Izibalo kumdlalo we-Dobble03
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Ngobusuku bethu bokugcina bomdlalo womndeni, izingane zaveza umdlalo ngentshiseko i-Dobble (uhlelo lukaHarry Potter). Ngemva kokuhlulwa imizuliswano emihlanu (ngaphandle kokuba ikhadi lami lifane nekhadi lokudlala), ngamangala ukuzwa ukuthi wonke umdlali angathola umdlalo kuyo yonke imizuliswano. Kodwa-ke, ukungakholwa kwami kwahlangatshezwa kuphela yimizuliswano eminingi elahlekile - izingane zazishesha nje.

Isizathu esanele sokubhekisisa umdlalo ngombono wezibalo. Okokuqala isimiso somdlalo: I-Dobble ingumdlalo wamakhadi olula onamakhadi ayindilinga $55$ , ngalinye libonisa izimpawu eziyisishiyagalombili ezihlukene. Wonke amakhadi aphathwa ngokushintshana, kusale kuphela ikhadi lokugcina phakathi netafula. Manje bonke abadlali kufanele baqhathanise kanyekanye izimpawu ezisekhadini nezimpawu ezisekhadini labo eliphezulu lamanje. Uma umdlali ethole uphawu olufanayo kuwo womabili amakhadi, angabeka ikhadi lakhe esitakini ngokuba oshesha ukubiza uphawu. Umdlali olahla wonke amakhadi akhe uyawina kuqala.

Kungenzeka kanjani ukuthi kunamakhadi anjalo angama $55$ akhiwe ngendlela yokuthi noma yimaphi amakhadi ama-2 abe nophawu olulodwa ncamashi afana ngalo? Ingakanani inombolo encane yezimpawu ezinjalo okufanele zisetshenziswe? Ingakanani inombolo enkulu yamakhadi anjalo?

Okokuqala, sakha lawa makhadi sisebenzisa lezi zinyathelo ezinengqondo ezilandelayo (wonke amakhadi akhiwa kamuva anendawo ahlelwa ngokulandelana okukhuphukayo): Ikhadi lokuqala kufanele libe nezimpawu ezihlukene eziyi-8, okungukuthi ukufunda:

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

Siwakha kanjani amakhadi alandelayo ukuze abe nophawu olulodwa olufana nekhadi lokuqala?:

Noma iyiphi inombolo yamakhadi anjalo isingakwazi ukwakhiwa lapha (uvele ugcwalise izindawo ngohlelo olunyukayo, uqale ngo $9$ ). Leli cala elincane alinambitheki, nokho, njengoba sinentshisekelo kusethi enenombolo encane yezimpawu (kanye nenani eliphakeme lamakhadi). Manje sicabangela uphawu lwesibili $ x_{l.2} $ lwekhadi ngalinye, okusobala ukuthi okulandelayo kufanele kusebenze kulo: $ x_{1.2} \neq x_{2.2} \neq x_{3.2} \neq \ldots \neq x_{k.2} $ . Ngakho-ke sethule $ k $ izimpawu ezintsha. Kodwa manje $ k \leq 8-1 = 7 $ , njengoba lungekho uphawu $ 7 $ $ x_{1.2},\, x_{1.3},\, x_{1.4},\, x_{1.5},\, x_{1.6},\, x_{1.7},\, x_{1.8} $ (ekhadini elingakwesokunxele) lingase lifane nophawu lwesibili lwekhadi ngalinye (ngaphandle kwalokho kuzoba nezimpawu ezimbili ezifanayo ).

