{"id":67,"date":"2022-01-19T13:12:00","date_gmt":"2022-01-19T12:12:00","guid":{"rendered":""},"modified":"2023-08-04T22:56:47","modified_gmt":"2023-08-04T20:56:47","slug":"dirichletov-princip","status":"publish","type":"post","link":"https:\/\/www.jankafialka.sk\/?p=67","title":{"rendered":"Dirichletov princ\u00edp"},"content":{"rendered":"\n

Je tak\u00e1 \u00faloha: ak m\u00e1m v z\u00e1suvke kopu \u010diernych a bielych pono\u017eiek, ko\u013eko mus\u00edm minim\u00e1lne vytiahnu\u0165, aby som mala istotu, \u017ee budem ma\u0165 dve pono\u017eky rovnakej farby? To je \u00faplne typick\u00e1 \u00faloha pre Dirichletov princ\u00edp.<\/p>\n\n\n\n

\"\"
Pono\u017eky v z\u00e1suvke<\/figcaption><\/figure>\n\n\n\n

T\u00e1to \u00faloha sa v r\u00f4znych obmen\u00e1ch pravidelne vyskytuje v matematick\u00fdch s\u00fa\u0165a\u017eiach. Pritom jej rie\u0161enie je a\u017e zar\u00e1\u017eaj\u00faco \u013eahk\u00e9. Sta\u010dia tri. Dve by toti\u017e e\u0161te mohli by\u0165 r\u00f4znej farby.<\/p>\n\n\n\n

Princ\u00edp, ktor\u00fd za rie\u0161en\u00edm stoj\u00ed, je v\u0161ak ve\u013emi pekn\u00fd a u\u017eito\u010dn\u00fd. Naz\u00fdva sa aj z\u00e1suvkov\u00fd princ\u00edp. Dirichletov\u00fdm sa naz\u00fdva po nemeckom matematikovi, ktor\u00fd ho v dev\u00e4tn\u00e1stom storo\u010d\u00ed presne sformuloval. V\u0161eobecne, ak m\u00e1me m<\/i> mno\u017e\u00edn, v ktor\u00fdch je aspo\u0148 m<\/i>+1 prvkov, tak ur\u010dite aspo\u0148 v jednej mno\u017eine musia by\u0165 aspo\u0148 dva prvky. V horeuvedenej \u00falohe s\u00fa dve mno\u017einy: biele a \u010dierne. Preto sta\u010d\u00ed, aby sme do nich vlo\u017eili tri prvky a zaru\u010dene v jednej z nich bud\u00fa aspo\u0148 dve pono\u017eky. \ud83d\ude0a Ak by sme mali pono\u017eky piatich farieb, museli by sme vytiahnu\u0165 aspo\u0148 \u0161es\u0165.<\/p>\n\n\n\n

Existuje e\u0161te jedna pekn\u00e1 \u00faloha, ktor\u00e1 ilustruje e\u0161te v\u0161eobecnej\u0161ie pou\u017eitie Dirichletovho princ\u00edpu. Je trochu vtipn\u00e1, ale ja som ju nevymyslela. Takto si ju pam\u00e4t\u00e1m:<\/p>\n\n\n\n

Muchy na \u0161tvorcovom stole<\/h2>\n\n\n\n

„Na \u0161tvorcovom stole s rozmermi 1×1 meter sed\u00ed 51 m\u00fach. \ud83e\udd23 Dok\u00e1\u017ete, \u017ee je mo\u017en\u00e9 trafi\u0165 aspo\u0148 tri muchy jednou ranou s hrncom, ktor\u00fd m\u00e1 priemer podstavy 30cm.“<\/p>\n\n\n\n

Kto to kedy videl, zab\u00edja\u0165 muchy hrncom? Ve\u010f v\u0161etky uletia. Muchy c\u00edtia tlak vzduchu, ktor\u00fd by ten hrniec tla\u010dil pred sebou. Preto maj\u00fa mucholapky dierky. Ja by som teda v zadan\u00ed ur\u010dite zmenila hrniec za kruhov\u00fa mucholapku. Ale t\u00e1to \u00faloha u\u017e neznie tak jednoducho, ako t\u00e1 pono\u017ekov\u00e1. Preto sa mi p\u00e1\u010di viac.<\/p>\n\n\n\n

Po\u010fme ju teda vyrie\u0161i\u0165. Ak by sme nikdy nepo\u010duli o Dirichletovom princ\u00edpe, asi by sme v\u00f4bec nevedeli, ako za\u010da\u0165. Rozhoduj\u00faca je \u00favaha: ak s\u00fa tie muchy niekde na kope, \u013eahko ich zabi\u0165 ve\u013ea naraz. Jedine, ak by boli na stole rozlezen\u00e9 viacmenej rovnomerne, mohol by to by\u0165 probl\u00e9m. Tak\u017ee uva\u017eujme, \u017ee s\u00fa rozmiestnen\u00e9 rovnomerne. St\u00f4l rozdel\u00edme na nieko\u013eko men\u0161\u00edch rovnak\u00fdch \u010dast\u00ed, do ktor\u00fdch umiestnime 51 m\u00fach tak, \u017ee v ka\u017edej \u010dasti bud\u00fa dve, len v jednej bude o jednu muchu viac. To by bolo to najhor\u0161ie rozmiestnenie. Ka\u017ed\u00e9 in\u00e9 je lep\u0161ie. 51 je dvakr\u00e1t 25 plus jedna. Tak\u017ee \u0161tvorcov\u00fd st\u00f4l m\u00f4\u017eeme pohodlne rozdeli\u0165 na 5×5 men\u0161\u00edch \u0161tvorcov. Cel\u00fd mal stranu d\u013a\u017eky jeden meter. P\u00e4tina bude teda dvadsa\u0165 centimetrov. Zost\u00e1va u\u017e len overi\u0165, \u010di kruhov\u00e1 mucholapka pokryje cel\u00fd jeden \u0161tvorec. To u\u017e je \u00faloha \u010disto geometrick\u00e1: zmest\u00ed sa \u0161tvorec so stranou d\u013a\u017eky 20 cm do kruhu s priemerom 30 cm? \u00daloha je to pomerne \u013eahk\u00e1, ke\u010f si uvedom\u00edme, \u017ee uhloprie\u010dka toho \u0161tvorca mus\u00ed by\u0165 men\u0161ia ako t\u00fdch 30 cm, \u010do je (vypo\u010d\u00edtan\u00e9 pomocou Pytagorovej vety), tak\u017ee naozaj aj pri fakt najhor\u0161om mo\u017enom rozmiestnen\u00ed m\u00fach na stole, v\u017edy bude existova\u0165 miesto, kam s mucholapkou \u0165apn\u00fa\u0165, aby sme zabili aspo\u0148 tri muchy.<\/p>\n\n\n\n

