figural Series (Şekil Dizileri) Work on TR-YÖS

In the abstract and spatial reasoning portion of the TR-YÖS Sayısal Yetenek Testi, figural series test your ability to extract deterministic mathematical laws from visual sequences. Each frame $F_n$ (for $n = 1, 2, 3, 4, \dots$) is a composite state vector made of 2 to 5 distinct visual sub-elements:

$$F_n = \big(e_n^{(1)},\; e_n^{(2)},\; e_n^{(3)},\; e_n^{(4)}\big)$$

Because ÖSYM constructs distractor options by keeping two or three sub-elements correct while corrupting just one, trying to "see" the whole picture moving at once leads to visual overload and trap selection. By treating each component $e_n^{(k)}$ as an independent sequence with its own governing rule, you transform an intimidating graphic puzzle into simple arithmetic.

Single-Element vs. Multi-Element Independent Tracking (Bağımsız Eleman Takibi)

When approaching any Şekil Dizileri item, decompose the figure into separated structural layers and track them one by one:

1. Layer 1 — Primary Reference Frame or Outer Polygon: Does the outer boundary rotate, change its number of sides, or remain static as a coordinate frame?
2. Layer 2 — Directional Pointer or Axis Element: Track the tip of an arrow, a clock-like hand, or an asymmetric hook using exact compass angles ($0^\circ, 45^\circ, 90^\circ, \dots$).
3. Layer 3 — Peripheral or Corner Markers: Track small dots, stars, or triangles stepping along the vertices, edges, or grid cells of the boundary.
4. Layer 4 — Shading and Fill Parity: Track whether sub-regions alternate between black, white, and hatched states, or whether the shaded region itself rotates independently of the boundary.

In practice, tracking Layer 2 alone typically eliminates 2 or 3 of the 5 options (A–E). Tracking Layer 3 resolves the remaining tie in under 20 seconds without ever needing to analyze the remaining decorative features.

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Exact Angular Rotation Mathematics

• Clockwise (Saat Yönü / CW): Following the hands of a standard analog clock (top $\to$ right $\to$ bottom $\to$ left).
• Counterclockwise (Saat Yönünün Tersi / CCW): Moving against the clock hands (top $\to$ left $\to$ bottom $\to$ right).

The Standard 8-Point Compass Grid ($45^\circ$ Increments)

Use the angles supplied by the question; the following eight-point system is useful for multiples of $45^\circ$. Visualizing a square frame as an 8-point compass makes angle tracking instantaneous:

• $45^\circ$ (One-Eighth Turn / Half-Side Step): Moves a pointer from a cardinal direction (such as North / top midpoint) to the adjacent diagonal corner (Northeast / top-right corner). Full period $= 8$ steps.
• $90^\circ$ (One-Quarter Turn / Right Angle): Moves a pointer from North to East (or from one corner of a square to the next adjacent corner). Full period $= 4$ steps.
• $135^\circ$ (Three-Eighths Turn): Equivalent to $90^\circ + 45^\circ$ in the same direction, or $1 + \frac{1}{2}$ sides of a square frame. Full period $= 8$ steps.
• $180^\circ$ (Half Turn / Point Inversion): Sends any element directly to its diametrically opposite position (North $\leftrightarrow$ South, Northeast $\leftrightarrow$ Southwest). Note that $+180^\circ$ CW and $-180^\circ$ CCW produce the exact same state. Full period $= 2$ steps.

Modular Cyclic Periodicity vs. Progressive Rotation

For a constant angular velocity series with initial orientation $\theta_1$ in Frame 1 and constant step $\Delta\theta$ per transition, the orientation in Frame $n$ is governed by modular arithmetic:

$$\theta_n = \big(\theta_1 + (n - 1)\cdot \Delta\theta\big) \pmod{360^\circ}$$

Remember that a rotation of $270^\circ$ CW is identical in final position to $90^\circ$ CCW ($-90^\circ \equiv +270^\circ \pmod{360^\circ}$), and $315^\circ$ CW is identical to $45^\circ$ CCW.