Sithole ubuningi bamakhadi amasha angu-7:

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

Ngempikiswano efanayo manje sakha amamephu $7$ alandelayo (owokuqala kulawa mamephu kufanele angqubuzane nemephu yethu yokuqala, hhayi no- $1$ , ngaphandle kwalokho bekuzoba ne- $7$ ngaphambilini. thola amamephu):

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

Le mpikiswano ingaqhubeka kumakhadi alandelayo e $7$ ngendlela efanayo; inani eliphelele le $8-2 = 6$ ngaphezulu. Ngakho-ke amakhadi okugcina e $7$:

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

Uma manje ubufuna ukwakha elinye ikhadi $$\left(\begin{array}{c} 9 \\ x_{57.2} \\ x_{57.3} \\ x_{57.4} \\ x_{57.5} \\ x_{57.6} \\ x_{57.7} \\ x_{57.8} \end{array}\right)$$ nophawu olujwayelekile nekhadi lokuqala. Lokhu kusho ukuthi sakhe umkhawulo wamakhadi $1 + 8 \cdot 7 = 57$ wamakhadi. Umgomo wethu manje uwukwakha amaningi nje.

Ukuze senze lokhu, sibheka amakhadi amasha ayi-7 okuqala atholiwe bese sifinyelela esiphethweni sokuthi sidinga ngokuphelele $7 \cdot 7$ izimpawu ezintsha (ayikho ikhadi elingase libe nophawu oluyimpinda futhi uphawu ngalunye oluzonikezwa lungase lungabonakali kabili ngoba $1$ isivele kabili):

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

Lokhu kusho ukuthi sidinga ubuncane bezimpawu $8 + (7 \cdot 7) = 57$ (okungukuthi izimpawu eziningi njengamakhadi!). Manje sizama ukudlula ngale nombolo futhi sithole ukucaciswa kwedizayini yazo zonke ezinye izakhi. Ukwenza lokhu, sakha i-dobble encane kancane ephethe kuphela izimpawu $3$ ngekhadi futhi ithole njengekhadi lokuqala.

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

kanye namanye amakhadi

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

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

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

nesamba $1 + 3 \cdot 2 = 7$ amakhadi kanye $ 3 + (2 \cdot 2) = 7$ izimpawu. Ngokuzama okuncane kanye nephutha (kanye nokusebenzisa izimpawu esezinikeziwe) uthola i-dobble elandelayo:

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

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

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

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

Ingabe lokhu nakho kungatholakala ngokuhlelekile? Ukwenza lokhu, sifaka izimpawu ezisanda kunikezwa $4, 5, 6, 7$ ku-matrix yesikwele.:

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

Manje sicabanga ngamakhadi amabili okuqala (kusukela ngezimpawu zokuqala \ $4$ kanye $5$ ) imigqa yokuxhuma eqondile ezimpawu eziphansi $6$ kanye $7$:

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

Njengoba le migqa ingaphambanisi, sithola (ngokuhlela izimpawu emigqeni yokuxhuma umugqa ngomugqa) amakhadi aseduze avumelekile.:

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

Okokugcina, sicabanga ukuxhuma imigqa ngomthambeka ohlukile (kulokhu ngomthambeka $1$ ):

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

Ulayini wokuxhuma wesibili (phakathi kuka $5$ kanye no $6$ ) ushiya i-matrix emaphethelweni angakwesokudla bese ungena kabusha onqenqemeni lwesokunxele. Ngokukhetha ngobuhlakani i-gradient, siqinisekisa ukuthi imigqa yokuxhuma ayiphambanisi, kodwa futhi ukuthi imigqa yokuxhuma yangaphambilini (emile) ayiphambani. Lo mbono wokuklama ekugcineni uholela kufomula yokuklama elandelayo:

I-dobble ethi $k \in \mathbb{N} \, | \, (k-1) \text{ prim} $ ine $1+(k \cdot (k-1)) = k^2-k+1 = k + (k-1)(k-1)$ amakhadi nezimpawu. $K_x$ ethi $x \in \mathbb{N}$ kanye $0 \leq x \leq (k-1) \cdot k$ iyasebenza:

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

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

Kukhona $(k-1)\cdot k + 1 = k + (k-1)(k-1)$ izingcezu zalawa makhadi. Manje kusasele kuphela ukukhombisa:

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

Icala lokuqala: $ x_1 = 0 $
- Icala 1a: $ 0 < x_2 < k $
  - Ku $y_1 = 1$ kanye $y_2 = 1$ :
    $f(x_1, y_1) = f(0, 1) = 1$
    $f(x_2, y_2) = f(x_2, 1) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$ .
  - Ku $y_1 \neq 1$ kanye $y_2 = 1$ :
    $f(x_1, y_1) = f(0, y_1) = y_1 \neq 1$
    $f(x_2, y_2) = f(x_2, y_2) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$
  - Ku $y_1 = 1$ kanye $y_2 \neq 1$ :
    $f(x_1, y_1) = f(0, 1) = 1$
    $f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) =$
    $(k+1)(x-1) + (k-1) + y \geq (k+1)(x-1)+y > 1$
  - Ngokuba $y_1 \neq 1$ kanye $y_2 \neq 1$ ithi:
    $f(x_1, y_1) = f(0, y_1) = y_1 \leq k$
    $f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) > k$
- Icala 1b: $ x_2 \geq k $
  - Okwe $y_1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$ kanye $y_2 = 1$ sine:
    $f(x_1, y_1) = f(0, \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
  - Okwe $y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$ kanye $y_2 = 1$ ithi:
    $f(x_1, y_1) = f(0, y_1) = y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$
  - Ngokuba $y_2 \neq 1$ ithi:
    $f(x_1, y_1) = f(0, y_1) = y_1 \leq k$
    $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y_2-2) \geq (k+1)+(k-1)(y_2-2) > k $
Icala lesibili: $ 0 < x_1 < k $
- Icala 2a: $ 0 < x_2 < k $
  - Ku $y_1 = 1$ kanye $y_2 = 1$ :
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$
  - Ku $y_1 \neq 1$ kanye $y_2 = 1$ :
    $f(x_1, y_1) = f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) > 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$
  - Ku $y_1 = 1$ kanye $y_2 \neq 1$ :
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) > 1$
  - Ngokuba $y_1 \neq 1$ kanye $y_2 \neq 1$ ithi:
    $f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) \leq$
    $(k+1)+(k-1)(x_1-1)+(k-2)$
    $f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) \geq$
    $(k+1)+(k-1)((x_1+1)-1)+(y_2-2) =$
    $(k+1)+(k-1)(x_1-1) + (k-1) + (y_2-2) \geq$
    $(k+1)+(k-1)(x_1-1) + (k-1) + (2-2) \geq$
    $(k+1)+(k-1)(x_1-1) + (k-1) > (k+1)+(k-1)(x_1-1) + (k-2)$
- Icala 2b: $ x_2 \geq k $
  - Ku $y_1 = 1$ kanye $y_2 = 1$ :
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 \geq \left\lfloor \frac{k-1}{k-1} \right\rfloor + 1 = 2 > 1$
  - Ku $y_1 = 1$ kanye $y_2 \neq 1$ :
    $f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$
    $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y_2-2) \geq (k+1) + (k-1)(y_2-2) > 1$
  - Ku $y_1 \neq 1$ kanye $y_2 = 1$ :
    $f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y_1-2) \geq (k+1) + (k-1)(y_1-2) > 1$
    $f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$
  - Ngoba $y_1 \neq 1$ futhi $y_2 \neq 1$ kuyinto:
    $(k+1) + (k-1)(x_1-1) + (y_1-2) =$
    $\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$
    $+ (k+1) + (k-1)(y-2)$
    $\Leftrightarrow y_1 = (k-1)y_2 - (k-1)(x_1+1) +$
    $\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) $
    Ngoba $y_2 = x_1+1$ nge $ 2 \leq y_2 \leq k$ kuyinto
    $y_1 = 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right)$ nge $ 2 \leq y_1 \leq k$.
    Kunesixazululo esisodwa kuphela lapha $ (y_1, y_2) $.
    Ngoba siyakhetha $y^*_2=y_2-1$ njengoba value, kuyinto $y^*_1 = y_1-(k-1) < 2$.
    Ngaphezu kwalokho, for $y^*_2*=y_2+1$ ke $y^*_1 = y_1+(k-1) > k$.
