Habeenkii ugu dambeeyay ee ciyaarteenna qoyska, carruurtu si xamaasad leh ayay u soo saareen ciyaarta Dobble (daabacaadda Harry Potter). Ka dib markii laga badiyay shan wareeg (iyada oo aan kaarkaygu si muuqata ula mid ahayn kaarka ciyaarta), waxaan la yaabay markii aan ogaaday in ciyaaryahan kasta uu had iyo jeer heli karo ciyaar wareeg kasta. Si kastaba ha ahaatee, aaminaad la'aantayda waxaa la kulmay wareegyo badan oo lumay - carruurtu si fudud ayay u dhaqso badnaayeen.
Sababta ku filan inaad si dhow uga fiirsato ciyaarta dhinaca xisaabta. Marka hore mabda'a ciyaarta: Dobble waa ciyaar kaar fudud oo leh \(55\) kaararka wareega, mid kastaa wuxuu muujinayaa siddeed calaamadood oo kala duwan. Dhammaan kaadhadhka markooda ayaa loo kala qaybiyaa, kaarka ugu dambeeya oo keliya ayaa miiska dhexdiisa ku yaal. Hadda ciyaartoyda oo dhan waa inay isku mar is barbar dhigaan calaamadaha kaadhka iyo calaamadaha ku yaal kaadhka sare ee hadda. Haddii ciyaaryahan uu ka helay calaamad isku mid ah labada kaarar, wuxuu dhigi karaa kaarkiisa xirmada isagoo ah kan ugu dhaqsaha badan ee magaca calaamadda. Ciyaaryahanka tuura dhammaan kaararkooda marka hore wuu guuleystaa.
Sidee bay ku noqon kartaa in ay jiraan \(55\) kaarar oo sidan oo kale ah oo loo dhisay si ay 2 kaarba u leeyihiin hal calaamad oo sax ah? Waa imisa tirada ugu yar ee calaamadaha noocaas ah ee ay tahay in la isticmaalo? Waa imisa tirada ugu badan ee kaararka noocaas ah?
Marka hore, waxaan ku dhiseynaa kaadhadhkan anagoo adeegsanayna tillaabooyinka macquulka ah ee soo socda (dhammaan kaadhadhka la dhisay waxay leeyihiin hanti ay u kala soocaan siday u korayaan): Kaadhka kowaad waa inuu lahaadaa 8 calaamadood oo kala duwan, tusaale ahaan akhrinta:
$$\left(\begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}\right)$$
Sideen u dhisnaa kaararka soo socda si ay u yeeshaan hal calaamad oo la mid ah kaarka koowaad?:
$$\left(\begin{array}{c} 1 \\ x_{1.2} \\ x_{1.3} \\ x_{1.4} \\ x_{1.5} \\ x_{1.6} \\ x_{1.7} \\ x_{1.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{2.2} \\ x_{2.3} \\ x_{2.4} \\ x_{2.5} \\ x_{2.6} \\ x_{2.7} \\ x_{2.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{3.2} \\ x_{3.3} \\ x_{3.4} \\ x_{3.5} \\ x_{3.6} \\ x_{3.7} \\ x_{3.8} \end{array}\right), \ldots, \left(\begin{array}{c} 1 \\ x_{k.2} \\ x_{k.3} \\ x_{k.4} \\ x_{k.5} \\ x_{k.6} \\ x_{k.7} \\ x_{k.8} \end{array}\right)$$
Tiro kasta oo kaararka noocaas ah mar hore ayaa laga dhisi karaa halkan (waxaad si fudud u buuxinaysaa meelaha sida ay u korayaan, laga bilaabo \(9\) ). Si kastaba ha ahaatee, kiiskan fudud waa mid aan xiiso lahayn, si kastaba ha ahaatee, tan iyo markii aan xiisaynayso set leh tirada ugu yar ee calaamadaha (iyo tirada ugu badan ee kaararka). Waxaan hadda tixgelineynaa calaamadda labaad \( x_{l.2} \) ee kaar kasta, taaso iska cad kuwa soo socda waa in lagu dabaqo: \( x_{1.2} \neq x_{2.2} \neq x_{3.2} \neq \ldots \neq x_{k.2} \) . Sidaa darteed waxaan si lama huraan ah u soo saarnay \( k \) calaamado cusub. Laakin hadda \( k \leq 8-1 = 7 \) , mar haddii calaamadihii \( \( 7 \) \) midkoodna \( x_{1.2},\, x_{1.3},\, x_{1.4},\, x_{1.5},\, x_{1.6},\, x_{1.7},\, x_{1.8} \) (kaarka bidix ee ugu dambeeya) waxa laga yaabaa inuu u dhigmo calaamadda labaad ee mid kasta oo ka mid ah kaararka kale (hadii kale waxaa jiri doona laba calaamadood oo isku mid ah). ).
