From: Martin Mares <mj@ucw.cz>
Date: Mon, 4 Apr 2011 18:49:15 +0000 (+0200)
Subject: Historicke poznamky presunuty do adresare 2009
X-Git-Url: http://mj.ucw.cz/gitweb/?a=commitdiff_plain;h=93fbc92df9ef7a07f747f7889abe7a89eee4dbaf;p=ads1.git

Historicke poznamky presunuty do adresare 2009
---

diff --git a/11-stromy/11-stromy.tex b/11-stromy/11-stromy.tex
deleted file mode 100644
index 173c36c..0000000
--- a/11-stromy/11-stromy.tex
+++ /dev/null
@@ -1,116 +0,0 @@
-\input ../lecnotes.tex
-
-% Vkladani obrazku
-\input ../mjipe.tex
-\def\treepic#1{
-\medskip
-\IpeInput{treepic/t#1.ipe}
-\medskip
-}
-
-\prednaska{11}{Vyhledávací stromy}{}
-
-Pøedstavme si následující problém: potøebujeme si udr¾ovat urèitá setøídìná data, napøíklad slovník. Polo¾kami jsou uspoøádané dvojice (klíè, hodnota). Klíèe jsou unikátní a jsou to prvky nìjakého lineárnì uspoøádaného universa. Hodnoty mohou být libovolné.
-
-Na na¹ich datech budeme chtít provádìt následující operace: 
-\itemize\ibull
-\:{\it Insert} -- vlo¾it novou polo¾ku
-\:{\it Delete} -- smazat polo¾ku
-\:{\it Find} -- najít polo¾ku
-\:{\it Max $\&$ Min} -- vybrat polo¾ku s~nejvìt¹ím, resp. nejmen¹ím klíèem
-\:{\it Pred $\&$ Succ} -- vybrat polo¾ku s~klíèem nejbli¾¹ím men¹ím, resp. vìt¹ím
-\endlist
-
-Jak by vypadalo nejjednodu¹¹í øe¹ení? Staticky bychom data mohli udr¾ovat v~setøídìném poli. Takové pole se vyrobí v~èase $\Theta(n \cdot \log{n})$. Operace {\it Find} by trvala $\Theta(\log{n})$ (vyhledávali bychom samozøejmì binárnì). Ale problém by nastal s~operacemi {\it Insert} a {\it Delete}, které by se takto implementovat nedaly, nebo by trvaly hodnì dlouho (napø. v~pøípadì {\it Insertu} bychom museli celé pole pøestavìt).
-
-\s{Pozorování:} Proces binárního vyhledávání v~setøídìném poli se dá reprezentovat binárním vyhledávacím stromem.
-
-\s{Definice:} {\I Binární strom:} Strom je binární, pokud je zakoøenìný a ka¾dý vrchol má nejvý¹e dva syny, u nich¾ rozli¹ujeme, který je levý a který pravý.
-
-\s{Definice:} Pro vrchol $v$ znaèíme:
-\itemize\ibull
-\:$l(v)$ a $p(v)$ -- levý a pravý syn vrcholu $v$
-\:$L(v)$ a $P(v)$ -- levý a pravý podstrom vrcholu $v$
-\:$S(v)$ -- pøíslu¹ný podstrom s~koøenem $v$
-\:$h(v)$ -- hloubka stromu $S(v)$, neboli délka nejdel¹í cesty z koøene do listu
-\endlist
-
-\s{Definice:} {\I Binární vyhledávací strom} (BVS): Binární strom je vyhledávací, pokud v~ka¾dém vrcholu je ulo¾ena dvojice (klíè, hodnota) [ztoto¾níme vrchol s~klíèem] a pro v¹echny vrcholy platí: $\left ( \forall u \in L(v) : u < v \right ) $ $ \&$ $ \left ( \forall u \in P(v) : u > v \right )$.
-
-\s{Pøíklady binárních vyhledávacích stromù:}
-
-\treepic{2}
-
-\break
- 
-Jak budou tedy vypadat operace {\it Find}, {\it Insert} a {\it Delete} na~binárním vyhledávacím stromu?
-
-{\bo Find$(v,x)$:}
-\algo
-\:Pokud $v = \emptyset \Rightarrow$ vrátíme $\emptyset$.
-\:Pokud $v = x \Rightarrow$ vrátíme $v$.
-\:Pokud $v < x \Rightarrow$ vrátíme $\<Find>(p(v),x)$.
-\:Pokud $v > x \Rightarrow$ vrátíme $\<Find>(l(v),x)$.
-\endalgo
-
-{\bo Insert$(v,x)$:}
-\algo
-\:Jako \<Find> a na~konci pøidáme list.
-\endalgo
-
-{\bo Delete$(v)$:}
-\algo
-\:Pokud $v$ je list $\Rightarrow$ jednodu¹e list utrhneme.
-\:Pokud $v$ má jednoho syna $\Rightarrow$ vrchol \uv{vyøízneme}.
-\:Jinak má $v$ oba syny $\Rightarrow$ do vrcholu vlo¾íme minimum z $P(v)$, co¾ bude list, a ten utrhneme.
-\endalgo
-
-\s{Poznámka:} Pokud má vrchol $v$ pøi operaci {\it Delete} oba syny, je vlo¾ení minima z $P(v)$ ekvivalentní s~vlo¾ením maxima z $L(v)$.
-
-\s{Pøíklady operací  {\it Insert} a {\it Delete} na~BVS:}
-\treepic{1}
-\treepic{3}
-
-\break
-Èasová slo¾itost v¹ech tøí operací je $\Theta(\<hloubka stromu>)$, co¾ mù¾e být $\Theta(n)$, kdy¾ budeme mít smùlu a strom bude (témìø) lineární spojový seznam, nebo $\Theta(\log{n})$ kdy¾ bude strom pìknì vyvá¾enì vystavìný. Vidíme tedy, ¾e slo¾itost operací stojí a padá s~hloubkou stromu. Proto by se nám líbilo, aby mìl ná¹ strom v¾dy hloubku $\Theta(\log{n})$. Podívejme se tedy, jak se dá navrhnout binární vyhledávací strom, aby tuto podmínku splòoval \dots
-
-\h{Vyvá¾ené binární vyhledávací stromy}
-
-\s{Definice:} {\I Dokonalá vyvá¾enost:} Strom je dokonale vyvá¾ený, pokud pro v¹echny jeho vrcholy platí: $\left \vert \vert L(v)\vert - \vert P(v)\vert \right \vert \leq 1 $.
-
-Takto definovaný binární strom bude mít urèitì logaritmickou hloubku. Jak takový strom ale konstruovat? To se nám podaøí buï staticky, nebo na~nìm budou operace dra¾¹í ne¾ $\Theta(\log{n})$.
-
-\s{Statická konstrukce dokonale vyvá¾eného BVS:} Vybereme prostøední prvek ze setøídìného pole (tedy medián posloupnosti) a dáme ho do~koøene stromu. Jeho syny pak vystavíme rekurzivnì z~levé a pravé pùlky pole. Celá konstrukce tedy trvá $\O(n)$.
-
-Vidíme tedy, ¾e to ná¹ problém pøíli¹ neøe¹í. Potøebovali bychom, aby se strom dal také efektivnì udr¾ovat. Zkusíme proto slab¹í podmínku:
-
-\s{Definice:} {\I Hloubková vyvá¾enost:} Strom je hloubkovì vyvá¾ený, pokud pro v¹echny jeho vrcholy platí: $\left \vert h(L(v)) - h(P(v)) \right \vert \leq 1 $.
-
-Stromùm s~hloubkovým vyvá¾ením se øíká {\I AVL stromy} (objeviteli je ru¹tí matematikové G. M. Adelson-Velsky a E. M. Landis, podle nich jsou také pojmenovány) a platí o~nich následující lemma:
-
-\s{Lemma:} AVL strom na~$n$ vrcholech má hloubku $ \Theta(\log{n}) $.
-
-\proof
-Uva¾me posloupnost $A_k = $ minimální poèet vrcholù AVL stromù hloubky $k$. Staèí ukázat, ¾e $A_k$ roste exponenciálnì. Podívejme se na minimální AVL stromy:
-$$\eqalign{
-A_0 &= 1 \cr
-A_1 &= 2 \cr
-A_2 &= 4 \cr
-A_3 &= 7 \cr
-&\vdots \cr
-A_k &= 1 + A_{k - 1} + A_{k - 2}. \cr
-}$$
-
-Rekurentní vzorec jsme dostali rekurzivním stavìním stromu hloubky $k$: nový koøen a 2 podstromy o~hloubkách $k - 1$ a $k - 2$.
-
-Vidíme tedy, ¾e $A_n = F_{n + 2} - 1$. (Mù¾eme dokázat napø. indukcí.)
-Teï nám ji¾ staèí dokázat, ¾e posloupnost $A_k$ roste exponenciálnì S výhodou mù¾eme vyu¾ít toho, ¾e na první pøedná¹ce jsme si ji¾ dokázali, ¾e Fibonacciho èísla rostou exponenciálnì. Nicménì pro zapomnìtlivé mù¾eme dùkaz ve struènosti zopakovat:
-
-Indukcí doká¾eme, ¾e $ A_k \geq 2^{k \over 2} $.
-První indukèní krok jsme si u¾ ukázali, teï pro $ k \geq 2 $ platí:
-$ A_k = 1 + A_{k - 1} + A_{k - 2} > 2^{{k - 1} \over 2} + 2^{{k - 2} \over 2} = 2^{k \over 2} \cdot (2^{-{1 \over 2}} + 2^{-1}) \cong 2^{k \over 2} \cdot 1.21 > 2^{k \over 2}.$
-
-Tímto jsme dokázali, ¾e na~ka¾dé hladinì je minimálnì exponenciálnì vrcholù, co¾ nám zaruèuje hloubku $ \Theta(\log{n})$.
-\qed
-
-\bye
diff --git a/11-stromy/Makefile b/11-stromy/Makefile
deleted file mode 100644
index 6f4f489..0000000
--- a/11-stromy/Makefile
+++ /dev/null
@@ -1,3 +0,0 @@
-P=11-stromy
-
-include ../Makerules
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diff --git a/12-hash/12-hash.tex b/12-hash/12-hash.tex
deleted file mode 100644
index f12db20..0000000
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-\input lecnotes.tex
-
-\prednaska{12}{Hashování}{}
-
-V pøedchozích kapitolách jsme se zabývali tøídìním prvkù a zjistili jsme, ¾e v obecném pøípadì (kdy¾ smíme prvky pouze porovnávat a prohazovat) nelze tøídit rychleji ne¾ v èase $\Theta(n\log n)$. Zároveò jsme do¹li k tomu, ¾e pokud víme o prvcích nìco více (napø. jejich rozsah), tak jsme v nìkterých pøípadech schopni tøídit i s lineární èasovou slo¾itostí. 
-
-V praxi nicménì vìt¹inou netøídíme samotná èísla, ale spí¹e nìjaké polo¾ky, napø. urèené uspoøádanou dvojicí (klíè, hodnota), kde klíèe jsou unikátní a existuje na nich lineární uspoøádání. Ukázali jsme si, jak takovouto mno¾inu udr¾ovat (aby se v ní dalo rychle vyhledávat, zaøazovat i mazat) ve vyhledávacích stromech. Dokázali jsme, ¾e pokud budeme udr¾ovat strom dostateènì vyvá¾ený, tak budou jednotlivé operace (\<Find>, \<Insert> a \<Delete>) trvat $\Theta(\log n)$.
-
-Ne¹lo by to ale rychleji? Odpovìï je \uv{jak kdy}. Podívejme se na dal¹í mo¾nosti. Nejdøíve si ale pøipomeòme základní pojmy, které budeme pou¾ívat.
-
-Klíèe budou prvky universa, co¾ je mno¾ina $\{ 0, \dots, U-1 \}$, kterou budeme oznaèovat $[U]$. Tuto mno¾inu budeme chtít udr¾ovat v nìjaké ¹ikovné datové struktuøe, abychom v ní mohli rychle vyhledávat, zaøazovat i mazat. Dále $n$ bude znaèit (maximální) poèet prvkù, které si budeme v jednu chvíli pamatovat.
-
-
-\h{1. Pole {\it indexované od $0$ do $U-1$}}
-
-Pokud bude velikost universa dostateènì malá, tak si mù¾eme polo¾ky pamatovat v poli o velikosti universa. Operace \<Find>, \<Insert> i \<Delete> pracují v~èase $\O(1)$, ale platíme za~to pamì»ovou slo¾itostí $\Theta(U)$. Kdy¾ uvá¾íme, ¾e si budeme pamatovat mnohem ménì prvkù, ne¾ je velikost universa, tak je to velké plýtvání.
-
-\h{2. Èíslicový strom}
-
-Zvolíme vhodný základ soustavy $b$, klíè zapí¹eme v soustavì o základu $b$ a zápis ulo¾íme do stromu. Tímto nám vznikne \uv{$b$-ární strom}, tedy strom, jeho¾ ka¾dý vrchol má a¾ $b$ synù.
-
-Na první hladinì se vìtvíme podle první èíslice, na druhé podle druhé, atd. Na poslední hladinì bude tedy $n$ listù. Hloubka tohoto stromu je $\lceil\log_b U\rceil$.
-
-Operace \<Find> bude trvat $O(\log_b U)$ (najití jednoho klíèe odpovídá projití jedné cesty ve stromì od koøene do listu). Stejný èas budou trvat i operace \<Insert> a \<Delete>. (Pøi vkládání zakládáme vrcholy, které odpovídají na¹emu klíèi a které je¹tì nejsou zalo¾ené. Pøi mazání sma¾eme list a následnì vrcholy, ze kterých nic nevede. Pro tuto operaci je výhodné si ve vrcholech udr¾ovat jejich stupnì.) Strom zabere prostor velikosti $O(nb\log_b U)$. 
-
-
-Èím bude základ $b$ men¹í, tím bude lep¹í pamì»ová a hor¹í èasová slo¾itost
-a obrácenì. Èíslicový strom tedy vy¾aduje peèlivé nastavení parametrù.
-
-\s{Pøíklad:} Pøedstavme si, ¾e klíèe budou IP-adresy, tedy 32-bitová èísla. Jak zvolit $b$, aby byla èasová i pamì»ová slo¾itost pøijatelná? Zkusme rozdìlit IP-adresu na 4 kusy, $b$ bude tedy 256. Strom bude mít hloubku 4. Èas bude v poøádku ($\log_b U = 4$), ale po¾adavky na prostor jsou dost vysoké($nb\log_b U = n \cdot 256\cdot 4 = 1024$). Výhodnìj¹í bude rozdìlit IP-adresu na 8 èástí, $b$ bude tedy 16, hloubka stromu 8 a nároky na èas i pamì» jsou pøijatelné, nebo» $\log_b U = 8$ a $nb\log_b U = n\cdot 16 \cdot 8 = 128$.
-
-
-\h{4. Hashování }
-
-Mìjme $m$ pøihrádek a v ka¾dé z nich spojový seznam. Dále $h$ bude hashovací funkce, která libovolnému prvku z universa pøiøadí právì jednu pøihrádku. 
-
-Kdyby $h$ fungovala \uv{rovnomìrnì } (ka¾dému prvku by pøiøadila jinou pøihrádku, tedy ka¾dý spojový seznam by mìl jen jeden prvek), tak bychom umìli vyhledávat, pøidávat i mazat prvky v konstantním èase. To je ov¹em samozøejmì pøíli¹ krásné na to, aby to byla pravda.
-
-Problém je, ¾e pokud bude $m$ výraznì men¹í ne¾ $U$ (co¾ chceme, jinak se jedná o normální (obrovské)  pole a funkce je identita -- tento pøípad jsme ji¾ vyøe¹ili), tak urèitì budou existovat prvky, kterým funkce pøiøadí stejnou hodnotu. Takovému pøípadu se øíká {\it kolize}. Prvky se pak zaøadí do spojového seznamu. Kdyby byla naopak funkce hodnì ne¹ikovná a namapovala by nám v¹echny prvky do jedné pøihrádky, tak budeme v¾dy muset projít (v nejhor¹ím pøípadì celý) dlouhý spojový seznam.
-
-Jak se dá tedy kolizím bránit? Nejvýhodnìj¹í by bylo mít dobrou hashovací funkci, která by mapovala prvky pokud mo¾no rovnomìrnì. Ale nìkdy staèí i prùmìrnì dobrá funkce a dostateènì náhodné klíèe.
-
-
-\s{Vsuvka:} Jaká je pravdìpodobnost, ¾e 2 z k lidí mají narozeniny ve stejný den? Staèí 23 lidí, aby nastala kolize s pravdìpodobností vìt¹í ne¾ jedna polovina. Aby nenastala, potøebovali bychom vìt¹í poèet pøihrádek: $m \approx n^2$.
-
-\h{Praktické hashovací funkce}
-
-\s{První pøíklad}
-
-Tato hashovací funkce $h: [U] \rightarrow [m]$ je definovaná pøedpisem: $h(x) = x \mod m$. Pro dostateènì náhodné hodnoty bude fungovat, ale v mnoha pøípadech mù¾e být nevhodná. Napøíklad je zøejmé, ¾e pro sudé $m$ zachovává paritu. V pøípadì ¾e sudých, resp. lichých klíèù bude mnohem více, vznikne mnoho zbyteèných kolizí. Pro takovouto hashovací funkci je nejlep¹í zvolit za $m$ nìjaké prvoèíslo.
-
-\s{Druhý pøíklad}
-
-Zvolme si iracionální $\alpha$ z intervalu $(0,1)$. Funkce bude definovaná následovnì: $$h(x) = \lfloor m (x \cdot \alpha \mod 1) \rfloor.$$ Tato funkce by mìla na $x$ záviset velmi málo, ale nebudeme si to dokazovat.
-
-\s{Implementace:} Reálná èísla se samozøejmì implementují dosti nároènì (resp. to vùbec nejde), tak¾e v praxi se tento postup aproximuje celoèíselnì. Místo reálné $\alpha$ si zvolme $A = \lfloor \alpha \cdot 2^w \rfloor$, kde $2^w$ je ¹íøka slova ($w$ je napø. poèet bitù integeru). Potom místo výrazu $(x \cdot \alpha \mod 1)$, který je z intervalu $(0,1)$, budeme poèítat s výrazem $(x \cdot A \mod 2^w)$. Potom na¹e funkce vyjde následovnì:
-
-$$h(x) = \left \lfloor { m(x \cdot A \mod 2^w) \over 2^w } \right \rfloor$$
-
-
-Zbývá jen otázka, jak volit $\alpha$? Osvìèená hodnota je $1/\tau$, co¾ je pøibli¾nì 0,618. ($\tau$ je hodnota zlatého øezu.) Takto zvolené $\alpha$ dobøe rozbíjí aritmetické posloupnosti, ale i toto je pouze pøedstava a nebudeme si to dokazovat.
-
-\h{Hashování øetìzcù}
-
-Vzhledem k tomu, ¾e hashování nijak speciálnì nepotøebuje, aby byly prvky èísla, tak se mohou hashovat i øetìzce. Uka¾me si jeden pøíklad funkce pro hashování øetìzcù.
-Prázdnému øetìzci pøiøaïme nulu: $h(\varepsilon) = 0$. Øetìzci znakù $x_1, \dots, x_n$ pøiøaïme hodnotu:
-$$h(x_1, \dots, x_n) = (h(x_1, \dots, x_{n-1}) \cdot + x_n) \mod m.$$
-$A$ je zde nìjaká konstanta. Aby se vyu¾ily v¹echny pøihrádky, tak je vhodné zvlit $A$ tak, aby $A^{{\bb E}[\<délka øetìzce>]} \gg m$. Tato funkce vyu¾ívá podobných vlastností jako lineární konkurenèní generátor náhodných èísel, který pomocí vhodných konstant $A$ a $B$ a ¹íøky slova $w$ vyrábí nové náhodné èíslo $x'$ z minulého èísla $x$ jako $$x' = (x \cdot A + B) \mod 2^w.$$
-
-
-\s{Vìta} Pokud jsou $h(x_1), \dots, h(x_n)$ rovnomìrnì náhodné, pak $$\forall i: {\bb E}[\<poèet prvkù v i-té pøihrádce>] = {n \over m}.$$
-
-\proof
-Poèet prvkù v $i$-té pøihrádce si oznaème $N_i$. Definujme si indikátor $P_{ij}$, který je roven 1, pokud $x_j$ padlo do $i$-té pøihrádky, jinak je roven 0.
-Vidíme, ¾e poèet prvkù v $i$-té pøihrádce je souèet $P_{ij}$ pøes v¹echna $j$.
-$$N_i = \sum_{j} P_{i,j}.$$
-Dále vyu¾ijeme mimo jiné vìtu o linearitì støední hodnoty.
-$${\bb E}[P_{ij}] = {1 \over m} \cdot 1 + \left(1 - {1 \over m} \right) \cdot 0 = {1 \over m}.$$
-$${\bb E}[N_i] = {\bb E}\left[\sum_{j} P_{i,j}\right] = \sum_{j} {\bb E}[P_{i,j}] = {n \over m}.$$
-\qed
-
-\s{Dùsledek:} Pokud bude $n$ pøibli¾nì stejnì velké jako $m$, tak v ka¾dé pøihrádce bude prùmìrnì jeden prvek.
-
-\bye
diff --git a/12-hash/Makefile b/12-hash/Makefile
deleted file mode 100644
index e0cfe0f..0000000
--- a/12-hash/Makefile
+++ /dev/null
@@ -1,3 +0,0 @@
-P=12-hash
-
-include ../Makerules
diff --git a/2-ram/2-ram.tex b/2-ram/2-ram.tex
deleted file mode 100644
index 71c3a25..0000000
--- a/2-ram/2-ram.tex
+++ /dev/null
@@ -1,245 +0,0 @@
-\input ../lecnotes.tex
-
-\prednaska{2}{Slo¾itost, grafové algoritmy}
-{(zapsal Martin Koutecký)}
-
-\h{Model {\sc Ram}}
-
-Pøi analýze algoritmu bychom chtìli nìjak popsat jeho slo¾itost. Abychom mohli
- udìlat toto, potøebujeme nejprve definovat výpoèetní model. Výpoèetních modelù
-je více, my vybereme jeden pomìrnì blízký skuteèným poèítaèùm:
-
-\s{Definice:} Random Access Machine ({\sc Ram})
-
-{\sc Ram} poèítá jen s celými èísly -- znaky, stringy a podobnì reprezentujeme
-èísly, jejich posloupnostmi atd. Pamì» je tvoøena buòkami, které obsahují
-èísla. Pamì»ové buòky jsou adresované takté¾ èísly. A program samotný je
-koneèná posloupnost instrukcí následujících druhù:
-\itemize\ibull
-\:Aritmetické a logické:
-$X$ |<-| $Y \oplus Z, \oplus\in\{|+|, |-|, |*|, |div|, |mod|, \&,
-{\tt\char124}, |<<|, |>>|\}$
-\:Øídící: |goto| \<label>, |halt|
-\:Podmínky: pro libovolnou nepodmínìnou instrukci mù¾u pou¾ít \hfil \break
-if~$X$~|<|~$Y$~|==>|~instrukce % Tady to prosím je¹tì zkontroluj. Myslím, ¾e
-% zápis je správný, ale sází se to divnì a vidím èerný obdélníèek na konci
-% øádku. Díky.
-\endlist
-
-\s{Poznámka} (operandy):
-\itemize\ibull
-\:Konstanty (1, 2, \dots)
-\:Adresované pøímo -- M[konst.] -- budeme pou¾ívat písmena {\tt A-Z} jako aliasy
-pro
-buòky pamìti $-1$ a¾ $-26$, které nazýváme registry.
-(tedy A={\tt M[-1]})
-\:Adresované nepøímo -- {\tt M[M[konst.]]} -- budeme pou¾ívat zkratku [[konst.]]
-\endlist
-
-Samotný výpoèet probíhá takto:
-\algo
-\:Do smluvených bunìk umístíme vstup, obsah zbylých pamì»ových bunìk není
-definován.
-\:Provádíme program postupnì po instrukcích, dokud nedojdeme k haltu nebo konci
-programu.
-\:Pokud se program nezacyklil, tedy pokud skonèil, ze smluvených bunìk pøeèteme
-výstup.
-\endalgo
-
-
-\h{Slo¾itost}
-\> Jak dobøe popsat slo¾itost?
-\numlist\ndotted
-\:{\I {\sc Ram} s jednotkovou cenou}: èas $\approx$ \#instrukcí, prostor
-$\approx$
-maximální èíslo buòky minus minimální èíslo buòky pou¾ité pøi výpoètu.
-Toto není moc dobrý nápad, proto¾e není nijak penalizována napøíklad práce s
-velmi dlouhými èísly -- poøád je to jedna instrukce, tak¾e cena je stejná, ale
-poèítaèe se tak pøece nechovají. Velikost èísel ale omezit nesmíme, proto¾e
-bychom omezili pamì» (èísly ji adresujeme).
-\:{\I{\sc Ram} s logaritmickou cenou}: cena instrukce $\approx$ \#bitù
-zpracovávaných èísel,
-prostor $\approx$ \# bitù v¹ech pou¾itých bunìk. To je teoreticky pøesné, ale
-dost nepraktické (ve v¹ech slo¾itostech by byly logaritmy).
-\:{\I{\sc Ram} s omezenými èísly}: jednotková cena instrukcí, ale èísla omezíme
-nìjakým polynomem $P(n)$. Tím zmizí paradoxy prvního modelu, ale
-mù¾eme adresovat jen polynomiální prostor (to nám ov¹em obvykle nevadí).
-\endlist
-
-Nadále tedy budeme pøedpokládat tøetí zmínìný model.
-
-% Z minulých zápiskù.
-\s{Definice:}
-\itemize\ibull
-\:{\I Èas bìhu algoritmu} $t(x)$ pro vstup~$x$ mìøíme jako sumu èasù provedených
-operací, které program provedl pøi zpracování vstupu
-$x$.
-\:{\I Prostor bìhu algoritmu} $s(x)$ je analogicky poèet pamì»ových
-bunìk spotøebovaných pøi výpoètu se vstupem~$x$.
