4.3 Magic Squares, Latin Grids, Sudoku Variants & Relational Matrix Tables

Magic Squares, Latin Grids, Sudoku Variants & Relational Matrix Tables

Quick Strategy: For these independent grid exercises, begin with the requested target cells; solve additional cells if the target requires them. For Latin Squares and Sudoku variants, cross out symbols along the row and column intersecting the target cell $X$. For $3 \times 3$ Magic Squares (Sihirli Kareler), memorize two algebraic identities that bypass trial-and-error completely: the magic sum is always three times the center cell ($M = 3c$), and every corner cell is the average of the two non-adjacent edge cells ($\text{Corner} = \frac{\text{Edge}_1 + \text{Edge}_2}{2}$).

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Latin Squares & Shape/Number Sudoku ($4 \times 4$, $5 \times 5$, $6 \times 6$)

An $n \times n$ Latin Square is a grid filled with $n$ distinct symbols—typically digits $\{1, 2, \dots, n\}$, letters $\{A, B, \dots\}$, or geometric shapes $\{\triangle, \square, \bigcirc, \lozenge, \star\}$—such that every symbol appears exactly once in each row and once in each column.

In Sudoku Variants, the grid imposes one additional regional constraint:

• $4 \times 4$ Mini-Sudoku: Partitioned into four bold $2 \times 2$ sub-boxes, each containing $\{1, 2, 3, 4\}$.
• $6 \times 6$ Semi-Sudoku: Partitioned into six bold $2 \times 3$ rectangular blocks, each containing $\{1, 2, 3, 4, 5, 6\}$.
• Diagonal Latin Square (Köşegen Sudoku): Both main diagonals must also contain all $n$ symbols without repetition.

Target-Focused Deduction Techniques

Because the exam only asks for the value of one or two labeled cells ($X$ or $X + Y$), filling every empty square wastes valuable time. Instead, use two rapid elimination methods:

1. Intersection Elimination (Naked Single / Kesişim Elemesi): To find the symbol in cell $(r, c)$, take the union of all symbols already visible in Row $r$, Column $c$, and (if applicable) its bold sub-block. Whenever $n - 1$ distinct symbols appear in that combined peer group, the single missing symbol is forced into $(r, c)$.
2. Hidden Single (Gizli Tekli) via Column/Row Blocking: Suppose you need cell $(r, c)$, and Row $r$ still has three empty cells: $(r, c_1), (r, c_2), (r, c_3)$. If symbol $S$ already appears in Column $c_1$ and in Column $c_2$, then $S$ is blocked from both $(r, c_1)$ and $(r, c_2)$ and must go into $(r, c_3)$—even if you have no idea what the other two cells in Row $r$ contain!
3. The Line-Sum Invariant: In any numerical $n \times n$ Latin square using $\{1, 2, \dots, n\}$, the sum of every complete row, column, or sub-block is fixed:

$$S_n = \frac{n(n+1)}{2} \implies S_4 = 10,\quad S_5 = 15,\quad S_6 = 21$$

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$3 \times 3$ Arithmetic Magic Squares (Sihirli Kareler)

A $3 \times 3$ Magic Square is a numerical grid in which all 3 rows, all 3 columns, and both 3-cell main diagonals sum to the exact same Magic Constant $M$ (Sihirli Toplam):

$$\begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & c & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{bmatrix}$$

Instead of setting up nine messy linear equations, use two golden theorems that simplify a specified 3-by-3 magic-square problem.

