Received: (qmail 8230 invoked from network); 26 May 2000 08:05:17 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 26 May 2000 08:05:17 -0000 Received: from pascal.fmph.uniba.sk (158.195.17.101) by cyril.fmph.uniba.sk (MX Fri, 26 May 2000 10:06:39 +0200 Resent-Date: Fri, 26 May 2000 10:04:18 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 58; Fri, 26 May 2000 10:04:07 +0200 Date: Fri, 26 May 2000 10:04:06 +0200 Subject: Skuska DM 26.5. Status: RO Lines: 29 Ahojte, takze tu su niektore (vsetky) z dnesnych prikladov na skuske z DM, vidno, ze staci mat precitane skripta a prepocitane priklady z papierov: 1.Nech x1,..,xn su realne cisla |xi|>1. Dokazte, ze v lubovolnom intervale dlzky 2 mame nie viac ako ( n ) suctov tvaru n ( |_ n/2 _| ) --- \ E . x ,kde E = + 1 /__ k k k - k=1 2.Kazda postupnost n^2+1 roznych prirodzenych cisel obsahuje alebo rastucu alebo klesajucu podpostupnost n+1 prvkov. 3.Uvedte a zdovodnite algoritmus pre najdenie systemu roznych reprezentantov. 4.Dokazte, ze prienik rozkladov (particii) cisla n na navzajom rozne scitance je rovny prieniku rozkladov cisla n na neparne scitance, t.j. kazdy scitanec je neparne cislo. (Uvazujeme neusporiadane particie) 5.Uloha o nepriatelskych dvojiciach. Kolkymi sposobmi mozeme posadit za okruhly stol n nepriatelskych dvojic tak, aby ziadna dvojica nepriatelov nesedela vedla seba ? (stolicky su ocislovane) 6.Zostavte a dokazte Cantorovu vetu. Zdovodnite jej dosledky. Takze vela stastia, hadam vam to na nieco bude Pa Laci Received: (qmail 6268 invoked from network); 30 May 2000 12:57:26 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 30 May 2000 12:57:26 -0000 Received: from turing.fmph.uniba.sk (158.195.16.207) by cyril.fmph.uniba.sk (MX Tue, 30 May 2000 14:58:51 +0200 Resent-Date: Tue, 30 May 2000 14:56:34 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 257; Tue, 30 May 2000 14:55:32 +0200 Date: Tue, 30 May 2000 14:55:31 +0200 Subject: DM 30.5.2000 Status: RO Lines: 33 Takze zdrawim : priklady: 1. Nech A_n je pocet Spernerovych systemov v n-prvkovej mnozine. Potom plati: (2^T_n) 2^T_n < A_n < ( nad ) ( T_n ) 2. Nech M=(S_1,S_2, ... , S_m) je system konecnych neprazdnych mnozin. Pozadujeme aby kazda mnozina mala viacej ako jedneho reprezentanta (r>1) pozadujeme roznost reprezentantov. Vyslovte a dokazte nutnu a postacujucu podmienku existencie roznych reprezentantov. 3. Dokazte ze plati : (n+r-1)! n(n+r-3)! n(n-1) (n+r-5)! n!(n-1)! ------- - -------- + ------.------- - ... + ... = ------- r! 1 (r-2)! 1.2 (r-4)! r!(n-r)! 4. F_0=0 F_(n+2)=F_(n+1) +F_n (klasicka fibonnaciovka) F_1=1 dokazte ze plati F_n=1/5^(1/2)[ ...] ten nerekurzivny vzorcek 5. Odvodte vzorec pre vypocet A'(r) 6. Vyslovte a dokazte Ramseyovu vetu. ------------ He no a potom uz len cakat a cakat .. 'Ste na hrane, musite zabojovat, Odvodte mi E(m,n,r) a F(m,n,r)' trosku taktickeho zdrzovania a .. 'Tak dajte sem index ..' tada sice zatri ale perla se vyplati :))) guud luck hoshi Wol Received: (qmail 6917 invoked from network); 30 May 2000 13:17:56 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 30 May 2000 13:17:56 -0000 Received: from turing.fmph.uniba.sk (158.195.16.207) by cyril.fmph.uniba.sk (MX Tue, 30 May 2000 15:19:20 +0200 Resent-Date: Tue, 30 May 2000 15:17:00 +0200 Received: from 158 (158.195.97.152) by turing.fmph.uniba.sk (MX V5.1 AnBn) with Subject: DM 30.5.2000 Date: Tue, 30 May 2000 15:24:01 +0200 Status: RO Lines: 17 Zdravim! No, pisomku uz mate (aspon ju nemusim pisat ... :-)) ). Takze na ustnej mi Edko najprv dal ulohu o nepriatelskych dvojiciach (stolicky su ocislovane) a potom mi dal zratat horny odhad v prvom priklade (v tomto sa mi ho este nepodarilo obabrat a uznal mi iba dolny). No, nejak haluzne som to odhadol, no poslednu nerovnost, ktoru som dostal som ani za toho ...