Konvey zanjirband etilgan o'q yozuvlari, matematik tomonidan yaratilgan Jon Xorton Konvey, nihoyatda aniq ifoda etish vositasidir katta raqamlar.[1] Bu shunchaki cheklangan ketma-ketlikdir musbat tamsayılar o'ng o'qlar bilan ajratilgan, masalan.
.
Ko'pchilik kabi kombinatorial yozuvlari, ta'rifi rekursiv. Bunday holda, yozuv oxir-oqibat ba'zi bir (odatda juda katta) butun sonli kuchga ko'tarilgan eng chap raqamga aylanadi.
Ta'rif va umumiy nuqtai
"Konvey zanjiri" quyidagicha ta'riflanadi:
- Har qanday musbat butun uzunlik zanjiri
. - Uzunlik zanjiri n, so'ngra o'ng o'q → va musbat butun son bilan birga uzunlik zanjiri hosil bo'ladi
.
Quyidagi beshta (texnik jihatdan to'rtta) qoidaga muvofiq har qanday zanjir butun sonni ifodalaydi. Ikkita zanjir bir xil sonni ifodalasa, ularga teng deyiladi.
Agar
,
va
musbat butun sonlar va
subchain, keyin:
- Bo'sh zanjir (yoki 0 uzunlikdagi zanjir) ga teng
va zanjir
raqamni ifodalaydi
.
ga teng
.
ga teng
. (qarang Knutning yuqoriga qarab o'qi )
ga teng
(bilan
nusxalari
,
nusxalari
va
juft qavslar; uchun murojaat qiladi
> 0).- Chunki
ga teng
, (2-qoida bo'yicha) va shuningdek
=
, (3-qoida bo'yicha) biz belgilashimiz mumkin
tenglashtirish ![p ^ {q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75ead10b72d59f44fbc277dbd66b6797dcbbf634)
E'tibor bering, to'rtinchi qoidani oldini olish uchun takroriy ikkita qoidani qo'llash bilan almashtirish mumkin ellipslar:
- 4a.
ga teng ![X](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
- 4b.
ga teng ![{ displaystyle X to (X to p to (q + 1)) to q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c817ff0cfbadb8f7d73364d65ef153699dc23b67)
Xususiyatlari
- Zanjir birinchi raqamning mukammal kuchiga baho beradi
- Shuning uchun,
ga teng ![1](https://wikimedia.org/api/rest_v1/media/math/render/svg/92d98b82a3778f043108d4e20960a9193df57cbf)
ga teng ![X](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
ga teng ![4](https://wikimedia.org/api/rest_v1/media/math/render/svg/295b4bf1de7cd3500e740e0f4f0635db22d87b42)
ga teng
(bilan aralashmaslik kerak
)
Tafsir
O'q zanjirini davolash uchun ehtiyot bo'lish kerak bir butun sifatida. Ok zanjirlari ikkilik operatorning takrorlanadigan dasturini tavsiflamaydi. Boshqa qo'shilgan belgilar zanjirlari (masalan, 3 + 4 + 5 + 6 + 7) ko'pincha ma'no o'zgarmasdan (masalan, (3 + 4) + 5 + (6 + 7)) qismlarda ko'rib chiqilishi mumkin (qarang. assotsiativlik ), yoki hech bo'lmaganda belgilangan tartibda bosqichma-bosqich baholanishi mumkin, masalan. 34567 o'ngdan chapga, bu Konveyning o'q zanjirlarida unday emas.
Masalan:
![$ 2 rightarrow3 rightarrow2 = 2 uparrow uparrow3 = 2 ^ {2 ^ 2} = 16](https://wikimedia.org/api/rest_v1/media/math/render/svg/3aa8384f1bb4f7fc9ca91f09ebc096769e6e7ed8)
![2 rightarrow left (3 rightarrow 2 right) = 2 ^ {{(3 ^ {2})}} = 2 ^ {{3 ^ {2}}} = 512](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ae9db644dcfcbbd4b2d75055af9074e97c06af4)
![chap (2 o'ng chiziq 3 o'ng) o'ng chiziq 2 = chap (2 ^ {3} o'ng) ^ {2} = 64](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c875876d5316a1ecfac4e18c28d5058dae4524f)
To'rtinchi qoida - yadro: 2 yoki undan yuqori bilan tugaydigan 4 va undan ortiq elementlardan iborat zanjir oldingi (odatda juda katta) oshirilgan oldingi element bilan bir xil uzunlikdagi zanjirga aylanadi. Ammo uning yakuniy element kamayadi, natijada zanjirni qisqartirish uchun ikkinchi qoidaga ruxsat beriladi. So'ngra, iborani o'zgartirish uchun Knuth, "juda batafsil", zanjir uchta elementga qisqartiriladi va uchinchi qoida rekursiyani tugatadi.