Golden Formula 1: The Center Cell & Opposite-Pair Law ($M = 3c$)

Add the four lines that pass directly through the center cell $c = a_{2,2}$ (Row 2, Column 2, Main Diagonal, and Anti-Diagonal):

$$\text{Row}_2 + \text{Col}_2 + \text{Diag}_1 + \text{Diag}_2 = 4M$$

Every outer cell is included exactly once in this sum, while the center cell $c$ is included four times (three extra times beyond the sum of all 9 cells, which equals $3M$):

$$3M + 3c = 4M \implies M = 3c \quad \text{and} \quad c = \frac{M}{3}$$

• Total Grid Sum: The sum of all 9 numbers in the magic square is always $3M = 9c$.
• Symmetric Opposite Pairs: Since every line through the center sums to $M = 3c$, any two cells symmetrically opposite across the center sum to $2c$ (so $c$ is the exact arithmetic mean of every opposite pair):

$$a_{1,1} + a_{3,3} = a_{1,3} + a_{3,1} = a_{1,2} + a_{3,2} = a_{2,1} + a_{2,3} = 2c$$

Golden Formula 2: The Corner-Edge Mean Law (Köşe-Kenar Ortalaması)

What if a TR-YÖS problem gives you a sparse magic square where neither $M$ nor the center cell $c$ is known? Use the Corner-Edge Mean Law:

Every corner cell equals the arithmetic mean of the two non-adjacent edge midpoints (the two edge cells that do not share a row or column with that corner):

$$a_{1,1} = \frac{a_{3,2} + a_{2,3}}{2}, \quad a_{1,3} = \frac{a_{3,2} + a_{2,1}}{2}, \quad a_{3,1} = \frac{a_{1,2} + a_{2,3}}{2}, \quad a_{3,3} = \frac{a_{1,2} + a_{2,1}}{2}$$

• Quick Proof: In Row 3 and Column 3, $a_{3,1} + a_{3,2} + a_{3,3} = M$ and $a_{1,3} + a_{2,3} + a_{3,3} = M$. Adding these gives $(a_{3,1} + a_{1,3}) + a_{3,2} + a_{2,3} + 2a_{3,3} = 2M$. Since the anti-diagonal gives $a_{3,1} + a_{1,3} = 2c$ and $a_{1,1} + a_{3,3} = 2c \implies a_{3,3} = 2c - a_{1,1}$, substituting and using $2M = 6c$ yields:

$$2c + a_{3,2} + a_{2,3} + 4c - 2a_{1,1} = 6c \implies 2a_{1,1} = a_{3,2} + a_{2,3}$$

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Skyscraper (Apartmanlar) & Marginal Sum-Clue Grids

In a Skyscraper (Apartmanlar) puzzle, an $N \times N$ Latin square contains building heights $\{1, 2, \dots, N\}$ in every row and column. Numbers placed outside the border indicate how many buildings are visible when looking along that row or column from that side (taller buildings hide all shorter buildings behind them).

• Clue $= 1$: The tallest building $N$ must stand in the very first cell immediately adjacent to the clue (blocking all $N - 1$ buildings behind it).
• Clue $= N$: Every single building is visible, forcing the strictly ascending sequence $(1, 2, 3, \dots, N)$ starting from the clue.
• Clue $= 2$ with $N$ at the Far End: If building $N$ is in the last cell of a row, then whichever building stands in the first cell must hide every intermediate building before $N$, forcing $N - 1$ into the first cell.
• Extreme Pair $(1, 2)$ Across a Row: If the left clue is $1$ and the right clue is $2$, then $N$ is in the leftmost cell and $N - 1$ is in the rightmost cell!

In Marginal Sum-Clue / Mine-Count Grids, numbers along the top and left borders specify either the sum of values inside shaded cells or the count of black squares/stars in that row or column. Always start with rows/columns whose clue is $0$ (cross out the entire line) or equals the full line capacity, then propagate forced cells into intersecting lines.

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Grid Constraint & Magic Square Reference Table

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Worked Examples

Worked Example 1: Solving a Sparse $3 \times 3$ Magic Square in 10 Seconds

Problem: In a $3 \times 3$ arithmetic magic square where every row, column, and main diagonal has the same sum $M$, only three entries are given:

• Top-Middle Edge $a_{1,2} = 19$
• Middle-Left Edge $a_{2,1} = 17$
• Bottom-Left Corner $a_{3,1} = 16$

Find the Bottom-Right Corner $a_{3,3}$, the Center Cell $c = a_{2,2}$, and the Magic Constant $M$.

