# IMathematika kumdlalo weDobble0322

Ngobusuku bokugqibela bentsapho, umdlalo we- Dobble (kwi-Harry Potter Edition) waziswa ngenzondelelo etafileni ngabantwana. Emva kwe-5th elahlekileyo ngeenxa zonke (ngaphandle kwe-hit ebonakalayo yekhadi lam kunye nekhadi lokudlala) Ndaxelelwa, ngokumangala kwam, ukuba wonke umdlali unokuhlala efumana i-hit kuwo wonke umjikelezo. Kodwa ukungakholelwa kwam kwavunywa kuphela ngokulahleka okungaphezulu - abantwana babekhawuleza.

Isizathu esaneleyo sokujonga ngakumbi umdlalo ngokwembono yezibalo. Okokuqala umgaqo womdlalo: IDobble ngumdlalo wekhadi olula onamakhadi arawndi $$55$$ , ngalinye libonisa iisimboli ezisibhozo ezahlukeneyo. Onke amakhadi aphathwa ngokulandelelana, kushiya kuphela ikhadi lokugqibela phakathi kwetafile. Ngoku bonke abadlali kufuneka ngaxeshanye bathelekise iisimboli ekhadini kunye neempawu kwikhadi labo eliphezulu langoku. Ukuba umdlali ufumene isimboli efanayo kuwo omabini amakhadi, unokubeka ikhadi lakhe kwisitaki ngokuba ngoyena ukhawulezayo ukubiza isimboli. Umdlali olahla onke amakhadi abo kuqala uphumelele.

Kwenzeka njani ukuba kukho $$55$$ amakhadi anjalo akhiwe ngendlela yokuba nawaphi na amakhadi ama-2 abe nesimboli enye kanye efanayo? Leliphi inani elincinane leesimboli ekufuneka zisetyenziswe? Lithini inani eliphezulu lamakhadi anjalo?

Okokuqala, sakha la makhadi sisebenzisa la manyathelo alandelayo anengqiqo (onke amakhadi akhiwe emva koko anepropathi ehlelwe ngokomyalelo onyukayo): Ikhadi lokuqala kufuneka libe neesimboli ezisi-8 ezahlukeneyo, o.k.:

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

Ngoku sakha la makhadi alandelayo ngendlela yokuba abe nesimboli enye kanye efanayo nekhadi lokuqala:

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

Naliphi na inani lalo makhadi asenokwakhiwa apha (ugcwalisa ngokulula iindawo ngolandelelwano olunyukayo, ukuqala ngo $$9$$ ). Eli tyala lincinci alinamdla, nangona kunjalo, kuba sinomdla kwiseti enenani elincinci leempawu (kunye nenani eliphezulu lamakhadi). Ngoku siqwalasela isimboli sesibini $$x_{l.2}$$ yekhadi ngalinye, ngokucacileyo oku kulandelayo kufuneka kusebenze: $$x_{1.2} \neq x_{2.2} \neq x_{3.2} \neq \ldots \neq x_{k.2}$$ Ke ngoko kufuneka sazise $$k$$ iisimboli ezintsha. Kodwa ngoku $$k \leq 8-1 = 7$$ , ekubeni akukho nanye $$7$$ $$x_{1.2},\, x_{1.3},\, x_{1.4},\, x_{1.5},\, x_{1.6},\, x_{1.7},\, x_{1.8}$$ (kwelona khadi lisekhohlo) linokutshatisa isimboli sesibini sekhadi ngalinye (kungenjalo kuya kubakho iisimboli ezimbini ezifanayo ).

Sifumene ubuninzi kula makhadi asi-7 matsha:

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

Kwale ngxoxo inye ngoku sakha iimephu ezilandelayo $$7$$ (eyokuqala kwezi mephu kufuneka ingqubane nemephu yethu yokuqalisa, kwaye hayi nge $$1$$ , kungenjalo iyakuba nge $$7$$ ngaphambili. ufumene iimephu):

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

Le ngxoxo inokuqhubekeka kumakhadi alandelayo $$7$$ ; Iyonke $$8-2 = 6$$ amaxesha amaninzi. Amakhadi okugqibela $$7$$ anjalo:

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

Ukuba ubunokongeza elinye ikhadi $$\left(\begin{array}{c} 9 \\ x_{57.2} \\ x_{57.3} \\ x_{57.4} \\ x_{57.5} \\ x_{57.6} \\ x_{57.7} \\ x_{57.8} \end{array}\right)$$ iya kusilela kuba eli khadi alinabelani ngesimboli kunye nekhadi lokuqalisa. Siye sakha ubuninzi bemephu $$1 + 8 \cdot 7 = 57$$ yeemephu. Injongo yethu ngoku kukwakha nokuba zininzi.

