# 游戏中的数学 Dobble0322

$$\left(\begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ x_{1.2} \\ x_{1.3} \\ x_{1.4} \\ x_{1.5} \\ x_{1.6} \\ x_{1.7} \\ x_{1.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{2.2} \\ x_{2.3} \\ x_{2.4} \\ x_{2.5} \\ x_{2.6} \\ x_{2.7} \\ x_{2.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{3.2} \\ x_{3.3} \\ x_{3.4} \\ x_{3.5} \\ x_{3.6} \\ x_{3.7} \\ x_{3.8} \end{array}\right), \ldots, \left(\begin{array}{c} 1 \\ x_{k.2} \\ x_{k.3} \\ x_{k.4} \\ x_{k.5} \\ x_{k.6} \\ x_{k.7} \\ x_{k.8} \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ x_{1.2} \\ x_{1.3} \\ x_{1.4} \\ x_{1.5} \\ x_{1.6} \\ x_{1.7} \\ x_{1.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{2.2} \\ x_{2.3} \\ x_{2.4} \\ x_{2.5} \\ x_{2.6} \\ x_{2.7} \\ x_{2.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{3.2} \\ x_{3.3} \\ x_{3.4} \\ x_{3.5} \\ x_{3.6} \\ x_{3.7} \\ x_{3.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{4.2} \\ x_{4.3} \\ x_{4.4} \\ x_{4.5} \\ x_{4.6} \\ x_{4.7} \\ x_{4.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{5.2} \\ x_{5.3} \\ x_{5.4} \\ x_{5.5} \\ x_{5.6} \\ x_{5.7} \\ x_{5.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{6.2} \\ x_{6.3} \\ x_{6.4} \\ x_{6.5} \\ x_{6.6} \\ x_{6.7} \\ x_{6.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{7.2} \\ x_{7.3} \\ x_{7.4} \\ x_{7.5} \\ x_{7.6} \\ x_{7.7} \\ x_{7.8} \end{array}\right)$$

$$\left(\begin{array}{c} 2 \\ x_{8.2} \\ x_{8.3} \\ x_{8.4} \\ x_{8.5} \\ x_{8.6} \\ x_{8.7} \\ x_{8.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{9.2} \\ x_{9.3} \\ x_{9.4} \\ x_{9.5} \\ x_{9.6} \\ x_{9.7} \\ x_{9.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{10.2} \\ x_{10.3} \\ x_{10.4} \\ x_{10.5} \\ x_{10.6} \\ x_{10.7} \\ x_{10.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{11.2} \\ x_{11.3} \\ x_{11.4} \\ x_{11.5} \\ x_{11.6} \\ x_{11.7} \\ x_{11.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{12.2} \\ x_{12.3} \\ x_{12.4} \\ x_{12.5} \\ x_{12.6} \\ x_{12.7} \\ x_{12.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{13.2} \\ x_{13.3} \\ x_{13.4} \\ x_{13.5} \\ x_{13.6} \\ x_{13.7} \\ x_{13.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{14.2} \\ x_{14.3} \\ x_{14.4} \\ x_{14.5} \\ x_{14.6} \\ x_{14.7} \\ x_{14.8} \end{array}\right)$$

$$\left(\begin{array}{c} 8 \\ x_{50.2} \\ x_{50.3} \\ x_{50.4} \\ x_{50.5} \\ x_{50.6} \\ x_{50.7} \\ x_{50.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{51.2} \\ x_{51.3} \\ x_{51.4} \\ x_{51.5} \\ x_{51.6} \\ x_{51.7} \\ x_{51.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{52.2} \\ x_{52.3} \\ x_{52.4} \\ x_{52.5} \\ x_{52.6} \\ x_{52.7} \\ x_{52.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{53.2} \\ x_{53.3} \\ x_{53.4} \\ x_{53.5} \\ x_{53.6} \\ x_{53.7} \\ x_{53.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{54.2} \\ x_{54.3} \\ x_{54.4} \\ x_{54.5} \\ x_{54.6} \\ x_{54.7} \\ x_{54.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{55.2} \\ x_{55.3} \\ x_{55.4} \\ x_{55.5} \\ x_{55.6} \\ x_{55.7} \\ x_{55.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{56.2} \\ x_{56.3} \\ x_{56.4} \\ x_{56.5} \\ x_{56.6} \\ x_{56.7} \\ x_{56.8} \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ 9 \\ 10 \\ 11 \\ 12 \\ 13 \\ 14 \\ 15 \end{array}\right), \left(\begin{array}{c} 1 \\ 16 \\ 17 \\ 18 \\ 19 \\ 20 \\ 21 \\ 22 \end{array}\right), \left(\begin{array}{c} 1 \\ 23 \\ 24 \\ 25 \\ 26 \\ 27 \\ 28 \\ 29 \end{array}\right), \left(\begin{array}{c} 1 \\ 30 \\ 31 \\ 32 \\ 33 \\ 34 \\ 35 \\ 36 \end{array}\right), \left(\begin{array}{c} 1 \\ 37 \\ 38 \\ 39 \\ 40 \\ 41 \\ 42 \\ 43 \end{array}\right), \left(\begin{array}{c} 1 \\ 44 \\ 45 \\ 46 \\ 47 \\ 48 \\ 49 \\ 50 \end{array}\right), \left(\begin{array}{c} 1 \\ 51 \\ 52 \\ 53 \\ 54 \\ 55 \\ 56 \\ 57 \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ 4 \\ 5 \end{array}\right), \left(\begin{array}{c} 1 \\ 6 \\ 7 \end{array}\right)$$