Misollar
Misollar tezda juda murakkablashadi. Mana ba'zi kichik misollar:
![n](https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b)
(1-qoida bo'yicha)
![{ displaystyle p dan q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fccb3827df1efe7930f9d1febd15d2359971b93)
(5-qoida bo'yicha)- Shunday qilib,
![{ displaystyle 3 dan 4 = 3 ^ {4} = 81} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fa08e81c568ef11592c24fcbf47b96dc125339b)
![{ displaystyle 4 dan 3 dan 2} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/88618a1c84b39128f3822985c6806499bb6b6452)
(3-qoida bo'yicha)![{ displaystyle = 4 uparrow (4 uparrow 4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8304e91f4eb7d9f1e6404c26f57f7b95e62b2c8d)
![{ displaystyle = 4 uparrow 256}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99bb3852d9cdaf296fb57ee687ae67656f43912b)
![{ displaystyle = 4 ^ {256}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96a9414c3f9c6114bb2431e458fdb10edb73ff3f)
![{ displaystyle 546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,096}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94716d4e7cbeeaa588f4e71f603fb4ad3c74a1a9)
![{ displaystyle taxminan 1.34 * 10 ^ {154}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b3b0feb5ef130fb1cd17971a4294b8202147f0f)
![{ displaystyle 2 dan 2 gacha a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c316bdf912fa665d2656211e0aeb6b8545849144)
(3-qoida bo'yicha)
(qarang Knut yuqoriga o'q o'qi )
![{ displaystyle 2 dan 4 gacha 3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60e529d982d572b6514402bf47534cd59a47b035)
(3-qoida bo'yicha)![{ displaystyle = 2 uparrow uparrow (2 uparrow uparrow (2 uparrow uparrow 2))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79f7b1eaf598090c4c021f74ad3d16b930517965)
![{ displaystyle = 2 uparrow uparrow (2 uparrow uparrow 4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f51fc90e34868974a160967a88524a8969995cc3)
![{ displaystyle = 2 uparrow uparrow (2 uparrow (2 uparrow (2 uparrow 2)))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cd58c398f7064b011fa05da88aea74de47ada26)
![{ displaystyle = 2 uparrow uparrow (2 uparrow (2 uparrow 4))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9dc7535f38bd59ebece3c29e99fc4e20fd442421)
![{ displaystyle = 2 uparrow uparrow (2 uparrow 16)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4109f93b6eeab5cfeef14be92fda09e2ede4999a)
![{ displaystyle = 2 uparrow uparrow (65536)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a5d1b4af4007ef724a02e87afd101560677222e)
(qarang tebranish )
![{ displaystyle 2 dan 3 dan 2 dan 2} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d5069c4a054afdcc20e8dcd9e522b339f38b377)
(4-qoida bo'yicha)
(5-qoida bo'yicha)
(2-qoida bo'yicha)
(3-qoida bo'yicha)- = oldingi raqamdan ancha katta
![{ displaystyle 3 to 2 to 2 to 2} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb22674ce8d465ebdd0d3a0fee99a77b08c8e323)
(4-qoida bo'yicha)
(5-qoida bo'yicha)
(2-qoida bo'yicha)
(3-qoida bo'yicha)- = oldingi raqamdan ancha katta
Tizimli misollar
To'rt atamadan iborat bo'lgan eng oddiy holatlar (2 dan kam bo'lmagan tamsayılardan iborat):
![a to b dan 2 ga 2 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/503a9391fe1e16897b44d29a4b904d74d0704ea4)
![= a to b dan 2 ga (1 + 1)](https://wikimedia.org/api/rest_v1/media/math/render/svg/468296bcf60aba1ed32ad5f29737789540cdf279)
![= a dan b ga (a dan b) ga 1 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/afa2d5a7191f2b9418c860b1346b6817d3d1ff67)