-\:{\I Èasová slo¾itost} (v~nejhor¹ím pøípadì) je:
-$$T(n) := \max \{t(x) ; \hbox{$x$ je vstup délky $n$}\}.$$
-\:{\I Prostorová slo¾itost} (v~nejhor¹ím pøípadì) je:
-$$S(n) := \max \{s(x) ; \hbox{$x$ je vstup délky $n$}\}.$$
-\endlist
-
-Nyní zkusíme zanalyzovat nìjaký konkrétní algoritmus. Vezmìme napøíklad øazení
-pomocí pøímeho výbìru (selection sort). Na vstupu dostaneme poèet èísel $n$ (v
-registru {\tt N}), v buòkách $1,\dots, n$ je nesetøídìná posloupnost èísel. Ta
-pak
-tøídíme následujícím algoritmem zapsaným v pseudokódu:
-\algo
-\:Pro $i=1$ do $n$:
-\::$j\leftarrow i$
-\::Pro $k=i$ do $n$:
-\:::Je-li $[k]<[j]\Rightarrow j\leftarrow k$
-\::$[i]$ prohodíme s $[j]$.
-\endalgo
-
-Jak by takový algoritmus vypadal zapsaný v instrukcích {\sc Ram}? Budeme muset
-pou¾ít návì¹tí a goto místo cyklù, jména registrù místo promìnných a
-tøeba prohození musíme provést pøes tøetí promìnnou. Nìjak takto:
-
-\verbatim{	I <- 1
-LOOP:	J <- I
-	M <- I
-MIN:	IF [J]<[M] ==> M <- J
-	J <- J+1
-	IF J<=N ==> GOTO MIN
-	X <- [I]
-	[I] <- [M]
-	[M] <- X
-	I <- I+1
-	IF I<=N ==> GOTO LOOP
-}
-
-Pojïme se podívat, jaká je èasová slo¾itost jednotlivých èástí algoritmu.
-Cyklus |MIN| provede $3\cdot (N-I+1)$ instrukcí. Mimo cyklu |MIN| je v |LOOP|
-je¹tì 7 instrukcí, tedy celý |LOOP| provede $3\cdot (N-I+1)+7$ instrukcí.
-Celkovì se dostáváme k souètu
-$$1+3\cdot N+7+3\cdot (N-1)+7+3\cdot (N-2)+7+3\cdot (N-3)+\dots +3\cdot 1+7 =$$
-$$1+7\cdot N+3\cdot {{N(N+1)}\over{2}} = {{3}\over{2}}N^2 + 8{,}5N + 1$$
-
-Na multiplikativních konstantách ale nezále¾í -- zaprvé se na reálných strojích
-ceny jednotlivých (pro nás jednotkových) instrukcí stejnì li¹í, zadruhé
-asymptoticky pomalej¹í funkce nakonec pro velké $N$ v¾dy prohraje, tak¾e nemá
-cenu
-(alespoò pøi prvním pøiblí¾ení k problému) multiplikativními konstantami se
-zabývat. Tím pádem nezále¾í ani na èlenech ni¾¹í øádù:
-$$1{,}5N^2 + 8{,}5N + 1 \leq 1{,}5N^2 + 8{,}5N^2 + N^2 = 11N^2\approx N^2$$
-Kdy¾ u¾ toto víme, mù¾eme zanedbávat konstanty prùbì¾nì: $N$ cyklù po
-${}\approx~N$~krocích $\Rightarrow~\approx~N^2$ krokù. To nás vede k zavedení
-tzv.
-{\I asymptotické notace:}
-
-\h{Asymptotická notace}
-\s{Definice:} Pro funkce $f,g: {\bb N} \rightarrow {\bb R}^+$ øekneme,
-¾e $f$ je $\O(g)$ právì tehdy, kdy¾ $\exists c>0: \forall ^{*} n \in {\bb N}:
-f(n) \leq c \cdot g(n)$.
-Zde $\forall^* n \in {\bb N}$ je zkratka za \uv{$\exists n_0 \in {\bb N}:
-\forall n \geq n_0$}, tedy
-\uv{pro v¹echna~$n$ a¾ na~koneènì mnoho výjimek.}
-
-\s{Poznámka:} $\O$-notace tedy vyjadøuje, ¾e funkce~$f$ je skoro v¹ude men¹í
-nebo nejvý¹e rovná nìjakému reálnému násobku funkce~$g$. Aèkoliv zápis vypadá
-jako rovnost, rozhodnì není symetrický: napøíklad platí $\log n=\O(n)$, ale
-neplatí $n=\O(\log n)$. Formálnì by bylo lep¹í pova¾ovat $\O(g)$ za tøídu
-funkcí, pro které platí, ¾e se dají shora odhadnout kladným násobkem funkce~$g$,
-a~psát tedy~$f\in\O(g)$, ale zvyk je bohu¾el ¾elezná ko¹ile.
-
-\s{Pøíklady:} $2{,}5n^{2} = \O(n^{2})$, $2{,}5n^{2}+30n = \O(n^{2})$.
-
-\>Také platí:
-$$
-\O(f)+\O(g) \in \O(f+g),
-$$
-èím¾ myslíme, ¾e pokud vezmeme libovolnou $f'=O(f)$ a $g'=O(g)$, bude
-$f'+g'=O(f+g)$.
-To platí, jeliko¾ skoro v¹ude je $f' \leq cf$, $g'\leq dg$, a~tedy $f'+g' \le
-cf+dg \le (c+d)(f+g)$.
-
-\s{Cvièení:} Uka¾te, ¾e:
-\itemize\ibull
-\:$\O(f) \cdot \O(g)=\O(f \cdot g)$,
-\:$\O(f+g)=\O(\max(f,g))$,
-\:$\O(n^{2})+\O(n)=\O(n^{2}+n)=\O(n^{2})$.
-\endlist
-
-$\O$-notace popisuje horní odhad asymptotického chování funkce. Mnohdy v¹ak
-potøebujeme také odhadnout funkci zespodu (chceme-li øíci, ¾e algoritmus
-potøebuje {\I alespoò} nìjaké mno¾ství èasu nebo pamìti), pøípadnì z~obou stran:
-
-\s{Definice:}
-
-\itemize\ibull
-\:$f=\Omega(g) \equiv \exists c>0:\forall^* n \in {\bb N}: f(n) \geq c\cdot
-g(n)$.
-
-$\Omega$-notace tedy øíká, ¾e hodnota funkce $f$ je v¾dy stejná nebo vy¹¹í ne¾
-nìjaký $c$-násobek funkce $g$, a tedy $g=\O(f)$.
-\:$f=\Theta(g) \equiv f=\O(g) \wedge f=\Omega(g)$
-
-nebo výøeènìji:
-
-$f=\Theta(g) \equiv \exists$ $c_{1},c_{2} > 0: \forall^* n \in {\bb
-N}: c_{1}\cdot g(n) \leq f(n) \leq
-c_{2}\cdot g(n)$ To znamená, ¾e existují nezáporné reálné konstanty
-$c_{1},c_{2}$ takové, ¾e se funkce $f(n)$ dá ohranièit $c_{1}$- a
-$c_{2}$-násobkem funkce $g(n)$.
-\endlist
-
-\s{Porovnání rùstu funkcí:} (aneb jak moc máme algoritmy rádi podle jejich
-chování od~nejlep¹ích k~nejhor¹ím)
-
-\itemize\ibull
-\:$\Theta(1) \ldots$ funkce zespoda i shora ohranièené konstantami
-\:$\Theta(\log{( \log{n} )})$
-\:$\Theta(\log{n})$ \dots\ logaritmická
-\:$\Theta(n^{\varepsilon}), \varepsilon \in (0,1)$ \dots\ sublineární
-\:$\Theta(n)$ \dots\ lineární
-\:$\Theta(n^{2})$ \dots\ kvadratická
-\:$\Theta(n^{k})$ \dots\ polynomiální
-\:$\Theta(2^{n})$ \dots\ exponenciální pøi základu $2$
-\:$\Theta(3^{n})$ \dots\ exponenciální pøi základu $3$
-\:$\Theta(k^{n})$ \dots\  exponenciální pøi základu $k>1$
-\:$\Theta(n!)$ \dots\ faktoriálová
-\:$\Theta(n^{n})$
-\:\dots\ nekoneènì mnoho dal¹ích tøíd (i mezi tìmi vý¹e uvedenými)
-\endlist
-
-\>{\sl Poznámka:} Pokud se v~odhadu slo¾itosti vyskytne logaritmus (jinde ne¾
-v~exponentu), nezále¾í na tom, jaký má základ, proto¾e platí:
-$$
-\log_k{n}={{\log_c{n}}\over{\log_c{k}}}={{1}\over{\log_c{k}}}\cdot \log_c{n},
-$$
-kde $1/\log_c{k}$ je jen konstanta, tak¾e ji mù¾eme \uv{schovat do~$\O$}.
-
-\s{Pøíklad:} Select sort (rozebraný vý¹e):
-Kdy¾ jej pustíme na $n$ èísel, pak èasová slo¾itost je $T(n) = \Theta(n^2)$ a
-prostorová $S(n) = \Theta(n)$.
-
-\h{Úvod do grafových algoritmù}
-Dal¹í dùle¾itou a zajímavou kapitolou jsou grafové algoritmy. Napøíklad
-následující pøíklady lze (i kdy¾ to tak obèas na první pohled nevypadá) øe¹it
-nìjakým grafovým algoritmem:
-\itemize\ibull
-\:Mám mapku silnièní sítì, v ní místa (vrcholy) oznaèená \uv{Doma} a \uv{©kola}.
-Dostanu se do
-¹koly (le¾í ve stejné komponentì souvislosti)? Dostanu se do ¹koly, kdy¾ v zimì
-napadne hodnì snìhu a nìkteré cesty budou neprùjezdné? A jaký nejkrat¹í úsek
-cest musí silnièáøi prohrnout, aby byla v¹echna místa na mapì dostupná?
-\:Mìjme hlavolam \uv{Lloydova devítka} -- krabièku $3\times3$ se ètvereèky
-oznaèenými èísly od jedné do osmi a jednou mezerou, ètvereèky jsou zamíchané a
-na¹ím úkolem je správnì je seøadit pomocí pøesouvání ètvereèkù sousedících s
-mezerou do této mezery. Jak to udìlat? Kolik nejménì krokù nám na
-to staèí? Jde to vùbec se zadáním, které jsme dostali?
-\:Jaké je nejkrat¹í (kladné, celé) èíslo v desítkové soustavì zapsané jen
-èíslicemi 1, 0, které je dìlitelné tøinácti? Nakreslíme orientovaný graf s
-vrcholy 0 a¾ 12
-a hranami $(x,y),$ $y=10\cdot x \mod 13$ a $y=(10\cdot x + 1) \mod 13$
-(z~ka¾dého vrcholu vychází jedna hrana za pøidání èíslice 1 a dal¹í za èíslici
-0). Hledané èíslo existuje právì tehdy, kdy¾ graf obsahuje orientovaný sled
-z~1~do~0. Jakým algoritmem takový sled najdeme?
-\endlist
-Podobné a dal¹í úlohy budeme øe¹it v následujících kapitolách.
-
-\bye
diff --git a/2-ram/Makefile b/2-ram/Makefile
deleted file mode 100644
index 382372f..0000000
--- a/2-ram/Makefile
+++ /dev/null
@@ -1,3 +0,0 @@
-P=2-ram
-
-include ../Makerules
diff --git a/2009/11-stromy/11-stromy.tex b/2009/11-stromy/11-stromy.tex
new file mode 100644
index 0000000..173c36c
--- /dev/null
+++ b/2009/11-stromy/11-stromy.tex
@@ -0,0 +1,116 @@
+\input ../lecnotes.tex
+
+% Vkladani obrazku
+\input ../mjipe.tex
+\def\treepic#1{
+\medskip
+\IpeInput{treepic/t#1.ipe}
+\medskip
+}
+
+\prednaska{11}{Vyhledávací stromy}{}
+
+Pøedstavme si následující problém: potøebujeme si udr¾ovat urèitá setøídìná data, napøíklad slovník. Polo¾kami jsou uspoøádané dvojice (klíè, hodnota). Klíèe jsou unikátní a jsou to prvky nìjakého lineárnì uspoøádaného universa. Hodnoty mohou být libovolné.
+
+Na na¹ich datech budeme chtít provádìt následující operace: 
+\itemize\ibull
+\:{\it Insert} -- vlo¾it novou polo¾ku
+\:{\it Delete} -- smazat polo¾ku
+\:{\it Find} -- najít polo¾ku
+\:{\it Max $\&$ Min} -- vybrat polo¾ku s~nejvìt¹ím, resp. nejmen¹ím klíèem
+\:{\it Pred $\&$ Succ} -- vybrat polo¾ku s~klíèem nejbli¾¹ím men¹ím, resp. vìt¹ím
+\endlist
+
+Jak by vypadalo nejjednodu¹¹í øe¹ení? Staticky bychom data mohli udr¾ovat v~setøídìném poli. Takové pole se vyrobí v~èase $\Theta(n \cdot \log{n})$. Operace {\it Find} by trvala $\Theta(\log{n})$ (vyhledávali bychom samozøejmì binárnì). Ale problém by nastal s~operacemi {\it Insert} a {\it Delete}, které by se takto implementovat nedaly, nebo by trvaly hodnì dlouho (napø. v~pøípadì {\it Insertu} bychom museli celé pole pøestavìt).
+
+\s{Pozorování:} Proces binárního vyhledávání v~setøídìném poli se dá reprezentovat binárním vyhledávacím stromem.
+
+\s{Definice:} {\I Binární strom:} Strom je binární, pokud je zakoøenìný a ka¾dý vrchol má nejvý¹e dva syny, u nich¾ rozli¹ujeme, který je levý a který pravý.
+
+\s{Definice:} Pro vrchol $v$ znaèíme:
+\itemize\ibull
+\:$l(v)$ a $p(v)$ -- levý a pravý syn vrcholu $v$
+\:$L(v)$ a $P(v)$ -- levý a pravý podstrom vrcholu $v$
+\:$S(v)$ -- pøíslu¹ný podstrom s~koøenem $v$
+\:$h(v)$ -- hloubka stromu $S(v)$, neboli délka nejdel¹í cesty z koøene do listu
+\endlist
+
+\s{Definice:} {\I Binární vyhledávací strom} (BVS): Binární strom je vyhledávací, pokud v~ka¾dém vrcholu je ulo¾ena dvojice (klíè, hodnota) [ztoto¾níme vrchol s~klíèem] a pro v¹echny vrcholy platí: $\left ( \forall u \in L(v) : u < v \right ) $ $ \&$ $ \left ( \forall u \in P(v) : u > v \right )$.
+
+\s{Pøíklady binárních vyhledávacích stromù:}
+
+\treepic{2}
+
+\break
+ 
+Jak budou tedy vypadat operace {\it Find}, {\it Insert} a {\it Delete} na~binárním vyhledávacím stromu?
+
+{\bo Find$(v,x)$:}
+\algo
+\:Pokud $v = \emptyset \Rightarrow$ vrátíme $\emptyset$.
+\:Pokud $v = x \Rightarrow$ vrátíme $v$.
+\:Pokud $v < x \Rightarrow$ vrátíme $\<Find>(p(v),x)$.
+\:Pokud $v > x \Rightarrow$ vrátíme $\<Find>(l(v),x)$.
+\endalgo
+
+{\bo Insert$(v,x)$:}
+\algo
+\:Jako \<Find> a na~konci pøidáme list.
+\endalgo
+
+{\bo Delete$(v)$:}
+\algo
+\:Pokud $v$ je list $\Rightarrow$ jednodu¹e list utrhneme.
+\:Pokud $v$ má jednoho syna $\Rightarrow$ vrchol \uv{vyøízneme}.
+\:Jinak má $v$ oba syny $\Rightarrow$ do vrcholu vlo¾íme minimum z $P(v)$, co¾ bude list, a ten utrhneme.
+\endalgo
+
+\s{Poznámka:} Pokud má vrchol $v$ pøi operaci {\it Delete} oba syny, je vlo¾ení minima z $P(v)$ ekvivalentní s~vlo¾ením maxima z $L(v)$.
+
+\s{Pøíklady operací  {\it Insert} a {\it Delete} na~BVS:}
+\treepic{1}
+\treepic{3}
+
+\break
+Èasová slo¾itost v¹ech tøí operací je $\Theta(\<hloubka stromu>)$, co¾ mù¾e být $\Theta(n)$, kdy¾ budeme mít smùlu a strom bude (témìø) lineární spojový seznam, nebo $\Theta(\log{n})$ kdy¾ bude strom pìknì vyvá¾enì vystavìný. Vidíme tedy, ¾e slo¾itost operací stojí a padá s~hloubkou stromu. Proto by se nám líbilo, aby mìl ná¹ strom v¾dy hloubku $\Theta(\log{n})$. Podívejme se tedy, jak se dá navrhnout binární vyhledávací strom, aby tuto podmínku splòoval \dots
+
+\h{Vyvá¾ené binární vyhledávací stromy}
+
+\s{Definice:} {\I Dokonalá vyvá¾enost:} Strom je dokonale vyvá¾ený, pokud pro v¹echny jeho vrcholy platí: $\left \vert \vert L(v)\vert - \vert P(v)\vert \right \vert \leq 1 $.
+
+Takto definovaný binární strom bude mít urèitì logaritmickou hloubku. Jak takový strom ale konstruovat? To se nám podaøí buï staticky, nebo na~nìm budou operace dra¾¹í ne¾ $\Theta(\log{n})$.
+
+\s{Statická konstrukce dokonale vyvá¾eného BVS:} Vybereme prostøední prvek ze setøídìného pole (tedy medián posloupnosti) a dáme ho do~koøene stromu. Jeho syny pak vystavíme rekurzivnì z~levé a pravé pùlky pole. Celá konstrukce tedy trvá $\O(n)$.
+
+Vidíme tedy, ¾e to ná¹ problém pøíli¹ neøe¹í. Potøebovali bychom, aby se strom dal také efektivnì udr¾ovat. Zkusíme proto slab¹í podmínku:
+
+\s{Definice:} {\I Hloubková vyvá¾enost:} Strom je hloubkovì vyvá¾ený, pokud pro v¹echny jeho vrcholy platí: $\left \vert h(L(v)) - h(P(v)) \right \vert \leq 1 $.
+
+Stromùm s~hloubkovým vyvá¾ením se øíká {\I AVL stromy} (objeviteli je ru¹tí matematikové G. M. Adelson-Velsky a E. M. Landis, podle nich jsou také pojmenovány) a platí o~nich následující lemma:
+
+\s{Lemma:} AVL strom na~$n$ vrcholech má hloubku $ \Theta(\log{n}) $.
+
+\proof
+Uva¾me posloupnost $A_k = $ minimální poèet vrcholù AVL stromù hloubky $k$. Staèí ukázat, ¾e $A_k$ roste exponenciálnì. Podívejme se na minimální AVL stromy:
+$$\eqalign{
+A_0 &= 1 \cr
+A_1 &= 2 \cr
+A_2 &= 4 \cr
+A_3 &= 7 \cr
+&\vdots \cr
+A_k &= 1 + A_{k - 1} + A_{k - 2}. \cr
+}$$
+
+Rekurentní vzorec jsme dostali rekurzivním stavìním stromu hloubky $k$: nový koøen a 2 podstromy o~hloubkách $k - 1$ a $k - 2$.
+
+Vidíme tedy, ¾e $A_n = F_{n + 2} - 1$. (Mù¾eme dokázat napø. indukcí.)
+Teï nám ji¾ staèí dokázat, ¾e posloupnost $A_k$ roste exponenciálnì S výhodou mù¾eme vyu¾ít toho, ¾e na první pøedná¹ce jsme si ji¾ dokázali, ¾e Fibonacciho èísla rostou exponenciálnì. Nicménì pro zapomnìtlivé mù¾eme dùkaz ve struènosti zopakovat:
+
+Indukcí doká¾eme, ¾e $ A_k \geq 2^{k \over 2} $.
+První indukèní krok jsme si u¾ ukázali, teï pro $ k \geq 2 $ platí:
+$ A_k = 1 + A_{k - 1} + A_{k - 2} > 2^{{k - 1} \over 2} + 2^{{k - 2} \over 2} = 2^{k \over 2} \cdot (2^{-{1 \over 2}} + 2^{-1}) \cong 2^{k \over 2} \cdot 1.21 > 2^{k \over 2}.$
+
+Tímto jsme dokázali, ¾e na~ka¾dé hladinì je minimálnì exponenciálnì vrcholù, co¾ nám zaruèuje hloubku $ \Theta(\log{n})$.
+\qed
+
+\bye
diff --git a/2009/11-stromy/Makefile b/2009/11-stromy/Makefile
new file mode 100644
index 0000000..6f4f489
--- /dev/null
+++ b/2009/11-stromy/Makefile
@@ -0,0 +1,3 @@
+P=11-stromy
+
+include ../Makerules
diff --git a/2009/11-stromy/treepic/t1.ipe b/2009/11-stromy/treepic/t1.ipe
new file mode 100644
index 0000000..84e02ad
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@@ -0,0 +1,299 @@
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+end %% of Ipe figure
diff --git a/2009/12-hash/12-hash.tex b/2009/12-hash/12-hash.tex
new file mode 100644
index 0000000..f12db20
--- /dev/null
+++ b/2009/12-hash/12-hash.tex
@@ -0,0 +1,84 @@
+\input lecnotes.tex
+
+\prednaska{12}{Hashování}{}
+
+V pøedchozích kapitolách jsme se zabývali tøídìním prvkù a zjistili jsme, ¾e v obecném pøípadì (kdy¾ smíme prvky pouze porovnávat a prohazovat) nelze tøídit rychleji ne¾ v èase $\Theta(n\log n)$. Zároveò jsme do¹li k tomu, ¾e pokud víme o prvcích nìco více (napø. jejich rozsah), tak jsme v nìkterých pøípadech schopni tøídit i s lineární èasovou slo¾itostí. 
+
+V praxi nicménì vìt¹inou netøídíme samotná èísla, ale spí¹e nìjaké polo¾ky, napø. urèené uspoøádanou dvojicí (klíè, hodnota), kde klíèe jsou unikátní a existuje na nich lineární uspoøádání. Ukázali jsme si, jak takovouto mno¾inu udr¾ovat (aby se v ní dalo rychle vyhledávat, zaøazovat i mazat) ve vyhledávacích stromech. Dokázali jsme, ¾e pokud budeme udr¾ovat strom dostateènì vyvá¾ený, tak budou jednotlivé operace (\<Find>, \<Insert> a \<Delete>) trvat $\Theta(\log n)$.
+
+Ne¹lo by to ale rychleji? Odpovìï je \uv{jak kdy}. Podívejme se na dal¹í mo¾nosti. Nejdøíve si ale pøipomeòme základní pojmy, které budeme pou¾ívat.
+
+Klíèe budou prvky universa, co¾ je mno¾ina $\{ 0, \dots, U-1 \}$, kterou budeme oznaèovat $[U]$. Tuto mno¾inu budeme chtít udr¾ovat v nìjaké ¹ikovné datové struktuøe, abychom v ní mohli rychle vyhledávat, zaøazovat i mazat. Dále $n$ bude znaèit (maximální) poèet prvkù, které si budeme v jednu chvíli pamatovat.
+
+
+\h{1. Pole {\it indexované od $0$ do $U-1$}}
+
+Pokud bude velikost universa dostateènì malá, tak si mù¾eme polo¾ky pamatovat v poli o velikosti universa. Operace \<Find>, \<Insert> i \<Delete> pracují v~èase $\O(1)$, ale platíme za~to pamì»ovou slo¾itostí $\Theta(U)$. Kdy¾ uvá¾íme, ¾e si budeme pamatovat mnohem ménì prvkù, ne¾ je velikost universa, tak je to velké plýtvání.
+
+\h{2. Èíslicový strom}
+
+Zvolíme vhodný základ soustavy $b$, klíè zapí¹eme v soustavì o základu $b$ a zápis ulo¾íme do stromu. Tímto nám vznikne \uv{$b$-ární strom}, tedy strom, jeho¾ ka¾dý vrchol má a¾ $b$ synù.
+
+Na první hladinì se vìtvíme podle první èíslice, na druhé podle druhé, atd. Na poslední hladinì bude tedy $n$ listù. Hloubka tohoto stromu je $\lceil\log_b U\rceil$.
+
+Operace \<Find> bude trvat $O(\log_b U)$ (najití jednoho klíèe odpovídá projití jedné cesty ve stromì od koøene do listu). Stejný èas budou trvat i operace \<Insert> a \<Delete>. (Pøi vkládání zakládáme vrcholy, které odpovídají na¹emu klíèi a které je¹tì nejsou zalo¾ené. Pøi mazání sma¾eme list a následnì vrcholy, ze kterých nic nevede. Pro tuto operaci je výhodné si ve vrcholech udr¾ovat jejich stupnì.) Strom zabere prostor velikosti $O(nb\log_b U)$. 
+
+
+Èím bude základ $b$ men¹í, tím bude lep¹í pamì»ová a hor¹í èasová slo¾itost
+a obrácenì. Èíslicový strom tedy vy¾aduje peèlivé nastavení parametrù.
+
+\s{Pøíklad:} Pøedstavme si, ¾e klíèe budou IP-adresy, tedy 32-bitová èísla. Jak zvolit $b$, aby byla èasová i pamì»ová slo¾itost pøijatelná? Zkusme rozdìlit IP-adresu na 4 kusy, $b$ bude tedy 256. Strom bude mít hloubku 4. Èas bude v poøádku ($\log_b U = 4$), ale po¾adavky na prostor jsou dost vysoké($nb\log_b U = n \cdot 256\cdot 4 = 1024$). Výhodnìj¹í bude rozdìlit IP-adresu na 8 èástí, $b$ bude tedy 16, hloubka stromu 8 a nároky na èas i pamì» jsou pøijatelné, nebo» $\log_b U = 8$ a $nb\log_b U = n\cdot 16 \cdot 8 = 128$.
