4.2 Pattern & Grid Completion (Şekil Tamamlama), Quarter-Tile Symmetry & Tessellations

Pattern & Grid Completion (Şekil Tamamlama), Quarter-Tile Symmetry & Tessellations

Quick Strategy: Never start a Şekil Tamamlama question by trying to draw the entire missing quadrant. Instead, look at the single corner of the missing box that touches the exact center of the main figure (usually the top-left corner of the bottom-right quadrant). Checking the corner can eliminate incompatible choices before you inspect the remaining tile features.

In a pattern-completion exercise, a tile is missing from a larger design. Identify whether the displayed pattern uses reflection, rotation or translation, and compare the touching edges before selecting a tile. The examples below choose the bottom-right quadrant for convenience; that position is not an official format rule.

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Four Symmetry Models

Let the complete square figure be partitioned by a vertical axis ($x = 0$) and a horizontal axis ($y = 0$) into four quadrants: Top-Left ($Q_{\text{TL}}$), Top-Right ($Q_{\text{TR}}$), Bottom-Left ($Q_{\text{BL}}$), and the missing Bottom-Right ($Q_{\text{BR}}$).

1. Bilateral Axis Symmetry (Dikey ve Yatay Eksen Simetrisi)

In a bilaterally symmetric pattern, the vertical centerline acts as a mirror between the left and right halves, and the horizontal centerline acts as a mirror between the top and bottom halves:

$$Q_{\text{TR}} = \text{Flip}_H(Q_{\text{TL}}), \quad Q_{\text{BL}} = \text{Flip}_V(Q_{\text{TL}})$$

Consequently, the missing bottom-right quadrant $Q_{\text{BR}}$ satisfies three equivalent relationships simultaneously:

$$Q_{\text{BR}} = \text{Flip}_H(Q_{\text{BL}}) = \text{Flip}_V(Q_{\text{TR}}) = \text{Rot}_{180^\circ}(Q_{\text{TL}})$$

• Radial Distance Invariance: Any feature located at distance $(d_x, d_y)$ outward from the center origin $(0,0)$ in $Q_{\text{TL}}$ appears at the exact same radial distances $(d_x, d_y)$ outward from the center in $Q_{\text{BR}}$. Thus, whatever touches the bottom-right corner of $Q_{\text{TL}}$ (at the center of the whole figure) must touch the top-left corner of $Q_{\text{BR}}$!
• Slope Reversal Rule: When reflecting $Q_{\text{BL}}$ horizontally into $Q_{\text{BR}}$ (or $Q_{\text{TR}}$ vertically into $Q_{\text{BR}}$), every diagonal line segment of slope $m$ reverses sign to $-m$ (/ becomes \ and \ becomes /), while horizontal lines remain horizontal and vertical lines remain vertical.

2. 4-Fold Rotational Pinwheel Symmetry ($90^\circ$ CW Cycle / Dönel Simetri)

In a rotational (pinwheel or vortex) figure, there are no mirror axes. Instead, each quadrant is generated by rotating the preceding quadrant $90^\circ$ clockwise around the central point:

$$Q_{\text{TL}} \xrightarrow{90^\circ\text{ CW}} Q_{\text{TR}} \xrightarrow{90^\circ\text{ CW}} Q_{\text{BR}} \xrightarrow{90^\circ\text{ CW}} Q_{\text{BL}} \xrightarrow{90^\circ\text{ CW}} Q_{\text{TL}}$$

Reflection Trap: A rotational pinwheel can have no mirror axes. Reflecting a neighbouring tile reverses handedness, whereas rotating it preserves handedness. Check an asymmetric feature and apply the actual transformation before selecting a tile; no particular distractor is guaranteed to appear.

• How to Diagnose Rotational vs. Bilateral Symmetry in 2 Seconds: Compare $Q_{\text{TL}}$ and $Q_{\text{TR}}$ before looking at the missing quadrant. Find an asymmetric element (such as a shaded triangle, a hook, or parallel stripes). If horizontal stripes in $Q_{\text{TL}}$ stay horizontal in $Q_{\text{TR}}$, the figure is Bilateral. If horizontal stripes in $Q_{\text{TL}}$ turn vertical in $Q_{\text{TR}}$, the figure is Rotational ($90^\circ$ CW), and $Q_{\text{BR}}$ must be obtained by rotating $Q_{\text{TR}}$ $90^\circ$ clockwise (or rotating $Q_{\text{TL}}$ by $180^\circ$).