In harder TR-YÖS items, the rotation is progressive (accelerating or decelerating) rather than constant. The angular increment $\Delta\theta_k$ between Frame $k$ and Frame $k+1$ follows an arithmetic sequence:

• Linearly Increasing Step: $\Delta\theta_1 = +45^\circ,\; \Delta\theta_2 = +90^\circ,\; \Delta\theta_3 = +135^\circ,\; \Delta\theta_4 = +180^\circ$ (all CW).
• Alternating Direction Step: $\Delta\theta_1 = +90^\circ\text{ (CW)},\; \Delta\theta_2 = -45^\circ\text{ (CCW)},\; \Delta\theta_3 = +90^\circ\text{ (CW)},\; \Delta\theta_4 = -45^\circ\text{ (CCW)}$.

Symmetry Trap: If a symmetric shape (such as a regular square or an equilateral cross) rotates by $90^\circ$, it looks unchanged unless it carries an asymmetric marker. If a plain line segment rotates by $45^\circ$ per frame, its visual period is $180^\circ$ (4 frames) rather than $360^\circ$ (8 frames) because its two ends are indistinguishable.

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Perimeter & Grid Translation Tracks

Instead of rotating around a central pivot, discrete elements often translate (slide) along a polygon's perimeter or across the cells of a matrix grid.

1. Polygon Perimeter Stepping

Suppose a marker moves along the boundary of a regular $N$-gon (such as a pentagon with $N = 5$ vertices or a hexagon with $N = 6$ vertices):

• Vertex-to-Vertex Stepping: Label the vertices $0, 1, 2, \dots, N-1$ in clockwise order. If a dot advances $k$ vertices CW per frame from starting vertex $p_1$, its vertex index at Frame $n$ is:

$$p_n = \big(p_1 + (n - 1)\cdot k\big) \pmod{N}$$

• Half-Side (Vertex-to-Midpoint) Stepping: On a square (4 vertices + 4 edge midpoints $= 8$ perimeter stations), a marker moving $\frac{1}{2}$ side length per frame is equivalent to stepping $1$ station on an $8$-station cycle. Moving $1.5$ sides CW per frame equals $+3$ stations per step modulo $8$.

2. Matrix Grid Translation: Wrap-Around vs. Bounce-Back

When a marker moves inside a $3 \times 3$ or $4 \times 4$ grid along a horizontal row, vertical column, or main diagonal, you must distinguish between two boundary rules as soon as the marker hits an outer wall:

• Cyclic Wrap-Around (Toroidal Track): After reaching the last cell of a row or column—say $(r, 4)$ in a $4 \times 4$ grid—the next step wraps immediately back to the first cell $(r, 1)$ of the same row (or snakes into $(r+1, 1)$ of the next row).
• Reflective Bounce-Back (Pendulum Track): Upon hitting the boundary cell $(r, 4)$, the marker reverses direction like a billiard ball, moving to $(r, 3)$ on the next step, then $(r, 2)$, then $(r, 1)$, before bouncing forward again. In a 4-cell row, a 1-cell bounce-back track follows the 6-step periodic cycle: $1 \to 2 \to 3 \to 4 \to 3 \to 2 \to 1$.

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Quantitative Figural Growth & Shading Parity

Many TR-YÖS series combine spatial movement with arithmetic counting rules:

1. Polygon Vertex and Side Counts: The number of sides $S_n$ may increase arithmetically (triangle $3 \to$ quadrilateral $4 \to$ pentagon $5 \to$ hexagon $6 \to$ heptagon $7$) or follow coupled opposite progressions (outer polygon gains $+1$ side per frame while inner polygon loses $-1$ side per frame).
2. Line-Segment Additions & Erasures: A figure may gain one new chord or external tick per frame at a specific clockwise position, or alternate between adding 2 segments and erasing 1 segment ($L_n: 2, 4, 3, 5, 4, 6$).
3. Intersection-Point Counts (Kesişim Noktası): Even when line orientations look chaotic, count the exact number of interior crossings between lines ($I_1 = 1, I_2 = 2, I_3 = 3, I_4 = 4 \implies I_5 = 5$).
4. Shading Parity Alternation: Track fill states numerically:
   • 2-State Toggle (Period 2): Black $\leftrightarrow$ White ($B, W, B, W, B$).
   • 3-State Cyclic Shift (Period 3): Black $\to$ White $\to$ Striped $\to$ Black ($B, W, S, B, W$).

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Worked Step-by-Step Examples

Worked Example 1: Progressive Rotation + Alternating Half-Side Perimeter Walk

Problem Setup: Each frame consists of a stationary outer square, an internal arrow pivoted at the center, and a small black circle on the square's perimeter.