+
+
+\h{4. Hashování }
+
+Mìjme $m$ pøihrádek a v ka¾dé z nich spojový seznam. Dále $h$ bude hashovací funkce, která libovolnému prvku z universa pøiøadí právì jednu pøihrádku. 
+
+Kdyby $h$ fungovala \uv{rovnomìrnì } (ka¾dému prvku by pøiøadila jinou pøihrádku, tedy ka¾dý spojový seznam by mìl jen jeden prvek), tak bychom umìli vyhledávat, pøidávat i mazat prvky v konstantním èase. To je ov¹em samozøejmì pøíli¹ krásné na to, aby to byla pravda.
+
+Problém je, ¾e pokud bude $m$ výraznì men¹í ne¾ $U$ (co¾ chceme, jinak se jedná o normální (obrovské)  pole a funkce je identita -- tento pøípad jsme ji¾ vyøe¹ili), tak urèitì budou existovat prvky, kterým funkce pøiøadí stejnou hodnotu. Takovému pøípadu se øíká {\it kolize}. Prvky se pak zaøadí do spojového seznamu. Kdyby byla naopak funkce hodnì ne¹ikovná a namapovala by nám v¹echny prvky do jedné pøihrádky, tak budeme v¾dy muset projít (v nejhor¹ím pøípadì celý) dlouhý spojový seznam.
+
+Jak se dá tedy kolizím bránit? Nejvýhodnìj¹í by bylo mít dobrou hashovací funkci, která by mapovala prvky pokud mo¾no rovnomìrnì. Ale nìkdy staèí i prùmìrnì dobrá funkce a dostateènì náhodné klíèe.
+
+
+\s{Vsuvka:} Jaká je pravdìpodobnost, ¾e 2 z k lidí mají narozeniny ve stejný den? Staèí 23 lidí, aby nastala kolize s pravdìpodobností vìt¹í ne¾ jedna polovina. Aby nenastala, potøebovali bychom vìt¹í poèet pøihrádek: $m \approx n^2$.
+
+\h{Praktické hashovací funkce}
+
+\s{První pøíklad}
+
+Tato hashovací funkce $h: [U] \rightarrow [m]$ je definovaná pøedpisem: $h(x) = x \mod m$. Pro dostateènì náhodné hodnoty bude fungovat, ale v mnoha pøípadech mù¾e být nevhodná. Napøíklad je zøejmé, ¾e pro sudé $m$ zachovává paritu. V pøípadì ¾e sudých, resp. lichých klíèù bude mnohem více, vznikne mnoho zbyteèných kolizí. Pro takovouto hashovací funkci je nejlep¹í zvolit za $m$ nìjaké prvoèíslo.
+
+\s{Druhý pøíklad}
+
+Zvolme si iracionální $\alpha$ z intervalu $(0,1)$. Funkce bude definovaná následovnì: $$h(x) = \lfloor m (x \cdot \alpha \mod 1) \rfloor.$$ Tato funkce by mìla na $x$ záviset velmi málo, ale nebudeme si to dokazovat.
+
+\s{Implementace:} Reálná èísla se samozøejmì implementují dosti nároènì (resp. to vùbec nejde), tak¾e v praxi se tento postup aproximuje celoèíselnì. Místo reálné $\alpha$ si zvolme $A = \lfloor \alpha \cdot 2^w \rfloor$, kde $2^w$ je ¹íøka slova ($w$ je napø. poèet bitù integeru). Potom místo výrazu $(x \cdot \alpha \mod 1)$, který je z intervalu $(0,1)$, budeme poèítat s výrazem $(x \cdot A \mod 2^w)$. Potom na¹e funkce vyjde následovnì:
+
+$$h(x) = \left \lfloor { m(x \cdot A \mod 2^w) \over 2^w } \right \rfloor$$
+
+
+Zbývá jen otázka, jak volit $\alpha$? Osvìèená hodnota je $1/\tau$, co¾ je pøibli¾nì 0,618. ($\tau$ je hodnota zlatého øezu.) Takto zvolené $\alpha$ dobøe rozbíjí aritmetické posloupnosti, ale i toto je pouze pøedstava a nebudeme si to dokazovat.
+
+\h{Hashování øetìzcù}
+
+Vzhledem k tomu, ¾e hashování nijak speciálnì nepotøebuje, aby byly prvky èísla, tak se mohou hashovat i øetìzce. Uka¾me si jeden pøíklad funkce pro hashování øetìzcù.
+Prázdnému øetìzci pøiøaïme nulu: $h(\varepsilon) = 0$. Øetìzci znakù $x_1, \dots, x_n$ pøiøaïme hodnotu:
+$$h(x_1, \dots, x_n) = (h(x_1, \dots, x_{n-1}) \cdot + x_n) \mod m.$$
+$A$ je zde nìjaká konstanta. Aby se vyu¾ily v¹echny pøihrádky, tak je vhodné zvlit $A$ tak, aby $A^{{\bb E}[\<délka øetìzce>]} \gg m$. Tato funkce vyu¾ívá podobných vlastností jako lineární konkurenèní generátor náhodných èísel, který pomocí vhodných konstant $A$ a $B$ a ¹íøky slova $w$ vyrábí nové náhodné èíslo $x'$ z minulého èísla $x$ jako $$x' = (x \cdot A + B) \mod 2^w.$$
+
+
+\s{Vìta} Pokud jsou $h(x_1), \dots, h(x_n)$ rovnomìrnì náhodné, pak $$\forall i: {\bb E}[\<poèet prvkù v i-té pøihrádce>] = {n \over m}.$$
+
+\proof
+Poèet prvkù v $i$-té pøihrádce si oznaème $N_i$. Definujme si indikátor $P_{ij}$, který je roven 1, pokud $x_j$ padlo do $i$-té pøihrádky, jinak je roven 0.
+Vidíme, ¾e poèet prvkù v $i$-té pøihrádce je souèet $P_{ij}$ pøes v¹echna $j$.
+$$N_i = \sum_{j} P_{i,j}.$$
+Dále vyu¾ijeme mimo jiné vìtu o linearitì støední hodnoty.
+$${\bb E}[P_{ij}] = {1 \over m} \cdot 1 + \left(1 - {1 \over m} \right) \cdot 0 = {1 \over m}.$$
+$${\bb E}[N_i] = {\bb E}\left[\sum_{j} P_{i,j}\right] = \sum_{j} {\bb E}[P_{i,j}] = {n \over m}.$$
+\qed
+
+\s{Dùsledek:} Pokud bude $n$ pøibli¾nì stejnì velké jako $m$, tak v ka¾dé pøihrádce bude prùmìrnì jeden prvek.
+
+\bye
diff --git a/2009/12-hash/Makefile b/2009/12-hash/Makefile
new file mode 100644
index 0000000..e0cfe0f
--- /dev/null
+++ b/2009/12-hash/Makefile
@@ -0,0 +1,3 @@
+P=12-hash
+
+include ../Makerules
diff --git a/2009/2-ram/2-ram.tex b/2009/2-ram/2-ram.tex
new file mode 100644
index 0000000..71c3a25
--- /dev/null
+++ b/2009/2-ram/2-ram.tex
@@ -0,0 +1,245 @@
+\input ../lecnotes.tex
+
+\prednaska{2}{Slo¾itost, grafové algoritmy}
+{(zapsal Martin Koutecký)}
+
+\h{Model {\sc Ram}}
+
+Pøi analýze algoritmu bychom chtìli nìjak popsat jeho slo¾itost. Abychom mohli
+ udìlat toto, potøebujeme nejprve definovat výpoèetní model. Výpoèetních modelù
+je více, my vybereme jeden pomìrnì blízký skuteèným poèítaèùm:
+
+\s{Definice:} Random Access Machine ({\sc Ram})
+
+{\sc Ram} poèítá jen s celými èísly -- znaky, stringy a podobnì reprezentujeme
+èísly, jejich posloupnostmi atd. Pamì» je tvoøena buòkami, které obsahují
+èísla. Pamì»ové buòky jsou adresované takté¾ èísly. A program samotný je
+koneèná posloupnost instrukcí následujících druhù:
+\itemize\ibull
+\:Aritmetické a logické:
+$X$ |<-| $Y \oplus Z, \oplus\in\{|+|, |-|, |*|, |div|, |mod|, \&,
+{\tt\char124}, |<<|, |>>|\}$
+\:Øídící: |goto| \<label>, |halt|
+\:Podmínky: pro libovolnou nepodmínìnou instrukci mù¾u pou¾ít \hfil \break
+if~$X$~|<|~$Y$~|==>|~instrukce % Tady to prosím je¹tì zkontroluj. Myslím, ¾e
+% zápis je správný, ale sází se to divnì a vidím èerný obdélníèek na konci
+% øádku. Díky.
+\endlist
+
+\s{Poznámka} (operandy):
+\itemize\ibull
+\:Konstanty (1, 2, \dots)
+\:Adresované pøímo -- M[konst.] -- budeme pou¾ívat písmena {\tt A-Z} jako aliasy
+pro
+buòky pamìti $-1$ a¾ $-26$, které nazýváme registry.
+(tedy A={\tt M[-1]})
+\:Adresované nepøímo -- {\tt M[M[konst.]]} -- budeme pou¾ívat zkratku [[konst.]]
+\endlist
+
+Samotný výpoèet probíhá takto:
+\algo
+\:Do smluvených bunìk umístíme vstup, obsah zbylých pamì»ových bunìk není
+definován.
+\:Provádíme program postupnì po instrukcích, dokud nedojdeme k haltu nebo konci
+programu.
+\:Pokud se program nezacyklil, tedy pokud skonèil, ze smluvených bunìk pøeèteme
+výstup.
+\endalgo
+
+
+\h{Slo¾itost}
+\> Jak dobøe popsat slo¾itost?
+\numlist\ndotted
+\:{\I {\sc Ram} s jednotkovou cenou}: èas $\approx$ \#instrukcí, prostor
+$\approx$
+maximální èíslo buòky minus minimální èíslo buòky pou¾ité pøi výpoètu.
+Toto není moc dobrý nápad, proto¾e není nijak penalizována napøíklad práce s
+velmi dlouhými èísly -- poøád je to jedna instrukce, tak¾e cena je stejná, ale
+poèítaèe se tak pøece nechovají. Velikost èísel ale omezit nesmíme, proto¾e
+bychom omezili pamì» (èísly ji adresujeme).
+\:{\I{\sc Ram} s logaritmickou cenou}: cena instrukce $\approx$ \#bitù
+zpracovávaných èísel,
+prostor $\approx$ \# bitù v¹ech pou¾itých bunìk. To je teoreticky pøesné, ale
+dost nepraktické (ve v¹ech slo¾itostech by byly logaritmy).
+\:{\I{\sc Ram} s omezenými èísly}: jednotková cena instrukcí, ale èísla omezíme
+nìjakým polynomem $P(n)$. Tím zmizí paradoxy prvního modelu, ale
+mù¾eme adresovat jen polynomiální prostor (to nám ov¹em obvykle nevadí).
+\endlist
+
+Nadále tedy budeme pøedpokládat tøetí zmínìný model.
+
+% Z minulých zápiskù.
+\s{Definice:}
+\itemize\ibull
+\:{\I Èas bìhu algoritmu} $t(x)$ pro vstup~$x$ mìøíme jako sumu èasù provedených
+operací, které program provedl pøi zpracování vstupu
+$x$.
+\:{\I Prostor bìhu algoritmu} $s(x)$ je analogicky poèet pamì»ových
+bunìk spotøebovaných pøi výpoètu se vstupem~$x$.
+\:{\I Èasová slo¾itost} (v~nejhor¹ím pøípadì) je:
+$$T(n) := \max \{t(x) ; \hbox{$x$ je vstup délky $n$}\}.$$
+\:{\I Prostorová slo¾itost} (v~nejhor¹ím pøípadì) je:
+$$S(n) := \max \{s(x) ; \hbox{$x$ je vstup délky $n$}\}.$$
+\endlist
+
+Nyní zkusíme zanalyzovat nìjaký konkrétní algoritmus. Vezmìme napøíklad øazení
+pomocí pøímeho výbìru (selection sort). Na vstupu dostaneme poèet èísel $n$ (v
+registru {\tt N}), v buòkách $1,\dots, n$ je nesetøídìná posloupnost èísel. Ta
+pak
+tøídíme následujícím algoritmem zapsaným v pseudokódu:
+\algo
+\:Pro $i=1$ do $n$:
+\::$j\leftarrow i$
+\::Pro $k=i$ do $n$:
+\:::Je-li $[k]<[j]\Rightarrow j\leftarrow k$
+\::$[i]$ prohodíme s $[j]$.
+\endalgo
+
+Jak by takový algoritmus vypadal zapsaný v instrukcích {\sc Ram}? Budeme muset
+pou¾ít návì¹tí a goto místo cyklù, jména registrù místo promìnných a
+tøeba prohození musíme provést pøes tøetí promìnnou. Nìjak takto:
+
+\verbatim{	I <- 1
+LOOP:	J <- I
+	M <- I
+MIN:	IF [J]<[M] ==> M <- J
+	J <- J+1
+	IF J<=N ==> GOTO MIN
+	X <- [I]
+	[I] <- [M]
+	[M] <- X
+	I <- I+1
+	IF I<=N ==> GOTO LOOP
+}
+
+Pojïme se podívat, jaká je èasová slo¾itost jednotlivých èástí algoritmu.
+Cyklus |MIN| provede $3\cdot (N-I+1)$ instrukcí. Mimo cyklu |MIN| je v |LOOP|
+je¹tì 7 instrukcí, tedy celý |LOOP| provede $3\cdot (N-I+1)+7$ instrukcí.
+Celkovì se dostáváme k souètu
+$$1+3\cdot N+7+3\cdot (N-1)+7+3\cdot (N-2)+7+3\cdot (N-3)+\dots +3\cdot 1+7 =$$
+$$1+7\cdot N+3\cdot {{N(N+1)}\over{2}} = {{3}\over{2}}N^2 + 8{,}5N + 1$$
+
+Na multiplikativních konstantách ale nezále¾í -- zaprvé se na reálných strojích
+ceny jednotlivých (pro nás jednotkových) instrukcí stejnì li¹í, zadruhé
+asymptoticky pomalej¹í funkce nakonec pro velké $N$ v¾dy prohraje, tak¾e nemá
+cenu
+(alespoò pøi prvním pøiblí¾ení k problému) multiplikativními konstantami se
+zabývat. Tím pádem nezále¾í ani na èlenech ni¾¹í øádù:
+$$1{,}5N^2 + 8{,}5N + 1 \leq 1{,}5N^2 + 8{,}5N^2 + N^2 = 11N^2\approx N^2$$
+Kdy¾ u¾ toto víme, mù¾eme zanedbávat konstanty prùbì¾nì: $N$ cyklù po
+${}\approx~N$~krocích $\Rightarrow~\approx~N^2$ krokù. To nás vede k zavedení
+tzv.
+{\I asymptotické notace:}
+
+\h{Asymptotická notace}
+\s{Definice:} Pro funkce $f,g: {\bb N} \rightarrow {\bb R}^+$ øekneme,
+¾e $f$ je $\O(g)$ právì tehdy, kdy¾ $\exists c>0: \forall ^{*} n \in {\bb N}:
+f(n) \leq c \cdot g(n)$.
+Zde $\forall^* n \in {\bb N}$ je zkratka za \uv{$\exists n_0 \in {\bb N}:
+\forall n \geq n_0$}, tedy
+\uv{pro v¹echna~$n$ a¾ na~koneènì mnoho výjimek.}
+
+\s{Poznámka:} $\O$-notace tedy vyjadøuje, ¾e funkce~$f$ je skoro v¹ude men¹í
+nebo nejvý¹e rovná nìjakému reálnému násobku funkce~$g$. Aèkoliv zápis vypadá
+jako rovnost, rozhodnì není symetrický: napøíklad platí $\log n=\O(n)$, ale
+neplatí $n=\O(\log n)$. Formálnì by bylo lep¹í pova¾ovat $\O(g)$ za tøídu
+funkcí, pro které platí, ¾e se dají shora odhadnout kladným násobkem funkce~$g$,
+a~psát tedy~$f\in\O(g)$, ale zvyk je bohu¾el ¾elezná ko¹ile.
+
+\s{Pøíklady:} $2{,}5n^{2} = \O(n^{2})$, $2{,}5n^{2}+30n = \O(n^{2})$.
+
+\>Také platí:
+$$
+\O(f)+\O(g) \in \O(f+g),
+$$
+èím¾ myslíme, ¾e pokud vezmeme libovolnou $f'=O(f)$ a $g'=O(g)$, bude
+$f'+g'=O(f+g)$.
+To platí, jeliko¾ skoro v¹ude je $f' \leq cf$, $g'\leq dg$, a~tedy $f'+g' \le
+cf+dg \le (c+d)(f+g)$.
+
+\s{Cvièení:} Uka¾te, ¾e:
+\itemize\ibull
+\:$\O(f) \cdot \O(g)=\O(f \cdot g)$,
+\:$\O(f+g)=\O(\max(f,g))$,
+\:$\O(n^{2})+\O(n)=\O(n^{2}+n)=\O(n^{2})$.
+\endlist
+
+$\O$-notace popisuje horní odhad asymptotického chování funkce. Mnohdy v¹ak
+potøebujeme také odhadnout funkci zespodu (chceme-li øíci, ¾e algoritmus
+potøebuje {\I alespoò} nìjaké mno¾ství èasu nebo pamìti), pøípadnì z~obou stran:
+
+\s{Definice:}
+
+\itemize\ibull
+\:$f=\Omega(g) \equiv \exists c>0:\forall^* n \in {\bb N}: f(n) \geq c\cdot
+g(n)$.
+
+$\Omega$-notace tedy øíká, ¾e hodnota funkce $f$ je v¾dy stejná nebo vy¹¹í ne¾
+nìjaký $c$-násobek funkce $g$, a tedy $g=\O(f)$.
+\:$f=\Theta(g) \equiv f=\O(g) \wedge f=\Omega(g)$
+
+nebo výøeènìji:
+
+$f=\Theta(g) \equiv \exists$ $c_{1},c_{2} > 0: \forall^* n \in {\bb
+N}: c_{1}\cdot g(n) \leq f(n) \leq
+c_{2}\cdot g(n)$ To znamená, ¾e existují nezáporné reálné konstanty
+$c_{1},c_{2}$ takové, ¾e se funkce $f(n)$ dá ohranièit $c_{1}$- a
+$c_{2}$-násobkem funkce $g(n)$.
+\endlist
+
+\s{Porovnání rùstu funkcí:} (aneb jak moc máme algoritmy rádi podle jejich
+chování od~nejlep¹ích k~nejhor¹ím)
+
+\itemize\ibull
+\:$\Theta(1) \ldots$ funkce zespoda i shora ohranièené konstantami
+\:$\Theta(\log{( \log{n} )})$
+\:$\Theta(\log{n})$ \dots\ logaritmická
+\:$\Theta(n^{\varepsilon}), \varepsilon \in (0,1)$ \dots\ sublineární
+\:$\Theta(n)$ \dots\ lineární
+\:$\Theta(n^{2})$ \dots\ kvadratická
+\:$\Theta(n^{k})$ \dots\ polynomiální
+\:$\Theta(2^{n})$ \dots\ exponenciální pøi základu $2$
+\:$\Theta(3^{n})$ \dots\ exponenciální pøi základu $3$
+\:$\Theta(k^{n})$ \dots\  exponenciální pøi základu $k>1$
+\:$\Theta(n!)$ \dots\ faktoriálová
+\:$\Theta(n^{n})$
+\:\dots\ nekoneènì mnoho dal¹ích tøíd (i mezi tìmi vý¹e uvedenými)
+\endlist
+
+\>{\sl Poznámka:} Pokud se v~odhadu slo¾itosti vyskytne logaritmus (jinde ne¾
+v~exponentu), nezále¾í na tom, jaký má základ, proto¾e platí:
+$$
+\log_k{n}={{\log_c{n}}\over{\log_c{k}}}={{1}\over{\log_c{k}}}\cdot \log_c{n},
+$$
+kde $1/\log_c{k}$ je jen konstanta, tak¾e ji mù¾eme \uv{schovat do~$\O$}.
+
+\s{Pøíklad:} Select sort (rozebraný vý¹e):
+Kdy¾ jej pustíme na $n$ èísel, pak èasová slo¾itost je $T(n) = \Theta(n^2)$ a
+prostorová $S(n) = \Theta(n)$.
+
+\h{Úvod do grafových algoritmù}
+Dal¹í dùle¾itou a zajímavou kapitolou jsou grafové algoritmy. Napøíklad
+následující pøíklady lze (i kdy¾ to tak obèas na první pohled nevypadá) øe¹it
+nìjakým grafovým algoritmem:
+\itemize\ibull
+\:Mám mapku silnièní sítì, v ní místa (vrcholy) oznaèená \uv{Doma} a \uv{©kola}.
+Dostanu se do
+¹koly (le¾í ve stejné komponentì souvislosti)? Dostanu se do ¹koly, kdy¾ v zimì
+napadne hodnì snìhu a nìkteré cesty budou neprùjezdné? A jaký nejkrat¹í úsek
+cest musí silnièáøi prohrnout, aby byla v¹echna místa na mapì dostupná?
+\:Mìjme hlavolam \uv{Lloydova devítka} -- krabièku $3\times3$ se ètvereèky
+oznaèenými èísly od jedné do osmi a jednou mezerou, ètvereèky jsou zamíchané a
+na¹ím úkolem je správnì je seøadit pomocí pøesouvání ètvereèkù sousedících s
+mezerou do této mezery. Jak to udìlat? Kolik nejménì krokù nám na
+to staèí? Jde to vùbec se zadáním, které jsme dostali?
+\:Jaké je nejkrat¹í (kladné, celé) èíslo v desítkové soustavì zapsané jen
+èíslicemi 1, 0, které je dìlitelné tøinácti? Nakreslíme orientovaný graf s
+vrcholy 0 a¾ 12
+a hranami $(x,y),$ $y=10\cdot x \mod 13$ a $y=(10\cdot x + 1) \mod 13$
+(z~ka¾dého vrcholu vychází jedna hrana za pøidání èíslice 1 a dal¹í za èíslici
+0). Hledané èíslo existuje právì tehdy, kdy¾ graf obsahuje orientovaný sled
+z~1~do~0. Jakým algoritmem takový sled najdeme?
+\endlist
+Podobné a dal¹í úlohy budeme øe¹it v následujících kapitolách.
+
+\bye
diff --git a/2009/2-ram/Makefile b/2009/2-ram/Makefile
new file mode 100644
index 0000000..382372f
--- /dev/null
+++ b/2009/2-ram/Makefile
@@ -0,0 +1,3 @@
+P=2-ram
+
+include ../Makerules
diff --git a/2009/3-dfs/3-dfs.tex b/2009/3-dfs/3-dfs.tex
new file mode 100644
index 0000000..b183ac4
--- /dev/null
+++ b/2009/3-dfs/3-dfs.tex
@@ -0,0 +1,249 @@
+\input ../lecnotes.tex
+
+\prednaska{3}{Prohledání do~¹íøky a do~hloubky}{()}
+
+\h{Prohledání do~¹íøky (BFS) {\I Breadth-First Search} }
+
+Jde o grafový algoritmus, který postupnì prochází v¹echny vrcholy v~dané komponentì souvislosti.
+Algoritmus nejprve projde v¹echny sousedy poèáteèního vrcholu, poté sousedy sousedù, atd\dots
+Díky tomuto zpùsobu procházení se nìkdy té¾ nazývá \uv{\I algoritmus vlny }, nebo» se z~poèáteèního vrcholu ¹íøí pomyslná vlna, která v~ka¾dém kroku nalezne v¹echny uzly, které mají od~poèáteèního vrcholu stejnou vzdálenost. Algoritmus se tedy skvìle hodí napøíklad pro hledání nejkra¹í cesty mezi dvìma vrcholy v~grafu.
+\figure{praseci-graf.eps}{Praseèí graf}{55mm}
+
+
+\s{Popis algoritmu:}
+Na zaèátku vlo¾íme do~fronty $Q$ poèáteèní vrchol $v_0$. Dále si v~poli $Z$ budeme pro ka¾dý vrchol pamatovat znaèku, zda jsme ho ji¾ nav¹tívili, èi nikoli. Pro vrchol $v_0$ si tedy dosazením jednièky zapamatujeme, ¾e je ji¾ nav¹tívený. V~dal¹ím kroku pak zkoumáme frontu $Q$: pokud není prázdná, vezmeme z~ní první vrchol a podíváme se na~v¹echny jeho sousedy $w$. Pokud je¹tì nejsou oznaèené (tedy $Z[w]=0$), tak je oznaèíme (zapamatujeme si, ¾e je pøedáváme ke zpracování a u¾ je nemáme znovu nav¹tìvovat) a pøidáme je do~fronty k~následnému zpracování. Takto cyklus opakujeme, dokud není fronta prázdná.
+
+\s{Poznámka:}
+Nejprve se podívejme, jak algoritmus pracuje na orientovaném grafu. Pro neorientovaný graf funguje algoritmus analogicky, co¾ si uká¾eme pozdìji.
+
+\s{Algoritmus:}
+
+\algo
+\:$Q \leftarrow \{v_0\}$.
+\:$Z[*] \leftarrow 0, Z[v_0] \leftarrow 1$.
+\:Dokud $Q \not= \emptyset $ opakujeme:
+\::Vyzvedneme vrcholy $u$ z~$Q$.
+\::$\forall w: (u,w)\in E$:
+\:::Je-li $Z[w]=0 \Rightarrow Z[w] \leftarrow 1$, pøidáme $w$ do~$Q$.
+\endalgo
+
+
+\>{\I Pozorování:} {\I BFS} se zastaví.
+
+\proof Zpracováváme jen vrcholy, které byly ve~frontì. Ka¾dý vrchol se dostane do~fronty maximálnì jednou. (Ka¾dý je oznaèen max. jednou, znaèky neodstraòujeme.)