3. Main & Anti-Diagonal Reflection Symmetry (Köşegen Simetrisi)

Some TR-YÖS grids lack vertical and horizontal mirror symmetry but fold cleanly across one or both $45^\circ$ diagonals:

• Main Diagonal ($\text{NW} \to \text{SE}$ axis): Folding across the main diagonal maps $Q_{\text{TR}} \leftrightarrow Q_{\text{BL}}$ and maps $Q_{\text{BR}}$ onto itself across its own internal $\text{NW} \to \text{SE}$ diagonal.
• Coordinate Transposition Law: Reflecting across the NW-to-SE array diagonal swaps $(r,c)$ with $(c,r)$. Reflection across the other diagonal of an $n\times n$ array maps $(r,c)$ to $(n+1-c,n+1-r)$, turning horizontal lines into vertical lines and vertical lines into horizontal lines.

4. Translational Wallpaper & Tessellation Grids (Öteleme ve Döşeme Örüntüleri)

In a periodic wallpaper or carpet tessellation, the full square is simply a $2 \times 2$ or $3 \times 3$ repetition of an identical unit tile shifted horizontally and vertically without rotation or reflection:

$$Q_{\text{TL}} = Q_{\text{TR}} = Q_{\text{BL}} = Q_{\text{BR}}$$

Alternatively, the grid may follow a checkerboard translation where $Q_{\text{BR}} = Q_{\text{TL}}$ and $Q_{\text{TR}} = Q_{\text{BL}}$ with inverted black/white coloring.

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The 3-Step Corner-First & Boundary-Continuity Technique

When solving any quarter-tile completion item where $Q_{\text{BR}}$ is missing, apply these three steps in strict order:

1. Step 1 — The Center-Corner Anchor (Merkez Köşe Kontrolü): Look at the exact center point of the large composite figure, where the bottom-right corner of $Q_{\text{TL}}$, the bottom-left corner of $Q_{\text{TR}}$, and the top-right corner of $Q_{\text{BL}}$ meet. The top-left corner of the missing tile $Q_{\text{BR}}$ must complete that central junction (e.g., completing a central black diamond or white circle). Scan the top-left corner of all options—this single glance routinely can eliminate incompatible options.
2. Step 2 — Cut-Edge Boundary Continuity (Kenar Sürekliliği): Trace the lines, arcs, and shaded regions that hit the North cut edge (the bottom border of $Q_{\text{TR}}$) and the West cut edge (the right border of $Q_{\text{BL}}$):
   • Position Matching: If a line from $Q_{\text{TR}}$ hits the North cut edge at $1/3$ of the distance from left to right, the matching option must have a line starting on its top edge at exactly $1/3$ from the left.
   • Tangent & Color Continuity: A circular arc crossing the West border from $Q_{\text{BL}}$ must continue smoothly without a sharp kink unless the border itself is a mirror axis.
3. Step 3 — Outer-Corner & Internal Slope Check: Compare the two remaining options at the bottom-right outer corner and verify whether an internal diagonal stroke should slant / or \.

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Quadrant Transformation Comparison Table

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Worked Examples

Worked Example 1: Avoiding the Reflection Trap in a $90^\circ$ Rotational Grid

Problem: A $6 \times 6$ grid is divided into four $3 \times 3$ quadrants ($Q_{\text{TL}}, Q_{\text{TR}}, Q_{\text{BL}}, Q_{\text{BR}}$). Inside $Q_{\text{TL}}$, using local coordinates $(r, c)$ from $(1,1)$ at top-left to $(3,3)$ at bottom-right:

• Cell $(3,3)$ (touching the center of the main grid) contains a Black Circle.
• Cell $(1,2)$ contains a Upward Arrow ($\uparrow$).
• Cell $(2,1)$ is Shaded Gray.

In $Q_{\text{TR}}$, the corresponding features are located at:

• Local cell $(3,1)$ (touching the center of the main grid): Black Circle.
• Local cell $(2,3)$: Rightward Arrow ($\rightarrow$).
• Local cell $(1,2)$: Shaded Gray.

