Matématika ing game Dobble

Ing wayah sore kulawarga pungkasan, game Dobble (ing Edisi Harry Potter) kanthi antusias digawa menyang meja dening bocah-bocah. Sawise 5. babak ilang (karo ora katon hit saka kertu karo kertu playing) Aku iki marang, kaget, sing saben pemain tansah bisa nemokake hit ing saben babak. Nanging rasa ora percayaku mung diakoni kanthi puteran sing ilang - bocah-bocah mung 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.

Carane bisa sing ana \(55\) kertu sing padha dibangun ing kuwi cara sing sembarang 2 SIM duwe persis siji simbol ing umum? 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.:

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

Saiki kita mbangun kertu ing ngisor iki kanthi cara sing padha duwe persis siji simbol sing padha karo kertu pisanan:

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

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):

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

Argumentasi iki uga bisa diterusake kanggo kertu \(7\) sabanjure; Gunggunge \(8-2 = 6\) kaping pindho. Kertu \(7\) pungkasan sing cocog:

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

Yen sampeyan nambahake kertu liyane $$\left(\begin{array}{c} 9 \\ x_{57.2} \\ x_{57.3} \\ x_{57.4} \\ x_{57.5} \\ x_{57.6} \\ x_{57.7} \\ x_{57.8} \end{array}\right)$$ bakal gagal amarga kertu iki ora nuduhake simbol karo kertu wiwitan. Kita wis mbangun maksimum \(1 + 8 \cdot 7 = 57\) maps. Tujuane saiki yaiku mbangun paling ora akeh.

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

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

Dadi kita kudu minimal \(8 + (7 \cdot 7) = 57\) simbol (supaya akeh simbol minangka kertu!). Kita saiki nyoba kanggo njaluk dening nomer iki lan nemokake aturan construction kanggo kabeh unsur liyane. Kanggo nindakake iki, kita mbangun dobble rada cilik sing mung nduweni simbol \(3\) saben kertu lan nampa minangka 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 skilfully milih slope, kita mesthekake ing tangan siji sing garis nyambungake ora intersect karo saben liyane, nanging uga sing sadurunge (vertikal) garis nyambungake 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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