+
+\s{Lemma:} BFS($v_0$) oznaèí $v$ právì tehdy, kdy¾ existuje cesta z~$v_0$ do~$v$.
+
+\proof 
+\uv{$\Longrightarrow$}: 
+Platí jako invariant po celou dobu bìhu algoritmu. To doká¾eme indukcí dle doby bìhu algoritmu:
+
+První krok indukce je triviální, nebo» cesta z~$v_0$ do~$v_0$ existuje v¾dy. Nyní si pøedstavme, ¾e oznaèujeme vrchol~$v$ pøes hranu~$uv$. To znamená, ¾e vrchol~$u$ ji¾ musel být oznaèený. Dle indukèního pøedpokladu tedy existuje cesta z~$v_0$ do~$u$, a tudí¾ pokud k~této cestì \uv{pøilepíme} hranu~$uv$, dostáváme sled z~$v_0$ do~$v$, který lze v¾dy zredukovat na cestu.
+
+\uv{$\Longleftarrow$} Sporem: Nech» existuje neoznaèený vrchol $v$ dosa¾itelný po nìjaké cestì z~$v_0$. Uva¾me nejkrat¹í cestu $(v_0, v)$: $v_0, v_1 \dots, v_k = v$ a vezmìme minimální~$i$ takové, ¾e~$v_i$ není oznaèený. Víme, ¾e~$i>0$ (nebo»~$v_0$ je urèitì oznaèen u¾ na zaèátku algoritmu). Tudí¾~$v_{i-1}$ je oznaèený. Museli jsme tedy pou¾ít hranu~$v_{i-1}~v_i$ a oznaèit vrchol~$v_i$, co¾ je SPOR. \qed
+
+Nyní tedy víme, ¾e je algoritmus správný, a máme pøedstavu o tom, jak funguje. Podíváme-li se na~nìj podrobnìji, zjistíme, ¾e je hodnì závislý na~tom, jak si budeme graf pamatovat. Zanedlouho zároveò zjistíme, ¾e nám reprezentace grafu v~pamìti znatelnì ovlivní èasovou (i pamì»ovou) slo¾itost celého algoritmu.
+
+\h{Reprezentace grafu v~pamìti}
+
+Oznaème vrcholy grafu na~následujícím obrázku písmeny A, B, C, D.
+Pokud bychom chtìli tento graf uchovat v~pamìti poèítaèe, máme na~výbìr
+hned nìkolik zpùsobù, jak to udìlat.
+\figure{imgn_o4.eps}{}{\epsfxsize}
+\s{1. matice sousednosti}
+
+Matice sousednosti pro graf $G$ na~$n$ vrcholech je ètvercové pole $A$ o velikosti $n \times n$, jeho¾ prvky na
+souøadnicích $i, j$ jsou dány následujícím pøedpisem:
+
+$$ A_{i,j} = \left\{ \matrix {1 \Leftrightarrow \{i,j\} \in E  \cr
+				0 \Leftrightarrow \{i,j\} \notin E \cr            			
+				}
+\right.$$
+
+Na~pozicích $i,j$ je jednièka, pokud v~grafu $G$ vede hrana z~vrcholu~$i$ do~vrcholu~$j$, jinak to je nula.
+Ná¹ graf z~obrázku vý¹e by tedy v~maticové reprezentaci vypadal takto:
+
+$$\bordermatrix{
+  & A & B & C & D\cr
+A & 0 & 1 & 1 & 0\cr
+B & 0 & 0 & 0 & 1\cr
+C & 0 & 0 & 0 & 1\cr
+D & 1 & 1 & 0 & 0\cr
+}$$
+
+S touto maticí se pracuje velmi snadno, napø. v¹echny sousedy $i$-tého vrcholu
+zjistíme jednodu¹e tak, ¾e projdeme $i$-tý øádek matice.
+Má ov¹em dvì zøejmé nevýhody: èasovou a pamì»ovou slo¾itost. Projití sousedù jednoho vrcholu trvá v¾dy $\Theta(n)$, projití sousedù pro v¹echny vrcholy (co¾ potøebujeme v~BFS) pak trvá $\Theta(n^2)$. Velikost matice je v¾dy $n \times n$, bez ohledu na~to, jak \uv{øídký} je graf. U grafu s mnoha vrcholy, ale s malým poètem hran, tedy budeme zbyteènì plýtvat místem v~pamìti. Tato reprezentace je tedy nevýhodná pøedev¹ím pro tøídy grafù jako jsou stromy, které mají $n-1$ hran, nebo rovinné grafy, které mají nejvý¹e $3n-6$ hran.
+
+\s{Pozorování:} BFS s reprezentací maticí sousednosti bì¾í v~èase: $\Theta(n^2)$.
+
+\proof
+U¾ jsme si uvìdomili, ¾e ka¾dý vrchol se dostane do~fronty $Q$ nejvý¹e jednou. Pro ka¾dý vrchol ve~frontì potøebujeme projít jeho sousedy, co¾ nám trvá s~reprezentací maticí sousednosti $\Theta(n)$. Vrcholù je celkem $n$, tedy èasová slo¾itost je $\Theta(n^2)$.
+\qed
+
+\s{2. seznam sousedù}
+
+V~matici sousednosti jsme museli procházet jak hrany, tak nehrany, co¾ bylo zbyteèné. Bylo by tedy výhodnìj¹í pamatovat si pro ka¾dý vrchol pouze jeho sousedy. To mù¾eme zaøídit napøíklad jedním ze~dvou následujících zpùsobù:
+
+Uchovávejme pole indexované vrcholy tak, ¾e v~ka¾dém prvku pole je ukazatel na~spojový seznam sousedù tohoto vrcholu. Tedy $L(v)=\{w: vw \in E(G)\}$.
+
+Pokud se nám nebude chtít pracovat se spojovými seznamy, mù¾eme vyu¾ít reprezentaci pomocí dvou polí. První pole~$V$ bude opìt indexované vrcholy. V~druhém poli~$E$ budou pro ka¾dý vrchol ulo¾eni jeho sousedé. V~poli~$V$ si pamatujeme pro ka¾dý vrchol~$i$ index do~pole~$E$, kde zaèínají jeho sousedé. K sousedùm vrcholu~$i$ se tedy dostaneme ji¾ snadno - nalezneme je na pozicích $V[i]~\dots~V[i+1]-1$.
+\figure{imgn_nei.eps}{Znázornìní polí seznamu sousedù.}{\epsfxsize}
+
+
+Na tuto reprezentaci u¾ staèí prostor $O(n + m)$, co¾ u¾ je, na~rozdíl od~pøedchozího kvadratického prostoru, docela pøíjemné.
+
+\s{Pozorování:} BFS bì¾í v~èase: $\Theta(n+m)$.
+
+\proof
+Algoritmus vezme ka¾dý vrchol i ka¾dou hranu do~ruky nejvý¹e jednou. Èasová slo¾itost bude tedy:
+$$\Theta\left(n+\sum_{v\in V(G)} {\rm deg}(v)\right) = \Theta(n+m).$$
+\qed
+
+\s{3. orákulum}
+
+Dal¹í mo¾ností reprezentace je pak jakési orákulum, které nám øekne (spoèítá), kam vedou hrany z~daného vrcholu\dots To se hodí napøíklad tehdy, pokud si nepotøebujeme pamatovat celý graf, ale staèí nám naleznout sousedy nìjakého vrcholu, které orákulum jednoznaènì výpoètem doká¾e urèit. Napøíklad pøi øe¹ení známé úlohy odmìøení daného mno¾ství vody za pomocí nádob rùzných objemù mù¾eme tento zpùsob reprezentace grafu pou¾ít. Problém pøevedeme na prohledávání grafu do~¹íøky, kde vrcholùm odpovídají jednotlivé stavy. Stav øíká, kolik vody je zrovna ve které nádobì. Potom nám ji¾ staèí ze zadaného poèáteèního vrcholu (stavu) najít cestu do~cílového. Orákulum je zde vlastnì funkce, které pøedáme vrchol a ona nám vrátí v¹echny vrcholy sousední -- tedy takové stavy, ke kterým dojde, kdy¾ vyzkou¹í v¹echny mo¾nosti pøelití kapaliny z jedné nádoby do~druhé.
+
+\h{Roz¹íøení algoritmu:}
+
+Abychom mohli vyu¾ít toho, ¾e algoritmus prochází vrcholy grafu ve~vlnì, a jiných hezkých vlastností, tak si dodefinujeme následující oznaèení:
+
+V~poli $D$ bude pro ka¾dý vrchol ulo¾ena vzdálenost od~poèáteèního vrcholu.
+V~poli $P$ si budeme pro ka¾dý vrchol pamatovat jeho pøedchùdce. Dále budeme vyu¾ívat fáze bìhu algoritmu, které budou simulovat onu vlnu:
+
+\s{Definice: {\I Fáze bìhu algoritmu}:} Ve~fázi $F_0$ je zpracováván vrchol $v_0$. Ve~fázi $F_{i+1}$ jsou zpracovávány vrcholy ulo¾ené do~fronty $Q$ bìhem fáze $F_i$.
+
+\s{Roz¹íøený algoritmus:}
+\algo
+\:$Q \leftarrow \{v_0\}$.
+\:$Z[*] \leftarrow 0, Z[v_0] \leftarrow 1$.
+\:$D[*] \leftarrow \infty, D[v_0] \leftarrow 0$.
+\:Dokud $Q \not= \emptyset $ opakujeme:
+\::Vyzvedneme vrchol $u$ z~$Q$.
+\::Pro ka¾dý vrchol $w$, který je sousedem vrcholu $u$:
+\:::Je-li $Z[w]=0 \Rightarrow Z[w] \leftarrow 1, D[w] \leftarrow D[u]+1, P[w] \leftarrow u$
+\::::Pøidáme $w$ do~$Q$.
+\endalgo
+
+\s{Lemma:} Na~konci BFS pro v¹echny vrcholy dosa¾itelné z~$v_0$ platí, ¾e vrchol $v$ byl zpracován ve~fázi $F_i$ právì tehdy, kdy¾ vzdálenost $v_0$ a $v$ (délka nejkrat¹í cesty z~$v_0$ do~$v$) je rovna $i$. Formálnì zapsáno: $v \in F_i \Leftrightarrow d(v_0,v) = i$. (Kde $d(x,y)$ je délka nejkrat¹í cesty z~$x$ do~$y$).
+
+\proof
+\uv{$\Longrightarrow$}: 
+Dùkaz provedeme indukcí podle $i$ (èísla fáze bìhu algoritmu).
+
+První krok indukce je triviální, nebo» ve~fázi $F_0$ je oznaèen (dle definice) pouze vrchol $v_0$ a to je jediný vrchol vzdálený~0 od~$v_0$. 
+
+Pokud je vrchol $v$ zpracováván ve~fázi $F_i$, pak musel být zaøazen do~fronty bìhem fáze $F_{i-1}$ jako soused nìjakého vrcholu $u$. Pro vrchol $u$ mù¾eme pou¾ít indukèní pøedpoklad, tedy ¾e délka nejkrat¹í cesty z~$v_0$ do~$u$ je $d(v_0,u)=i-1$. Pak tedy $d(v_0,v)\leq i$, nebo» je¹tì nevíme, zda cesta $v_0, \dots, u, v$ je nejkrat¹í. Kdyby ale existovala nìjaká krat¹í, tedy délky nejvý¹e $i-1$, tak by byl vrchol $v$ objeven u¾ døíve ne¾ ve fázi $F_i$. Proto $d(v_0,v) = i$.
+
+
+\uv{$\Longleftarrow$}: Ka¾dý dosa¾itelný vrchol padne do~nìjaké fáze (viz. minulé lemma).
+\qed
+
+Ji¾ tedy víme, ¾e vrchol $v_i$, jeho¾ vzdálenost od~vrcholu $v_0$ je $i$, bude zpracován v~$i$-té fázi. Jak ale po~skonèení algoritmu zjistíme, ve které fázi byl zpracován, neboli jak je vzdálený od~startovního vrcholu? Tato informace je právì ulo¾ena v~poli $D$ s indexem $i$ (v~$D[i]$).
+
+Zároveò nás mù¾e zajímat, jak bychom nejkrat¹í cestu z~$v_0$ do~$v_i$ rekonstruovali. Pro tento úèel jsme si zavedli pole $P$. Nejkrat¹í cesta z~$v_0$ do~$v_i$ bude v~obráceném poøadí vypadat: $v_i, P[v_i], P[P[v_i]], P[P[P[v_i]]], \dots, v_0$.
+
+
+\s{Pozorování:} Pokud víme, ¾e $v_0, v_1, \dots, v_{k-1}, v_k$ je nejkrat¹í cesta z~$v_0$ do~$v_k$, pak $v_0,v_1,\dots,v_{k-1}$ je nejkrat¹í cesta z~$v_0$ do~$v_{k-1}$. Neboli prefix nejkrat¹í cesty je nejkrat¹í cesta.
+
+\proof
+Kdyby existovala krat¹í cesta z~$v_0$ do~$v_{k-1}$, pak bychom mohli zkrátit i cestu z~$v_0$ do~$v_k$.
+
+
+\s{Pozorování:} BFS u~neorientovaného grafu projde celou komponentu souvislosti.
+
+\proof
+Víme, ¾e BFS($v_0$) oznaèí $v$ právì tehdy, kdy¾ existuje cesta z~$v_0$ do~$v$. V~neorientovaném grafu existuje cesta z~$v_0$ do~právì v¹ech vrcholù, které jsou ve~stejné komponentì souvislosti jako $v_0$. Pokud tedy spustíme BFS na~$v_0$, tak se postupnì projdou v¹echny vrcholy této komponenty souvislosti.
+\qed
+
+\s{Pozorování:} Pokud BFS postupnì spou¹tíme na~dosud neobarvené vrcholy v~ neorientovaném grafu, nalezneme nakonec v~èase $\Theta(n+m)$ v¹echny komponenty souvislosti.
+
+\s{Algoritmus:}
+\algo
+\:Pro v¹echny vrcholy $v \in V(G)$ opakuj:
+\::Pokud je vrchol $v$ neobarvený $ \Rightarrow \<BFS(v)>$.
+\endalgo
+
+\proof
+Ka¾dým spu¹tìním na~dosud neobarvený vrchol neorientovaného grafu obarvíme právì jednu komponentu souvislosti (tu, ve~které je tento vrchol). 
+Postupnì projdeme v¹echny vrcholy od prvního k poslednímu a v¾dy pokud je vrchol neobarvený, spustíme na nìj BFS. Tak nakonec obarvíme v¹echny komponenty souvislosti. Èasová slo¾itost bude stejná jako u~samotného BFS, tedy $\Theta(n + m)$.
+\qed
+
+\s{Vìta:} $BFS(v_0)$ v~èase $\Theta(n + m)$ spoète:
+\itemize\ibull
+\:vrcholy dosa¾itelné z~$v_0$
+\:vzdálenosti tìchto vrcholù od~$v_0$
+\:strom nejkrat¹ích cest z~$v_0$
+\endlist
+
+\s{Poznámka:} Algoritmus na prohledávání do~¹íøky bude fungovat i na neorientovaném grafu. Mù¾eme si jednodu¹e pøedstavit, ¾e je to orientovaný graf, kde ka¾dá hrana má oboustrannou orientaci.
+
+
+Prohledávání do~¹íøky ale není jediný algoritmus, který nìjak systematicky prochází graf. Jak u¾ název kapitoly napovídá, budeme se zabývat je¹tì druhým algoritmem, prohledáváním do~hloubky. Podívejme se, jak bude vypadat \dots
+
+\h{Prohledávání do~hloubky (DFS) {\I Depth-First Search} }
+
+Tento algoritmus neprochází graf ve~vlnì jako BFS, ale prochází ho rekurzivnì. V¾dy se zanoøí co nejhloubìji a¾ do~listu a pak se o~kus vrátí a opìt se sna¾í zanoøit. Vrcholy, ve kterých u¾ byl, ignoruje.
+
+Opìt uva¾me nejdøíve graf orientovaný. Následnì si uká¾eme, ¾e v~neorientovaném grafu budou pouze malé zmìny.
+
+Budeme pou¾ívat podobné znaèení jako u~BFS. V~poli $Z$ si budeme pamatovat, zda jsme vrchol ji¾ nav¹tívili (hodnota 1), nebo ne (hodnota 0). Navíc promìnná~$T$ bude znaèit dobu bìhu algoritmu - tedy jakési \uv{hodiny}. Pøi ka¾dém nalezení nového vrcholu èi jeho opu¹tìní tuto promìnnou zvý¹íme o~1. Do~polí $\<in>$ a $\<out>$ si budeme ukládat èas (prvního) nalezení a opu¹tìní vrcholu.
+
+\s{Algoritmus:}
+
+\algo
+\: inicializace: $Z[*] \leftarrow 0, T \leftarrow 1, \<in>[*] \leftarrow ?, \<out>[*] \leftarrow ?$
+\: $DFS(v): Z[v] \leftarrow 1, \<in>[v] \leftarrow T|++|$
+\:: Pro $w$: $vw \in E(G)$:
+\::: Pokud $Z[w]=0 \Rightarrow DFS(w)$
+\:: $out[v] \leftarrow T|++|$
+\endalgo
+
+\s {Vìta:} DFS($v_0$) v~èase $\Theta(m+n)$ oznaèí právì v¹echny vrcholy dosa¾itelné z~$v_0$.
+
+\proof
+Nejdøíve je potøeba dokázat, ¾e pokud je vrchol $v$ dosa¾itelný z~vrcholu $v_0$, tak jej DFS oznaèí. Dùkaz bude podobný jako u~BFS.
+
+V analýze èasové slo¾itosti si pak opìt uvìdomíme, ¾e algoritmus vezme ka¾dý vrchol i hranu do~ruky právì jednou, tak¾e èasová slo¾itost bude $\Theta(n + m)$.
+\qed
+
+\figure{img5_dfso.eps}{Graf a znázornìní prùbìhu DFS s~jednotlivými hranami:}{\epsfxsize}
+
+Mù¾eme si v¹imnout, ¾e jak DFS prochází graf, rozdìluje hrany do~4 skupin: hrany {\I stromové, zpìtné, dopøedné a pøíèné}.
+
+Jedinì {\I stromové} hrany jsou takové, ¾e se po~nich DFS opravdu vydá. Vedou toti¾ do~vrcholu, který nebyl dosud objeven. V~ukázkovém grafu to jsou hrany: $\{(A \rightarrow B), (B \rightarrow C), (B \rightarrow D)\}$.
+
+Pokud algoritmus objeví z~vrcholu $v$ hranu do~ji¾ døíve nav¹tíveného vrcholu $w$ a zároveò platí, ¾e $w$ je ve~stejném podstromu jako $v$, tak nazveme hranu $vw$ {\I zpìtnou}. Pro rozpoznání je dùle¾ité, ¾e vrchol $w$ byl ji¾ objeven, ale je¹tì ne opu¹tìn. V~ukázkovém grafu se jedná o hranu: $(C \rightarrow A)$.
+
+Kdy¾ pøi prohledávání sousedù vrcholu $v$ narazíme na~vrchol $w$, který jsme ji¾ nav¹tívili, a to v~podstromì vrcholu $v$, tak nazveme hranu $vw$ {\I dopøednou}, nebo» vede z~$v$ do~jeho potomka. Platí tedy, ¾e jsme nejdøíve objevili vrchol $v$, potom vrchol $w$, pak jsme vrchol $w$ opustili a nyní jsme na~nìj znovu narazili po~dopøedné hranì. V~ukázkovém grafu hrana: $(A \rightarrow D)$.
+
+Posledním typem hran je {\I pøíèná} hrana. Ta vede do~vrcholu v~sousedním podstromì zprava doleva. V~tomto pøípadì jsme tedy nejdøíve objevili vrchol $w$, ten jsme následnì opustili a a¾ pak jsme objevili vrchol $v$. V~ukázkovém grafu to je hrana: $(D \rightarrow C)$.
+
+\s{K zamy¹lení:} Proè nemohou vést pøíèné hrany také zleva doprava?
+
+K~rozpoznávání typù hran se nám tedy velmi hodí pole $\<in>$ a $\<out>$, ve~kterých si pamatujeme èasy objevení a opu¹tìní vrcholu. Podle toho, jak se intervaly objevení a opu¹tìní obou vrcholù pøekrývají, mù¾eme jednoznaènì rozhodnout, o jaký typ hrany se jedná:
+
+U~zpìtných hran je poøadí: $\<in>(w)$, $\<in>(v)$, $\<out>(v)$, $\<out>(w)$. Intervaly do~sebe budou zanoøené takto: $<<>_v>_w$.
+
+U~dopøedných hran je poøadí: $\<in>(v)$, $\<in>(w)$, $\<out>(w)$, $\<out>(v)$. Intervaly do~sebe budou zanoøené takto: $<<>_w>_v$.
+
+U~pøíèných hran je poøadí: $\<in>(w)$, $\<out>(w)$, $\<in>(v)$, $\<out>(v)$. Intervaly do~sebe budou zanoøené takto: $<>_w<>_v$.
+
+Pozn: Pou¾íváme zde toto znaèení: $<>_v = <in(v), out(v)>$. Jedná se o interval objevení a opu¹tìní vrcholu $v$.
+
+\s{Typy hran ($v \rightarrow w$):}
+
+\itemize\ibull
+\:Stromové hrany \dots po nich DFS pro¹lo. $\{(A \rightarrow B), (B \rightarrow C), (B \rightarrow D)\}$
+\:Zpìtné hrany $<<>_v>_w$\dots vedou do~pøedchùdce $v$ ve~stromu. $\{(C \rightarrow A)\}$
+\:Dopøedné hrany $<<>_w>_v$\dots vedou do~potomka. $v$ $\{(A \rightarrow D)\}$
+\:Pøíèné hrany $<>_w<>_v$\dots vedou do~vrcholu $v$ v~sousedním podstromì, v¾dy zprava doleva. $\{(D \rightarrow A)\}$
+\endlist
+
+\s{Pozorování:} Hrany, po~kterých DFS pro¹lo, tvoøí DFS strom.
+
+\s{Pozorování:} Intervaly ($\<in>(v)$, $\<out>(v)$) $\forall v \in V(G) $ tvoøí dobré uzávorkování. (Intervaly synù disjunktnì vyplòují otce $\Rightarrow$ intervaly se nemohou køí¾it).
+
+Nakonec si je¹tì uvìdomme, jak bude vypadat prohledávání do~hloubky na~neorientovaném grafu. Algortimus bude úplnì stejný, jenom se nám zredukuje poèet typù hran na~dvì: {\I stromové} a~{\I zpìtné}. Ani {\I dopøedné} ani {\I pøíèné} nebudou existovat, nebo» se z~{\I pøíèných} stanou {\I stromové} a z~{\I dopøedných zpìtné}.
+
+\bye
diff --git a/2009/3-dfs/Makefile b/2009/3-dfs/Makefile
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index 0000000..62ff1bc
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+
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diff --git a/2009/3-dfs/praseci-graf.eps b/2009/3-dfs/praseci-graf.eps
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diff --git a/2009/6-kostry/6-kostry.tex b/2009/6-kostry/6-kostry.tex
new file mode 100644
index 0000000..3c1a111
--- /dev/null
+++ b/2009/6-kostry/6-kostry.tex
@@ -0,0 +1,192 @@
+\input lecnotes.tex
+
+\prednaska{6}{Problém minimální kostry}{}
+
+\s{Zadání úlohy:} Pro neorientovaný graf $G$ s~ohodnocením hran {\I váhami} $w: E(G) \rightarrow \bb R$,
+chceme najít kostru $T$ s minimálním ohodnocením $w(T)=\sum_{e\in E(T)} w(e)$.
+
+\s{Navíc pøedpokládáme:} (bez újmy na~obecnosti)
+\itemize\ibull
+\:Graf $G$ je souvislý (jinak ho nejprve rozlo¾íme na komponenty).
+\:$\forall e,f \in E(G$) : $e\neq f \Rightarrow w(e)\neq w(f)$.
+\endlist
+
+Nyní si uká¾eme tøi algoritmy pro hledání minimální kostry, konkrétnì se jedná
+o~Jarníkùv, Borùvkùv a Kruskalùv algoritmus.
+
+\h{Jarníkùv algoritmus}
+
+\s{Algoritmus:}
+
+\algo
+\algin Graf~$G$ s~ohodnocením~$w$.
+\:Zvolíme libovolný vrchol $v_0\in V(G)$.
+\:$T\leftarrow(\left\{v_0\right\},\emptyset)$ (zatím jednovrcholový strom)
+\:Dokud $\vert V(T) \vert \neq n$:
+\::Vybereme hranu $uv\in E(G) : u\in V(T), v\notin V(T)$ tak, aby $w(uv)$ byla minimálni.
+\::$T\leftarrow T+uv$.
+\algout Minimální kostra~$T$.
+\endalgo
+
+\s{Vìta:} Jarníkùv algoritmus se zastaví po maximálnì $n$ iteracích a vydá minimální kostru grafu $G$.
+
+\proof
+Pøi ka¾dé iteraci algoritmus pøidá jeden vrchol do~$T$, a~proto se po~maximálnì $n$ iteracích zastaví.
+Vydaný graf je strom, proto¾e se stále pøidává list k ji¾ existujícímu stromu, a~jeliko¾ má $n$~vrcholù,
+je to kostra. Zbývá nám u¾ jen dokázat, ¾e nalezená kostra je minimální. K~tomu pomu¾e následující lemma:
+
+{\narrower
+
+\s{Definice:} {\I Øez} v~grafu $G=(V,E)$ je mno¾ina hran $F\subseteq E$ taková, ¾e $\exists  U\subset V$ :
+$F=\left\{uv\in E:u\in U, v\notin U \right\}$.
+
+\s{Lemma (øezové):} Pokud $G$ je graf, $w$ jeho prosté ohodnocení, $F$ je øez v
+grafu $G$ a $f$ je nejlehèí hrana v øezu $F$, pak pro ka¾dou minimální kostru
+$T$ grafu $G$ je $f\in E(T)$.
+
+\proof
+Buï $T$ kostra a $f=uv\notin E(T)$. Pak existuje cesta $P\subseteq T$ spojující $u$ a $v$.