• Frame 1: Arrow points North ($0^\circ$); circle is at the top-left corner.
• Frame 2: Arrow points Northeast ($45^\circ$ CW); circle is at the top-right corner.
• Frame 3: Arrow points Southeast ($135^\circ$ CW); circle is at the right-edge midpoint.
• Frame 4: Arrow points West ($270^\circ$ CW); circle is at the bottom-edge midpoint.
• Question: Find the state of Frame 5.

Step-by-Step Solution:

1. Isolate Layer 1 (Central Arrow): Write the angles in degrees clockwise from North: $\theta_1 = 0^\circ,\; \theta_2 = 45^\circ,\; \theta_3 = 135^\circ,\; \theta_4 = 270^\circ$. Compute the differences: $\Delta\theta_1 = +45^\circ$, $\Delta\theta_2 = +90^\circ$, $\Delta\theta_3 = +135^\circ$. This is a progressive rotation increasing by $+45^\circ$ each transition. Therefore, $\Delta\theta_4 = +180^\circ$. In Frame 5, the arrow points at $\theta_5 = (270^\circ + 180^\circ) \pmod{360^\circ} = 450^\circ \pmod{360^\circ} = 90^\circ$ (East).
2. Isolate Layer 2 (Perimeter Circle): Label the 8 perimeter stations (4 corners + 4 midpoints) clockwise starting from top-left corner $= 0$, top midpoint $= 1$, top-right corner $= 2$, right midpoint $= 3$, bottom-right corner $= 4$, bottom midpoint $= 5$, bottom-left corner $= 6$, left midpoint $= 7$.
   • Positions across frames: $s_1 = 0,\; s_2 = 2,\; s_3 = 3,\; s_4 = 5$.
   • Step differences: $0 \to 2$ is $+2$ stations ($1$ full side CW); $2 \to 3$ is $+1$ station ($0.5$ side CW); $3 \to 5$ is $+2$ stations ($1$ full side CW). The perimeter step alternates $+2, +1, +2, +1$ stations clockwise. Thus, from Frame 4 ($s_4 = 5$) to Frame 5, the circle advances $+1$ station to $s_5 = 6$ (bottom-left corner). Combining Arrow = East ($90^\circ$) and Circle = Bottom-Left Corner uniquely identifies Frame 5.

Worked Example 2: Multi-Layer Polygon Growth + Shading Cycle

Problem Setup: A sequence of nested polygons with an interior dot counter:

• Frame 1: A black 3-sided triangle containing 6 white dots.
• Frame 2: A white 4-sided square containing 5 black dots.
• Frame 3: A striped 5-sided pentagon containing 4 white dots.
• Frame 4: A black 6-sided hexagon containing 3 black dots.
• Question: Determine Frame 5.

Step-by-Step Solution:

1. Polygon Side Count ($S_n$): $3, 4, 5, 6 \implies S_5 = 7$ (heptagon).
2. Polygon Shading State: Cycles through a 3-state sequence: $\text{Black} \to \text{White} \to \text{Striped} \to \text{Black} \implies$ Frame 5 polygon must be White.
3. Dot Count ($D_n$): Decreases by 1 each frame: $6, 5, 4, 3 \implies D_5 = 2$ dots (preserving the invariant $S_n + D_n = 9$ in every frame).
4. Dot Color Parity: Alternates with period 2: $\text{White} \to \text{Black} \to \text{White} \to \text{Black} \implies$ Frame 5 dots must be White.

• Result: A white heptagon (7 sides) containing 2 white dots outlined in black so they remain visible.

Original Visual Practice: Two Independent Tracks

Read each cell as a pointer direction and a marked corner; the corner labels refer to the stationary square frame. The circle alternates filled and hollow.

The pointer rotates 90 degrees clockwise on each transition, so the missing pointer is ←. The marker moves one corner clockwise, reaching SW, and its alternating fill makes it hollow: ○ SW. Verify both tracks independently. A choice showing ← with a filled SW marker would satisfy direction and position but fail the colour rule. All figures and rules here are original practice.

Review the Method

Quantitative figural progressions govern polygon side counts, line-segment additions, intersection-point totals, and 2-state or 3-state shading parity cycles