Matématika ing game Dobble03
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Ing wengi dolanan kulawarga pungkasan, bocah-bocah kanthi semangat ngetokake dolanan Dobble (edisi Harry Potter). Sawise kalah limang babak (kartuku ora katon cocog karo kertu dolanane), aku kaget ngerti yen saben pemain mesthi bisa nemokake jodhone ing saben babak. Nanging, rasa ora percayaku mung ketemu karo babak sing luwih akeh sing kalah - bocah-bocah luwih cepet.

Alesan cukup kanggo njupuk dipikir nyedhaki ing game saka titik matématika tampilan. Pisanan prinsip game: Dobble punika game kertu prasaja karo $55$ kertu babak, saben nuduhake wolung simbol beda. Kabeh kertu sing urusan ing siji, ninggalake mung kertu pungkasan ing tengah meja. Saiki kabeh pemain kudu bebarengan mbandhingaké simbol ing kertu karo simbol ing kertu ndhuwur saiki. Yen pemain wis ketemu simbol padha ing loro SIM, kang bisa nyelehake kertu ing tumpukan dening paling cepet kanggo jeneng simbol. Pamuter sing mbuwang kabeh kertu pisanan menang.

Kepiye carane bisa ana $55$ kertu sing dibangun kanthi cara sing 2 kertu duwe siji simbol sing padha? Apa nomer minimal simbol kasebut sing kudu digunakake? Apa nomer maksimum kertu kuwi?

Kaping pisanan, kita mbangun kertu kasebut kanthi nggunakake langkah-langkah logis ing ngisor iki (kabeh kertu sing dibangun sabanjure duwe properti sing diurutake kanthi urutan munggah): Kertu pisanan kudu duwe 8 simbol sing beda, yaiku maca.:

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

Kepriye carane nyusun kertu-kertu ing ngisor iki supaya nduweni persis siji simbol sing padha karo kertu sing pertama?:

Sembarang nomer kertu kuwi wis bisa dibangun ing kene (sampeyan mung ngisi panggonan ing urutan munggah, miwiti karo $9$ ). Kasus sepele iki ora menarik, Nanging, amarga kita kasengsem ing pesawat kanthi jumlah minimal simbol (lan jumlah maksimum kertu). Saiki kita nimbang simbol kaloro $ x_{l.2} $ saben kertu, sing jelas kudu ditrapake: $ x_{1.2} \neq x_{2.2} \neq x_{3.2} \neq \ldots \neq x_{k.2} $ . Mulane kita kudu ngenalaken $ k $ simbol anyar. Nanging saiki $ k \leq 8-1 = 7 $ , amarga ora ana simbol \ $ 7 $ $ x_{1.2},\, x_{1.3},\, x_{1.4},\, x_{1.5},\, x_{1.6},\, x_{1.7},\, x_{1.8} $ (saka kertu paling kiwa) bisa cocog karo simbol kaloro saben kertu liyane (yen ora ana rong simbol sing padha. ).

Kita wis ketemu maksimum iki 7 kertu anyar:

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

Kanthi argumentasi sing padha, saiki kita mbangun peta $7$ sabanjure (pisanan peta iki kudu tabrakan karo peta wiwitan, lan ora karo $1$ , yen ora bakal karo $7$ sadurunge. ketemu peta):

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

Argumen iki bisa diterusake kanggo kertu $7$ sabanjure kanthi cara sing padha; total $8-2 = 6$ kaping luwih akeh. Mulane, kertu $7$ pungkasan yaiku:

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

Yen saiki sampeyan pengin nggawe kertu liyane $$\left(\begin{array}{c} 9 \\ x_{57.2} \\ x_{57.3} \\ x_{57.4} \\ x_{57.5} \\ x_{57.6} \\ x_{57.7} \\ x_{57.8} \end{array}\right)$$ duwe simbol umum karo kertu wiwitan. Iki tegese kita wis dibangun maksimum $1 + 8 \cdot 7 = 57$ kertu. Tujuan kita saiki yaiku mbangun kaya akeh.

Kanggo nindakake iki, kita katon ing pisanan 7 kertu anyar ketemu lan teka menyang kesimpulan sing pancene kudu $7 \cdot 7$ simbol anyar (ora ana kertu bisa duwe simbol duplikat lan saben simbol bakal dianugerahi ora katon kaping pindho amarga $1$ wis pindho).:

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

Iki tegese kita kudu minimal $8 + (7 \cdot 7) = 57$ simbol (i.e. minangka akeh simbol minangka kertu!). Kita saiki nyoba kanggo njaluk dening nomer iki lan nemokake specification desain kanggo kabeh unsur liyane. Kanggo nindakake iki, kita mbangun dobble sing rada cilik sing mung nggawa simbol $3$ saben kertu lan dadi kertu wiwitan.