![= a to b ga ^ {b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0094c520d14938b357886e55172363b94def2e8)
![= a [a ^ {b} +2] b](https://wikimedia.org/api/rest_v1/media/math/render/svg/5122c5c8b65604a751640ca32b6fa474637eaeec)
- (oxirgi ko'rsatilgan mulkka teng)
![a to b dan 3 ga 2 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/11156b7d320be94a196fc292b572034fe2067d12)
![= a to b dan 3 ga (1 + 1)](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bec29bc73e0eeed4b5249572e03cc8c8abcb19d)
![= a to b to (a to b to (a to b) to 1) to 1](https://wikimedia.org/api/rest_v1/media/math/render/svg/311e4bec4fa82d025ca429f4daa1a7e6824b320e)
![= a to b to (a to b to a ^ {b})](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebb34957cf5d0feba97bd727fd2888977e4f1f51)
![= a [a to b to 2 to 2 + 2] b](https://wikimedia.org/api/rest_v1/media/math/render/svg/53ae38786c37933b56883d852a17a2b8e4f1d222)
![a to b dan 4 ga 2 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/b562f60620783d8bfc61aad74cc5f7864fa5d247)
![= a to b to (a to b to (a to b to a ^ {b})))](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf02fe36909b537b3b67913c3d5eadb602a080d6)
![= a [a to b to 3 to 2 + 2] b](https://wikimedia.org/api/rest_v1/media/math/render/svg/41077be13272dbc2d9a89dd71889a505af53c6c1)
Bu erda naqshni ko'rishimiz mumkin. Agar biron bir zanjir uchun bo'lsa
, biz ruxsat berdik
keyin
(qarang funktsional kuchlar ).
Buni qo'llash
, keyin
va ![a to b to p to 2 = a [a to b to (p-1) to 2 + 2] b = f ^ {p} (1)](https://wikimedia.org/api/rest_v1/media/math/render/svg/8462b1d161e5a1235b8676f01d91a95614413f6c)
Shunday qilib, masalan,
.
Davom etmoq:
![a to b dan 2 dan 3 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/39bb65239fe28cc8f026f612900443e23b6d7680)
![= a to b dan 2 ga (2 + 1)](https://wikimedia.org/api/rest_v1/media/math/render/svg/05dd09c3157020b08a8a68f93f82b94d54acfa0f)
![= a dan b ga (a dan b) ga 2 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fa10b0ad746a86e3d39fa8681f2b909ab94f1ff)
![= a to b ga ^ {b} dan 2 gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/2140e0b51ed47462875b53c1b789b813b679ca3f)
![= f ^ {{a ^ {b}}} (1)](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7705e04fc1f6878aef5642cc6ef908317f6280b)
Yana biz umumlashtirishimiz mumkin. Biz yozganimizda
bizda ... bor
, anavi,
. Yuqoridagi holatda,
va
, shuning uchun ![a to b dan 2 ga 3 = g_ {3} (2) = g_ {2} ^ {2} (1) = g_ {2} (g_ {2} (1)) = f ^ {{f (1)}} (1) = f ^ {{a ^ {b}}} (1)](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd22bef2cad35e8a4b260bc4c63c71f60c3e9acb)
Ackermann funktsiyasi
The Ackermann funktsiyasi Conway zanjirli o'q belgisi yordamida ifodalanishi mumkin:
uchun
(Beri
yilda giperoperatsiya )
shu sababli
uchun ![n> 2](https://wikimedia.org/api/rest_v1/media/math/render/svg/44e71ac55b9fbf1e9f341b946cda63d61d3ef2cd)
- (
va
bilan mos keladi
va
, bu mantiqan qo'shilishi mumkin).
Gremning raqami
Gremning raqami
o'zini Konveyning zanjirli o'q yozuvida aniq ifodalash mumkin emas, lekin u quyidagilar bilan chegaralangan:
![{ displaystyle 3 rightarrow 3 rightarrow 64 rightarrow 2 <G <3 rightarrow 3 rightarrow 65 rightarrow 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a5cf29133e49988d93d0c5cb21fdbdba1913594)
Isbot: Avval oraliq funktsiyani aniqlaymiz
, bu orqali Graham raqamini quyidagicha aniqlash mumkin
. (64-ustki belgi a ni bildiradi funktsional quvvat.)
