### 数学代写|组合数学代写Combinatorial mathematics代考|MATH418

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## 数学代写|组合数学代写Combinatorial mathematics代考|Solutions to Exercises

2.1. The area of an $\mathrm{L}$-tromino (see Figure 1.6) is equal to 3 . The area of the $4 \times 5$ rectangle is equal to 20 . Hence, in order to tile this rectangle by L-trominoes, it is necessary to use $\frac{20}{3}$ L-trominoes, which is impossible.

Generally, if an $a \times b$ rectangle can be tiled by L-trominoes, then $a b$ is divisible by 3 . This is a necessary condition for the existence of such a tiling.
2.2. Due to Exercise $2.1$, if an $a \times a$ square can be tiled by Ltrominoes, then $a$ is divisible by 3 . The $6 \times 6$ square can be decomposed into $2 \times 3$ blocks (Figure $2.3$ ), and consequently, the $6 \times 6$ square can be tiled by L-trominoes (see Figure 2.2). So, it remains to be seen whether the $3 \times 3$ square can be tiled by L-trominoes. The answer is “no.” Indeed, all the ways to cover the upper-left corner are shown in Figure 2.4.

In Figure 2.4(a), it is impossible to cover the upper right corner. In Figure 2.4(b), it is impossible to cover the lower left corner. Finally, in Figure 2.4(c), the only ways to cover the upper right corner is shown

in Figure 2.5. But then the three bottom squares remain uncovered. Thus the investigation of all possibilities shows that the $3 \times 3$ square cannot be tiled by L-trominoes. Consequently, the $6 \times 6$ square is the smallest tileable square.
2.3. If $b$ is divisible by 3 , then a $2 \times b$ rectangle can be decomposed into $2 \times 3$ blocks (see Figure 2.6) and, consequently, this rectangle can be tiled by L-trominoes. The reasoning in the solution to Exercise $2.1$ shows that there are no other possibilities. So, the $2 \times b$ rectangle can be tiled by L-trominoes if and only if $b$ is divisible by 3 .

## 数学代写|组合数学代写Combinatorial mathematics代考|Tetrominoes and Chromatic Reasoning

There exist five types of tetrominoes (Figure 3.1). We will call them O-, Z-, L-, T-, and I-tetrominoes. In this section we consider the problem of tiling a rectangle by O-, Z-, L-, and T-tetrominoes. For I-tetrominoes, the problem will be generalized and solved in the next section. In order to solve these problems we will often use “color” reasoning that was mentioned in Section 1.

Exercise 3.1. Prove that the $m \times n$ rectangle can be tiled with Otetrominoes if and only if $m, n$ are even.

Exercise 3.2. Prove that no rectangle can be tiled with Z-tetrominoes. The above exercises do not require the use of “color” reasoning. But such reasoning will be very useful for other types of tetrominoes.
Exercise 3.3. Prove that if all squares of an $a \times b$ chess-board are colored in two colors black and white (Figure 3.2), then the difference between the number of black squares and the number of white squares is equal to either 0 or $\pm 1$.

Exercise 3.4. Prove that if an $m \times n$ rectangle is colored in two colors, black and white, by columns (Figure 3.3), then the difference between the number of black squares and the number of white squares is equal to either 0 or $\pm m$.

Fxercise 3.5. Prove that if an $a \times b$ rectangle is tiled with I.tetrominoes, then the number of tiles is even.

This exercise shows that the area of a rectangle tiled by $\mathrm{L}$ tetrominoes must be divisible by 8 .

It is obvious that the $2 \times 4$ rectangle can be tiled by two Ltetrominoes (Figure 3.4). The following problem arises naturally: What other rectangles can be tiled by tetrominoes of this type? Can you solve this problem on your own? If not, try to solve the following exercises to pave the way for a successful train of thought.

# 组合数学代写

## 数学代写|组合数学代写Combinatorial mathematics代考|Solutions to Exercises

2.1. 的面积大号-tromino（见图 1.6）等于 3 。的面积4×5矩形等于 20 。因此，为了用 L-trominoes 平铺这个矩形，有必要使用203L-trominoes，这是不可能的。

2.2. 由于运动2.1， 如果一种×一种正方形可以用 Ltrominoes 平铺，然后一种能被 3 整除。这6×6正方形可以分解为2×3块（图2.3), 因此,6×6正方形可以用 L-trominoes 平铺（见图 2.2）。因此，是否有待观察3×3正方形可以用 L-trominoes 平铺。答案是不。” 事实上，所有覆盖左上角的方法如图 2.4 所示。

2.3. 如果b能被 3 整除，那么 a2×b矩形可以分解为2×3块（见图 2.6），因此，这个矩形可以用 L-trominoes 平铺。Exercise 解决方案中的推理2.1表明没有其他可能性。所以2×b矩形可以被 L-trominoes 平铺当且仅当b能被 3 整除。

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## MATLAB代写

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