Ukwenza oku, sijonga amakhadi amatsha angama-7 okuqala afunyenweyo kwaye sifikelela kwisigqibo sokuba sifuna ngokupheleleyo iisimboli ezintsha apha (akukho khadi linokuba nesimboli kabini kwaye isimboli ngasinye $$7 \cdot 7$$ kufuneka singaveli. kabini, ekubeni yeyiphi $$1$$ sele iphindwe kabini):

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

Ngoko ke sifuna kancinci $$8 + (7 \cdot 7) = 57$$ iisimboli (kangangoko iisimboli ezininzi njengamakhadi!). Ngoku sizama ukuphumelela ngeli nani kwaye sifumane umthetho wokwakha kuzo zonke ezinye izinto. Ukwenza oku, sakha idobble encinci encinci eneempawu kuphela $$3$$ ngekhadi ngalinye kwaye siyifumane njengekhadi lokuqala.

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

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

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

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

ngetotali $$1 + 3 \cdot 2 = 7$$ amakhadi kunye $$3 + (2 \cdot 2) = 7$$ iisimboli. Ngolingo oluncinci kunye nempazamo (kunye nokusebenzisa iisimboli esele zabelwe) ufumana le dobble ilandelayo:

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

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

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

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

Ngaba oku kunokufumaneka ngokucwangcisiweyo? Ukwenza oku, singenisa iisimboli ezisanda kwabelwe $$4, 5, 6, 7$$ kwisikwere sematriki.:

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

Ngoku sicingela amakhadi amabini okuqala (ukuqala ngeesimboli zokuqala \ $$4$$ kunye $$5$$ ) imigca eqhagamshelayo ethe nkqo kwiisimboli ezisezantsi $$6$$ kunye $$7$$:

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

Ekubeni le migca ingadibanisi, sifumana (ngokucwangcisa iisimboli kumgca wokudibanisa ngomgca) awona makhadi asondeleyo asemthethweni.:

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

Okokugqibela, sicinga ukudibanisa imigca kunye nethambeka elahlukileyo (kulo mzekelo kunye nethambeka $$1$$ ):

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

Umgca wokudibanisa wesibini (phakathi $$5$$ kunye $$6$$ ) ushiya i-matrix kumda wasekunene kwaye uphinde ungene kumda wasekhohlo. Ngokukhetha ngobuchule i-slope, siqinisekisa kwelinye icala ukuba imigca yokudibanisa ayiphambanisi, kodwa kwakhona ukuba imigca yangaphambili (ethe nkqo) yokudibanisa ayiphambanisi. Le ngcamango yoyilo ekugqibeleni ikhokelela kwifomula yoyilo ilandelayo:

Idoboli eline $$k \in \mathbb{N} \, | \, (k-1) \text{ prim}$$ ine $$1+(k \cdot (k-1)) = k^2-k+1 = k + (k-1)(k-1)$$ amakhadi kunye neempawu. $$K_x$$ nge $$x \in \mathbb{N}$$ kunye $$0 \leq x \leq (k-1) \cdot k$$ iyasebenza:

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

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

Kukho $$(k-1)\cdot k + 1 = k + (k-1)(k-1)$$ amaqhekeza ala makhadi. Ngoku kushiyeke kuphela ukubonisa:

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

• Ityala lokuqala: $$x_1 = 0$$
• Ityala 1a: $$0 < x_2 < k$$
• Kuba $$y_1 = 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(0, 1) = 1$$
$$f(x_2, y_2) = f(x_2, 1) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$$ .
• Ku- $$y_1 \neq 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(0, y_1) = y_1 \neq 1$$
$$f(x_2, y_2) = f(x_2, y_2) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1$$
• Kwi $$y_1 = 1$$ kunye $$y_2 \neq 1$$ :
$$f(x_1, y_1) = f(0, 1) = 1$$
$$f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) =$$
$$(k+1)(x-1) + (k-1) + y \geq (k+1)(x-1)+y > 1$$
• Kuba $$y_1 \neq 1$$ kunye $$y_2 \neq 1$$ ithi:
$$f(x_1, y_1) = f(0, y_1) = y_1 \leq k$$
$$f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) > k$$
• Ityala 1b: $$x_2 \geq k$$
• Ku $$y_1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(0, \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
• Kuba $$y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$ kunye $$y_2 = 1$$ yile:
$$f(x_1, y_1) = f(0, y_1) = y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1$$
• Ye $$y_2 \neq 1$$ ithi:
$$f(x_1, y_1) = f(0, y_1) = y_1 \leq k$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y_2-2) \geq (k+1)+(k-1)(y_2-2) > k$$
• Ityala lesibini: $$0 < x_1 < k$$
• Ityala 2a: $$0 < x_2 < k$$
• Kuba $$y_1 = 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$$
• Ku- $$y_1 \neq 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) > 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$$
• Kwi $$y_1 = 1$$ kunye $$y_2 \neq 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) > 1$$
• Kuba $$y_1 \neq 1$$ kunye $$y_2 \neq 1$$ ithi:
$$f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) \leq$$
$$(k+1)+(k-1)(x_1-1)+(k-2)$$
$$f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) \geq$$
$$(k+1)+(k-1)((x_1+1)-1)+(y_2-2) =$$
$$(k+1)+(k-1)(x_1-1) + (k-1) + (y_2-2) \geq$$
$$(k+1)+(k-1)(x_1-1) + (k-1) + (2-2) \geq$$
$$(k+1)+(k-1)(x_1-1) + (k-1) > (k+1)+(k-1)(x_1-1) + (k-2)$$
• Ityala 2b: $$x_2 \geq k$$
• Kuba $$y_1 = 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 \geq \left\lfloor \frac{k-1}{k-1} \right\rfloor + 1 = 2 > 1$$
• Kwi $$y_1 = 1$$ kunye $$y_2 \neq 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y_2-2) \geq (k+1) + (k-1)(y_2-2) > 1$$
• Ku- $$y_1 \neq 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y_1-2) \geq (k+1) + (k-1)(y_1-2) > 1$$
$$f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1$$
• Kuba $$y_1 \neq 1$$ kwaye $$y_2 \neq 1$$ yi:
$$(k+1) + (k-1)(x_1-1) + (y_1-2) =$$
$$\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)$$
$$+ (k+1) + (k-1)(y-2)$$
$$\Leftrightarrow y_1 = (k-1)y_2 - (k-1)(x_1+1) +$$
$$\left( 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right) \right)$$
Kuba $$y_2 = x_1+1$$ kunye $$2 \leq y_2 \leq k$$ yi
$$y_1 = 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right)$$ kunye $$2 \leq y_1 \leq k$$.
Sinye kuphela isisombululo apha $$(y_1, y_2)$$.
Ngenxa yokuba sikhetha $$y^*_2=y_2-1$$ njengexabiso, liyi $$y^*_1 = y_1-(k-1) < 2$$.
Ukongeza, kuba $$y^*_2*=y_2+1$$ ngoko $$y^*_1 = y_1+(k-1) > k$$.
• 3. Ityala: $$x_1 \geq k$$
• Ityala 3a: $$x_2 \geq k$$
• Ityala 3a': $$m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$$
• Kuba $$y_1 = 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = m_1$$
$$f(x_2, y_2) = f(x_2, 1) = m_2 = m_1$$
• Kwi $$y_1 = 1$$ kunye $$y_2 \neq 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$$
$$\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) \geq$$
$$(k+1) + (k-1)(y_2-2) \geq (k+1) > k$$
• Ku- $$y_1 \neq 1$$ kunye $$y_2 = 1$$ :
Bona $$y_1 = 1$$ kunye $$y_2 \neq 1$$ .
• Kuba $$y_1 \neq 1$$ kwaye $$y_2 \neq 1$$ yi:
$$f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$$
Emva koko $$f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$$
$$L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 - L_2 = (k-1)(y_2-y_1)$$
Kuba $$y_1 \neq y_2$$ yi $$L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$$.
Kuba $$y_1 = y_2$$ yi $$L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$$ kwaye
$$\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$$
$$\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$$
$$x_1 = x_2 + (k-1)\cdot l$$ ngokuchaseneyo ne $$m_1 = m_2$$.
• Icala 3a'': $$m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2$$
• Kuba $$y_1 = 1$$ kunye $$y_2 = 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = m_1$$
$$f(x_2, y_2) = f(x_2, 1) = m_2 \neq m_1$$
• Kwi $$y_1 = 1$$ kunye $$y_2 \neq 1$$ :
$$f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =$$
$$\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) \geq$$
$$(k+1) + (k-1)(y_2-2) \geq (k+1) > k$$
• Ku- $$y_1 \neq 1$$ kunye $$y_2 = 1$$ :
Bona $$y_1 = 1$$ kunye $$y_2 \neq 1$$ .
• Kuba $$y_1 \neq 1$$ kwaye $$y_2 \neq 1$$ yi:
$$f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)$$
$$f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%$$
$$(k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)$$
Emva koko $$f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow$$
$$L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow$$
$$L_1 - L_2 = (k-1)(y_2-y_1)$$
Kuba $$y_1 \neq y_2$$ yi $$L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)$$.
Kuba $$y_1 = y_2$$ yi $$L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2$$ kwaye
$$\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =$$
$$\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow$$
$$y = \frac{(k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)}{m_2 - m_1}$$
Kulungile apho $$2 \leq y \leq k$$ rhoqo a $$l \in \mathbb{N}_0$$, lo nto
$$m_2 - m_1 \mid (k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)$$.
Bungqina: khona $$(k-1)$$ yiprime, yi (ngenxa ye-lemma ka-Bézout)
$$(k-1)\cdot l \equiv -\left( (3-k)(m_2-m_1) + (x_1-x_2) \right) \, \mod (m_2-m_1)$$
inokusombulula, kuba $$\text{ggT}\left((k-1),(m_2-m_1)\right) = 1$$ Ukwahlula $$-\left( (3-k)(m_2-m_1) + (x_1-x_2) \right)$$.
Emva koko oku kuphela kwesisombululo $$l_1$$, ngenxa yokuba enye
$$l_2 = l_1 + (m_2-m_1)$$ yi $$y_2 = y_1 + (k-1) > k$$.

Unokufumana kwakhona ulwazi olunomdla lwemvelaphi kwisihloko se-dobble kunye nemathematika apha okanye apha . Kolu shicilelo lulandelayo ungabona ifomula engqiniweyo yangaphambili isebenza: IiDobbles (ye $$(k-1)$$ prim) zinokuveliswa ngokucofa iqhosha.:

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