Dirichletov princ\u00edp v be\u017enom \u017eivote<\/h2>\n\n\n\n

Dirichletov princ\u00edp m\u00f4\u017eeme pou\u017ei\u0165 aj na celkom re\u00e1lne ot\u00e1zky. Napr\u00edklad, v\u0161imli sme si, \u017ee v ka\u017edej triede existuje dvojica \u017eiakov, ktor\u00ed maj\u00fa narodeniny v tom istom mesiaci. Za mesiac si m\u00f4\u017eete doplni\u0165 znamenie zverokruhu, ak chcete. Je to pravidlo, alebo n\u00e1hoda? Aplikujeme Dirichletov princ\u00edp a hne\u010f vieme, \u017ee sta\u010d\u00ed, aby v triede bolo aspo\u0148 trin\u00e1s\u0165 \u017eiakov a je jasn\u00e9, \u017ee aspo\u0148 dvaja bud\u00fa ma\u0165 narodeniny v tom istom mesiaci. Ak je v triede aspo\u0148 25 \u017eiakov, tak tam mus\u00ed by\u0165 ur\u010dite aspo\u0148 trojica tak\u00fdchto \u017eiakov.<\/p>\n\n\n\n

Podobne, ak \u017eijem v meste\u010dku s 10000 obyvate\u013emi, m\u00f4\u017eem uva\u017eova\u0165, \u010di je medzi nimi niekto, kto m\u00e1 narodeniny v rovnak\u00fd de\u0148, ako ja. Toto ale Dirichletov princ\u00edp nevyrie\u0161i. M\u00f4\u017eu to by\u0165 v\u0161etci aj nikto. Vyrie\u0161i ale ot\u00e1zku, minim\u00e1lne ak\u00e1 ve\u013ek\u00e1 skupina \u013eud\u00ed v tomto meste oslavuje narodeniny v tom istom dni. Nevieme, v ktorom konkr\u00e9tne, ale \u017ee tak\u00fd de\u0148 existuje. Rok ma 365 dn\u00ed, niekedy 366. Uva\u017eujme teda o 366 z\u00e1suvk\u00e1ch. 10000:366= 27, zv. 118. Tak\u017ee m\u00f4\u017eem rozhodi\u0165 t\u00fdch \u013eud\u00ed po dvadsiatich siedmych do jednotliv\u00fdch dn\u00ed a \u010fal\u0161\u00edch 118 mus\u00edm niekde e\u0161te prida\u0165. Z\u00e1ver: v meste s 10000 obyvate\u013emi zaru\u010dene existuje skupina aspo\u0148 28 \u013eud\u00ed, ktor\u00ed maj\u00fa narodeniny v tom istom dni.<\/p>\n\n\n\n

Je mo\u017en\u00e9, \u017ee by sme podobne mohli zd\u00f4vodni\u0165 aj existenciu dvojn\u00edkov. Sta\u010dilo by spo\u010d\u00edta\u0165, ko\u013eko mo\u017enost\u00ed pre \u010drty tv\u00e1re dok\u00e1\u017eeme vo\u013en\u00fdm okom rozl\u00ed\u0161i\u0165 a ak to bude menej ako osem mili\u00e1rd, tak m\u00e1me d\u00f4kaz, \u017ee musia existova\u0165 aspo\u0148 dvaja \u013eudia, ktor\u00ed vyzeraj\u00fa \u00faplne rovnako. Nie je to n\u00e1hoda, ale nutnos\u0165.<\/p>\n","protected":false},"excerpt":{"rendered":"

Je tak\u00e1 \u00faloha: ak m\u00e1m v z\u00e1suvke kopu \u010diernych a bielych pono\u017eiek, ko\u013eko mus\u00edm minim\u00e1lne vytiahnu\u0165, aby som mala istotu, \u017ee budem ma\u0165 dve pono\u017eky rovnakej farby? To je \u00faplne typick\u00e1 \u00faloha pre Dirichletov princ\u00edp. T\u00e1to \u00faloha sa v r\u00f4znych obmen\u00e1ch pravidelne vyskytuje v matematick\u00fdch s\u00fa\u0165a\u017eiach. Pritom jej rie\u0161enie je a\u017e zar\u00e1\u017eaj\u00faco \u013eahk\u00e9. Sta\u010dia tri. […]<\/p>\n","protected":false},"author":2,"featured_media":1965,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-67","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nezaradene"],"yoast_head":"\nDirichletov princ\u00edp - jankafialka<\/title>\n<meta name=\"description\" content=\"Dirichletov princ\u00edp na pr\u00edklade pomie\u0161an\u00fdch pono\u017eiek. 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