3. Icala: $ x_1 \geq k $
- Icala 3a: $ x_2 \geq k $
  - Icala 3a': $m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$
    - Ku $y_1 = 1$ kanye $y_2 = 1$ :
      $f(x_1, y_1) = f(x_1, 1) = m_1$
      $f(x_2, y_2) = f(x_2, 1) = m_2 = m_1$
    - Ku $y_1 = 1$ kanye $y_2 \neq 1$ :
      $f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$
      $\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) \geq$
      $(k+1) + (k-1)(y_2-2) \geq (k+1) > k$
    - Ku $y_1 \neq 1$ kanye $y_2 = 1$ :
      Bona $y_1 = 1$ kanye $y_2 \neq 1$ .
    - Ngoba $y_1 \neq 1$ futhi $y_2 \neq 1$ kuyinto:
      $f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$
      Ngemuva kwalokho $f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$
      $L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 - L_2 = (k-1)(y_2-y_1)$
      Ngoba $y_1 \neq y_2$ kuyinto $L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$.
      Ngoba $y_1 = y_2$ kuyinto $L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$ futhi
      $\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$
      $\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$
      $x_1 = x_2 + (k-1)\cdot l$ ngokuphikisana ne $m_1 = m_2$.
  - Icala 3a'': $m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$
    - Ku $y_1 = 1$ kanye $y_2 = 1$ :
      $f(x_1, y_1) = f(x_1, 1) = m_1$
      $f(x_2, y_2) = f(x_2, 1) = m_2 \neq m_1$
    - Ku $y_1 = 1$ kanye $y_2 \neq 1$ :
      $f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$
      $\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) \geq$
      $(k+1) + (k-1)(y_2-2) \geq (k+1) > k$
    - Ku $y_1 \neq 1$ kanye $y_2 = 1$ :
      Bona $y_1 = 1$ kanye $y_2 \neq 1$ .
    - Ngoba $y_1 \neq 1$ futhi $y_2 \neq 1$ kuyinto:
      $f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$
      $f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$
      $(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$
      Ngemuva kwalokho $f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$
      $L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$
      $L_1 - L_2 = (k-1)(y_2-y_1)$
      Ngoba $y_1 \neq y_2$ kuyinto $L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$.
      Ngoba $y_1 = y_2$ kuyinto $L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$ futhi
      $\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$
      $\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$
      $y = \frac{(k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)}{m_2 - m_1}$
      Well lapho for $2 \leq y \leq k$ njalo a $l \in \mathbb{N}_0$, ukuze
      $m_2 - m_1 \mid (k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)$.
      Ubufakazi: lapho $(k-1)$ iyinhloko, yi (ngenxa ye-lemma ka-Bézout)
      $(k-1)\cdot l \equiv -\left( (3-k)(m_2-m_1) + (x_1-x_2) \right) \, \mod (m_2-m_1)$
      kuxazululeka, ngoba $\text{ggT}\left((k-1),(m_2-m_1)\right) = 1$ Ukuhlukana $-\left( (3-k)(m_2-m_1) + (x_1-x_2) \right)$.
      Khona-ke yilona kuphela ikhambi $l_1$, ngoba munye
      $l_2 = l_1 + (m_2-m_1)$ kuyinto $ y_2 = y_1 + (k-1) > k$.

Ungathola futhi ulwazi lwangemuva oluthakazelisayo esihlokweni se-dobble nezibalo lapha noma lapha . Kuskripthi esilandelayo ungabona ifomula efakazelwe ngaphambilini isebenza: Ama-Dobbles (ye- $(k-1)$ prim) angakhiqizwa ngokucindezela inkinobho.:

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