+Cesta musí øez alespoò jednou pøekroèit. Proto existuje $e\in P \cap F$ a navíc víme, ¾e $w(e) > w(f)$. Uva¾me $T'=T-e+f$.
+Tento graf je rovnì¾ kostra grafu $G$, proto¾e odebraním hrany $e$ se graf rozpadne na dvì komponenty a pøidáním
+hrany $f$ se tyto komponenty opìt spojí. Navíc $w(T')=w(T)-w(e)+w(f)<w(T)$.
+\qed
+
+}
+
+\>V~dùkazu korektnosti Jarníkova algoritmu toto lemma vyu¾ijeme tak, ¾e si v¹imneme, ¾e hrany mezi
+vrcholy stromu~$T$ a zbytkem grafu tvoøí øez a algoritmus nejlehèí hranu tohoto øezu pøidá
+do~$T$. Podle lemmatu tedy v¹echny hrany~$T$ musí být souèástí ka¾dé minimální kostry a jeliko¾~$T$ je strom,
+musí být minimální kostrou.
+
+\qed
+
+
+\s{Dùsledky:} Graf $G$ s prostým ohodnocením má pravì jednu minimální kostru. Minimální kostra je
+jednoznaènì urèená lineárním uspoøádáním hran.
+
+\s{Implementace:}
+\itemize\ibull
+\:Pøímoèará: pamatujeme si, které vrcholy a hrany jsou v kostøe $T$ a které ne. Èasová slo¾itost je $\O(nm)$.
+\:Chytøej¹í: Pro $v\notin V(T)$ si pamatujeme $D(v)=\min\left\{w(uv):u\in T\right\}$. Pøi ka¾dém
+prùchodu hlavním cyklem pak procházíme v¹echna~$D(v)$ (to v¾dy trvá $\O(n)$) a pøi pøidání vrcholu do~$T$ kontrolujeme
+okolní~$D(w)$ pro $vw\in E$ a pøípadnì je sni¾ujeme (za~ka¾dou hranu~$\O(1)$). Èasovou slo¾itost tím celkovì
+zlep¹íme na~$\O(n^2+m)=\O(n^2)$.
+\:Také se dá pou¾ít halda pro uchovávání hran nebo hodnot~$D(v)$.
+\endlist
+
+\h{Borùvkùv algoritmus}
+
+\s{Algoritmus:}
+
+\algo
+\algin Graf~$G$ s~ohodnocením~$w$.
+\:$F\leftarrow(V(G),\emptyset)$
+\:Dokud $F$ má alespoò dvì komponenty:
+\::Pro ka¾dou komponentu $T_i$ grafu $F$ vybereme nejlehèí incidentní hranu $t_i$.
+\::V¹echny hrany $t_i$ pøidáme do $F$.
+\algout Minimální kostra~$F$.
+\endalgo
+
+\s{Vìta:} Borùvkùv algoritmus se zastaví po $\left\lceil \log_2 n\right\rceil$ iteracích a vydá minimální kostru grafu $G$.
+
+\proof
+V¹imnìme si nejprve, ¾e po~$k$ iteracích mají v¹echny komponenty grafu~$F$ minimálnì $2^k$ vrcholù
+(indukcí -- na~poèátku jsou v¹echny komponenty jednovrcholové, v~ka¾dé dal¹í
+iteraci se komponenty sluèují do~vìt¹ích,
+ka¾dá s~alespoò jednou sousední, tak¾e se velikosti komponent minimálnì zdvojnásobí).
+Proto nejpozdìji po~$\left\lceil \log_2 n\right\rceil$ iteracích u¾~velikost komponenty dosáhne poètu v¹ech vrcholù a algoritmus
+se zastaví.
+
+Hrany mezi ka¾dou komponentou a~zbytkem grafu tvoøí øez, tak¾e podle øezového
+lemmatu v¹echny hrany pøidané do~$F$ musí být souèástí (jednoznaènì
+urèené) minimální kostry. Graf $F\subseteq G$ je tedy v¾dy les a a¾ se
+algoritmus zastaví, bude roven minimální kostøe.
+\qed
+
+\s{Implementace:}
+\itemize\ibull
+\:Inicializace pøímoèará.
+\:Pomocí DFS rozlo¾íme les na komponenty. Pro ka¾dý vrchol si pamatujeme èíslo komponenty.
+\:Pro ka¾dou hranu zjistíme, do které komponenty patøí, a pro ka¾dou komponentu
+si uchováme nejlehèí hranu.
+\endlist
+
+\>Takto doká¾eme ka¾dou iteraci provést v~èase $\O(m)$ a celý algoritmus dobìhne v~$\O(m\log n)$.
+
+\h{Kruskalùv neboli hladový algoritmus}
+
+\s{Algoritmus:}
+
+\algo
+\algin Graf~$G$ s~ohodnocením~$w$.
+\:Setøídíme v¹echny hrany z $E(G)$ tak, aby: $w(e_1)<...<w(e_m)$.
+\:$F\leftarrow (V(G),\emptyset)$.
+\:Pro $i=1$ do $m$:
+\::Pokud $F+e_i$ je acyklický, provedeme $F\leftarrow F+e_i$.
+\algout Minimální kostra~$F$.
+\endalgo
+
+\s{Vìta:} Kruskalùv algoritmus se zastaví po~$m$ iteracích a vydá minimální kostru.
+
+\proof
+Ka¾dá iterace algoritmu zpracovává jednu hranu, tak¾e iterací je~$m$. Indukcí doká¾eme,
+¾e~$F$ je v¾dy podgrafem minimální kostry: prázdné poèáteèní~$F$ je podgrafem èehokoliv,
+ka¾dá hrana, kterou pak pøidáme, je minimální v~øezu oddìlujícím nìjakou
+komponentu~$F$ od~zbytku grafu (ostatní hrany tohoto øezu je¹tì nebyly
+zpracovány, a~tudí¾ jsou tì¾¹í). Naopak ¾ádná hrana, kterou jsme se rozhodli
+do~$F$ nepøidat, nemù¾e být souèástí minimální kostry, jeliko¾ s~hranami,
+o~kterých ji¾ víme, ¾e v~minimální kostøe le¾í, tvoøí kru¾nici. \qed
+
+% Toto jsme na pøedná¹ce dìlali sporem a mì se to líbí více, ale oba dùkazy
+% urèitì fungují.
+
+\s{Implementace:}
+\itemize\ibull
+\:Setøídìní v èase $\O(m\log m)=\O(m\log n)$.
+\:Pak potøebujeme udr¾ovat komponenty souvislosti grafu~$F$, abychom umìli rychle
+  urèit, jestli právì zpracovávaná hrana vytvoøí kru¾nici. Potøebujeme tedy strukturu
+  pro udr¾ování komponent souvislosti, které se $m$-krát zeptáme, zda dva vrcholy
+  le¾í v~té¾e komponentì (tomu budeme øíkat operace \<Find>), a~$(n-1)$-krát spojíme
+  dvì komponenty do jedné (\<Union>).
+\endlist
+
+\>Kruskalùv algoritmus tedy pobì¾í v~èase $\O(m\log n + mT_f + nT_u)$, kde~$T_u$ je
+èas na~provedení jedné operace \<Union> a $T_f$ na~operaci \<Find>.
+
+\s{Jednoduchá struktura pro komponenty:}
+Budeme si pamatovat v~poli èísla komponent, ve~kterých le¾í jednotlivé
+vrcholy. \<Find> zvládneme v~èase $\O(1)$, ale \<Union> bude stát $\O(n)$. Celý
+algoritmus pak pobì¾í v~èase $\O(m\log n+ m + n^2) = \O(m\log n+n^2)$.
+
+\s{Chytøej¹í struktura:} Ka¾dou komponentou si ulo¾íme jako strom orientovaný smìrem ke koøeni
+-- ka¾dý vrchol si pamatuje svého otce, navíc ka¾dý koøen si pamatuje velikost
+komponenty. 
+%Hloubku podstromu? Zaøazujeme mìlèí pod hlub¹í, ne nutnì men¹í pod vìt¹í.
+%Myslím, ¾e s hloubkami to funguje lépe, ovìøit.
+Operace \<Find> vystoupá z~obou vrcholù ke~koøeni a koøeny porovná. \<Union>
+rovnì¾ najde koøeny a pøipojí koøen men¹í komponenty pod koøen té vìt¹í. Obojí
+zvládneme v~èase lineárním v~hloubce stromu a jak si uká¾eme, tato hloubka je
+v¾dy nejvý¹e logaritmická, a proto celý Kruskalùv algoritmus pobì¾í v~èase
+$\O(m\log n + m\log n + n\log n) = \O(m\log n)$.\foot{%
+Drobnou úpravou bychom mohli dosáhnout daleko efektivnìj¹í struktury, ale tu
+bychom neupotøebili,
+jeliko¾ by nás stejnì brzdilo tøídìní, a analýza slo¾itosti by byla \dots\ inu, slo¾itìj¹í.}
+
+% V originále je jen "strom hloubky..." a to pøece pro obecný (napø.
+% zdegenerovaný) strom neplatí.
+
+% Mimochodem, zde se u¾ mluví o spojování podle hloubky, tak¾e to nahoøe je asi
+% vá¾nì chyba
+\s{Lemma:} \<Union-Find> strom hloubky $h$ má alespoò $2^h$ prvkù.
+
+\proof
+Indukcí: Pokud \<Union> spojí strom s~hloubkou $h$ s jiným s~hloubkou men¹í ne¾ $h$, pak hloubka výsledného
+stromu zùstává~$h$. Pokud spojuje dva stromy stejné hloubky~$h$, pak má výsledný strom hloubku $h+1$.
+Z~indukèního pøedpokladu víme, ¾e strom hloubky $h$ má minimálnì $2^h$ vrcholù,
+a~tedy výsledný strom hloubky $h+1$ má alespoò $2^{h+1}$ vrcholù. \qed
+
+% Union-find je jsme mìli na pøedná¹ce detailnìji rozebraný (i s pseudokódem),
+% není to nutnost, ale mù¾u ho klidnì rozepsat trochu výøeènìji, aby mìl ètenáø
+% vìt¹í jistotu. Zále¾í na tobì.
+
+\bye
diff --git a/2009/6-kostry/Makefile b/2009/6-kostry/Makefile
new file mode 100644
index 0000000..b86ea6f
--- /dev/null
+++ b/2009/6-kostry/Makefile
@@ -0,0 +1,3 @@
+P=6-kostry
+
+include ../Makerules
diff --git a/2009/8-rozdel/8-rozdel.tex b/2009/8-rozdel/8-rozdel.tex
new file mode 100644
index 0000000..6103891
--- /dev/null
+++ b/2009/8-rozdel/8-rozdel.tex
@@ -0,0 +1,408 @@
+\input ../lecnotes.tex
+
+\prednaska{8}{Rozdìl a~panuj}{}
+
+Známá strategie {\sl Divide et Impera} (nìkdy pøekládáno spí¹e jako \uv {Roze¹tvi a~panuj}) pochází z dob antického Øíma\foot{Aè se touto strategií øím¹tí panovníci øídili, není nalezen ¾ádný antický zdroj výroku; ten je pøipisován a¾ renesanènímu N. Machiavellimu.
+}, kdy panovník zasel nesvár mezi místní kmeny (èím¾ je rozdìlil a~roze¹tval) a~následnì pøi¹el, urovnal napìtí, aby je mohl opìt ovládnout a~panovat a~aby byli v¹ichni spokojení.
+
+Tato strategie ov¹em pøetvrává (aè tøeba V~jiných oblastech) a~pomáhá øe¹it ty problémy, které se dají rozdìlit na~men¹í jednodu¹¹í podproblémky.
+
+Jak tedy algoritmus typu {\it rozdìl a~panuj} pracuje? Mìjme problém, který má tu vlastnost, ¾e kdy¾ jej rozdìlíme na~vhodné podproblémy, které mají stejný charakter, a~ty vyøe¹íme, slo¾ením jejich øe¹ení mù¾eme získat øe¹ení pùvodního problému. Algoritmus tedy bude rekurzivnì volat sám sebe, ne¾ se dostane k~podproblému
+nìjaké konstantní velikosti, který u¾ umí vyøe¹it triviálnì, a~pak se zaène z~rekurze vracet a~skládat jednotlivá dílèí øe¹ení.
+
+\h{Pøíklad 1 -- MergeSort:}
+
+Tento tøídící algoritmus pracuje na~principu, ¾e vstup rozdìlíme na~dvì (skoro) stejnì velké èásti, které rekurzivním voláním setøídíme, a~nakonec výsledné dvì posloupnosti slijeme do~jedné. 
+
+\s{Algoritmus:}
+
+\algo
+\algin posloupnost $x_1,\dots, x_n$.
+\:Pokud $n \leq 1 \Rightarrow$ vrátíme vstup.
+\:$y_1,\dots,y_{\lfloor n/2 \rfloor} \leftarrow$ \<MergeSort> $(x_1,\dots,x_{\lfloor n/2 \rfloor})$.
+\:$z_1,\dots,z_{\lceil n/2 \rceil} \leftarrow$ \<MergeSort> $(x_{\lfloor n/2 \rfloor + 1},\dots,x_n)$.
+\:Vrátíme $\<Merge>(y_1,\dots,y_{\lfloor n/2 \rfloor}; z_1,\dots,z_{\lceil n/2 \rceil})$.
+\endalgo
+
+\noindent
+Na slití dvou setøídìných posloupností do~jedné pou¾íváme funkci Merge:
+
+{\bo $\<Merge>(y_1, \dots,y_a;z_1, \dots,z_b)$:}
+\algo
+\:$i \leftarrow 1, j \leftarrow 1, k \leftarrow 1$.
+\:Dokud $k \leq a+b$:
+\::Je-li $(j>b)$ nebo $(i \leq a) \& (y_i < z_j) \Rightarrow x_k \leftarrow y_i, k|++|, i|++|$.
+\::Jinak $\Rightarrow x_k \leftarrow z_j, k|++|, j|++|$.
+\:Vrátíme $x_1, \dots,x_n$.
+\endalgo
+
+\s{Pozorování:}
+\<Merge> trvá $\Theta (n)$, nebo» ka¾dou ze~slévaných posloupností projdeme právì jednou.
+
+\s{Èasová slo¾itost MergeSortu:}
+
+Rozdìlování a~slévání nám trvá lineárnì dlouho, tak¾e pro èasovou slo¾itost MergeSortu platí tato rekurentní rovnice: $$T(n)= 2 \cdot T(n/2) + c \cdot n.$$ 
+Pøitom $T(n)$ zde znaèí èas strávený na~vstupu délky $n$ a~$c$ je nìjaká vhodná konstanta. Pro jednoduchost BÚNO pøedpokládejme, ¾e $n$ je mocnina dvojky. Zároveò si jednotky zvolme tak, ¾e bude platit $T(1)=1$. Mù¾eme si tedy rekurentní vztah rozepsat následovnì:
+
+$$\eqalign{
+T(n) &= 2 \cdot (2T(n/4) + c \cdot n/2) + c \cdot n = 4T(n/4) + 2cn = \cr
+		 &= 4(2T(n/8) + c(n/4)) + 2cn = 8T(n/8) + 3cn = \dots\cr
+		 &= 2^k T(n/2^k) + kcn.
+}$$
+
+Pokud zvolíme $k = \log_2{n}$, získáme:
+$$T(n) = 2^{\log_2{n}} \cdot T(n/2^{\log_2{n}}) + \log_2{n} \cdot c \cdot n = n \cdot T(1) + c \cdot n \cdot \log_2{n} = \Theta(n \log{n}). $$
+
+Ke stejnému výsledku mù¾eme ale dojít také úplnì jinou cestou. Pøedstavme si strom rekurzivních volání. Ka¾dý vrchol má dva syny (dìlíme vstup na~dvì èásti), v~nich¾ jsou vstupy polovièní velikosti. V~ka¾dém vrcholu trávíme èas lineární s~velikostí jeho vstupu, souèet velikostí vstupù pøes ka¾dou hladinu je~$n$ a~hloubka stromu musí být $\Theta(\log n)$. Vyjde nám tedy, ¾e $T(n)=\Theta(n\log n)$.\foot{Po pozornìj¹ím zamy¹lení si ètenáø mù¾e uvìdomit, ¾e se jedná vlastnì pouze o~jiný pohled na~stejný dùkaz jako rozepisování rekurentního vzorce.}
+
+\s{Pamì»ová slo¾itost MergeSortu:}
+
+$M(n) = d \cdot n + M(n/2) = d \cdot n + d \cdot n/2 + d \cdot n/4 + \dots \leq 2d n = \Theta(n).$
+
+Tento vztah platí pro nìjakou vhodnou konstantu $d$. Mù¾eme ho opìt nahlédnout napøíklad ze stromu rekurzivních volání. Podívejme se na~libovolnou cestu od~koøene do~listu. V~jednotlivých vrcholech potøebujeme pamìti pøesnì $d \cdot n/2^k$, kde $k$ je èíslo hladiny. Kdy¾ tyto hodnoty seèteme pøes v¹echny vrcholy na~této cestì, výsledek bude konvergovat k~$2dn$, co¾ dává pamì»ovou slo¾itost $\Theta(n)$.
+
+\s{Závìr:}
+Mergesort bì¾í V~èase $\Theta(n \log{n})$ a~pamìti $\Theta(n)$. Lineární pamì»ová slo¾itost není výhodná, ale na~druhou stranu se tento algoritmus velmi hodí napøíklad na~tøídìní lineárních spojových seznamù.
+
+\h{Pøiklad 2 -- Násobení èísel:} 
+Pokud násobíme dvì èísla $X$ a~$Y$ (obì délky $n$; pokud bylo jedno krat¹í, tak ho doplníme nulami zleva tak, aby byla obì èísla stejnì dlouhá) zpùsobem, který nás uèili na~základní ¹kole, výsledný algoritmus má èasovou slo¾itost $\Theta(n^2)$. Proto¾e se jedná o~dost èastou operaci, zamysleme se, zda by ne¹la zrychlit. Nasmìrujme na¹e úvahy na~postup {\it rozdìl a~panuj}. Rozdìlíme ka¾dého èinitele na~dvì stejnì dlouhé èásti. Pro jednoduchost pøedpokládejme, ¾e toto roz¹tìpení èinitele probìhne v¾dy bez zbytku:
+$$
+X=A \cdot 10^{{n}/2}+B, \qquad Y=C \cdot10^{{n}/{2}}+D.
+$$
+Zde $A, B, C, D$ jsou u¾ jen $(n/2)$-ciferná èísla. Pùvodní souèin získáme jako:
+$$
+XY=(A\cdot 10^{{n}/{2}}+B) (C\cdot 10^{{n}/{2}}+D)=AC \cdot 10^{n}+(AD+BC)\cdot 10^{{n}/{2}}+BD.
+$$
+Nyní, jak vidíme, staèí spoèítat souèin ètyø $(n/2)$-ciferných èísel a~pak výsledky spolu seèíst. Uva¾me,
+jakou bude mít tento algoritmus èasovou slo¾itost:
+$$T(n) = 4T(n/2) + cn.$$
+Toto platí pro nìjakou vhodnou konstantu $c$ (výraz $cn$ je re¾ie na~sèítání).  Jednotky si zvolme tak, aby platilo: $$T(1)=1.$$
+Jak takovou rekurenci vyøe¹íme? Máme opìt dvì mo¾nosti:
+
+\>{\sl 1. zpùsob: Øe¹ení rozepsáním rekurentního vztahu:}
+$$\eqalign{
+T(n)&= 4T(n/2)+cn = \cr
+    &= 4\cdot (4T(n/4)+cn/2)+cn = 4^2T(n/4)+2cn+cn = 4^2T(n/4)+3cn = \cr
+    &= 4^2\cdot (2T(n/8)+cn/4)+3cn = 4^3T(n/8)+4cn+3cn = 4^3T(n/8)+7cn = \cr
+    &\dots\cr
+}$$
+Odtud snadno vypozorujeme, ¾e jednotlivé vztahy se vyvíjejí podle vzorce
+$T(n)=4^kT(n/2^k) + (2^k-1)cn.$ Pro $k=\lceil\log_2 n\rceil$ je ov¹em
+$2^k\le 1$, tak¾e $T(n/2^k)=\Theta(1)$ a~dostaneme (horní celou èást zanedbáme,
+ta ovlivní jen konstanty):
+$$
+T(n) = 4^{\log_2 n}\Theta(1) + (2^{\log_2 n}-1)cn = n^2\Theta(1) + (n-1)cn = \Theta(n^2).
+$$
+
+\>{\sl 2. zpùsob: Úvaha o~stromu:} Nakreslíme si strom rekurzivních volání
+na¹eho algoritmu:
+\fig{figure.eps}{4in}
+Na~$i$-té hladinì stromu le¾í $4^i$ vrcholù, v~nich jsou vstupy velikosti
+$n/2^i$, tak¾e na~celé hladinì trávíme èas celkem $\Theta(4^i\cdot n/2^i)
+= \Theta(2^in)$. Velikosti vstupù klesají exponenciálnì, tak¾e celý strom
+je hluboký $k=\log_2 n$ (opìt si dovolíme zapomenout na~horní celou èást).
+Celkem tedy spotøebujeme èas $\sum_{i=0}^{k}\Theta(2^in) = \Theta(n\cdot\sum_{i=0}^k 2^i) = \Theta(n^2)$.
+
+Oba zpùsoby analýzy se tedy shodují na~tom, ¾e ná¹ algoritmus má kvadratickou èasovou
+slo¾itost a~¾e jsme si oproti klasickému algoritmu nikterak nepomohli.
+Podívejme se je¹tì jednou na~to, jak se ná¹ algoritmus vìtví:
+$$\vbox{\halign{\hfil#\hfil \quad & \hfil#\hfil \quad &\hfil#\hfil\cr
+hloubka & poèet úloh & velikost podúlohy\cr
+\noalign{\smallskip\hrule\medskip}
+0 & $4^{0}$ & ${n}/{2^{0}}$\cr
+1 & $4^{1}$ & ${n}/{2^{1}}$\cr
+2 & $4^{2}$ & ${n}/{2^{2}}$\cr
+3 & $4^{3}$ & ${n}/{2^{3}}$\cr
+\vdots & \vdots & \vdots\cr
+$k$ & $4^{k}$ & ${n}/{2^{k}}$\cr}}$$
+Naskýtá se otázka, jestli bychom nemohli èasovou slo¾itost zlep¹it. Toho bychom
+mohli dosáhnout napøíklad zlep¹ením èlenu $cn$ v~na¹í rekurenci, èili zefektivnìním
+spojování podúloh. To ov¹em není pøíli¹ nadìjné (pokud ètenáø nevìøí, mù¾e si to dokázat),
+tak¾e místo toho vyu¾ijeme druhou ¹anci a~to omezení vìtvení ze~ètyø vìtví na~tøi.
+Pøipomeòme si, ¾e potøebujeme spoèítat:
+$$
+XY=AC\cdot 10^{n}+(AD+BC)\cdot 10^{n/2}+BD.
+$$
+Pøitom ale nepotøebujeme znát souèiny $AD$ ani $BC$ samostatnì, nám staèí zjistit hodnotu celého výrazu $AD+BC$. Kdy¾ budeme znát hodnotu výrazù: $AC$, $BD$ a~$(A + B)(C + D)$ (k tomu nám staèí 3 násobení a~2 sèítání), tak mù¾eme výraz $AD + BC$ získat následovnì:
+$$(A + B)(C + D) - AC - BD = AC + AD + BC + BD - AC - BD = AD + BC$$
+Nyní nám ji¾ staèí jen tøi násobení, ale potøebujeme dvì sèítání a~jedno odèítání navíc. Nicménì tato komplikace je zanedbatelná oproti práci u¹etøené men¹ím vìtvením. (Nová dvì sèítání a~jedno odèítání se V~èasové slo¾itosti schová do~$cn$.) Podívejme se opìt na~tabulku:
+$$\vbox{\halign{\hfil#\hfil \quad & \hfil#\hfil \quad &\hfil#\hfil\cr
+hloubka & poèet úloh & velikost podúlohy\cr
+\noalign{\smallskip\hrule\medskip}
+0 & $3^{0}$ & ${n}/{2^{0}}$\cr
+1 & $3^{1}$ & ${n}/{2^{1}}$\cr
+2& $3^{2}$ & ${n}/{2^{2}}$\cr
+3 & $3^{3}$ & ${n}/{2^{3}}$\cr
+\vdots & \vdots & \vdots\cr
+$k$ & $3^{k}$ & ${n}/{2^{k}}$\cr}}$$
+Ná¹ rekurentní vztah po~zbavení se jednoho násobení vypadá následovnì:
+$$T(n) = 3T(n/2)+ cn.$$
+
+Opìt uva¾me, kolik práce spotøebujeme v~souètu pøes v¹echny hladiny (hloubka stromu
+$k$ je opìt $\lceil\log_2 n\rceil$ a~horní celou èást zanedbáme):
+$$\sum_{i=0}^{k}3^{i}\cdot {{n}\over{2^{i}}}=\sum_{i=0}^{k} \left( {{3}\over{2}} \right) ^{i}\cdot n=n\cdot \sum_{i=0}^{k} \left( {{3}\over{2}} \right) ^{i}=n\cdot {{ \left( {{3}\over{2}} \right) ^{k+1}-1}\over{{{3}\over{2}}-1}}=
+$$
+$$
+=n\cdot {{ \left( {3}\over{2} \right) ^{k+1}-1}\over{{{1}\over{2}}}}=2\cdot n\cdot  \left[ \left( {{3}\over{2}} \right) ^{k+1}-1 \right] = \Theta \left( n\cdot  \left( {{3}\over{2}} \right) ^{\log_2{n}} \right) =
+$$
+$$
+=\Theta \left( n\cdot {{3^{\log_2{n}}}\over{2^{\log_2{n}}}} \right)=\Theta \left( n\cdot {{3^{\log_2{n}}}\over{n}} \right)=\Theta \left( 3^{\log_2{n}} \right)=\Theta \left( (2^{\log_2{3}})^{\log_2{n}} \right)=
+$$
+$$
+=\Theta \left( 2^{(\log_2{n}) \cdot \log_2{3}} \right)=\Theta \left( (2^{\log_2{n}})^{\log_2{3}} \right)=\Theta \left( n^{\log_2{3}} \right) =\Theta \left( n^{1.585} \right).
