Constructions and Tilings | IT

Question 38

Is an m×nm \times n grid tileable with 2×12 \times 1 tiles, if both mm and nn are even? If yes, come up with a general strategy to tile it.

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Solution
Understand the Question
  • A single 2×12 \times 1 tile covers 22 unit squares.
  • An m×nm \times n grid has a total of m×nm \times n unit squares.
  • If both mm and nn are even, the grid can be decomposed into smaller 2×22 \times 2 squares, each of which can be tiled by two 2×12 \times 1 tiles.

Step 1 · Check Total Number of Squares

Let m=2km = 2k and n=2jn = 2j, where kk and jj are positive integers since mm and nn are both even.Diagram 1

Total squares=m×n=(2×k)×(2×j)=4×k×j\begin{aligned} \text{Total squares} &= m \times n \\ &= (2 \times k) \times (2 \times j) \\ &= 4 \times k \times j \end{aligned}

Since the total number of squares is a multiple of 44 (and therefore divisible by 22), the total area can potentially be covered by 2×12 \times 1 tiles.

Step 2 · Formulate Tiling Strategy

Since mm and nn are both even, we can partition the m×nm \times n grid into m2×n2\dfrac{m}{2} \times \dfrac{n}{2} smaller blocks of size 2×22 \times 2.Diagram 2

Each 2×22 \times 2 square consists of 44 unit squares and can be completely covered by placing two 2×12 \times 1 tiles horizontally (or vertically).

Repeating this placement for all m2×n2\dfrac{m}{2} \times \dfrac{n}{2} blocks tiles the entire m×nm \times n grid completely without any gaps or overlaps.

Answer

Yes, an m×nm \times n grid is tileable.

Strategy: Divide the m×nm \times n grid into m2×n2\dfrac{m}{2} \times \dfrac{n}{2} blocks of size 2×22 \times 2, and cover each 2×22 \times 2 block with two 2×12 \times 1 tiles placed horizontally.

Common Mistakes
  • Assuming Area Divisibility is Enough: While having an even total area is necessary to tile with 2×12 \times 1 tiles, a valid tiling requires a concrete placement strategy without gaps or overlaps.
  • Overcomplicating the Pattern: Since both dimensions are even, simply subdividing into 2×22 \times 2 blocks (or filling row-by-row) is sufficient rather than trying alternating or complex checkerboard patterns.

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