Waxaan helnay ugu badnaan 7-daan kaarar oo cusub:
$$\left(\begin{array}{c} 1 \\ x_{1.2} \\ x_{1.3} \\ x_{1.4} \\ x_{1.5} \\ x_{1.6} \\ x_{1.7} \\ x_{1.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{2.2} \\ x_{2.3} \\ x_{2.4} \\ x_{2.5} \\ x_{2.6} \\ x_{2.7} \\ x_{2.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{3.2} \\ x_{3.3} \\ x_{3.4} \\ x_{3.5} \\ x_{3.6} \\ x_{3.7} \\ x_{3.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{4.2} \\ x_{4.3} \\ x_{4.4} \\ x_{4.5} \\ x_{4.6} \\ x_{4.7} \\ x_{4.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{5.2} \\ x_{5.3} \\ x_{5.4} \\ x_{5.5} \\ x_{5.6} \\ x_{5.7} \\ x_{5.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{6.2} \\ x_{6.3} \\ x_{6.4} \\ x_{6.5} \\ x_{6.6} \\ x_{6.7} \\ x_{6.8} \end{array}\right), \left(\begin{array}{c} 1 \\ x_{7.2} \\ x_{7.3} \\ x_{7.4} \\ x_{7.5} \\ x_{7.6} \\ x_{7.7} \\ x_{7.8} \end{array}\right)$$
Isla doodda ayaanu hadda dhisaynaa khariidadaha soo socda ee \(7\) (ka ugu horreeya ee khariidadahani waa inay ku dhacaan khariidadda bilawga ah, ee maaha inay ku dhacaan \(1\) , haddii kale waxay la mid tahay \(7\) hore. Maab la helay):
$$\left(\begin{array}{c} 2 \\ x_{8.2} \\ x_{8.3} \\ x_{8.4} \\ x_{8.5} \\ x_{8.6} \\ x_{8.7} \\ x_{8.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{9.2} \\ x_{9.3} \\ x_{9.4} \\ x_{9.5} \\ x_{9.6} \\ x_{9.7} \\ x_{9.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{10.2} \\ x_{10.3} \\ x_{10.4} \\ x_{10.5} \\ x_{10.6} \\ x_{10.7} \\ x_{10.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{11.2} \\ x_{11.3} \\ x_{11.4} \\ x_{11.5} \\ x_{11.6} \\ x_{11.7} \\ x_{11.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{12.2} \\ x_{12.3} \\ x_{12.4} \\ x_{12.5} \\ x_{12.6} \\ x_{12.7} \\ x_{12.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{13.2} \\ x_{13.3} \\ x_{13.4} \\ x_{13.5} \\ x_{13.6} \\ x_{13.7} \\ x_{13.8} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{14.2} \\ x_{14.3} \\ x_{14.4} \\ x_{14.5} \\ x_{14.6} \\ x_{14.7} \\ x_{14.8} \end{array}\right)$$
Dooddan waxaa loo sii wadi karaa kaararka \(7\) ee xiga si la mid ah; wadarta guud ee \(8-2 = 6\) jeer ka badan. Sidaa darteed, kaararka \(7\) ee ugu dambeeya waa:
$$\left(\begin{array}{c} 8 \\ x_{50.2} \\ x_{50.3} \\ x_{50.4} \\ x_{50.5} \\ x_{50.6} \\ x_{50.7} \\ x_{50.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{51.2} \\ x_{51.3} \\ x_{51.4} \\ x_{51.5} \\ x_{51.6} \\ x_{51.7} \\ x_{51.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{52.2} \\ x_{52.3} \\ x_{52.4} \\ x_{52.5} \\ x_{52.6} \\ x_{52.7} \\ x_{52.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{53.2} \\ x_{53.3} \\ x_{53.4} \\ x_{53.5} \\ x_{53.6} \\ x_{53.7} \\ x_{53.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{54.2} \\ x_{54.3} \\ x_{54.4} \\ x_{54.5} \\ x_{54.6} \\ x_{54.7} \\ x_{54.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{55.2} \\ x_{55.3} \\ x_{55.4} \\ x_{55.5} \\ x_{55.6} \\ x_{55.7} \\ x_{55.8} \end{array}\right), \left(\begin{array}{c} 8 \\ x_{56.2} \\ x_{56.3} \\ x_{56.4} \\ x_{56.5} \\ x_{56.6} \\ x_{56.7} \\ x_{56.8} \end{array}\right)$$