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

lan kertu liyane

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

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

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

karo total $1 + 3 \cdot 2 = 7$ kertu lan $ 3 + (2 \cdot 2) = 7$ simbol. Kanthi nyoba lan kesalahan sethithik (lan nggunakake simbol sing wis ditemtokake) sampeyan entuk dobble ing ngisor iki:

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

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

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

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

Apa iki uga bisa ditemokake kanthi sistematis? Kanggo nindakake iki, kita ngetik simbol sing mentas ditugasake $4, 5, 6, 7$ ing matriks persegi.:

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

Saiki kita mbayangno kanggo rong kertu pisanan (wiwit simbol wiwitan \ $4$ lan $5$ ) garis nyambungake vertikal menyang simbol ngisor $6$ lan $7$:

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

Wiwit garis-garis iki ora intersect, kita njaluk (kanthi ngrancang simbol ing garis nyambungake baris dening baris) kertu sah paling cedhak.:

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

Pungkasan, kita mbayangno garis sing nyambungake kanthi slope sing beda (ing kasus iki karo slope $1$ ):

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

Garis panyambung kapindho (antarane $5$ lan $6$ ) ninggalake matriks ing pinggir tengen lan mlebu maneh ing pinggir kiwa. Kanthi pinter milih gradien, kita mesthekake yen garis nyambungake ora intersect saben liyane, nanging uga sing sadurunge (vertikal) nyambungake garis ora intersect. Ide desain iki pungkasane ndadékaké rumus desain ing ngisor iki:

Dobel kanthi $k \in \mathbb{N} \, | \, (k-1) \text{ prim} $ nduweni $1+(k \cdot (k-1)) = k^2-k+1 = k + (k-1)(k-1)$ kertu lan simbol. Kanggo peta $K_x$ nganggo $x \in \mathbb{N}$ lan $0 \leq x \leq (k-1) \cdot k$ ditrapake:

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

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

Ana $(k-1)\cdot k + 1 = k + (k-1)(k-1)$ bêsik kertu iki. Saiki mung tetep kanggo nuduhake:

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

Kasus kaping 1: $ x_1 = 0 $
- Kasus 1a: $ 0 < x_2 < k $
  - Kanggo $y_1 = 1$ lan $y_2 = 1$ duwe:
    $f(x_1, y_1) = f(0, 1) = 1$
    $f(x_2, y_2) = f(x_2, 1) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$ .
  - Kanggo $y_1 \neq 1$ lan $y_2 = 1$ duwe:
    $f(x_1, y_1) = f(0, y_1) = y_1 \neq 1$
    $f(x_2, y_2) = f(x_2, y_2) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$
  - Kanggo $y_1 = 1$ lan $y_2 \neq 1$ duwe:
    $f(x_1, y_1) = f(0, 1) = 1$
    $f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) =$
    $(k+1)(x-1) + (k-1) + y \geq (k+1)(x-1)+y > 1$
  - Kanggo $y_1 \neq 1$ lan $y_2 \neq 1$ yaiku:
    $f(x_1, y_1) = f(0, y_1) = y_1 \leq k$
    $f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) > k$
- Kasus 1b: $ x_2 \geq k $
  - Kanggo $y_1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$ lan $y_2 = 1$ kita duwe:
    $f(x_1, y_1) = f(0, \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
  - Kanggo $y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$ lan $y_2 = 1$ yaiku:
    $f(x_1, y_1) = f(0, y_1) = y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
  - Kanggo $y_2 \neq 1$ yaiku:
    $f(x_1, y_1) = f(0, y_1) = y_1 \leq k$
    $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y_2-2) \geq (k+1)+(k-1)(y_2-2) > k $
Kasus kaping 2: $ 0 < x_1 < k $
- Kasus 2a: $ 0 < x_2 < k $
  - Kanggo $y_1 = 1$ lan $y_2 = 1$ duwe:
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$
  - Kanggo $y_1 \neq 1$ lan $y_2 = 1$ duwe:
    $f(x_1, y_1) = f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) > 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$
  - Kanggo $y_1 = 1$ lan $y_2 \neq 1$ duwe:
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) > 1$
  - Kanggo $y_1 \neq 1$ lan $y_2 \neq 1$ yaiku:
    $f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) \leq$
    $(k+1)+(k-1)(x_1-1)+(k-2)$
    $f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) \geq$
    $(k+1)+(k-1)((x_1+1)-1)+(y_2-2) =$
    $(k+1)+(k-1)(x_1-1) + (k-1) + (y_2-2) \geq$
    $(k+1)+(k-1)(x_1-1) + (k-1) + (2-2) \geq$
    $(k+1)+(k-1)(x_1-1) + (k-1) > (k+1)+(k-1)(x_1-1) + (k-2)$
- Kasus 2b: $ x_2 \geq k $
  - Kanggo $y_1 = 1$ lan $y_2 = 1$ duwe:
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 \geq \left\lfloor \frac{k-1}{k-1} \right\rfloor + 1 = 2 > 1$
  - Kanggo $y_1 = 1$ lan $y_2 \neq 1$ duwe:
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y_2-2) \geq (k+1) + (k-1)(y_2-2) > 1$
  - Kanggo $y_1 \neq 1$ lan $y_2 = 1$ duwe:
    $f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y_1-2) \geq (k+1) + (k-1)(y_1-2) > 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$
  - Kanggo $y_1 \neq 1$ lan $y_2 \neq 1$ punika:
    $(k+1) + (k-1)(x_1-1) + (y_1-2) =$
    $\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y-2)$
    $\Leftrightarrow y_1 = (k-1)y_2 - (k-1)(x_1+1) +$
    $\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) $
    Kanggo $y_2 = x_1+1$ karo $ 2 \leq y_2 \leq k$ punika
    $y_1 = 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right)$ karo $ 2 \leq y_1 \leq k$.
    Mung ana siji solusi ing kene $ (y_1, y_2) $.
    Amarga kita milih $y^*_2=y_2-1$ minangka nilai, punika $y^*_1 = y_1-(k-1) < 2$.
    Kajaba iku, kanggo $y^*_2*=y_2+1$ banjur $y^*_1 = y_1+(k-1) > k$.
3. Kasus: $ x_1 \geq k $
- Kasus 3a: $ x_2 \geq k $
  - Kasus 3a: $m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$
    - Kanggo $y_1 = 1$ lan $y_2 = 1$ duwe:
      $f(x_1, y_1) = f(x_1, 1) = m_1$
      $f(x_2, y_2) = f(x_2, 1) = m_2 = m_1$
    - Kanggo $y_1 = 1$ lan $y_2 \neq 1$ duwe:
      $f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$
      $\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) \geq$
      $(k+1) + (k-1)(y_2-2) \geq (k+1) > k$
    - Kanggo $y_1 \neq 1$ lan $y_2 = 1$ duwe:
      Waca $y_1 = 1$ lan $y_2 \neq 1$ .
    - Kanggo $y_1 \neq 1$ lan $y_2 \neq 1$ punika:
      $f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$
      Banjur $f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$
      $L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 - L_2 = (k-1)(y_2-y_1)$
      Kanggo $y_1 \neq y_2$ punika $L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$.
      Kanggo $y_1 = y_2$ punika $L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$ lan
      $\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$
      $\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$
      $x_1 = x_2 + (k-1)\cdot l$ ing kontradiksi kanggo $m_1 = m_2$.
  - Kasus 3a: $m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$
    - Kanggo $y_1 = 1$ lan $y_2 = 1$ duwe:
      $f(x_1, y_1) = f(x_1, 1) = m_1$
      $f(x_2, y_2) = f(x_2, 1) = m_2 \neq m_1$
    - Kanggo $y_1 = 1$ lan $y_2 \neq 1$ duwe:
      $f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$
      $\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) \geq$
      $(k+1) + (k-1)(y_2-2) \geq (k+1) > k$
    - Kanggo $y_1 \neq 1$ lan $y_2 = 1$ duwe:
      Waca $y_1 = 1$ lan $y_2 \neq 1$ .
    - Kanggo $y_1 \neq 1$ lan $y_2 \neq 1$ punika:
      $f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$
      Banjur $f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$
      $L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 - L_2 = (k-1)(y_2-y_1)$
      Kanggo $y_1 \neq y_2$ punika $L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$.
      Kanggo $y_1 = y_2$ punika $L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$ lan
      $\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$
      $\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$
      $y = \frac{(k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)}{m_2 - m_1}$
      Uga ana kanggo $2 \leq y \leq k$ tansah a $l \in \mathbb{N}_0$, supaya
      $m_2 - m_1 \mid (k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)$.
      Bukti: ana $(k-1)$ punika prima, punika (amarga lemma Bézout)
      $(k-1)\cdot l \equiv -\left( (3-k)(m_2-m_1) + (x_1-x_2) \right) \, \mod (m_2-m_1)$
      solvable, amarga $\text{ggT}\left((k-1),(m_2-m_1)\right) = 1$ Pisah $-\left( (3-k)(m_2-m_1) + (x_1-x_2) \right)$.
      Banjur iki mung solusi $l_1$, amarga kanggo siji
      $l_2 = l_1 + (m_2-m_1)$ punika $ y_2 = y_1 + (k-1) > k$.

Sampeyan uga bisa nemokake informasi latar mburi menarik ing topik dobble lan matématika kene utawa kene . Ing script ing ngisor iki sampeyan bisa ndeleng rumus sadurunge buktiaken ing tumindak: Dobbles (kanggo $(k-1)$ prim) bisa kui kanthi mencet tombol:

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

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