+$$
+Upravený algoritmus má u¾ tedy lep¹í èasovou slo¾itost, konkrétnì $\Theta(n^{1.585})$.
+V~praxi bychom samozøejmì èinitele ne¹tìpili a¾ na~jednociferná èísla,
+ale zastavili se u~nìjaké dostateènì malé délky (øeknìme 50~cifer) a~tam
+pøepnuli na~kvadratický algoritmus, který má men¹í re¾ii.
+
+(Mimochodem, asymptoticky tato slo¾itost není nejlep¹í známá, pro násobení èísel existují efektívnìj¹í algoritmy, které
+dosahují èasové slo¾itosti $\Theta(n \log{n})$, ale jednak mají vysoké multiplikativní konstanty a~druhak jsou v~nich u¾ potøeba trochu
+pokroèilej¹í techniky, jako je diskrétní Fourierova transformace, tak¾e si je necháme na~pøí¹tí semestr.)
+
+\h{Kuchaøková vìta {\it (Master Theorem)}}
+
+Metody øe¹ení rekurentních rovnic z pøedchozích dvou pøíkladù
+by jistì fungovaly i na~jiné algoritmy, ale proè poka¾dé zbyteènì upravovat tolik  výrazù? Radìji si doká¾eme obecnou vìtu, která pùjde pou¾ít na~vìt¹inu
+takovýchto rekurencí. Øíká se jí {\it Master Theorem} nebo také (vzhledem k~tomu,
+jak se pou¾ívá) {\it Kuchaøková vìta}.
+
+\s{Vìta:} \>{\sl (Master Theorem)} 
+
+Pøedpokládejme, ¾e $T(1)=\Theta(1)$ a~$T(n)=a\cdot T(\lceil {{n}\over{b}} \rceil)+\Theta(n^d)$, kde $a \geq 1$, $b>1$, $d \geq 0$ a~$a,b \in \bb N$. Potom $T(n)$ je:
+
+\smallskip
+
+\halign{#&#&#\cr
+\indent & $\Theta(n^d)$ & kdy¾ $a<b^d$,\cr
+& $\Theta(n^d\cdot \log{n})$ & kdy¾ $a=b^d$,\cr
+& $\Theta(n^{\log_b{a}})$ & kdy¾ $a>b^d$.\cr}
+
+\proof Pøedpokládejme, ¾e $n=b^k, k \in \bb{N}$, aby platilo $\lceil
+{{n}\over{b}} \rceil = {{n}\over{b}}$. Pou¾ijeme opìt \uv{dùkaz stromem}.
+Strom rekurzivních volání se v¾dy vìtví na~stejný poèet vìtví, konkrétnì~$a$,
+a~velikosti vstupù klesají $b$-krát. Podívejme se na~tabulku:
+$$\vbox{\halign{\hfil#\hfil \quad & \hfil#\hfil \quad & \hfil#\hfil \quad & \hfil#\hfil \cr
+poèet vrcholù na~hladinì & velikost vstupu & èas ve vrcholu & èas na~hladinì \cr
+\noalign{\medskip\hrule\medskip}
+$1$ & $n$ & $\Theta(n^d)$ & $\Theta(n^d)$\cr
+$a$ & $n/{b^1}$ & $\Theta((n/b^1)^d)$ & ${\Theta(a^1 \cdot ({n/{b^1}})^d)}$\cr
+$a^2$ & $n/{b^2}$ & $\Theta((n/b^2)^d)$ & ${\Theta(a^2 \cdot ({n/{b^2}})^d)}$\cr
+$a^3$ & $n/{b^3}$ & $\Theta((n/b^3)^d)$ & ${\Theta(a^3 \cdot ({n/{b^3}})^d)}$\cr
+\vdots & \vdots  & \vdots & \vdots\cr
+$a^k$ & $n/{b^k}$  & $\Theta((n/b^k)^d)$ & ${\Theta(a^k \cdot ({n/{b^k}})^d)}$\cr}}$$
+
+\noindent
+Celkový èas potøebný na~vyøe¹ení v¹ech dílèích podúloh je následovný:
+$$
+T(n)=\sum_{i=0}^k\Theta \left( a^i \left( {n\over{b^i}} \right) ^d \right)=\sum_{i=0}^k\Theta \left( n^d \left( {a\over{b^d}} \right) ^i \right)=\Theta \left( n^d \cdot \sum_{i=0}^k \left( {a\over{b^d}} \right) ^i \right).
+$$
+V¹imnìme si sumy $\sum_{i=0}^k \left( {a\over{b^d}} \right) ^i$. Jedná se vlastnì o~geometrickou øadu s kvocientem $q={a\over{b^d}}$. Rozli¹me následující pøípady:
+
+\>{\I 1.} $q<1$: Práce na~jednotlivých hladinách exponenciálnì ubývá a~souèet øady (i~kdyby byla nekoneèná) se dá omezit nìjakou konstantou, tedy $T(n)=\Theta(n^d)$.
+
+\>{\I 2.} $q=1$: Práce na~jednotlivých hladinách je stejnì, to znamená, ¾e suma je právì $1 + \log_b n$, a~tedy $T(n) = \Theta(n^d \cdot \log{n})$.
+
+\>{\I 3.} $q>1$: Práce na~jednotlivých hladinách pøibývá, tak¾e musíme geometrickou øadu seèíst poctivì. Víme, ¾e souèet geometrické øady od~$0$ do~$k$ s kvocientem $q$ a~prvním èlenem $1$ je: ${q^{k+1} - 1}\over{q - 1}$, co¾ pøibli¾ne odpovídá $q^k$. Pak tedy platí: $$T(n) = \Theta(n^d \cdot q^{\log_b{n}}).$$ Tento výraz vypadá ponìkud o¹klivì, ale je¹tì ho trochu (alespoò kosmeticky) upravíme:
+$$
+\Theta\left(n^d \cdot q^{\log_b{n}}\right)=\Theta\left({ a^{\log_b{n}} \cdot n^d \over (b^d)^{\log_b{n}}}\right)=\Theta\left({\left(b^{\log_b{a}}\right)^{\log_b{n}} \cdot n^d \over{\left(b^d\right)^{\log_b{n}}}}\right)=
+$$
+$$
+=\Theta\left({\left(b^{\log_b{n}}\right)^{\log_b{a}} \cdot n^d \over{\left(b^{\log_b{n}}\right)^d}}\right)
+=\Theta\left({n^{\log_b{a}} \cdot n^d \over{n^d}}\right)
+=\Theta\left(n^{\log_b{a}}\right).
+$$
+Tyto tøi pøípady pøesnì odpovídají rozdìlení pøípadù v~tvrzení vìty.
+
+\noindent
+Vra»me se nyní k~mo¾nosti, kdy $n$ není mocnina~$b$. Pro obecné $n \geq 1$ zvolme: 
+
+\itemize\ibull
+\:$n^+ \dots$ nejbli¾¹í vy¹¹í mocnina $b$
+\:$n^-$ \dots nejbli¾¹í ni¾¹í mocnina $b$
+\endlist
+
+Platí tedy, ¾e ${n \over b} \leq n^- \leq n \leq n^+ \leq n \cdot b$. Zajisté i platí $T(n^-) \le T(n) \le T(n^+)$. Ale $T(n^-) = \Theta((n^-)^c) = \Theta(n^c)$ a~$T(n^+) = \Theta((n^+)^c) = \Theta(n^c)$. Pak tedy i $T(n) = \Theta(n^c)$. Vìta tedy platí i v~pøípadì, ¾e $n$ není mocnina~$b$.
+\qed
+
+\s{Pøíklad:}
+Porovnejme si nìkteré známé algoritmy a~jejich èasovou slo¾itost pomocí \>{\sl Master Theoremu}:
+$$\vbox{\halign{# \quad  \quad & # \quad  \quad & # \quad  \quad & # \quad  \quad & #\cr
+algoritmus & $a$ & $b$ & $d$ & èasová slo¾itost\cr
+\noalign{\smallskip\hrule\medskip}
+Mergesort & 2 & 2 & 1 & $\Theta({n \cdot \log{n}})$\cr
+Násobení I. & 4 & 2 & 1 & $\Theta(n^2)$\cr
+Násobení II. & 3 & 2 & 1 & $\Theta(n^{\log_2{3}})$\cr
+Binární vyhledávání & 1 & 2 & 0 & $\Theta(\log{n})$\cr}}$$
+
+
+\h{Hledání $k$-tého nejmen¹ího prvku (mediánu)}
+
+V tomto oddílu se budeme zabývat tím, jak co nejrychleji najít V~jakékoli posloupnosti $n$ èísel $k$-tý nejmen¹í prvek, popøípadì medián. Pro ty, kdo medián neznají, tu máme definici:
+
+\s{Definice:}
+{\I Medián} posloupnosti $a_1, a_2,\ldots , a_n$ je takové $a_i$, ¾e nejvý¹e $n/2$ prvkù je men¹ích ne¾ $a_i$ a~nejvý¹e $n/2$ prvkù je vìt¹ích ne¾ $a_i$. Platí tedy, ¾e medián je buï $\lfloor n/2\rfloor$-tý, nebo $\lceil n/2\rceil$-tý nejmen¹í prvek posloupnosti.
+
+Nejjednodu¹¹ím øe¹ením by urèitì bylo celou posloupnost nejdøíve setøídit a~pak u¾ jednodu¹e vybrat po¾adovaný prvek. To bychom dokázali V~celkem slu¹ném èase $\Theta(n\log n)$, ale u¾ teï mù¾eme prozradit, ¾e to jde V~èase $\Theta(n)$. Jak?
+
+Pou¾ijme metodu {\it rozdìl a~panuj}. Nìjakým zpùsobem si zvolíme jeden prvek posloupnosti a~nazveme ho {\it pivot}. Poté rozdìlíme zadanou posloupnost na~tøi disjunktní mno¾iny. do~první dáme v¹echny prvky men¹í ne¾ pivot, do~druhé rovné pivotu jako pivot a~do tøetí vìt¹í ne¾ pivot. Tímto máme zaji¹tìno, ¾e prvky z první mno¾iny jsou urèitì men¹í ne¾ prvky z druhé a~ty ne¾ prvky z tøetí.
+O tom, jak jsou prvky uspoøádány uvnitø tìchto mno¾in, ale nic nevíme.
+
+V posledním kroku na¹eho algoritmu se pak rozhodneme, na~kterou mno¾inu svùj algoritmus rekurzivnì zavoláme. Pokud je $k$ men¹í nebo rovno ne¾ velikost první mno¾iny, pokraèujeme V~první mno¾inì, pokud je $k$ men¹í nebo rovno ne¾ souèet velikostí první a~druhé mno¾iny, pak hledaným prvkem je právì vybraný pivot a~algoritmus skonèí, a~nakonec pokud ani jedna podmínka splnìna nebyla, pustíme se do~hledání ve tøetí mno¾inì, ov¹em u¾ nehledáme $k$-tý nejmen¹í prvek, ale $l$-tý, kde $l$ se rovná $k$ minus velikost prvních dvou mno¾in. Pro vìt¹í názornost zapí¹eme tento algoritmus formálnìji:
+
+\algo
+{\bo Select($k,X$):} (Hledání $k$-tého nejmen¹ího prvku V~mno¾ine $X$)
+\:Jestli¾e $\vert X\vert \le 1$, vyøe¹íme triviálnì.
+\:Zvolíme pivota $p \in X$.
+\:Rozdìlíme mno¾inu X na~tøi podmno¾iny: $L = \{x \in X; x < p\},$ $ S = \{x \in X; x = p\}, P = \{x \in X; x > p\}$.
+\:Jestli¾e $k \le \vert L\vert$, vrátíme výsledek funkce \<Select>($k$, $L$).
+\:Jestli¾e $ \vert L\vert < k \le \vert L\vert + \vert S\vert$, vrátíme pivota $p$.
+\:Jestli¾e $ \vert L\vert + \vert S\vert < k$, vrátíme výsledek funkce \<Select>($k - \vert L\vert - \vert S\vert, P$).
+\endalgo
+Na první pohled je vidìt, ¾e se algoritmus zastaví (vstup se v¾dy zmen¹í alespoò o~1) a~¾e vydá v¾dy správný výsledek. Jak je to ov¹em s èasovou slo¾itostí? Rozdìlení do~mno¾in a~podmínky V~druhém a~tøetím kroku mají lineární slo¾itost, èemu¾ se nevyhneme. Pøi ne¹»astné volbì pivota se nám mù¾e stát, ¾e poèet rekurentních volání mù¾e být a¾ $n$, tedy celková slo¾itost V~nejhor¹ím pøípadì je $\Theta(n^2)$, èím¾ jsme si oproti prostému setøídìní je¹tì pohor¹ili. Co s tím? Jak je vidìt, velmi dùle¾itá je volba pivota. Tu mù¾eme provést nìkolika zpùsoby:
+
+a) Pivot by se V~setøídìné posloupnosti vyskytoval uprostøed, vstup by se tedy stále pùlil. Èasovou slo¾itost vypoèteme z rekurentního zápisu:
+$$ T(n) = T\left({n \over 2}\right) + \Theta(n) = \Theta\left(n + {n \over 2} + {n \over 4} + \dots\right) = \Theta(n). $$
+To by bylo sice skvìlé, ale nalezení takového pivota je vlastnì vyøe¹ení úlohy hledání mediánu, o~co¾ se sna¾íme. Tedy jsme si vùbec nepomohli.
+
+b) Pivot by se V~setøídìné posloupnosti náchazel V~prostøedních dvou ètvrtinách (nazvìme tento prvek \uv{\I l¾imedián}). Tím bychom V~ka¾dém kroku urèitì odstranili mno¾inu velikosti alespoò ètvrtiny vstupu. Èasová slo¾itost tohoto øe¹ení by byla:
+
+$$ T(n) = T\left({3 \over 4}n\right) + \Theta(n) = \Theta\left(n + {3 \over 4}n + {9 \over 16}n + \dots\right) = \Theta(n). $$
+
+Tímto bychom tedy také dosáhli lineární èasové slo¾itosti. Ale jak vybrat pivota tak, aby se nacházel V~prostøedních dvou ètvrtinách a~aby nám nám to nepokazilo lineární slo¾itost?
+
+Zkusme vybrat pivota náhodnì. Pravdìpodobnost, ¾e vytáhneme zrovna l¾imedián, je alespoò $1/2$. (Pokud by se prvky nemohly opakovat a~byl jich sudý poèet, byla by to pøesnì $1/2$.) Tuto pravdìpodobnost si oznaème $p$. 
+
+\s{Volba l¾imediánu}
+\algo
+\:Vybereme rovnomìrnì náhodnì pivota z mno¾iny $X$.
+\:Otestujeme, zda je pivot l¾imedián.
+\:Pokud není $\Rightarrow$ pokraèujeme znovu od~zaèátku, jinak konec.
+\endalgo
+
+Pivota vybíráme rovnomìrnì náhodnì, tedy ka¾dý prvek posloupnosti má stejnou pravdìpodobnost, ¾e bude vybrán. 
+
+Oznaème si $T$ náhodnou velièinu znaèící dobu bìhu algoritmu. Potom støední hodnota této náhodné velièiny ${\bb E}[T] = \Theta(n) \cdot {\bb E}$[{\I poèet~prùchodù~cyklem}].
+
+Teï si doka¾me, ¾e budeme-li náhodnì vybírat pivota tak dlouho, a¾ se strefíme do~l¾imediánu, tak \uv{v prùmìru} budeme muset tahat jen $1/p$-krát.
+
+\s{Lemma: } {\I (O d¾bánu a~vodì)}
+Èekání na~náhodnou událost, která nastává s pravdìpodobností $p$, trvá V~prùmìru $1/p$.
+
+\proof
+Oznaème $N$ poèet pokusù. Potom støední hodnota poètu pokusù je:
+$$\eqalign{
+		{\bb E}[N] & = p \cdot 1 +  (1-p) \cdot (1 + {\bb E}[N])\cr
+		{\bb E}[N] & = p + 1 +  {\bb E}[N] - p -  p \cdot {\bb E}[N]\cr
+    {\bb E}[N] \cdot (1 - 1 + p) & = 1\cr
+    {\bb E}[N] & = 1/p\cr
+}$$
+S pravdìpodobností $p$ událost nastane a~s odvrácenou pravdìpodobností ($1-p$) jsme jeden pokus promarnili a~musíme celý proces opakovat. Jednoduchými úpravami nám vyjde, ¾e støední hodnota poètu pokusù je $1/p$.
+
+\noindent
+\uv{V prùmìru se tedy chodí $1/p$-krát se d¾bánem pro vodu, ne¾ se ucho utrhne...}
+\qed
+
+Z lemmatu tedy plyne, ¾e V~na¹em pøípadì ${\bb E}$[{\I poèet prùchodù cyklem}]$ \le 2 $. V~prùmìru tedy na~druhý pokus vytáhneme l¾imedián. Ten pak pou¾ijeme jako pivot. Tímto tudí¾ dosáhneme prùmìrné èasové slo¾itosti $\Theta(n)$.
+
+\s{Vìta: } Pravdìpodobnostním algoritmem lze najít $k$-tý nejmen¹í prvek z $n$ prvkù V~prùmìrném èase $\Theta(n)$.
+
+\s{Poznámka: } Kdy¾ budeme volit pivota náhodnì a~nebudeme se starat o~to, zda se jedná o~l¾imedián, èi ne, tak dostaneme také prùmìrný èas $\Theta(n)$.
+
+K volbì l¾imediánu jsme volili v¾dy náhodného pivota, jednalo se tedy o~randomizovaný algoritmus. Stejný výsledek nám ale vyjde i pøi deterministickém algoritmu, kdy budeme volit pivota v¾dy stejnì (napø. na~první pozici ve vstupní posloupnosti), ale budeme mít nìjak zaruèeno, ¾e vstupy budou dostateènì náhodné. U randomizovaného algoritmu se jednalo o~prùmìr pøes náhodná èísla, u deterministického algoritmu o~prùmìr pøes náhodné vstupy.
+
+U¾ tedy víme, ¾e kdy¾ budeme volit pivota hodnì ¹patnì, tak se mù¾eme dostat a¾ na~èasovou slo¾itost $\Theta(n^2)$. Kdy¾ budeme volit o~nìco lépe, tak se mù¾eme dosáhnout prùmìrné èasové slo¾itosti $\Theta(n)$. Existuje ale i algoritmus, který pracuje v¾dy (nejen V~prùmìrném pøípadì) V~èase $\Theta(n)$. Podívejme se, jak bude vypadat \dots
+
+\s{Volba pivota:}
+\algo
+\:Rozdìlíme vstup na~pìtice \dots $\Theta(n)$
+\:Spoèteme medián ka¾dé pìtice \dots $\Theta(n)$
+\:Spoèteme medián mediánù pìtic tak, ¾e rekurzivnì zavoláme tentý¾ algoritmus \<Select>($n / 10$, {mediány pìtic}). Tím získáme pivota.
+\endalgo
+
+Pro úplnost se podívejme, jak bude vypadat celý algoritmus na~hledání $k$-tého nejmen¹ího prvku posloupnosti za pou¾ití na¹í nové volby pivota:
+
+\s{Deterministický lineární algoritmus na~výbìr mediánu lineárnì}
+\algo
+\algin $X = x_1 \dots x_n$, $k$ ($1 \le k \le n$)
+\:Pokud $n \le 5 \Rightarrow$ vyøe¹íme hrubou silou.
+\:Vstup rozdìlíme na~pìtice $P_1 \dots P_{\lceil n/5 \rceil}$.
+\:$\forall i: m_i \leftarrow$ medián($P_i$).
+\:$p \leftarrow$ \<Select>($m_1 \dots m{\lceil n/5 \rceil}, \lceil n/10 \rceil$).
+\:Rozdìlíme $X$ na~$L,S,P$
+\:Rekurzivnì se zavoláme na~jednu z $L,S,P$ (tu, ve které se má vyskytovat hledaný prvek)
+\endalgo
+
+Ná¹ deterministický algoritmus tedy roz¹iøuje pùvodní algoritmus tak, ¾e v¾dy, kdy¾ je potøeba zvolit pivota z posloupnosti X, tak si posloupnost rozdìlí na~pìtice a~v ka¾dé spoète medián. Následnì potøebuje medián z tìchto mediánù, co¾ bude výsledný pivot -- medián V~pùvodní posloupnosti. Ten získá tak, ¾e se znovu zavolá na~\<Select> s parametry posloupnost mediánù a~èíslo $\lceil n/10 \rceil$, nebo» potøebuje právì prostøední prvek této poslounosti, která má délku $\lceil n/5 \rceil$. Kdy¾ máme koneènì výhodného pivota nalezeného, tak mù¾eme pokraèovat, ¾e si posloupnost opìt rozdìlíme na~tøi hromádky -- prvky men¹í ($L$), stejné ($S$) a~vìt¹í ($P$) ne¾ pivot. Následnì jen vybereme hromádku, která odpovídá umístìní $k$-tého nejmen¹ího prvku, a~na tu se rekurzivnì zavoláme.
+
+Abychom dokázali, ¾e tento algoritmus bude mít opravdu lineární èasovou slo¾itost, musíme si nejdøíve dokázat následující lemma:
+
+\s{Lemma:} V~ka¾dém kroku vypadne alespoò ${3/10}\cdot n$ prvkù.
+
+\proof 
+V dùkazu budeme trochu ètenáøe ¹idit tím, ¾e budeme pøedpokládat celoèíselné výsledky (napø. ¾e poèet prvkù posloupnosti, na~kterou se zavoláme, je v¾dy dìlitelný pìti). Ètenáø si mù¾e jako cvièení dokázat, ¾e kdyby výsledky nevycházely celoèíselnì, tak to nevadí.
+
+Pøedstavme si vybrané pìtice seøazené podle velikosti od~nejvìt¹ího prvku a~zakresleme je do~sloupcù. Jejich mediány tedy vyplòují prostøední øadu. Tyto pìtice pak seøaïme podle velikosti jejich mediánù (nejmen¹í vlevo)\foot{Algoritmus ov¹em nikde takto pìtice neøadí! Jen nám to pomù¾e v~dùkazu správnosti.}. Hledaný pivot se tedy nachází (pokud pøedpokládáme pro jednoduchost lichý poèet pìtic) pøesnì uprostøed. o~prvcích nad pivotem a~napravo od~nìj mù¾eme urèitì øíct, ¾e jsou vìt¹í nebo rovny pivotu, prvky pod ním a~nalevo od~nìj jsou zase urèitì men¹í nebo rovny pivotu.
+
+Podle konstrukce algoritmu tedy zaruèenì vypadne jedna nebo druhá skupina prvkù. Obì tyto skupiny pøitom obsahují, jak je vidìt z obrázku, alespoò $3/10 \cdot n$ prvkù.
+\qed
+
+\figure{petice.eps}{Pìtice}{125mm}
+
+Nyní u¾ se tedy mù¾eme pustit do~výpoètu èasové slo¾itosti. V~ka¾dém kroku funkce zavolá sama sebe nejdøíve na~vstup velikosti ${n \over 5}$ a~poté na~vstup velikosti nejvý¹e ${7 \over 10}n$. Ostatní operace zùstávají lineární. Èasovou slo¾itost V~nejhor¹ím pøípadì tedy mù¾eme zapsat rekurentním vzorcem:
+
+$$ T(n) = \Theta(n) + T\left({n \over 5}\right) + T\left({7 \over 10}n\right). $$
+
+Tento rekurentní vzorec ale zatím neumíme obecnì øe¹it. Mohli bychom ho postupnì rozepsat, ale trvalo by dlouho, ne¾ bychom z toho nìco vykoukali. Lep¹í bude rovnou dokázat, ¾e tento rekurentní vzorec odpovídá lineární slo¾itosti, zkusíme tedy dosadit $T(n) = cn$:
+
+$$ cn = n + {1 \over 5}cn + {7 \over 10}cn \quad\Longleftrightarrow\quad c = 10. $$
+
+Tímto jsme tedy dokázali, ¾e èasová slo¾itost tohoto algoritmu V~nejhor¹ím pøípadì je $\Theta(n)$.
+Jeliko¾ to rychleji urèitì není mo¾né, mù¾eme na~závìr této kapitoly zformulovat tuto vìtu:
+
+\s{Vìta:} Hledání $k$-tého nejmen¹ího prvku V~posloupnosti délky $n$ má èasovou slo¾itost V~nejhor¹ím pøípadì $\Theta(n)$. \qed
+
+\s{K zamy¹lení:} Proè jsme zvolili zrovna pìtice? Jak by to dopadlo pro trojice? a~jak pro sedmice? Fungoval by takový algoritmus? Byl by také lineární?
+
+\h{Quicksort}
+
+Ji¾ jsme se seznámili s tøídícím algoritmem {\it Mergesort}. Podívejme se teï na~algoritmus, který je také zalo¾en na~metodì {\it rozdìl a~panuj}, ale chová se V~podstatì opaènì. Namísto slévání men¹ích setøídìných posloupností posloupnost v¾dy rozdìlí na~dvì pokud mo¾no podobnì velké èásti $L$ a~$P$, kdy V~$L$ jsou èísla men¹í ne¾ nìjaký pivot a~v $P$ èísla vìt¹í. Pivot tvoøí samostatnou èást $S$. Obì èásti $L$ i $P$ rekurzivnì setøídí a~následnì vrátí posloupnost, kde budou setøídìné prvky z $L$, pak pivot z $S$ a~poté setøídìné prvky z $P$. Vyu¾ívá tedy podobnou my¹lenku jako hledání $k$-tého nejmen¹ího prvku.
+
+\s{Algoritmus:}
+{\bo \<QS>$(x = x_1, \dots,x_n)$:}
+\algo
+\:Pokud $n \leq 1 \Rightarrow$ vrátíme $x$.