Haddii aad hadda rabtay inaad dhisto kaar kale $$\left(\begin{array}{c} 9 \\ x_{57.2} \\ x_{57.3} \\ x_{57.4} \\ x_{57.5} \\ x_{57.6} \\ x_{57.7} \\ x_{57.8} \end{array}\right)$$ , kaarkani ma guuldarraystay leh calaamad guud oo leh kaadhka bilowga. Tani waxay ka dhigan tahay inaan dhisnay ugu badnaan kaararka \(1 + 8 \cdot 7 = 57\) . Hadafkayagu hadda waa inaan dhisno sida kuwa badan.
Si tan loo sameeyo, waxaan eegnaa 7 kaarka cusub ee ugu horreeya ee la helay oo aan gaadhno gabagabada inaan u baahanahay gabi ahaanba \(7 \cdot 7\) calaamado cusub (kaarna ma yeelan karo calaamad nuqul ah, calaamad kasta oo lagu abaalmariyo waxaa laga yaabaa inaysan u muuqan laba jeer sababtoo ah \(1\) waa horeba labanlaab):
$$\left(\begin{array}{c} 1 \\ 9 \\ 10 \\ 11 \\ 12 \\ 13 \\ 14 \\ 15 \end{array}\right), \left(\begin{array}{c} 1 \\ 16 \\ 17 \\ 18 \\ 19 \\ 20 \\ 21 \\ 22 \end{array}\right), \left(\begin{array}{c} 1 \\ 23 \\ 24 \\ 25 \\ 26 \\ 27 \\ 28 \\ 29 \end{array}\right), \left(\begin{array}{c} 1 \\ 30 \\ 31 \\ 32 \\ 33 \\ 34 \\ 35 \\ 36 \end{array}\right), \left(\begin{array}{c} 1 \\ 37 \\ 38 \\ 39 \\ 40 \\ 41 \\ 42 \\ 43 \end{array}\right), \left(\begin{array}{c} 1 \\ 44 \\ 45 \\ 46 \\ 47 \\ 48 \\ 49 \\ 50 \end{array}\right), \left(\begin{array}{c} 1 \\ 51 \\ 52 \\ 53 \\ 54 \\ 55 \\ 56 \\ 57 \end{array}\right)$$
Taas macnaheedu waa inaan u baahanahay ugu yaraan calaamadaha \(8 + (7 \cdot 7) = 57\) (tusaale ahaan, inta calaamad ee kaararka ah!). Waxaan hadda isku dayeynaa in aan ku helno lambarkan oo aan u helno qeexitaanka naqshadeynta dhammaan walxaha kale. Si tan loo sameeyo, waxaan dhiseynaa doble wax yar ka yar oo sita oo kaliya calaamadaha \(3\) kaarkiiba oo hel kaarka bilowga
$$\left(\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right)$$
iyo kaadhadhka kale
$$\left(\begin{array}{c} 1 \\ 4 \\ 5 \end{array}\right), \left(\begin{array}{c} 1 \\ 6 \\ 7 \end{array}\right)$$
$$\left(\begin{array}{c} 2 \\ x_{3.2} \\ x_{3.3} \end{array}\right), \left(\begin{array}{c} 2 \\ x_{4.2} \\ x_{4.3} \end{array}\right)$$
$$\left(\begin{array}{c} 3 \\ x_{5.2} \\ x_{5.3} \end{array}\right), \left(\begin{array}{c} 3 \\ x_{6.2} \\ x_{6.3} \end{array}\right)$$
wadar ahaan \(1 + 3 \cdot 2 = 7\) kaararka iyo \( 3 + (2 \cdot 2) = 7\) calaamadaha. Tijaabo yar iyo khalad (iyo adigoo isticmaalaya calaamadihii hore loo qoondeeyay) waxaad helaysaa dobble soo socda:
$$\left(\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right)$$
$$\left(\begin{array}{c} 1 \\ 4 \\ 5 \end{array}\right), \left(\begin{array}{c} 1 \\ 6 \\ 7 \end{array}\right)$$
$$\left(\begin{array}{c} 2 \\ 4 \\ 6 \end{array}\right), \left(\begin{array}{c} 2 \\ 5 \\ 7 \end{array}\right)$$
$$\left(\begin{array}{c} 3 \\ 4 \\ 7 \end{array}\right), \left(\begin{array}{c} 3 \\ 5 \\ 6 \end{array}\right)$$
Tan sidoo kale ma lagu heli karaa si habaysan? Si taas loo sameeyo, waxaanu geli calaamadaha cusub ee loo qoondeeyey \(4, 5, 6, 7\) ee matrix labajibbaaran.:
$$\begin{array}{ccc} 4 & & 5 \\ & & \\ 6 & & 7\end{array}$$
Hadda waxaan qiyaasi karnaa labada kaarar ee ugu horreeya (laga bilaabo calaamadaha bilowga \ \(4\) iyo \(5\) ) xadadka isku xirka tooska ah ee calaamadaha hoose \(6\) iyo \(7\):
$$\begin{array}{ccc} 4 & & 5 \\ \vdots & & \vdots \\ 6 & & 7\end{array}$$
Maaddaama xariiqyadani aysan isdhaafin, waxaan helnaa (adigoo ku sawiraya calaamadaha xariiqda isku xirka khadka) kaararka saxda ah ee ugu dhow.:
$$\left(\begin{array}{c} 2 \\ 4 \\ 6 \end{array}\right), \left(\begin{array}{c} 2 \\ 5 \\ 7 \end{array}\right)$$
Ugu dambeyntii, waxaan ku qiyaaseynaa isku xirka khadadka jiirada kale (kiiskan oo leh jiirada \(1\) ):
$$\begin{array}{ccccc} & 4 & & 5 & \\ \ddots & & \ddots & & \ddots \\ & 6 & & 7 &\end{array}$$
Xariiqda isku xirka labaad (inta u dhaxaysa \(5\) iyo \(6\) ) waxay ka tagtaa jaantuska cidhifka midig oo dib u galaa cidhifka bidix. Anaga oo si xariif ah u dooranayna gradient-ka, waxaanu hubinaynaa in xadhkaha isku xidha aanay isku xidhin midba midka kale, laakiin sidoo kale in xadhkaha isku xidha ee hore (tooska ah) aanay isku xidhin. Fikraddan naqshadeynta waxay ugu dambeyntii keentaa qaacidada naqshadeynta ee soo socota:
Dobble leh \(k \in \mathbb{N} \, | \, (k-1) \text{ prim} \) wuxuu leeyahay \(1+(k \cdot (k-1)) = k^2-k+1 = k + (k-1)(k-1)\) kaararka iyo calaamadaha. Khariidadda \(K_x\) ee leh \(x \in \mathbb{N}\) iyo \(0 \leq x \leq (k-1) \cdot k\) ayaa khuseysa:
$$K_x = \left(\begin{array}{c} f(x,1) \\ f(x,2) \\ \vdots \\ f(x,k) \end{array}\right), \,\, m = \left\lfloor \frac{x-1}{k-1} \right\rfloor + 1,$$
$$f(x,y) = \left\{\begin{array}{ll} y & \text{falls } x = 0 \\ \lfloor \frac{x-1}{k-1} \rfloor + 1, &\text{sonst falls } y = 1 \\ (k+1) + (k-1)(x-1) + (y-2), & \text{sonst falls } 0 < x < k \\ \left( \left((m-1)(k-1)+x\right)-1+ \left( (m-2)(y-2) \right) \right) \% (k-1) &\text{sonst} \\ + (k+1) + (k-1)(y-2)&\end{array}\right.$$
Waxaa jira qaybo ka mid ah kaararkan \((k-1)\cdot k + 1 = k + (k-1)(k-1)\) . Hadda waxa kaliya oo hadhay in la muujiyo:
$$ \forall x_1 < x_2 \in \{ 1, \ldots, k+(k-1)(k-1) \} \, \exists \, ! \, y_1, y_2 \in \{ 1, \ldots, k \}: f(x_1, y_1) = f(x_2, y_2) $$
- Kiiska 1aad: \( x_1 = 0 \)
- Kiis 1a: \( 0 < x_2 < k \)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(0, 1) = 1\)
\(f(x_2, y_2) = f(x_2, 1) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1\) . - Wixii \(y_1 \neq 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(0, y_1) = y_1 \neq 1\)
\(f(x_2, y_2) = f(x_2, y_2) = \lfloor \frac{x_2-1}{k-1} \rfloor + 1 = 1\) - Wixii \(y_1 = 1\) iyo \(y_2 \neq 1\) u haynaa:
\(f(x_1, y_1) = f(0, 1) = 1\)
\(f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) =\)
\((k+1)(x-1) + (k-1) + y \geq (k+1)(x-1)+y > 1\) - Waayo \(y_1 \neq 1\) iyo \(y_2 \neq 1\) waa:
\(f(x_1, y_1) = f(0, y_1) = y_1 \leq k\)
\(f(x_2, y_2) = f(x_2, y_2) = (k+1) + (k-1)(x-1) + (y-2) > k\)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
- Kiis 1b: \( x_2 \geq k \)
- Wixii \(y_1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\) iyo \(y_2 = 1\) waxaan leenahay:
\(f(x_1, y_1) = f(0, \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\)
\(f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\) - Wixii \(y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\) iyo \(y_2 = 1\) waa:
\(f(x_1, y_1) = f(0, y_1) = y_1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\)
\(f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\) - Waayo \(y_2 \neq 1\) waa:
\(f(x_1, y_1) = f(0, y_1) = y_1 \leq k\)
\(f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)\)
\(+ (k+1) + (k-1)(y_2-2) \geq (k+1)+(k-1)(y_2-2) > k \)
- Wixii \(y_1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1\) iyo \(y_2 = 1\) waxaan leenahay:
- Kiis 1a: \( 0 < x_2 < k \)
- Kiiska 2aad: \( 0 < x_1 < k \)
- Kiis 2a: \( 0 < x_2 < k \)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1\)
\(f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1\) - Wixii \(y_1 \neq 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) > 1\)
\(f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1\) - Wixii \(y_1 = 1\) iyo \(y_2 \neq 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1\)
\(f(x_2, y_2) = f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) > 1\) - Waayo \(y_1 \neq 1\) iyo \(y_2 \neq 1\) waa:
\(f(x_1, y_1) = (k+1)+(k-1)(x_1-1)+(y_1-2) \leq\)
\((k+1)+(k-1)(x_1-1)+(k-2)\)
\(f(x_2, y_2) = (k+1)+(k-1)(x_2-1)+(y_2-2) \geq\)
\((k+1)+(k-1)((x_1+1)-1)+(y_2-2) =\)
\((k+1)+(k-1)(x_1-1) + (k-1) + (y_2-2) \geq\)
\((k+1)+(k-1)(x_1-1) + (k-1) + (2-2) \geq\)
\((k+1)+(k-1)(x_1-1) + (k-1) > (k+1)+(k-1)(x_1-1) + (k-2)\)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
- Kiis 2b: \( x_2 \geq k \)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1\)