(krepeho) nevedel zbuchat. No, nakoniec som mu tam napisal nejaku indukciu (za ktoru by som ruku do ohna nedal, aj ked mozno sa dala dotiahnut) a on sa na nu ani nepozrel. Takze nezufajte a obabravajte ... :-))) P.S.:Ten odhad nemal na pisomke tusim nikto a na ustnej ho dal minimalne 3 ludom! Ivan Piliš 1i4 AD D59 Received: (qmail 3586 invoked from network); 13 Jun 2000 14:50:04 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 13 Jun 2000 14:50:04 -0000 Received: from pascal.fmph.uniba.sk (158.195.17.101) by cyril.fmph.uniba.sk (MX Tue, 13 Jun 2000 16:51:39 +0200 Resent-Date: Tue, 13 Jun 2000 16:49:27 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 76; Tue, 13 Jun 2000 16:49:09 +0200 Date: Tue, 13 Jun 2000 16:48:57 +0200 Subject: DM-13.6.2000 Status: RO Lines: 27 Dnes si Toman vymyslel toto: 1. uloha o manzelskych dvojiciach,mame ich usadit okolo okruhleho stola s 2n stolickami tak,aby ziadni manzelia nesedeli vedla seba a striedali sa muzi a zeny... (stolicky su ocislovane) 2. Hallova veta aj s dokazom 3. Nech m,n>2 a cisla R(m,n-1) a R(m-1,n) su parne.Potom plati: R(m,n) < R(m,n-1)+R(m-1,n) 4. Nech |A|=n a |B|=m. Potom: a) |A U B| = n + m ak A a B su disjunktne b) |A x B| = n . m c) |A ^ B| = n ^ m 5. Nech M je mnozina slov dlzky n v abecede A={a1,...ak} (k>=2). Kazde 2 slova mnoziny M sa lisia v aspon dvoch pismenach. Odhadnite zhora mohutnost mnoziny M. 6. Dokazte, ze |Z x <0,1)| = |R|, a potom aj |R|.|N| = |R|. P.S: Aj keby ste to mali dobre, je toho malo. Pridte nabuduce! (Toman) Received: (qmail 22589 invoked from network); 21 Jun 2000 12:23:04 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 21 Jun 2000 12:23:04 -0000 Received: from pascal.fmph.uniba.sk (158.195.17.101) by cyril.fmph.uniba.sk (MX Wed, 21 Jun 2000 14:24:34 +0200 Resent-Date: Wed, 21 Jun 2000 14:22:17 +0200 Received: from CYRIL (CYRIL::SYSTEM) by TURING (MX V5.1 AnBn) with ESMTP (DECnet); Wed, 21 Jun 2000 14:21:52 +0200 Received: from redbull.dcs.fmph.uniba.sk (158.195.18.166) by cyril.fmph.uniba.sk (MX V5.1-X AnBn) with SMTP for Received: (qmail 9450 invoked by uid 721); 21 Jun 2000 12:22:55 -0000 21 Jun 2000 12:22:55 -0000 Date: Wed, 21 Jun 2000 14:22:54 +0200 (CEST) Subject: DM 21.6.2000 Status: RO Lines: 37 oki, takze este nist nedoslo: tak to tu mate: pisomka: 1. konigova veta 2. ohranicet zhora mnozinu cisel ktore mozno vybrat z {1,...,2k+1} tak aby zjadne znix nebolo suctom dvox uz vybranyx (je to <=k+1) 3. nex su dane cele prirodzene cisla n,K a nezaporne cele cisla c1,..,ck. urcte pocet K-prvkovyx kombinacii s opakobanim z n prvkov, ktoryx sa pre kazde i z {1,...,n} poakuje prvok i-ty prvok najvjac ci-krat. ..iny grc: muoj wysledok (ET ho zozral, ale ja by som zan ruku do ohna nedal: n K k /n+K-1\ ---- k+1 /n\ ---- /n-K-SUM-l-1\ | ---- | | - \ (-1) * | | * \ | | | SUM= \ (cij+1) \ K / /___ \k/ /___ \ K-SUM-l / | /___ k = 1 l=SUM+1 j=1 co ine ako princip zapojenia a vypojenia :)) 4. A, B su konecne mneoziny |A|=|B| a exi stuju zobrazenja: f:A->B^k a g:B->A^k. treba dokazat ze ex. bij zobrazenja medzi A a B ktore su urcene f a g... (ET okolo toho dost kecal... ja som to moc nexcapal :)) 5. An - pocet Spernerovyx systemov n-prvkonej mnoziny, Tn = n nad |_ n/2 _|, dokazat ze: 2^Tn < An < 2^Tn nad Tn 6. Dokazat ze kazde prirodzene cislo sa da napisat ako sucet fibonacciho cisel, z ktoryx ani dve njesu "susedne". (induxia) Ustna: navyse Algoritmus... ruozni reprezentanti.. bla bla Zaklad: tvarit sa suverenne :))) c u when u get there ...tbc ________________________________________________________________________________ http://tbc.w3.to Received: (qmail 24305 invoked from network); 29 Jun 2000 11:48:30 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 29 Jun 2000 11:48:30 -0000 Received: from turing.fmph.uniba.sk (158.195.16.207) by cyril.fmph.uniba.sk (MX Thu, 29 Jun 2000 13:50:09 +0200 Resent-Date: Thu, 29 Jun 2000 13:48:12 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 163; Thu, 29 Jun 2000 13:47:53 +0200 Date: Thu, 29 Jun 2000 13:46:08 +0200 Subject: DM - 29. 