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Rado, na mojom blogu je teraz tiez mrtvo ;-) Ludia su asi v soku z padu investicnych bank, vyrkoov istych politikov, alebo jednoducho skoncili prazdniny.Ja osobne sa na tvoje prispevky vzdy tesim, odkedy som teda objavil QED, a mnohe ulohy ma pekne pobavia! Tato konkretna je naozaj pekna, velmi sa mi paci ze ta take elegantne napadaju!Zatial prezradim, ze riesenie je n^3, pre stranu sestuholnika rozdelenu na n casti. To som moc ale nepovedal, kedze to je lahke uhadnut z tvojich dvoch prikladov. Mne sa to podarilo dokazat pomocou indukcie [mam to tu nacarbane pred sebou na papieri]. Ked budes netrpezlivy s dalsim prikladom, na ktory sa tesim, napis do poznamky a ja sem s radostou napisem moje riesenie. Zaujimalo by ma potom, ci sa na to da ist aj inak! Zatial ma napada len toto jedno riesenie; take follow-your-nose...

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Okej tak predsa len ešte jeden človek sa našiel čo dal do rohov bielu farbu, Vladko Ujházy, takže nie si na svete sám Jaro :D.

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Wow, tak to je blbé, ja som ale použil biele rohy:)) Asi to svedčí o mojej duševnej nevyrovnanosti...

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Viacerí ste si všimli, že keď sa ten mnohouholník vyfarbí ako šachovnica, budú všetky políčka v rohoch rovnakej farby.... dosť by ma ale zaujímalo, prečo si zatiaľ //úplne// všetci riešitelia (a dokonca nezávisle na seba aj ja a Feráč) zvolili za túto farbu práve čiernu??? Že by nejaká drsná psychológia? :)

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Jo, to bola pointa. Akurát som mal tri rôzne sumácie, takže si proste musel trafiť...
Alfa spánok je skvelý na rátanie, kedysi na prípravku som tak zrátal jednu kombinatoriku;)

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(A predpokladam ze je to to iste co vymyslel Iustus a nevymyslel Kubo... ale az teraz mi to doslo.)

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Ach jaj, taky som hrdy na moj vzorak :), uplne je elegantny. Napadlo ma to vcera vecer ked som nemohol zaspat.... a zatial som to nevidel u ziadneho riesitela. Ale mozno sa este najde. Cez mrezove body. Skuste sa zamysliet.

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Myrec to asi skúšal vyriešiť teóriou katastrof:))

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Huh, čože, ako vyzerá mnohouholník v ktorom sa prekrývajú nejaké plochy? Či čo? :)

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uvazovali ste niekto aj taky utvar v ktorom sa prekryvaju nejake plochy aj viackrat(nie len dva krat;) ale aj veeeela krat).. a jednoducho sa to tam tak nejak zauzluje.. lebo fakt neviem si predstavit ze ako to overjete... :(
lebo to ma odradilo od toho to vypocitat cele..

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Hmm, Kubo, keby si to dotiahol, mohol si nám to natrieť ;)
Tvoja hypotéza nielenže platí a tvrdenie v úlohe z nej triviálne vyplýva, ale ešte aj je j dôkaz je jednoduchší a triviálnejší :)))

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nooo mne sa tento priklad pacil a myslim ze ho mam dobre a ani nie velmi zlozito :)

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.... Ale dosiel som na to, ze pocet vrcholov /4 = |pocet ciernych stvorcekov - pocet bielych|.

Predpokladám, že si sporom skúmal pravouholník s nepárnymi stranami, lebo všeobecne ten vzorec neplatí a často ani nedáva zmysel... Tak potom to je úplne super pozorovanie, aj keď som si jeho pravdivosť overil len náhmatkovo. Keďže v sebe zahŕňa tvrdenie, že taký útvar musí mať počet vrcholov deliteľný štvorkou, čo pre nepárnouholníky platí, tak by som mu celkom veril. Musím si to overiť, možno by sa to dalo použiť v jednom zovšeobecnení.
Každopádne, tá úloha má v sebe brutálny potenciál, dáva do súvislosti plošnú a dĺžkovú paritu a dá sa v nej kadečo dokázať ... a vôbec:)

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Skoro som to dal. Taky kusocek som bol od riesenia.... Ale dosiel som na to, ze pocet vrcholov /4 = |pocet ciernych stvorcekov - pocet bielych|. Len som to nevedel dokazat

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Tuším Ty, nie:)) Takže by som to upravil na "Čo si si navaril, ..."

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Uff to som zas dostal priklad :D, ale co sme si navarili.... priznajte sa, vyriesil to niekto?

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