\(f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 \geq \left\lfloor \frac{k-1}{k-1} \right\rfloor + 1 = 2 > 1\) - Wixii \(y_1 = 1\) iyo \(y_2 \neq 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 = 1\)
\(f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)\)
\(+ (k+1) + (k-1)(y_2-2) \geq (k+1) + (k-1)(y_2-2) > 1\) - Wixii \(y_1 \neq 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1)\)
\(+ (k+1) + (k-1)(y_1-2) \geq (k+1) + (k-1)(y_1-2) > 1\)
\(f(x_2, y_2) = f(x_2, 1) = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor + 1 = 1\) - Waayo \(y_1 \neq 1\) iyo \(y_2 \neq 1\) waa:
\((k+1) + (k-1)(x_1-1) + (y_1-2) =\)
\(\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1)\)
\(+ (k+1) + (k-1)(y-2)\)
\(\Leftrightarrow y_1 = (k-1)y_2 - (k-1)(x_1+1) +\)
\(\left( 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right) \right) \)
Waayo \(y_2 = x_1+1\) leh \( 2 \leq y_2 \leq k\) waa
\(y_1 = 2 + \left( \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \right)\) leh \( 2 \leq y_1 \leq k\).
Waxaa jira hal xal oo kaliya halkan \( (y_1, y_2) \).
Sababtoo ah waxaan dooranaa \(y^*_2=y_2-1\) sida qiimaha, waa \(y^*_1 = y_1-(k-1) < 2\).
Intaa waxaa dheer, loogu talagalay \(y^*_2*=y_2+1\) markaas \(y^*_1 = y_1+(k-1) > k\).
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
- Kiis 2a: \( 0 < x_2 < k \)
- 3. Kiis: \( x_1 \geq k \)
- Kiis 3a: \( x_2 \geq k \)
- Kiiska 3a': \(m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2\)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = m_1\)
\(f(x_2, y_2) = f(x_2, 1) = m_2 = m_1\) - Wixii \(y_1 = 1\) iyo \(y_2 \neq 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =\)
\(\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k\)
\(f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%\)
\((k-1) + (k+1) + (k-1)(y_2-2) \geq\)
\((k+1) + (k-1)(y_2-2) \geq (k+1) > k\) - Wixii \(y_1 \neq 1\) iyo \(y_2 = 1\) u haynaa:
Eeg \(y_1 = 1\) iyo \(y_2 \neq 1\) . - Waayo \(y_1 \neq 1\) iyo \(y_2 \neq 1\) waa:
\(f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%\)
\((k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)\)
\(f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%\)
\((k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)\)
Kadib \(f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow\)
\(L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow\)
\(L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow\)
\(L_1 - L_2 = (k-1)(y_2-y_1)\)
Waayo \(y_1 \neq y_2\) waa \(L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)\).
Waayo \(y_1 = y_2\) waa \(L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2\) iyo
\(\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =\)
\(\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow\)
\(x_1 = x_2 + (k-1)\cdot l\) lid ku ah \(m_1 = m_2\).