2-qoida va 4-qoidani orqaga qarab qo'llash orqali biz soddalashtiramiz:
![{ displaystyle f ^ {64} (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d9d62db12fe7163da673a8316a2718e2168029e)
(64 bilan
ning)![{ displaystyle = 3 rightarrow 3 rightarrow (3 rightarrow 3 rightarrow ( cdots (3 rightarrow 3 rightarrow (3 rightarrow 3) rightarrow 1) cdots) rightarrow 1) rightarrow 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8db08ceac0acb82912e2ba8eb714a40ec09d9bd)
![{ displaystyle = 3 rightarrow 3 rightarrow 64 rightarrow 2;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85cf8da9c12e843c3a4451aea27f09a9b642c78b)
![{ displaystyle left. { begin {matrix} = & 3 underbrace { uparrow uparrow cdots cdots cdots cdot uparrow} 3 & 3 underbrace { uparrow uparrow cdots cdots cdots uparrow} 3 & underbrace { qquad ; ; vdots qquad ; ;} & 3 underbrace { uparrow uparrow cdots cdot uparrow} 3 & 3 uparrow 3 end {matrix}} right } { text {64 qatlam}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53d5c0fde631fcf017a7f5b2dd81cd712101110f)
![{ displaystyle f ^ {64} (4) = G;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61f9d75494abb30ce3bfb756e17f49ba083e2814)
(64 bilan
ning)
![{ displaystyle left. { begin {matrix} = & 3 underbrace { uparrow uparrow cdots cdots cdots cdot uparrow} 3 & 3 underbrace { uparrow uparrow cdots cdots cdots uparrow} 3 & underbrace { qquad ; ; vdots qquad ; ;} & 3 underbrace { uparrow uparrow cdots cdot uparrow} 3 & 3 uparrow uparrow uparrow uparrow 3 end {matrix}} right } { text {64 qatlamlar}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93464984c9d1a5e5396dfe45509a9c7881cc32a8)
![{ displaystyle f ^ {64} (27)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e02319843b8d29e0265eddad8b4da66e873b8eb)
(64 bilan
ning)
(65 bilan
ning)
(yuqoridagi kabi hisoblash).![{ displaystyle = f ^ {65} (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/711ec7cea52950fa60c83384287e676a2d6dac34)
![{ displaystyle left. { begin {matrix} = & 3 underbrace { uparrow uparrow cdots cdots cdots cdot uparrow} 3 & 3 underbrace { uparrow uparrow cdots cdots cdots uparrow} 3 & underbrace { qquad ; ; vdots qquad ; ;} & 3 underbrace { uparrow uparrow cdots cdot uparrow} 3 & 3 uparrow 3 end {matrix}} right } { text {65 qatlam}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21a08110ba35326f58896a78ca19660b1e37acc4)
Beri f bu qat'iy ravishda ko'paymoqda,
![{ displaystyle f ^ {64} (1) <f ^ {64} (4) <f ^ {64} (27)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/121b887e5c368d4cb5f9741906b9d3f3fd5a4113)
berilgan tengsizlik.
Zanjirli o'qlar yordamida raqamni kattaroq qilib belgilash juda oson
, masalan,
.
![{ displaystyle 3 rightarrow 3 rightarrow 3 rightarrow 3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba8017fe2262ac3de792e70e4e53a6f408b367de)
![{ displaystyle = 3 rightarrow 3 rightarrow (3 rightarrow 3 rightarrow 27 rightarrow 2) rightarrow 2 ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fdda5bf60a2ca2f413d06917d0e8bd8638c1ded)
![{ displaystyle = f ^ {3 rightarrow 3 rightarrow 27 rightarrow 2} (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/588576db19618f0e6bc04aa1a8e36a4ed815a0a1)
![{ displaystyle = f ^ {f ^ {27} (1)} (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b0b485c557b9cdc4c4ff31399c76a0bd4c85c71)
![{ displaystyle left. { begin {matrix} = & 3 underbrace { uparrow uparrow cdots cdots cdots cdot cdot uparrow} 3 & 3 underbrace { uparrow uparrow cdots cdots cdots cdot uparrow} 3 & 3 underbrace { uparrow uparrow cdots cdots cdots uparrow} 3 & underbrace { qquad ; ; vdots qquad ; ;} & 3 underbrace { uparrow uparrow cdots cdot uparrow} 3 & 3 uparrow 3 end {matrix}} right } chap. { Begin {matrix} 3 underbrace { uparrow uparrow cdots cdots cdots cdot uparrow} 3 3 underbrace { uparrow uparrow cdots cdots cdots uparrow} 3 underbrace { qquad ; ; vdots qquad ; ; } 3 underbrace { uparrow uparrow cdots cdot uparrow} 3 3 uparrow 3 end {matrix}} right } 3 uparrow 3 = { text {27 qatlamlar}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9a73f48e8f8f8d3337799196d5d4f68727330a8)
bu Gremning sonidan ancha katta, chunki bu raqam
![{ displaystyle = f ^ {27} (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80193becc07359c0ca9ba05f2622d017847a6bec)
dan kattaroqdir
![{ displaystyle 65}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b29381eecdc70c14924bb25cdd343405f146ff4b)
.