+\:Vybereme pivota.
+\:Rozdìlíme posloupnost na~podposloupnosti $L,S,P$.
+\:$L \leftarrow$ \<QS($L$)>, $P \leftarrow$ \<QS($P$)>
+\:Vrátíme $L + S + P$.
+\endalgo
+
+\s{Pozorování:}
+
+Kdy¾ bude pivot v¾dy medián: $T(n) = \Theta(n) + 2T(n/2) \Rightarrow T(n) = \Theta(n \cdot\log{n})$
+
+Kdy¾ bude pivot v¾dy minimum: $T(n) = \Theta(n) + T(n-1) \Rightarrow T(n) = \Theta(n^2)$
+
+\s{Poznámka:} Hledání mediánu lineárnì je sice hezké a~asymptoticky lineární, ale má tak o¹klivé multiplikativní konstanty, ¾e se V~praxi nepou¾ívá. Doka¾me si tedy, ¾e Quicksort s náhodnou volbou pivota má prùmìrnou èasovou slo¾itost také $\Theta(n\cdot\log{n})$.
+
+\s{Vìta:} Quicksort s náhodnou volbou pivota provede ve støední hodnotì $\Theta(n\cdot\log{n})$ porovnání.
+
+\s{Dùsledek:} Quicksort má prùmìrnou èasovou slo¾itost  $\Theta(n\cdot\log{n})$.
+
+\proof
+Nech» $C$ znaèí poèet porovnání, $y_1, \dots, y_n$ je vstup V~setøídìném poøadí a~$C_{ij}$ znaèí poèet porovnání prvkù $y_i$ s $y_j$ pro $1 \leq i < j \leq n$. Uvìdomme si, ¾e dva prvky porovnává algoritmus nejvý¹e jednou, a~to pouze tehdy, kdy¾ je jeden z nich pivot. Proto $C_{ij}$ nabývá hodnot $0$ nebo $1$. $C_{ij}$ je tedy indikátor jevu, ¾e byly prvky $y_i$ a~$y_j$ porovnány. 
+
+Poèet v¹ech porovnání je pak souèet jednotlivých $C_{ij}$, èili $$C = \sum_{i,j} C_{ij}.$$ 
+Proto platí: $${\bb E}[C] = {\bb E}\left[\sum_{i,j} C_{ij}\right].$$ 
+Díky linearitì støední hodnoty mù¾eme tvrdit, ¾e: $${\bb E}[C] = \sum_{i,j} {\bb E}[C_{ij}].$$
+Støední hodnota indikátoru jevu, ¾e prvky  $y_i$ a~$y_j$ byly porovnány, je rovna pravdìpodobnosti, ¾e $C_{ij}$ je 1. $${\bb E}[C_{ij}] = P[C_{ij} = 1].$$
+Aby $C_{ij}$ se rovnalo $1$ pro prvky $y_i$ a~$y_j$, tak se musel pro podposloupnost $y_i, \dots, y_j$ stát pivotem jako první jeden z $y_i$ nebo $y_j$. Kdyby se tak nestalo, tak spolu pøi tomto rozdìlování prvky $y_i$ a~$y_j$ nebudou porovnány a~dostanou se ka¾dý do~jiné skupiny ($L$ a~$P$), tak¾e u¾ spolu porovnány nikdy ani nebudou.
+Pravdìpodobnost, ¾e $y_i$ nebo $y_j$ se staly pivotem pro posloupnost $y_i, \dots, y_j$  je rovna $2 \over {j-i+1}$ (nebo» pivota vybíráme rovnomìrnì náhodnì). Proto mù¾eme koneènì vyjádøit:
+
+$${\bb E}[C] = \sum_{1 \le i < j \le n} {2\over{j-i+1}} = \sum_{d=1}^{n-1} { \sum_{n=1}^{n-d} {2\over{d + 1}} }\le \sum_{d=1}^{n-1} {2n\over{d + 1}} = 2n\sum_{d=2}^n {1\over d} = \Theta(n \cdot\log{n}).$$
+
+\qed
+
+
+\bye
diff --git a/2009/8-rozdel/Makefile b/2009/8-rozdel/Makefile
new file mode 100644
index 0000000..4d9a469
--- /dev/null
+++ b/2009/8-rozdel/Makefile
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+P=8-rozdel
+
+include ../Makerules
diff --git a/2009/8-rozdel/figure.eps b/2009/8-rozdel/figure.eps
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+currentoverprint ne{setoverprint}{pop}ifelse}{1 eq setoverprint}ifelse dup _ eq
+{pop setprocesscolor_5 pop}{findcmykcustomcolor exch setcustomcolor}ifelse
+SepMode_5 0 eq{true}{GetGry 1 eq currentoverprint and not}ifelse}bd/colorimage
+where{pop/ColorImage{colorimage}def}{/ColorImage{/ncolors xd/$multi xd $multi
+true eq{ncolors 3 eq{/daqB xd/daqG xd/daqR xd pop pop exch pop abs{daqR pop
+daqG pop daqB pop}repeat}{/daqK xd/daqY xd/daqM xd/daqC xd pop pop exch pop abs
+{daqC pop daqM pop daqY pop daqK pop}repeat}ifelse}{/dataaq xd{dataaq ncolors
+dup 3 eq{/$dat xd 0 1 $dat length 3 div 1 sub{dup 3 mul $dat 1 index get 255
+div $dat 2 index 1 add get 255 div $dat 3 index 2 add get 255 div rgb2g 255 mul
+cvi exch pop $dat 3 1 roll put}for $dat 0 $dat length 3 idiv getinterval pop}{
+4 eq{/$dat xd 0 1 $dat length 4 div 1 sub{dup 4 mul $dat 1 index get 255 div
+$dat 2 index 1 add get 255 div $dat 3 index 2 add get 255 div $dat 4 index 3
+add get 255 div cmyk2rgb rgb2g 255 mul cvi exch pop $dat 3 1 roll put}for $dat
+0 $dat length ncolors idiv getinterval}if}ifelse}image}ifelse}bd}ifelse
+/setcmykcolor{1 5 1 roll _ currentoverprint @scc_5/$ffpnt xd}bd
+/currentcmykcolor{GetCmyk}bd/setrgbcolor{rgb2cmyk setcmykcolor}bd
+/currentrgbcolor{currentcmykcolor cmyk2rgb}bd/sethsbcolor{hsb2rgb setrgbcolor}
+bd/currenthsbcolor{currentrgbcolor rgb2hsb}bd/setgray{dup dup setrgbcolor}bd
+/currentgray{currentrgbcolor rgb2g}bd/InsideDCS false def/IMAGE/image ld/image
+{InsideDCS{IMAGE}{/EPSDict where{pop SepMode_5 0 eq{IMAGE}{dup type/dicttype eq
+{dup/ImageType get 1 ne{IMAGE}{dup dup/BitsPerComponent get 8 eq exch
+/BitsPerComponent get 1 eq or currentcolorspace 0 get/DeviceGray eq and{
+CurrentInkName_5(Black)eq{IMAGE}{dup/DataSource get/TCC xd/Height get abs{TCC
+pop}repeat}ifelse}{IMAGE}ifelse}ifelse}{2 index 1 ne{CurrentInkName_5(Black)eq
+{IMAGE}{/TCC xd pop pop exch pop abs{TCC pop}repeat}ifelse}{IMAGE}ifelse}
+ifelse}ifelse}{IMAGE}ifelse}ifelse}bd}ifelse/WaldoColor_5 true def/$fm 0 def
+/wfill{1 $fm eq{fill}{eofill}ifelse}bd/@Pf{@sv SepMode_5 0 eq $Psc 0 ne or
+$ink_5 3 eq or{0 J 0 j[]0 d $t $c $m $y $k $n $o @scc_5 pop $ctm setmatrix 72
+1000 div dup matrix scale dup concat dup Bburx exch Bbury exch itransform
+ceiling cvi/Bbury xd ceiling cvi/Bburx xd Bbllx exch Bblly exch itransform
+floor cvi/Bblly xd floor cvi/Bbllx xd $Prm aload pop $Psn load exec}{1 SetGry
+wfill}ifelse @rs @np}bd/F{matrix currentmatrix $sdf{$scf $sca $scp @ss}if $fil
+1 eq{CorelPtrnDoFill}{$fil 2 eq{@ff}{$fil 3 eq{@Pf}{$fil 4 eq
+{CorelShfillDoFill}{$t $c $m $y $k $n $o @scc_5{wfill}{@np}ifelse}ifelse}
+ifelse}ifelse}ifelse $sdf{$dsf $dsa $dsp @ss}if setmatrix}bd/f{@cp F}bd/S{
+matrix currentmatrix $ctm setmatrix $SDF{$SCF $SCA $SCP @ss}if $T $C $M $Y $K
+$N $O @scc_5{matrix currentmatrix $ptm concat stroke setmatrix}{@np}ifelse $SDF
+{$dsf $dsa $dsp @ss}if setmatrix}bd/s{@cp S}bd/B{@gs F @gr S}bd/b{@cp B}bd/_E{
+5 array astore exch cvlit xd}bd/@cc{currentfile $dat readhexstring pop}bd/@sm{
+/$ctm $ctm currentmatrix def}bd/@E{/Bbury xd/Bburx xd/Bblly xd/Bbllx xd}bd/@c{
+@cp}bd/@P{/$fil 3 def/$Psn xd/$Psc xd array astore/$Prm xd}bd/tcc{@cc}def/@B{
+@gs S @gr F}bd/@b{@cp @B}bd/@sep{CurrentInkName_5(Composite)eq{/$ink_5 -1 def}
+{CurrentInkName_5(Cyan)eq{/$ink_5 0 def}{CurrentInkName_5(Magenta)eq{/$ink_5 1
+def}{CurrentInkName_5(Yellow)eq{/$ink_5 2 def}{CurrentInkName_5(Black)eq
+{/$ink_5 3 def}{/$ink_5 4 def}ifelse}ifelse}ifelse}ifelse}ifelse}bd/@whi{@gs
+-72000 dup m -72000 72000 l 72000 dup l 72000 -72000 l @cp 1 SetGry fill @gr}
+bd/@neg{[{1 exch sub}/exec cvx currenttransfer/exec cvx]cvx settransfer @whi}
+bd/deflevel 0 def/@sax{/deflevel deflevel 1 add def}bd/@eax{/deflevel deflevel
+dup 0 gt{1 sub}if def deflevel 0 gt{/eax load}{eax}ifelse}bd/eax{{exec}forall}
+bd/@rax{deflevel 0 eq{@rs @sv}if}bd systemdict/pdfmark known not{/pdfmark
+/cleartomark ld}if/wclip{1 $fm eq{clip}{eoclip}ifelse}bd
+% Copyright (c)1992-2003 Corel Corporation
+% All rights reserved.     v12 r0.0
+/z{exch findfont exch scalefont setfont}bd/ZB{9 dict dup begin 4 1 roll
+/FontType 3 def/FontMatrix xd/FontBBox xd/Encoding 256 array def 0 1 255{
+Encoding exch/.notdef put}for/CharStrings 256 dict def CharStrings/.notdef{}
+put/Metrics 256 dict def Metrics/.notdef 3 -1 roll put/BuildChar{exch dup
+/$char exch/Encoding get 3 index get def dup/Metrics get $char get aload pop
+setcachedevice begin Encoding exch get CharStrings exch get end exec}def end
+definefont pop}bd/ZBAddChar{findfont begin dup 4 1 roll dup 6 1 roll Encoding 3
+1 roll put CharStrings 3 1 roll put Metrics 3 1 roll put end}bd/Z{findfont dup
+maxlength 2 add dict exch dup{1 index/FID ne{3 index 3 1 roll put}{pop pop}
+ifelse}forall pop dup dup/Encoding get 256 array copy dup/$fe xd/Encoding exch
+put dup/Fontname 3 index put 3 -1 roll dup length 0 ne{0 exch{dup type 0 type
+eq{exch pop}{$fe exch 2 index exch put 1 add}ifelse}forall pop}if dup 256 dict
+dup/$met xd/Metrics exch put dup/FontMatrix get 0 get 1000 mul 1 exch div 3
+index length 256 eq{0 1 255{dup $fe exch get dup/.notdef eq{pop pop}{5 index 3
+-1 roll get 2 index mul $met 3 1 roll put}ifelse}for}if pop definefont pop pop
+}bd/CorelIsValidCharpath{pathbbox 3 -1 roll sub abs 0.5 ge 3 1 roll sub abs 0.5
+ge and}bd/@ftx{{currentpoint 3 -1 roll(0)dup 3 -1 roll 0 exch put dup @gs true
+charpath $ctm setmatrix CorelIsValidCharpath{@@txt}if @gr @np stringwidth pop 3
+-1 roll add exch m}forall}bd/@ft{matrix currentmatrix exch $sdf{$scf $sca $scp
+@ss}if $fil 1 eq{/@@txt/@pf ld @ftx}{$fil 2 eq{/@@txt/@ff ld @ftx}{$fil 3 eq
+{/@@txt/@Pf ld @ftx}{$fil 4 eq{/@@txt/CorelShfillDoFill ld @ftx}{$t $c $m $y $k
+$n $o @scc_5{show}{pop}ifelse}ifelse}ifelse}ifelse}ifelse $sdf{$dsf $dsa $dsp
+@ss}if setmatrix}bd/@st{matrix currentmatrix exch $SDF{$SCF $SCA $SCP @ss}if $T
+$C $M $Y $K $N $O @scc_5{{currentpoint 3 -1 roll(0)dup 3 -1 roll 0 exch put dup
+@gs true charpath $ctm setmatrix $ptm concat stroke @gr @np stringwidth pop 3
+-1 roll add exch m}forall}{pop}ifelse $SDF{$dsf $dsa $dsp @ss}if setmatrix}bd
+/@te{@ft}bd/@tr{@st}bd/@ta{dup @gs @ft @gr @st}bd/@t@a{dup @gs @st @gr @ft}bd
+/@tm{@sm concat}bd/e{/t{@te}def}bd/r{/t{@tr}def}bd/o{/t{pop}def}bd/a{/t{@ta}
+def}bd/@a{/t{@t@a}def}bd/t{@te}def/T{@np $ctm setmatrix/$ttm matrix def}bd/ddt
+{t}def/@t{/$stm $stm currentmatrix def 3 1 roll m $ttm concat ddt $stm
+setmatrix}bd/@n{/$ttm exch matrix rotate def}bd/@s{}bd/@l{}bd/_lineorientation
+0 def/_bitfont null def/_bitlobyte 0 def/_bitkey null def/_bithibyte 0 def
+% Copyright (c)1992-2003 Corel Corporation
+% All rights reserved.     v12 r0.0
+/@ii{concat 3 index 3 index m 3 index 1 index l 2 copy l 1 index 3 index l 3
+index 3 index l clip pop pop pop pop}bd/@i{@sm @gs @ii 6 index 1 ne{/$frg true
+def pop pop}{1 eq{s1t s1c s1m s1y s1k s1n $O @scc_5/$frg xd}{/$frg false def}
+ifelse 1 eq{@gs $ctm setmatrix F @gr}if}ifelse @np/$ury xd/$urx xd/$lly xd
+/$llx xd/$bts xd/$hei xd/$wid xd/$dat $wid $bts mul 8 div ceiling cvi string
+def $bkg $frg or{$SDF{$SCF $SCA $SCP @ss}if $llx $lly Tl $urx $llx sub $ury
+$lly sub scale $bkg{$t $c $m $y $k $n $o @scc_5 pop}if $wid $hei abs $bts 1 eq
+{$bkg}{$bts}ifelse[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse]/tcc load $bts
+1 eq{imagemask}{image}ifelse $SDF{$dsf $dsa $dsp @ss}if}{$hei abs{tcc pop}
+repeat}ifelse @gr $ctm setmatrix}bd/@I{@sm @gs @ii @np/$ury xd/$urx xd/$lly xd
+/$llx xd/$ncl xd/$bts xd/$hei xd/$wid xd $ngx $llx $lly Tl $urx $llx sub $ury
+$lly sub scale $wid $hei abs $bts[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse
+]$msimage false eq $ncl 1 eq or{/$dat $wid $bts mul $ncl mul 8 div ceiling cvi
+string def/@cc load false $ncl ColorImage}{$wid $bts mul 8 div ceiling cvi $ncl
+3 eq{dup dup/$dat1 exch string def/$dat2 exch string def/$dat3 exch string def
+/@cc1 load/@cc2 load/@cc3 load}{dup dup dup/$dat1 exch string def/$dat2 exch
+string def/$dat3 exch string def/$dat4 exch string def/@cc1 load/@cc2 load
+/@cc3 load/@cc4 load}ifelse true $ncl ColorImage}ifelse $SDF{$dsf $dsa $dsp
+@ss}if @gr $ctm setmatrix}bd/@cc1{currentfile $dat1 readhexstring pop}bd/@cc2{
+currentfile $dat2 readhexstring pop}bd/@cc3{currentfile $dat3 readhexstring pop
+}bd/@cc4{currentfile $dat4 readhexstring pop}bd/$msimage false def/COMP 0 def
+/MaskedImage false def L2?{/@I_2{@sm @gs @ii @np/$ury xd/$urx xd/$lly xd/$llx
+xd/$ncl xd/$bts xd/$hei xd/$wid xd/$dat $wid $bts mul $ncl mul 8 div ceiling
+cvi string def $ngx $ncl 1 eq{/DeviceGray}{$ncl 3 eq{/DeviceRGB}{/DeviceCMYK}
+ifelse}ifelse setcolorspace $llx $lly Tl $urx $llx sub $ury $lly sub scale 8
+dict begin/ImageType 1 def/Width $wid def/Height $hei abs def/BitsPerComponent
+$bts def/Decode $ncl 1 eq{[0 1]}{$ncl 3 eq{[0 1 0 1 0 1]}{[0 1 0 1 0 1 0 1]}
+ifelse}ifelse def/ImageMatrix[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse]def
+/DataSource currentfile/ASCII85Decode filter COMP 1 eq{/DCTDecode filter}{COMP
+2 eq{/RunLengthDecode filter}if}ifelse def currentdict end image $SDF{$dsf $dsa
+$dsp @ss}if @gr $ctm setmatrix}bd}{/@I_2{}bd}ifelse/@I_3{@sm @gs @ii @np/$ury
+xd/$urx xd/$lly xd/$llx xd/$ncl xd/$bts xd/$hei xd/$wid xd/$dat $wid $bts mul
+$ncl mul 8 div ceiling cvi string def $ngx $ncl 1 eq{/DeviceGray}{$ncl 3 eq
+{/DeviceRGB}{/DeviceCMYK}ifelse}ifelse setcolorspace $llx $lly Tl $urx $llx sub
+$ury $lly sub scale/ImageDataDict 8 dict def ImageDataDict begin/ImageType 1
+def/Width $wid def/Height $hei abs def/BitsPerComponent $bts def/Decode $ncl 1
+eq{[0 1]}{$ncl 3 eq{[0 1 0 1 0 1]}{[0 1 0 1 0 1 0 1]}ifelse}ifelse def
+/ImageMatrix[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse]def/DataSource
+currentfile/ASCII85Decode filter COMP 1 eq{/DCTDecode filter}{COMP 2 eq{
+/RunLengthDecode filter}if}ifelse def end/MaskedImageDict 7 dict def
+MaskedImageDict begin/ImageType 3 def/InterleaveType 3 def/MaskDict
+ImageMaskDict def/DataDict ImageDataDict def end MaskedImageDict image $SDF
+{$dsf $dsa $dsp @ss}if @gr $ctm setmatrix}bd/@SetMask{/$mbts xd/$mhei xd/$mwid
+xd/ImageMaskDict 8 dict def ImageMaskDict begin/ImageType 1 def/Width $mwid def
+/Height $mhei abs def/BitsPerComponent $mbts def/DataSource maskstream def
+/ImageMatrix[$mwid 0 0 $mhei neg 0 $mhei 0 gt{$mhei}{0}ifelse]def/Decode[1 0]
+def end}bd/@daq{dup type/arraytype eq{{}forall}if}bd/@BMP{/@cc xd UseLevel 3 eq
+MaskedImage true eq and{7 -2 roll pop pop @I_3}{12 index 1 gt UseLevel 2 eq
+UseLevel 3 eq or and{7 -2 roll pop pop @I_2}{11 index 1 eq{12 -1 roll pop @i}{
+7 -2 roll pop pop @I}ifelse}ifelse}ifelse}bd
+end
+%%EndResource
+%%EndProlog
+%%BeginSetup
+wCorel12Dict begin
+@BeginSysCorelDict
+2.6131 setmiterlimit
+1.00 setflat
+/$fst 128 def
+%%BeginResource: font TimesNewRomanPSMT-NormalItalic
+%!FontType1-1.0: TimesNewRomanPSMT-NormalItalic 001.003
%%Creator: Corel PostScript Engine
10 dict begin
/FontName /TimesNewRomanPSMT-NormalItalic def
/PaintType 0 def
/FontType 1 def
/FontMatrix [0.001 0 0 0.001 0 0] readonly def
/Encoding 256 array 0 1 255 {1 index exch /.notdef put} for
dup 77 /M put
dup 101 /e put
dup 100 /d put
dup 105 /i put
dup 225 /aacute put
dup 110 /n put
dup 121 /y put
dup 112 /p put
dup 116 /t put
dup 99 /c put
dup 72 /H put
dup 108 /l put
dup 97 /a put
dup 253 /yacute put
dup 118 /v put
dup 111 /o put
readonly def
/FontBBox {0 0 0 0} readonly def
currentdict end
currentfile eexec

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diff --git a/3-dfs/3-dfs.tex b/3-dfs/3-dfs.tex
deleted file mode 100644
index b183ac4..0000000
--- a/3-dfs/3-dfs.tex
+++ /dev/null
@@ -1,249 +0,0 @@
-\input ../lecnotes.tex
-
-\prednaska{3}{Prohledání do~¹íøky a do~hloubky}{()}
-
-\h{Prohledání do~¹íøky (BFS) {\I Breadth-First Search} }
-
-Jde o grafový algoritmus, který postupnì prochází v¹echny vrcholy v~dané komponentì souvislosti.
-Algoritmus nejprve projde v¹echny sousedy poèáteèního vrcholu, poté sousedy sousedù, atd\dots
-Díky tomuto zpùsobu procházení se nìkdy té¾ nazývá \uv{\I algoritmus vlny }, nebo» se z~poèáteèního vrcholu ¹íøí pomyslná vlna, která v~ka¾dém kroku nalezne v¹echny uzly, které mají od~poèáteèního vrcholu stejnou vzdálenost. Algoritmus se tedy skvìle hodí napøíklad pro hledání nejkra¹í cesty mezi dvìma vrcholy v~grafu.
-\figure{praseci-graf.eps}{Praseèí graf}{55mm}
-
-
-\s{Popis algoritmu:}
-Na zaèátku vlo¾íme do~fronty $Q$ poèáteèní vrchol $v_0$. Dále si v~poli $Z$ budeme pro ka¾dý vrchol pamatovat znaèku, zda jsme ho ji¾ nav¹tívili, èi nikoli. Pro vrchol $v_0$ si tedy dosazením jednièky zapamatujeme, ¾e je ji¾ nav¹tívený. V~dal¹ím kroku pak zkoumáme frontu $Q$: pokud není prázdná, vezmeme z~ní první vrchol a podíváme se na~v¹echny jeho sousedy $w$. Pokud je¹tì nejsou oznaèené (tedy $Z[w]=0$), tak je oznaèíme (zapamatujeme si, ¾e je pøedáváme ke zpracování a u¾ je nemáme znovu nav¹tìvovat) a pøidáme je do~fronty k~následnému zpracování. Takto cyklus opakujeme, dokud není fronta prázdná.
-
-\s{Poznámka:}
-Nejprve se podívejme, jak algoritmus pracuje na orientovaném grafu. Pro neorientovaný graf funguje algoritmus analogicky, co¾ si uká¾eme pozdìji.
-
-\s{Algoritmus:}
-
-\algo
-\:$Q \leftarrow \{v_0\}$.
-\:$Z[*] \leftarrow 0, Z[v_0] \leftarrow 1$.
-\:Dokud $Q \not= \emptyset $ opakujeme:
-\::Vyzvedneme vrcholy $u$ z~$Q$.
-\::$\forall w: (u,w)\in E$:
-\:::Je-li $Z[w]=0 \Rightarrow Z[w] \leftarrow 1$, pøidáme $w$ do~$Q$.
-\endalgo
-
-
-\>{\I Pozorování:} {\I BFS} se zastaví.
-
-\proof Zpracováváme jen vrcholy, které byly ve~frontì. Ka¾dý vrchol se dostane do~fronty maximálnì jednou. (Ka¾dý je oznaèen max. jednou, znaèky neodstraòujeme.)
-
-\s{Lemma:} BFS($v_0$) oznaèí $v$ právì tehdy, kdy¾ existuje cesta z~$v_0$ do~$v$.
-
-\proof 
-\uv{$\Longrightarrow$}: 
-Platí jako invariant po celou dobu bìhu algoritmu. To doká¾eme indukcí dle doby bìhu algoritmu:
-
-První krok indukce je triviální, nebo» cesta z~$v_0$ do~$v_0$ existuje v¾dy. Nyní si pøedstavme, ¾e oznaèujeme vrchol~$v$ pøes hranu~$uv$. To znamená, ¾e vrchol~$u$ ji¾ musel být oznaèený. Dle indukèního pøedpokladu tedy existuje cesta z~$v_0$ do~$u$, a tudí¾ pokud k~této cestì \uv{pøilepíme} hranu~$uv$, dostáváme sled z~$v_0$ do~$v$, který lze v¾dy zredukovat na cestu.