$$\left(\begin{array}{c} 2 \\ x_{3.2} \\ x_{3.3} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{4.2} \\ x_{4.3} \end{array}\right)$$

$$\left(\begin{array}{c} 3 \\ x_{5.2} \\ x_{5.3} \end{array}\right), \left(\begin{array}{c} 3 \\ x_{6.2} \\ x_{6.3} \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right)$$

$$\left(\begin{array}{c} 1 \\ 4 \\ 5 \end{array}\right), \left(\begin{array}{c} 1 \\ 6 \\ 7 \end{array}\right)$$

$$\left(\begin{array}{c} 2 \\ 4 \\ 6 \end{array}\right), \left(\begin{array}{c} 2 \\ 5 \\ 7 \end{array}\right)$$

$$\left(\begin{array}{c} 3 \\ 4 \\ 7 \end{array}\right), \left(\begin{array}{c} 3 \\ 5 \\ 6 \end{array}\right)$$

$$\begin{array}{ccc} 4 & & 5 \\ & & \\ 6 & & 7\end{array}$$

$$\begin{array}{ccc} 4 & & 5 \\ \vdots & & \vdots \\ 6 & & 7\end{array}$$

$$\left(\begin{array}{c} 2 \\ 4 \\ 6 \end{array}\right), \left(\begin{array}{c} 2 \\ 5 \\ 7 \end{array}\right)$$

$$\begin{array}{ccccc} & 4 & & 5 & \\ \ddots & & \ddots & & \ddots \\ & 6 & & 7 &\end{array}$$

$$k \in \mathbb{N} \, | \, (k-1) \text{ prim}$$的双摆有$$1+(k \cdot (k-1)) = k^2-k+1 = k + (k-1)(k-1)$$卡片和符号。 对于具有$$x \in \mathbb{N}$$$$0 \leq x \leq (k-1) \cdot k$$的映射$$K_x$$ ) 适用:

$$K_x = \left(\begin{array}{c} f(x,1) \\ f(x,2) \\ \vdots \\ f(x,k) \end{array}\right), \,\, m = \left\lfloor \frac{x-1}{k-1} \right\rfloor + 1,$$

$$f(x,y) = \left\{\begin{array}{ll} y & \text{falls } x = 0 \\ \lfloor \frac{x-1}{k-1} \rfloor + 1, &\text{sonst falls } y = 1 \\ (k+1) + (k-1)(x-1) + (y-2), & \text{sonst falls } 0 < x < k \\ \left( \left((m-1)(k-1)+x\right)-1+ \left( (m-2)(y-2) \right) \right) \% (k-1) &\text{sonst} \\ + (k+1) + (k-1)(y-2)&\end{array}\right.$$

$$\forall x_1 < x_2 \in \{ 1, \ldots, k+(k-1)(k-1) \} \, \exists \, ! \, y_1, y_2 \in \{ 1, \ldots, k \}: f(x_1, y_1) = f(x_2, y_2)$$