6. 2000 ustna Status: RO Lines: 11 Cau vsetci Na ustnom som mal len jednu otazku: Nutna a postacujuca podmienka pre spocitatelnost mnoziny Toto je velmi lahke. Toman sa na to pozrel a povedal slaba trojka, dajte index. :-) Drzim vsetkym palce na skuskach. Patrik Received: (qmail 25850 invoked from network); 30 Aug 2000 10:09:30 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 30 Aug 2000 10:09:30 -0000 Received: from turing.fmph.uniba.sk (158.195.16.207) by cyril.fmph.uniba.sk (MX Wed, 30 Aug 2000 12:11:31 +0200 Resent-Date: Wed, 30 Aug 2000 12:06:40 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 24; Wed, 30 Aug 2000 12:06:20 +0200 Date: Wed, 30 Aug 2000 12:05:30 +0200 Subject: DM-30.6.2000-Toman Status: RO Lines: 23 Dnes mozete prejst vsetci,vsetko uz bolo... 1. Hallova veta + algoritmus na najdenie toho systemu (este neviem co tam po nas chcel,no uz viem-cely algoritmus,ha) 2. Eulerova veta o particiach. 3. Cantorova, dokaz + vsetky dosledky 4. x prirodzene, x>0 , nech p(x) je pocet prvocisel neprevysujucich x. Urcte p(n)-p(Sqrt(n)) , n je prir. cislo vacsie ako 1 Urcte pocet prvocisel neprevysujucich 36. 5. Dok, ze pre umocnovanie mohutnosti plati (|A|^|B|)^|C|=|A|^(|B|.|C|) 6. E^n={(alfa_1,...,alfa_n),alfa_i je 0 alebo 1 pre secky prislusne i} alfa~=(alfa_1,...,alfa_n), beta~=(beta_1,....,beta_n); alfa~ krutene_mensie rovne beta~ <=> alfa_i mensie rovne beta_i ;i=1,n (toto su porovnatelne prvky-ostatne neporovnatelne) (E^n,=<) je ciastocne usporiadanie mnoziny E^n, urcte maximalny pocet neporovnatelnych prvkov. Teraz sa uz len caka a caka, a caka... Nic tazke,len to vediet :) ((: No tak sa majte a vela zdaru :)) mayo Received: (qmail 27490 invoked from network); 30 Aug 2000 11:38:54 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 30 Aug 2000 11:38:54 -0000 Received: from pascal.fmph.uniba.sk (158.195.17.101) by cyril.fmph.uniba.sk (MX Wed, 30 Aug 2000 13:40:56 +0200 Resent-Date: Wed, 30 Aug 2000 13:36:00 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 65; Wed, 30 Aug 2000 13:35:45 +0200 Date: Wed, 30 Aug 2000 13:35:44 +0200 Subject: DM oral Status: RO Lines: 10 pisomnu cast uz mate tak teda co mi dal na oralnej malickost: napisat mu vsetko co viem o particiach teda vlastne cele scripta co on napisal myslim ze asi tyzden nebudem moct chytit pero do ruky ked som uz nevladal pisat tak som mu to odovzdal on sa len pozrel a dajte indeks mohol som tam popisat aj trebars ze toman je k... no co uz ale som rad ze to mam za sebou Received: (qmail 28985 invoked from network); 30 Aug 2000 12:57:11 -0000 Received: from cyril.fmph.uniba.sk (158.195.16.200) by hq.alert.sk with SMTP; 30 Aug 2000 12:57:11 -0000 Received: from pascal.fmph.uniba.sk (158.195.17.101) by cyril.fmph.uniba.sk (MX Wed, 30 Aug 2000 14:59:17 +0200 Resent-Date: Wed, 30 Aug 2000 14:54:26 +0200 Received: by st.fmph.uniba.sk (MX V5.1 AnBn) id 33; Wed, 30 Aug 2000 14:54:10 +0200 Date: Wed, 30 Aug 2000 14:53:19 +0200 Subject: DM-30.8.2000-Toman-ustna Status: RO Lines: 16 Este par veci z ustnej Konigova veta aj s tymi haluzami o bipartitnom grafe z jeho salabastrov Pocet permutacii n prvkov tak aby ziadny nebol na svojom mieste... K siestemu prikladu: najst, kedy plati rovnost m=n nad [n/2]. A S , potom aj rozklad m-prvkovej mnoziny na n casti. B Tak zatial vsetko. Statistika : dvaja leteli,ostatok za tri, mozno nejaka vynimka :)