Step-by-Step Solution:

1. Find Bottom-Right Corner $a_{3,3}$ via the Corner-Edge Mean Law: The two non-adjacent edge midpoints opposite to $a_{3,3}$ are $a_{1,2} = 19$ and $a_{2,1} = 17$:

$$a_{3,3} = \frac{a_{1,2} + a_{2,1}}{2} = \frac{19 + 17}{2} = 18$$

1. Find Middle-Right Edge $a_{2,3}$ via the Corner-Edge Mean Law on $a_{3,1}$: The corner $a_{3,1} = 16$ is the average of its two non-adjacent edges $a_{1,2} = 19$ and $a_{2,3}$:

$$16 = \frac{19 + a_{2,3}}{2} \implies 19 + a_{2,3} = 32 \implies a_{2,3} = 13$$

1. Find Center $c$ and Magic Constant $M$: Since $a_{2,1} = 17$ and $a_{2,3} = 13$ are opposite across the center:

$$2c = a_{2,1} + a_{2,3} = 17 + 13 = 30 \implies c = 15, \quad M = 3c = 45$$

(Indeed, the complete magic square of consecutive integers $11 \dots 19$ is $[12, 19, 14;\; 17, 15, 13;\; 16, 11, 18]$.)

Worked Example 2: $4 \times 4$ Skyscraper & Latin Square Deduction

Problem: A $4 \times 4$ Skyscraper grid uses building heights $\{1, 2, 3, 4\}$ in each row and column.

• The Top clues (looking down Columns 1 to 4) are: $\text{Col 1} = 4,\; \text{Col 2} = 1,\; \text{Col 3} = 2,\; \text{Col 4} = 2$.
• The Right clue looking left along Row 2 is $1$.

Determine the exact building height in cell $(2, 3)$ (Row 2, Column 3).

Step-by-Step Solution:

1. Apply the Top Clue $4$ on Column 1: A clue of $4$ forces Column 1 from top to bottom to be $(1, 2, 3, 4)^T$, so $a_{1,1} = 1,\; a_{2,1} = 2,\; a_{3,1} = 3,\; a_{4,1} = 4$.
2. Apply the Clue $1$s to Place All Four $4$s:
   • Top clue $\text{Col 2} = 1 \implies a_{1,2} = 4$.
   • Right clue on Row 2 $= 1 \implies a_{2,4} = 4$.
   • Since $4$ is already placed in Columns 1, 2, and 4 (at $a_{4,1}, a_{1,2}, a_{2,4}$), the remaining $4$ in Column 3 must sit in Row 3: $a_{3,3} = 4$.
3. Solve Row 2: In Row 2, we already have $a_{2,1} = 2$ and $a_{2,4} = 4$, leaving $\{1, 3\}$ for $a_{2,2}$ and $a_{2,3}$. In Row 1, since $a_{1,1} = 1$ and $a_{1,2} = 4$, the remaining cells $a_{1,3}$ and $a_{1,4}$ must be $\{2, 3\}$. Because the Top clue for Column 3 is $2$ and $a_{3,3} = 4$, cell $a_{2,3}$ cannot be taller than $a_{1,3}$ (otherwise $a_{1,3}, a_{2,3}, 4$ would all be visible from the top of Column 3, giving 3 visible buildings instead of 2). Choosing $a_{2,3} = 3$ would force $a_{1,3} = 2 < 3 < 4$, which violates the clue of $2$. Therefore, $a_{2,3} = 1$!

Review the Method

In $N \times N$ Skyscraper ('Apartmanlar') grids, a marginal clue of $1$ forces the tallest building $N$ into the immediately adjacent border cell, while a clue of $N$ forces the ascending sequence $(1, 2, \dots, N)$ along that line