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
- Kiiska 3a'': \(m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 \neq \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2\)
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = m_1\)
\(f(x_2, y_2) = f(x_2, 1) = m_2 \neq m_1\) - Wixii \(y_1 = 1\) iyo \(y_2 \neq 1\) u haynaa:
\(f(x_1, y_1) = f(x_1, 1) = m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor + 1 \leq \left\lfloor \frac{((k-1) \cdot k)-1}{k-1} \right\rfloor + 1 =\)
\(\left\lfloor k - \frac{1}{k-1} \right\rfloor + 1 = (k - 1) + 1 = k\)
\(f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%\)
\((k-1) + (k+1) + (k-1)(y_2-2) \geq\)
\((k+1) + (k-1)(y_2-2) \geq (k+1) > k\) - Wixii \(y_1 \neq 1\) iyo \(y_2 = 1\) u haynaa:
Eeg \(y_1 = 1\) iyo \(y_2 \neq 1\) . - Waayo \(y_1 \neq 1\) iyo \(y_2 \neq 1\) waa:
\(f(x_1, y_1) = \left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \%\)
\((k-1) + (k+1) + (k-1)(y_1-2) = L_1 + (k+1) + (k-1)(y_1-2)\)
\(f(x_2, y_2) = \left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \%\)
\((k-1) + (k+1) + (k-1)(y_2-2) = L_2 + (k+1) + (k-1)(y_2-2)\)
Kadib \(f(x_1, y_1) = f(x_2, y_2) \Leftrightarrow\)
\(L_1 + (k+1) + (k-1)(y_1-2) = L_2 + (k+1) + (k-1)(y_2-2) \Leftrightarrow\)
\(L_1 + (k-1)(y_1-2) = L_2 + (k-1)(y_2-2) \Leftrightarrow\)
\(L_1 - L_2 = (k-1)(y_2-y_1)\)
Waayo \(y_1 \neq y_2\) waa \(L_1-L_2 \leq (k-2 - 0) = k-2 < (k-1)(y_2-y_1)\).
Waayo \(y_1 = y_2\) waa \(L_1 - L_2 = 0 \Leftrightarrow L_1 = L_2\) iyo
\(\left( \left((m_1-1)(k-1)+x_1\right)-1+ \left( (m_1-2)(y_1-2) \right) \right) \% (k-1) =\)
\(\left( \left((m_2-1)(k-1)+x_2\right)-1+ \left( (m_2-2)(y_2-2) \right) \right) \% (k-1) \Leftrightarrow\)
\(y = \frac{(k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)}{m_2 - m_1}\)
Waa hagaag \(2 \leq y \leq k\) had iyo jeer a \(l \in \mathbb{N}_0\), sidaas darteed
\(m_2 - m_1 \mid (k-1)\cdot l + (3-k)(m_2 - m_1) + (x_1 - x_2)\).
Caddeyn: halkaas \((k-1)\) waa ra'iisul, waa (sababtoo ah Bézout's lemma)
\((k-1)\cdot l \equiv -\left( (3-k)(m_2-m_1) + (x_1-x_2) \right) \, \mod (m_2-m_1)\)
la xalin karo, sababtoo ah \(\text{ggT}\left((k-1),(m_2-m_1)\right) = 1\) Kala baxa \(-\left( (3-k)(m_2-m_1) + (x_1-x_2) \right)\).
Markaa tani waa xalka kaliya \(l_1\), sababtoo ah hal
\(l_2 = l_1 + (m_2-m_1)\) waa \( y_2 = y_1 + (k-1) > k\).
- Wixii \(y_1 = 1\) iyo \(y_2 = 1\) u haynaa:
- Kiiska 3a': \(m_1 = \left\lfloor \frac{x_1-1}{k-1} \right\rfloor +1 = \left\lfloor \frac{x_2-1}{k-1} \right\rfloor +1 = m_2\)
- Kiis 3a: \( x_2 \geq k \)
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