## 2 Polyomino Tiling Theory
: _According to condition 2 of Theorem 2, the smallest rectangle that can be tiled using 'T' trominoes is (2 \(\times\) 3). Let a = 4; then according to condition 2, the possible b values must be divisible by 3. Let's assume the dimension of the rectangle is 4 \(\times\) 6, Figure 3(a); then it is possible to decompose it to a 4 (2 \(\times\) 3) rectangles. Hence, a rectangle with a dimension was a \(>\) 3, and b is a multiple of 3, then the rectangle could be decomposed to n(2 \(\times\) 3) rectangles as in Figure 3(b), which can be tillable with 'T' trominoes. However, the rectangle with dimensions of 5 \(\times\) 3, 7 \(\times\) 3, and 9 \(\times\) 3 cannot be tiled using any arrangement of 'T' trominoes._  
Figure 3: Image Argument for Lemma 1. (a) a = 3 and b is even; (b) Decomposing a \(\times\) b rectangle into sub-rectangles. **Theorem 3**. : _A deficient \(n\) X \(n\) rectangle board can be tiled with the set of 'T' frominoes, if, and only if, either:_  
1. \(n\) _is odd, and_ \(>\)_5, and_ \(n^{2}-1\) _is divisible by 3._
2. \(n\) _is even, and_ \(>\)_1, and_ \(n^{2}-1\) _is divisible by 3._  
Also with Theorem 3, we are utilizing set 'T' fromino pieces in order to tile the given area. The term deficient in the theorem describes a square grid of side a and b with a single cell truncated. According to Theorem 3, if the sides of the squares are odd, it should be greater than 5, and to the square of that sides subtracted by 1 must be divisible by 3 to tile the space using 'T' frominoes. In order to tile an even-sided deficient square with 'T' frominoes, the value must be greater than 1 and, to the square of that value subtracted by 1 must be divisible by 3. The above notion regarding tiling the deficient square points to Conditions 1 and 2 of Theorem 3. Lemmas 3 and 4 validate Theorem 3 by proving Conditions 1 and 2. **Lemma 3**. : _Let \(M\) (a, b) be the modified rectangle and segmented into an (i, j) grids. Let's consider an \(M\) (7 \(\times\) 7) rectangle grid with a single (1, 1) square is removed Figure 5a Left. As shown in Figure 5a Right, the considered rectangle can be decomposed to a 2 \(\times\) 3, 3 \(\times\) 2, 5 \(\times\) 5, and three separate \(L\)-frominoes. According to Lemma 1, a 2 \(\times\) 3, and 3 \(\times\) 2, a rectangle can be easily tillable with 'T' fromino set. When it comes to a 5 \(\times\) 5 square, it is tileable with 'T' frominoes only if the corner cells are removed as shown is Figure 5b. Hence, the results show that an \(M\) (7 \(\times\) 7) rectangle with single cell removed is tillable with 'T' frominoes. It is also possible to tile an \(M\) (7 \(\times\) 7) rectangle when we delete (1, 4), (2, 3), (2, 4), or (4, 4) cells._  
Figure 4: Image Argument for Lemma 2. (**a**) A 4 \(\times\) 6 and 5 \(\times\) 6 Tiled rectangle; (**b**) Decomposition on a \(\times\) b rectangle according to Lemma 2. **Lemma 4**. : _Similarly, with condition 2 Theorem 3, we are considering an M (10 \(\times\) 10) rectangle. The considered rectangle can be split into a 7 \(\times\) 7, 6 \(\times\) 3, 3 \(\times\) 6, and 4 \(\times\) 4 sub-rectangles shown in Figure 6b. According to Lemma 3 a 7 \(\times\) 7 rectangle with one square removed can be tiled with 'T' tromioes. Similarly, with the help of Lemmas 1 & 2, we know that 6 \(\times\) 3, and 3 \(\times\) 6 can also be tiled with 'T' trominoes. For a 4 \(\times\) 4 square, it is proven that a \(2^{k}\times 2^{k}\) when \(k\geq 1\) can be tiled with 'T' trominoes as shown in Figure 6a. Hence, it is proved that a rectangle with even sided can be tiled using set of 'T' tromino pieces._  
Figure 5: Image Argument for Lemma 3. (**a**) Decomposing an a \(\times\) b rectangle according to Lemma 3; (**b**) A 5 \(\times\) 5 modified tiled rectangle. Figure 6: _Cont._  
**Theorem 4**. : _An Aztec Diamond, AZ(n) can be tiled using 'T' set of tromino tiling pieces if, and only if \(n\) (\(n+1\)) \(\equiv\) 0 mod 3, where \(n\) can be a positive integer._  
**Theorem 5**.