CG funktsiyasi
Konuey va Gay oddiy, bitta argumentli funktsiyani yaratdilar, bu butun yozuvlar bo'yicha diagonalizatsiya qiladi, quyidagicha tavsiflanadi:
![{ displaystyle cg (n) = underbrace {n rightarrow n rightarrow n rightarrow dots rightarrow n rightarrow n rightarrow n} _ {n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b767c26467944754e2b1f89a3ab13bfac0c9bd3a)
ketma-ketlikni anglatadi:
![{ displaystyle cg (1) = 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a74ab6fe3b328197885cf5e54ce936eb52fc3a1c)
![{ displaystyle cg (2) = 2 dan 2 = 2 ^ {2} = 4} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5bb0846d24af538b363a5c11c5826edde126e63)
![{ displaystyle cg (3) = 3 to 3 to 3 = 3 uparrow uparrow uparrow 3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8aad233ce6dfe9465866704b5b9e909120c1d160)
![{ displaystyle cg (4) = 4 to 4 to 4 to 4} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf5d170ae87bb8d5414205cdc1c481febe2b1ce9)
![{ displaystyle cg (5) = 5 dan 5 dan 5 dan 5 dan 5} gacha](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ae179a88b85029538b952effc6886b549320818)
...
Ushbu funktsiya, kutilganidek, juda tez o'sib boradi.
Piter Xurford tomonidan kengaytirilgan
Veb-dasturchi va statistik mutaxassis Piter Xurford ushbu yozuvning kengaytmasini aniqladi:
![{ displaystyle a rightarrow _ {b} c = underbrace {a rightarrow _ {b-1} a rightarrow _ {b-1} a rightarrow _ {b-1} dots rightarrow _ {b- 1} a rightarrow _ {b-1} a rightarrow _ {b-1} a} _ {c { text {arrow}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da5a7075d93adc33b680f192ba1b6eec6a979a35)
![{ displaystyle a rightarrow _ {1} b = a rightarrow b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b9a3fc2f414c6d87049accc4bdec8477f8a2cc1)
Aks holda barcha normal qoidalar o'zgarmagan.
allaqachon yuqorida aytib o'tilganlarga teng
va funktsiyasi
Conway va Guynikiga qaraganda ancha tez o'smoqda
.
Kabi iboralarga e'tibor bering
agar noqonuniy bo'lsa
va
turli xil raqamlar; bitta zanjirda faqat bitta turdagi o'ng o'q bo'lishi kerak.
Ammo, agar biz buni biroz o'zgartirsak:
![{ displaystyle a rightarrow _ {b} c rightarrow _ {d} e = a rightarrow _ {b} underbrace {c rightarrow _ {d-1} c rightarrow _ {d-1} c rightarrow _ {d-1} dots rightarrow _ {d-1} c rightarrow _ {d-1} c rightarrow _ {d-1} c} _ {e { text {arrow}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/042c21ad33a243f7cee75b9e0f88d07f1101d6ed)
keyin nafaqat qiladi
qonuniy holga keladi, lekin umuman belgi ancha kuchayadi.[2]
Shuningdek qarang
Adabiyotlar
Tashqi havolalar
|
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Birlamchi | |
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Chap dalil uchun teskari | |
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To'g'ri argument uchun teskari | |
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Tegishli maqolalar | |
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Misollar raqamli tartib | |
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Ifoda usullari | |
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Bog'liq maqolalar (alifbo tartibida)
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