• 第一种情况: $$x_1 = 0$$
• 案例1a: $$0 < x_2 < k$$
• 对于$$y_1 = 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(0, 1) = 1$$
$$f(x_2, y_2) = f(x_2, 1) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$$
• 对于$$y_1 \neq 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(0, y_1) = y_1 \neq 1$$
$$f(x_2, y_2) = f(x_2, y_2) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$$
• 对于$$y_1 = 1$$$$y_2 \neq 1$$有：
$$f(x_1, y_1) = f(0, 1) = 1$$
$$f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) =$$
$$(k+1)(x-1) + (k-1) + y \geq (k+1)(x-1)+y > 1$$
• 对于$$y_1 \neq 1$$$$y_2 \neq 1$$是：
$$f(x_1, y_1) = f(0, y_1) = y_1 \leq k$$
$$f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) > k$$
• 案例1b: $$x_2 \geq k$$
• 对于$$y_1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$$$y_2 = 1$$我们有：
$$f(x_1, y_1) = f(0, \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
• 对于$$y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$$$y_2 = 1$$是：
$$f(x_1, y_1) = f(0, y_1) = y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
• 对于$$y_2 \neq 1$$是：
$$f(x_1, y_1) = f(0, y_1) = y_1 \leq k$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y_2-2) \geq (k+1)+(k-1)(y_2-2) > k$$
• 第二例: $$0 < x_1 < k$$
• 案例2a: $$0 < x_2 < k$$
• 对于$$y_1 = 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$$
• 对于$$y_1 \neq 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) > 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$$
• 对于$$y_1 = 1$$$$y_2 \neq 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) > 1$$
• 对于$$y_1 \neq 1$$$$y_2 \neq 1$$是：
$$f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) \leq$$
$$(k+1)+(k-1)(x_1-1)+(k-2)$$
$$f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) \geq$$
$$(k+1)+(k-1)((x_1+1)-1)+(y_2-2) =$$
$$(k+1)+(k-1)(x_1-1) + (k-1) + (y_2-2) \geq$$
$$(k+1)+(k-1)(x_1-1) + (k-1) + (2-2) \geq$$
$$(k+1)+(k-1)(x_1-1) + (k-1) > (k+1)+(k-1)(x_1-1) + (k-2)$$
• 案例2b: $$x_2 \geq k$$
• 对于$$y_1 = 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 \geq \left\lfloor \frac{k-1}{k-1} \right\rfloor + 1 = 2 > 1$$
• 对于$$y_1 = 1$$$$y_2 \neq 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y_2-2) \geq (k+1) + (k-1)(y_2-2) > 1$$
• 对于$$y_1 \neq 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y_1-2) \geq (k+1) + (k-1)(y_1-2) > 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$$
• 为了 $$y_1 \neq 1$$$$y_2 \neq 1$$ 是:
$$(k+1) + (k-1)(x_1-1) + (y_1-2) =$$
$$\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y-2)$$
$$\Leftrightarrow y_1 = (k-1)y_2 - (k-1)(x_1+1) +$$
$$\left( 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right) \right)$$
为了 $$y_2 = x_1+1$$$$2 \leq y_2 \leq k$$
$$y_1 = 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right)$$$$2 \leq y_1 \leq k$$.
这里只有一种解决方案 $$(y_1, y_2)$$.
因为我们选择 $$y^*_2=y_2-1$$ 作为值，是 $$y^*_1 = y_1-(k-1) < 2$$.
此外，对于 $$y^*_2*=y_2+1$$ 然后 $$y^*_1 = y_1+(k-1) > k$$.
• 3.案例: $$x_1 \geq k$$
• 案例 3a: $$x_2 \geq k$$
• 案例 3a': $$m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$$
• 对于$$y_1 = 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = m_1$$
$$f(x_2, y_2) = f(x_2, 1) = m_2 = m_1$$
• 对于$$y_1 = 1$$$$y_2 \neq 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$$
$$\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) \geq$$
$$(k+1) + (k-1)(y_2-2) \geq (k+1) > k$$
• 对于$$y_1 \neq 1$$$$y_2 = 1$$有：
参见$$y_1 = 1$$$$y_2 \neq 1$$
• 为了 $$y_1 \neq 1$$$$y_2 \neq 1$$ 是:
$$f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$$
然后 $$f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$$
$$L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 - L_2 = (k-1)(y_2-y_1)$$
为了 $$y_1 \neq y_2$$$$L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$$.
为了 $$y_1 = y_2$$$$L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$$
$$\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$$
$$\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$$
$$x_1 = x_2 + (k-1)\cdot l$$$$m_1 = m_2$$.
• 案例3a'': $$m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$$
• 对于$$y_1 = 1$$$$y_2 = 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = m_1$$
$$f(x_2, y_2) = f(x_2, 1) = m_2 \neq m_1$$
• 对于$$y_1 = 1$$$$y_2 \neq 1$$有：
$$f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$$
$$\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) \geq$$
$$(k+1) + (k-1)(y_2-2) \geq (k+1) > k$$
• 对于$$y_1 \neq 1$$$$y_2 = 1$$有：
参见$$y_1 = 1$$$$y_2 \neq 1$$
• 为了 $$y_1 \neq 1$$$$y_2 \neq 1$$ 是:
$$f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$$
然后 $$f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$$
$$L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 - L_2 = (k-1)(y_2-y_1)$$
为了 $$y_1 \neq y_2$$$$L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$$.
为了 $$y_1 = y_2$$$$L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$$
$$\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$$
$$\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$$
$$y = \frac{(k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)}{m_2 - m_1}$$
好吧 $$2 \leq y \leq k$$ 总是一个 $$l \in \mathbb{N}_0$$, 以便
$$m_2 - m_1 \mid (k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)$$.
证明：有 $$(k-1)$$ 是素数，是（由于 Bézout 引理）
$$(k-1)\cdot l \equiv -\left( (3-k)(m_2-m_1) + (x_1-x_2) \right) \, \mod (m_2-m_1)$$
可解，因为 $$\text{ggT}\left((k-1),(m_2-m_1)\right) = 1$$ 分裂 $$-\left( (3-k)(m_2-m_1) + (x_1-x_2) \right)$$.
那么这是唯一的解决方案 $$l_1$$, 因为对于一个
$$l_2 = l_1 + (m_2-m_1)$$$$y_2 = y_1 + (k-1) > k$$.

See the Pen DOBBLE CREATOR by David Vielhuber (@vielhuber) on CodePen.