游戏中的数学 Dobble

在最后一个家庭之夜,孩子们热情地将游戏Dobble (在哈利波特版中)带到了餐桌上。 在第 5 轮失败后(我的牌与扑克牌没有明显的命中),令我惊讶的是,每个玩家总能在每一轮中找到命中。 但我的怀疑只有在进一步输掉几圈后才被承认——孩子们只是更快了。


有足够的理由从数学的角度仔细研究游戏。 首先游戏原理: Dobble 是一个简单的纸牌游戏,有\(55\)圆牌,每张牌显示八个不同的符号。 所有的牌都是轮流发的,只剩下最后一张牌在桌子中间。 现在所有玩家都必须同时将卡上的符号与他们当前顶卡上的符号进行比较。 如果玩家在两张牌上找到了相同的符号,他可以通过以最快的速度命名符号来将他的牌放入堆叠中。 首先丢弃所有牌的玩家获胜。

怎么会有\(55\)这样的牌以任何 2 张牌只有一个共同符号的方式构造? 必须使用的此类符号的最小数量是多少? 此类卡的最大数量是多少?

首先,我们使用以下逻辑步骤构建这些卡片(所有随后构建的卡片都具有按升序排序的属性):第一张卡片必须有 8 个不同的符号,即读取:

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

我们现在以这样的方式构造以下卡片,使它们与第一张卡片具有完全相同的一个符号:

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

此处已经可以构建任意数量的此类卡片(您只需按升序填写位置,从\(9\)开始)。 然而,这种简单的情况是无趣的,因为我们对具有最少符号数量(和最大数量卡片)的集合感兴趣。 我们现在考虑每张卡片的第二个符号\( x_{l.2} \) ,显然以下必须适用: \( x_{1.2} \neq x_{2.2} \neq x_{3.2} \neq \ldots \neq x_{k.2} \) 。 因此,我们必须引入\( k \)新符号。 但是现在\( k \leq 8-1 = 7 \) ,因为没有一个\( 7 \)符号\( x_{1.2},\, x_{1.3},\, x_{1.4},\, x_{1.5},\, x_{1.6},\, x_{1.7},\, x_{1.8} \) (最左边的牌)可以匹配其他每张牌的第二个符号(否则会有两个相同的符号)。

我们最多找到了这 7 张新卡:

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

使用相同的参数,我们现在构造下一个\(7\)映射(这些映射中的第一个必须与我们的起始映射碰撞,而不是与\(1\) ,否则它将与之前的\(7\)找到地图):

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

对于接下来的\(7\)牌,这个论点也可以继续; 总共\(8-2 = 6\)次。 最后的\(7\)卡相应地:

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

如果您要添加另一张卡片$$\left(\begin{array}{c} 9 \\ x_{57.2} \\ x_{57.3} \\ x_{57.4} \\ x_{57.5} \\ x_{57.6} \\ x_{57.7} \\ x_{57.8} \end{array}\right)$$将失败,因为这张牌与起始牌不共享符号。 我们已经构建了最多\(1 + 8 \cdot 7 = 57\)映射。 我们现在的目标是至少建造尽可能多的建筑。

为此,我们查看找到的前 7 张新牌,并得出结论,我们绝对需要在此处使用\(7 \cdot 7\)新符号(没有牌可能有两次符号,并且每个要分配的符号不得出现两次,因为其中\(1\)已经是两倍):

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

所以我们需要最少的\(8 + (7 \cdot 7) = 57\)符号(所以符号和卡片一样多!)。 我们现在正试图解决这个数字,并为所有其他元素找到一个构造规则。 为此,我们构建了一个稍微小一点的 dobble,每张牌只有\(3\)个符号,并将其作为起始牌接收

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

和其他卡

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

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

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

总共有\(1 + 3 \cdot 2 = 7\)卡片和\( 3 + (2 \cdot 2) = 7\)符号。 通过一些试验和错误(并使用已经分配的符号),您将获得以下 dobble:

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

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

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

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

这也能系统地找到吗? 为此,我们在方阵中输入新分配的符号\(4, 5, 6, 7\):

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

现在我们想象前两张牌(从起始符号 \ \(4\)\(5\)开始)垂直连接线到下面的符号\(6\)\(7\):

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

由于这些线不相交,我们得到(通过在连接线上逐行绘制符号)最接近的有效卡片:

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

最后,我们想象连接具有不同斜率的线(在这种情况下是斜率\(1\) ):

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

第二条连接线(在\(5\)\(6\)之间)在右边缘离开矩阵并在左边缘重新进入。 通过巧妙地选择坡度,我们一方面确保连接线不相交,而且确保之前的(垂直)连接线不相交。 这种设计思路最终导致了如下设计公式:

\(k \in \mathbb{N} \, | \, (k-1) \text{ prim} \)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:

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

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

这些卡片有\((k-1)\cdot k + 1 = k + (k-1)(k-1)\)块。 现在只剩下展示:

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

  • 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
          \(l_2 = l_1 + (m_2-m_1)\)\( y_2 = y_1 + (k-1) > k\).

您还可以在此处此处找到有关 Dobble 和数学主题的有趣背景信息。 在以下脚本中,您可以看到先前证明的公式在起作用: Dobbles (for \((k-1)\) prim) 可以通过按下按钮生成:

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

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