Where must these three features appear inside the missing quadrant $Q_{\text{BR}}$?

Step-by-Step Solution:

1. Diagnose the Regime ($Q_{\text{TL}} \to Q_{\text{TR}}$): Notice that the Upward Arrow ($\uparrow$) at $(1,2)$ in $Q_{\text{TL}}$ became a Rightward Arrow ($\rightarrow$) at $(2,3)$ in $Q_{\text{TR}}$, and the Gray cell at $(2,1)$ moved to $(1,2)$. In a $3 \times 3$ array, rotating $90^\circ$ clockwise maps $(r, c) \to (c, 4 - r)$:
   • $(1,2) \to (2, 4-1) = (2,3)$ and $\uparrow$ rotates $90^\circ$ CW to $\rightarrow$.
   • $(2,1) \to (1, 4-2) = (1,2)$.
   • $(3,3) \to (3, 4-3) = (3,1)$. This proves the figure is governed by 4-Fold Rotational Pinwheel Symmetry ($90^\circ$ CW), not bilateral mirror symmetry!
2. Apply $90^\circ$ CW Rotation to $Q_{\text{TR}}$ (or $180^\circ$ to $Q_{\text{TL}}$ via $(r,c) \to (4-r, 4-c)$):
   • Black Circle: $(3,3)$ in $Q_{\text{TL}} \to (4-3, 4-3) = (1,1)$ (the top-left center-touching corner of $Q_{\text{BR}}$).
   • Arrow: $(1,2)$ with $\uparrow$ in $Q_{\text{TL}} \to (4-1, 4-2) = (3,2)$ with a Downward Arrow ($\downarrow$).
   • Shaded Gray Cell: $(2,1)$ in $Q_{\text{TL}} \to (4-2, 4-1) = (2,3)$.
3. Conclusion: In $Q_{\text{BR}}$, the Black Circle is at $(1,1)$, the Gray cell is at $(2,3)$, and the Downward Arrow ($\downarrow$) is at $(3,2)$.

Worked Example 2: Corner-First & Boundary-Continuity Elimination

Problem: An ornamental carpet square has Bilateral Axis Symmetry (both horizontal and vertical mirror axes). In the Bottom-Left quadrant $Q_{\text{BL}}$:

• The top-right corner of $Q_{\text{BL}}$ (at the center of the carpet) is a Black Quarter-Disk.
• Along the right border of $Q_{\text{BL}}$ (the West cut edge of $Q_{\text{BR}}$), a bold diagonal line arrives from the interior of $Q_{\text{BL}}$ with slope $+1$ (rising /) and terminates at the midpoint of that vertical cut edge.
• The bottom-left outer corner of $Q_{\text{BL}}$ contains a white triangle pointing inward (/ hypotenuse).

Determine the exact features of the missing Bottom-Right tile $Q_{\text{BR}}$.

Step-by-Step Solution:

1. Step 1 (Center-Corner Anchor): Under bilateral symmetry, $Q_{\text{BR}} = \text{Flip}_H(Q_{\text{BL}})$. Flipping $Q_{\text{BL}}$ horizontally maps its top-right corner to the top-left corner of $Q_{\text{BR}}$. Thus, the top-left corner of $Q_{\text{BR}}$ must be a Black Quarter-Disk (completing the central black circle).
2. Step 2 (West Cut-Edge Continuity): The right border of $Q_{\text{BL}}$ is the shared mirror axis with the left border of $Q_{\text{BR}}$. Because the bold line in $Q_{\text{BL}}$ hits the midpoint of that vertical border with slope $+1$ (/), its horizontal reflection in $Q_{\text{BR}}$ must originate at the exact midpoint of the left border of $Q_{\text{BR}}$ and travel into $Q_{\text{BR}}$ with reversed slope $-1$ (\, heading down and right).
3. Step 3 (Outer Corner): The bottom-left outer corner of $Q_{\text{BL}}$ reflects horizontally to the bottom-right outer corner of $Q_{\text{BR}}$, reversing its hypotenuse from / to \.

Review the Method

Reflection across the NW-to-SE diagonal of an array swaps row and column indices; the other diagonal maps (r,c) to (n+1-c,n+1-r).