-
-\uv{$\Longleftarrow$} Sporem: Nech» existuje neoznaèený vrchol $v$ dosa¾itelný po nìjaké cestì z~$v_0$. Uva¾me nejkrat¹í cestu $(v_0, v)$: $v_0, v_1 \dots, v_k = v$ a vezmìme minimální~$i$ takové, ¾e~$v_i$ není oznaèený. Víme, ¾e~$i>0$ (nebo»~$v_0$ je urèitì oznaèen u¾ na zaèátku algoritmu). Tudí¾~$v_{i-1}$ je oznaèený. Museli jsme tedy pou¾ít hranu~$v_{i-1}~v_i$ a oznaèit vrchol~$v_i$, co¾ je SPOR. \qed
-
-Nyní tedy víme, ¾e je algoritmus správný, a máme pøedstavu o tom, jak funguje. Podíváme-li se na~nìj podrobnìji, zjistíme, ¾e je hodnì závislý na~tom, jak si budeme graf pamatovat. Zanedlouho zároveò zjistíme, ¾e nám reprezentace grafu v~pamìti znatelnì ovlivní èasovou (i pamì»ovou) slo¾itost celého algoritmu.
-
-\h{Reprezentace grafu v~pamìti}
-
-Oznaème vrcholy grafu na~následujícím obrázku písmeny A, B, C, D.
-Pokud bychom chtìli tento graf uchovat v~pamìti poèítaèe, máme na~výbìr
-hned nìkolik zpùsobù, jak to udìlat.
-\figure{imgn_o4.eps}{}{\epsfxsize}
-\s{1. matice sousednosti}
-
-Matice sousednosti pro graf $G$ na~$n$ vrcholech je ètvercové pole $A$ o velikosti $n \times n$, jeho¾ prvky na
-souøadnicích $i, j$ jsou dány následujícím pøedpisem:
-
-$$ A_{i,j} = \left\{ \matrix {1 \Leftrightarrow \{i,j\} \in E  \cr
-				0 \Leftrightarrow \{i,j\} \notin E \cr            			
-				}
-\right.$$
-
-Na~pozicích $i,j$ je jednièka, pokud v~grafu $G$ vede hrana z~vrcholu~$i$ do~vrcholu~$j$, jinak to je nula.
-Ná¹ graf z~obrázku vý¹e by tedy v~maticové reprezentaci vypadal takto:
-
-$$\bordermatrix{
-  & A & B & C & D\cr
-A & 0 & 1 & 1 & 0\cr
-B & 0 & 0 & 0 & 1\cr
-C & 0 & 0 & 0 & 1\cr
-D & 1 & 1 & 0 & 0\cr
-}$$
-
-S touto maticí se pracuje velmi snadno, napø. v¹echny sousedy $i$-tého vrcholu
-zjistíme jednodu¹e tak, ¾e projdeme $i$-tý øádek matice.
-Má ov¹em dvì zøejmé nevýhody: èasovou a pamì»ovou slo¾itost. Projití sousedù jednoho vrcholu trvá v¾dy $\Theta(n)$, projití sousedù pro v¹echny vrcholy (co¾ potøebujeme v~BFS) pak trvá $\Theta(n^2)$. Velikost matice je v¾dy $n \times n$, bez ohledu na~to, jak \uv{øídký} je graf. U grafu s mnoha vrcholy, ale s malým poètem hran, tedy budeme zbyteènì plýtvat místem v~pamìti. Tato reprezentace je tedy nevýhodná pøedev¹ím pro tøídy grafù jako jsou stromy, které mají $n-1$ hran, nebo rovinné grafy, které mají nejvý¹e $3n-6$ hran.
-
-\s{Pozorování:} BFS s reprezentací maticí sousednosti bì¾í v~èase: $\Theta(n^2)$.
-
-\proof
-U¾ jsme si uvìdomili, ¾e ka¾dý vrchol se dostane do~fronty $Q$ nejvý¹e jednou. Pro ka¾dý vrchol ve~frontì potøebujeme projít jeho sousedy, co¾ nám trvá s~reprezentací maticí sousednosti $\Theta(n)$. Vrcholù je celkem $n$, tedy èasová slo¾itost je $\Theta(n^2)$.
-\qed
-
-\s{2. seznam sousedù}
-
-V~matici sousednosti jsme museli procházet jak hrany, tak nehrany, co¾ bylo zbyteèné. Bylo by tedy výhodnìj¹í pamatovat si pro ka¾dý vrchol pouze jeho sousedy. To mù¾eme zaøídit napøíklad jedním ze~dvou následujících zpùsobù:
-
-Uchovávejme pole indexované vrcholy tak, ¾e v~ka¾dém prvku pole je ukazatel na~spojový seznam sousedù tohoto vrcholu. Tedy $L(v)=\{w: vw \in E(G)\}$.
-
-Pokud se nám nebude chtít pracovat se spojovými seznamy, mù¾eme vyu¾ít reprezentaci pomocí dvou polí. První pole~$V$ bude opìt indexované vrcholy. V~druhém poli~$E$ budou pro ka¾dý vrchol ulo¾eni jeho sousedé. V~poli~$V$ si pamatujeme pro ka¾dý vrchol~$i$ index do~pole~$E$, kde zaèínají jeho sousedé. K sousedùm vrcholu~$i$ se tedy dostaneme ji¾ snadno - nalezneme je na pozicích $V[i]~\dots~V[i+1]-1$.
-\figure{imgn_nei.eps}{Znázornìní polí seznamu sousedù.}{\epsfxsize}
-
-
-Na tuto reprezentaci u¾ staèí prostor $O(n + m)$, co¾ u¾ je, na~rozdíl od~pøedchozího kvadratického prostoru, docela pøíjemné.
-
-\s{Pozorování:} BFS bì¾í v~èase: $\Theta(n+m)$.
-
-\proof
-Algoritmus vezme ka¾dý vrchol i ka¾dou hranu do~ruky nejvý¹e jednou. Èasová slo¾itost bude tedy:
-$$\Theta\left(n+\sum_{v\in V(G)} {\rm deg}(v)\right) = \Theta(n+m).$$
-\qed
-
-\s{3. orákulum}
-
-Dal¹í mo¾ností reprezentace je pak jakési orákulum, které nám øekne (spoèítá), kam vedou hrany z~daného vrcholu\dots To se hodí napøíklad tehdy, pokud si nepotøebujeme pamatovat celý graf, ale staèí nám naleznout sousedy nìjakého vrcholu, které orákulum jednoznaènì výpoètem doká¾e urèit. Napøíklad pøi øe¹ení známé úlohy odmìøení daného mno¾ství vody za pomocí nádob rùzných objemù mù¾eme tento zpùsob reprezentace grafu pou¾ít. Problém pøevedeme na prohledávání grafu do~¹íøky, kde vrcholùm odpovídají jednotlivé stavy. Stav øíká, kolik vody je zrovna ve které nádobì. Potom nám ji¾ staèí ze zadaného poèáteèního vrcholu (stavu) najít cestu do~cílového. Orákulum je zde vlastnì funkce, které pøedáme vrchol a ona nám vrátí v¹echny vrcholy sousední -- tedy takové stavy, ke kterým dojde, kdy¾ vyzkou¹í v¹echny mo¾nosti pøelití kapaliny z jedné nádoby do~druhé.
-
-\h{Roz¹íøení algoritmu:}
-
-Abychom mohli vyu¾ít toho, ¾e algoritmus prochází vrcholy grafu ve~vlnì, a jiných hezkých vlastností, tak si dodefinujeme následující oznaèení:
-
-V~poli $D$ bude pro ka¾dý vrchol ulo¾ena vzdálenost od~poèáteèního vrcholu.
-V~poli $P$ si budeme pro ka¾dý vrchol pamatovat jeho pøedchùdce. Dále budeme vyu¾ívat fáze bìhu algoritmu, které budou simulovat onu vlnu:
-
-\s{Definice: {\I Fáze bìhu algoritmu}:} Ve~fázi $F_0$ je zpracováván vrchol $v_0$. Ve~fázi $F_{i+1}$ jsou zpracovávány vrcholy ulo¾ené do~fronty $Q$ bìhem fáze $F_i$.
-
-\s{Roz¹íøený algoritmus:}
-\algo
-\:$Q \leftarrow \{v_0\}$.
-\:$Z[*] \leftarrow 0, Z[v_0] \leftarrow 1$.
-\:$D[*] \leftarrow \infty, D[v_0] \leftarrow 0$.
-\:Dokud $Q \not= \emptyset $ opakujeme:
-\::Vyzvedneme vrchol $u$ z~$Q$.
-\::Pro ka¾dý vrchol $w$, který je sousedem vrcholu $u$:
-\:::Je-li $Z[w]=0 \Rightarrow Z[w] \leftarrow 1, D[w] \leftarrow D[u]+1, P[w] \leftarrow u$
-\::::Pøidáme $w$ do~$Q$.
-\endalgo
-
-\s{Lemma:} Na~konci BFS pro v¹echny vrcholy dosa¾itelné z~$v_0$ platí, ¾e vrchol $v$ byl zpracován ve~fázi $F_i$ právì tehdy, kdy¾ vzdálenost $v_0$ a $v$ (délka nejkrat¹í cesty z~$v_0$ do~$v$) je rovna $i$. Formálnì zapsáno: $v \in F_i \Leftrightarrow d(v_0,v) = i$. (Kde $d(x,y)$ je délka nejkrat¹í cesty z~$x$ do~$y$).
-
-\proof
-\uv{$\Longrightarrow$}: 
-Dùkaz provedeme indukcí podle $i$ (èísla fáze bìhu algoritmu).
-
-První krok indukce je triviální, nebo» ve~fázi $F_0$ je oznaèen (dle definice) pouze vrchol $v_0$ a to je jediný vrchol vzdálený~0 od~$v_0$. 
-
-Pokud je vrchol $v$ zpracováván ve~fázi $F_i$, pak musel být zaøazen do~fronty bìhem fáze $F_{i-1}$ jako soused nìjakého vrcholu $u$. Pro vrchol $u$ mù¾eme pou¾ít indukèní pøedpoklad, tedy ¾e délka nejkrat¹í cesty z~$v_0$ do~$u$ je $d(v_0,u)=i-1$. Pak tedy $d(v_0,v)\leq i$, nebo» je¹tì nevíme, zda cesta $v_0, \dots, u, v$ je nejkrat¹í. Kdyby ale existovala nìjaká krat¹í, tedy délky nejvý¹e $i-1$, tak by byl vrchol $v$ objeven u¾ døíve ne¾ ve fázi $F_i$. Proto $d(v_0,v) = i$.
-
-
-\uv{$\Longleftarrow$}: Ka¾dý dosa¾itelný vrchol padne do~nìjaké fáze (viz. minulé lemma).
-\qed
-
-Ji¾ tedy víme, ¾e vrchol $v_i$, jeho¾ vzdálenost od~vrcholu $v_0$ je $i$, bude zpracován v~$i$-té fázi. Jak ale po~skonèení algoritmu zjistíme, ve které fázi byl zpracován, neboli jak je vzdálený od~startovního vrcholu? Tato informace je právì ulo¾ena v~poli $D$ s indexem $i$ (v~$D[i]$).
-
-Zároveò nás mù¾e zajímat, jak bychom nejkrat¹í cestu z~$v_0$ do~$v_i$ rekonstruovali. Pro tento úèel jsme si zavedli pole $P$. Nejkrat¹í cesta z~$v_0$ do~$v_i$ bude v~obráceném poøadí vypadat: $v_i, P[v_i], P[P[v_i]], P[P[P[v_i]]], \dots, v_0$.
-
-
-\s{Pozorování:} Pokud víme, ¾e $v_0, v_1, \dots, v_{k-1}, v_k$ je nejkrat¹í cesta z~$v_0$ do~$v_k$, pak $v_0,v_1,\dots,v_{k-1}$ je nejkrat¹í cesta z~$v_0$ do~$v_{k-1}$. Neboli prefix nejkrat¹í cesty je nejkrat¹í cesta.
-
-\proof
-Kdyby existovala krat¹í cesta z~$v_0$ do~$v_{k-1}$, pak bychom mohli zkrátit i cestu z~$v_0$ do~$v_k$.
-
-
-\s{Pozorování:} BFS u~neorientovaného grafu projde celou komponentu souvislosti.
-
-\proof
-Víme, ¾e BFS($v_0$) oznaèí $v$ právì tehdy, kdy¾ existuje cesta z~$v_0$ do~$v$. V~neorientovaném grafu existuje cesta z~$v_0$ do~právì v¹ech vrcholù, které jsou ve~stejné komponentì souvislosti jako $v_0$. Pokud tedy spustíme BFS na~$v_0$, tak se postupnì projdou v¹echny vrcholy této komponenty souvislosti.
-\qed
-
-\s{Pozorování:} Pokud BFS postupnì spou¹tíme na~dosud neobarvené vrcholy v~ neorientovaném grafu, nalezneme nakonec v~èase $\Theta(n+m)$ v¹echny komponenty souvislosti.
-
-\s{Algoritmus:}
-\algo
-\:Pro v¹echny vrcholy $v \in V(G)$ opakuj:
-\::Pokud je vrchol $v$ neobarvený $ \Rightarrow \<BFS(v)>$.
-\endalgo
-
-\proof
-Ka¾dým spu¹tìním na~dosud neobarvený vrchol neorientovaného grafu obarvíme právì jednu komponentu souvislosti (tu, ve~které je tento vrchol). 
-Postupnì projdeme v¹echny vrcholy od prvního k poslednímu a v¾dy pokud je vrchol neobarvený, spustíme na nìj BFS. Tak nakonec obarvíme v¹echny komponenty souvislosti. Èasová slo¾itost bude stejná jako u~samotného BFS, tedy $\Theta(n + m)$.
-\qed
-
-\s{Vìta:} $BFS(v_0)$ v~èase $\Theta(n + m)$ spoète:
-\itemize\ibull
-\:vrcholy dosa¾itelné z~$v_0$
-\:vzdálenosti tìchto vrcholù od~$v_0$
-\:strom nejkrat¹ích cest z~$v_0$
-\endlist
-
-\s{Poznámka:} Algoritmus na prohledávání do~¹íøky bude fungovat i na neorientovaném grafu. Mù¾eme si jednodu¹e pøedstavit, ¾e je to orientovaný graf, kde ka¾dá hrana má oboustrannou orientaci.
-
-
-Prohledávání do~¹íøky ale není jediný algoritmus, který nìjak systematicky prochází graf. Jak u¾ název kapitoly napovídá, budeme se zabývat je¹tì druhým algoritmem, prohledáváním do~hloubky. Podívejme se, jak bude vypadat \dots
-
-\h{Prohledávání do~hloubky (DFS) {\I Depth-First Search} }
-
-Tento algoritmus neprochází graf ve~vlnì jako BFS, ale prochází ho rekurzivnì. V¾dy se zanoøí co nejhloubìji a¾ do~listu a pak se o~kus vrátí a opìt se sna¾í zanoøit. Vrcholy, ve kterých u¾ byl, ignoruje.
-
-Opìt uva¾me nejdøíve graf orientovaný. Následnì si uká¾eme, ¾e v~neorientovaném grafu budou pouze malé zmìny.
-
-Budeme pou¾ívat podobné znaèení jako u~BFS. V~poli $Z$ si budeme pamatovat, zda jsme vrchol ji¾ nav¹tívili (hodnota 1), nebo ne (hodnota 0). Navíc promìnná~$T$ bude znaèit dobu bìhu algoritmu - tedy jakési \uv{hodiny}. Pøi ka¾dém nalezení nového vrcholu èi jeho opu¹tìní tuto promìnnou zvý¹íme o~1. Do~polí $\<in>$ a $\<out>$ si budeme ukládat èas (prvního) nalezení a opu¹tìní vrcholu.
-
-\s{Algoritmus:}
-
-\algo
-\: inicializace: $Z[*] \leftarrow 0, T \leftarrow 1, \<in>[*] \leftarrow ?, \<out>[*] \leftarrow ?$
-\: $DFS(v): Z[v] \leftarrow 1, \<in>[v] \leftarrow T|++|$
-\:: Pro $w$: $vw \in E(G)$:
-\::: Pokud $Z[w]=0 \Rightarrow DFS(w)$
-\:: $out[v] \leftarrow T|++|$
-\endalgo
-
-\s {Vìta:} DFS($v_0$) v~èase $\Theta(m+n)$ oznaèí právì v¹echny vrcholy dosa¾itelné z~$v_0$.
-
-\proof
-Nejdøíve je potøeba dokázat, ¾e pokud je vrchol $v$ dosa¾itelný z~vrcholu $v_0$, tak jej DFS oznaèí. Dùkaz bude podobný jako u~BFS.
-
-V analýze èasové slo¾itosti si pak opìt uvìdomíme, ¾e algoritmus vezme ka¾dý vrchol i hranu do~ruky právì jednou, tak¾e èasová slo¾itost bude $\Theta(n + m)$.
-\qed
-
-\figure{img5_dfso.eps}{Graf a znázornìní prùbìhu DFS s~jednotlivými hranami:}{\epsfxsize}
-
-Mù¾eme si v¹imnout, ¾e jak DFS prochází graf, rozdìluje hrany do~4 skupin: hrany {\I stromové, zpìtné, dopøedné a pøíèné}.
-
-Jedinì {\I stromové} hrany jsou takové, ¾e se po~nich DFS opravdu vydá. Vedou toti¾ do~vrcholu, který nebyl dosud objeven. V~ukázkovém grafu to jsou hrany: $\{(A \rightarrow B), (B \rightarrow C), (B \rightarrow D)\}$.
-
-Pokud algoritmus objeví z~vrcholu $v$ hranu do~ji¾ døíve nav¹tíveného vrcholu $w$ a zároveò platí, ¾e $w$ je ve~stejném podstromu jako $v$, tak nazveme hranu $vw$ {\I zpìtnou}. Pro rozpoznání je dùle¾ité, ¾e vrchol $w$ byl ji¾ objeven, ale je¹tì ne opu¹tìn. V~ukázkovém grafu se jedná o hranu: $(C \rightarrow A)$.
-
-Kdy¾ pøi prohledávání sousedù vrcholu $v$ narazíme na~vrchol $w$, který jsme ji¾ nav¹tívili, a to v~podstromì vrcholu $v$, tak nazveme hranu $vw$ {\I dopøednou}, nebo» vede z~$v$ do~jeho potomka. Platí tedy, ¾e jsme nejdøíve objevili vrchol $v$, potom vrchol $w$, pak jsme vrchol $w$ opustili a nyní jsme na~nìj znovu narazili po~dopøedné hranì. V~ukázkovém grafu hrana: $(A \rightarrow D)$.
-
-Posledním typem hran je {\I pøíèná} hrana. Ta vede do~vrcholu v~sousedním podstromì zprava doleva. V~tomto pøípadì jsme tedy nejdøíve objevili vrchol $w$, ten jsme následnì opustili a a¾ pak jsme objevili vrchol $v$. V~ukázkovém grafu to je hrana: $(D \rightarrow C)$.
-
-\s{K zamy¹lení:} Proè nemohou vést pøíèné hrany také zleva doprava?
-
-K~rozpoznávání typù hran se nám tedy velmi hodí pole $\<in>$ a $\<out>$, ve~kterých si pamatujeme èasy objevení a opu¹tìní vrcholu. Podle toho, jak se intervaly objevení a opu¹tìní obou vrcholù pøekrývají, mù¾eme jednoznaènì rozhodnout, o jaký typ hrany se jedná:
-
-U~zpìtných hran je poøadí: $\<in>(w)$, $\<in>(v)$, $\<out>(v)$, $\<out>(w)$. Intervaly do~sebe budou zanoøené takto: $<<>_v>_w$.
-
-U~dopøedných hran je poøadí: $\<in>(v)$, $\<in>(w)$, $\<out>(w)$, $\<out>(v)$. Intervaly do~sebe budou zanoøené takto: $<<>_w>_v$.
-
-U~pøíèných hran je poøadí: $\<in>(w)$, $\<out>(w)$, $\<in>(v)$, $\<out>(v)$. Intervaly do~sebe budou zanoøené takto: $<>_w<>_v$.
-
-Pozn: Pou¾íváme zde toto znaèení: $<>_v = <in(v), out(v)>$. Jedná se o interval objevení a opu¹tìní vrcholu $v$.
-
-\s{Typy hran ($v \rightarrow w$):}
-
-\itemize\ibull
-\:Stromové hrany \dots po nich DFS pro¹lo. $\{(A \rightarrow B), (B \rightarrow C), (B \rightarrow D)\}$
-\:Zpìtné hrany $<<>_v>_w$\dots vedou do~pøedchùdce $v$ ve~stromu. $\{(C \rightarrow A)\}$
-\:Dopøedné hrany $<<>_w>_v$\dots vedou do~potomka. $v$ $\{(A \rightarrow D)\}$
-\:Pøíèné hrany $<>_w<>_v$\dots vedou do~vrcholu $v$ v~sousedním podstromì, v¾dy zprava doleva. $\{(D \rightarrow A)\}$
-\endlist
-
-\s{Pozorování:} Hrany, po~kterých DFS pro¹lo, tvoøí DFS strom.
-
-\s{Pozorování:} Intervaly ($\<in>(v)$, $\<out>(v)$) $\forall v \in V(G) $ tvoøí dobré uzávorkování. (Intervaly synù disjunktnì vyplòují otce $\Rightarrow$ intervaly se nemohou køí¾it).
-
-Nakonec si je¹tì uvìdomme, jak bude vypadat prohledávání do~hloubky na~neorientovaném grafu. Algortimus bude úplnì stejný, jenom se nám zredukuje poèet typù hran na~dvì: {\I stromové} a~{\I zpìtné}. Ani {\I dopøedné} ani {\I pøíèné} nebudou existovat, nebo» se z~{\I pøíèných} stanou {\I stromové} a z~{\I dopøedných zpìtné}.
-
-\bye
diff --git a/3-dfs/Makefile b/3-dfs/Makefile
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diff --git a/3-dfs/praseci-graf.eps b/3-dfs/praseci-graf.eps
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diff --git a/6-kostry/6-kostry.tex b/6-kostry/6-kostry.tex
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-\input lecnotes.tex
-
-\prednaska{6}{Problém minimální kostry}{}
-
-\s{Zadání úlohy:} Pro neorientovaný graf $G$ s~ohodnocením hran {\I váhami} $w: E(G) \rightarrow \bb R$,
-chceme najít kostru $T$ s minimálním ohodnocením $w(T)=\sum_{e\in E(T)} w(e)$.
-
-\s{Navíc pøedpokládáme:} (bez újmy na~obecnosti)
-\itemize\ibull
-\:Graf $G$ je souvislý (jinak ho nejprve rozlo¾íme na komponenty).
-\:$\forall e,f \in E(G$) : $e\neq f \Rightarrow w(e)\neq w(f)$.
-\endlist
-
-Nyní si uká¾eme tøi algoritmy pro hledání minimální kostry, konkrétnì se jedná
-o~Jarníkùv, Borùvkùv a Kruskalùv algoritmus.
-
-\h{Jarníkùv algoritmus}
-
-\s{Algoritmus:}
-
-\algo
-\algin Graf~$G$ s~ohodnocením~$w$.
-\:Zvolíme libovolný vrchol $v_0\in V(G)$.
-\:$T\leftarrow(\left\{v_0\right\},\emptyset)$ (zatím jednovrcholový strom)
-\:Dokud $\vert V(T) \vert \neq n$:
-\::Vybereme hranu $uv\in E(G) : u\in V(T), v\notin V(T)$ tak, aby $w(uv)$ byla minimálni.
-\::$T\leftarrow T+uv$.
-\algout Minimální kostra~$T$.
-\endalgo
-
-\s{Vìta:} Jarníkùv algoritmus se zastaví po maximálnì $n$ iteracích a vydá minimální kostru grafu $G$.
-
-\proof
-Pøi ka¾dé iteraci algoritmus pøidá jeden vrchol do~$T$, a~proto se po~maximálnì $n$ iteracích zastaví.
-Vydaný graf je strom, proto¾e se stále pøidává list k ji¾ existujícímu stromu, a~jeliko¾ má $n$~vrcholù,
-je to kostra. Zbývá nám u¾ jen dokázat, ¾e nalezená kostra je minimální. K~tomu pomu¾e následující lemma:
-
-{\narrower
-
-\s{Definice:} {\I Øez} v~grafu $G=(V,E)$ je mno¾ina hran $F\subseteq E$ taková, ¾e $\exists  U\subset V$ :
-$F=\left\{uv\in E:u\in U, v\notin U \right\}$.
-
-\s{Lemma (øezové):} Pokud $G$ je graf, $w$ jeho prosté ohodnocení, $F$ je øez v
-grafu $G$ a $f$ je nejlehèí hrana v øezu $F$, pak pro ka¾dou minimální kostru
-$T$ grafu $G$ je $f\in E(T)$.
-
-\proof
-Buï $T$ kostra a $f=uv\notin E(T)$. Pak existuje cesta $P\subseteq T$ spojující $u$ a $v$.
-Cesta musí øez alespoò jednou pøekroèit. Proto existuje $e\in P \cap F$ a navíc víme, ¾e $w(e) > w(f)$. Uva¾me $T'=T-e+f$.
-Tento graf je rovnì¾ kostra grafu $G$, proto¾e odebraním hrany $e$ se graf rozpadne na dvì komponenty a pøidáním
-hrany $f$ se tyto komponenty opìt spojí. Navíc $w(T')=w(T)-w(e)+w(f)<w(T)$.
-\qed
-
-}
-
-\>V~dùkazu korektnosti Jarníkova algoritmu toto lemma vyu¾ijeme tak, ¾e si v¹imneme, ¾e hrany mezi
-vrcholy stromu~$T$ a zbytkem grafu tvoøí øez a algoritmus nejlehèí hranu tohoto øezu pøidá
-do~$T$. Podle lemmatu tedy v¹echny hrany~$T$ musí být souèástí ka¾dé minimální kostry a jeliko¾~$T$ je strom,
-musí být minimální kostrou.
-
-\qed
-
-
-\s{Dùsledky:} Graf $G$ s prostým ohodnocením má pravì jednu minimální kostru. Minimální kostra je
-jednoznaènì urèená lineárním uspoøádáním hran.
-
-\s{Implementace:}
-\itemize\ibull
-\:Pøímoèará: pamatujeme si, které vrcholy a hrany jsou v kostøe $T$ a které ne. Èasová slo¾itost je $\O(nm)$.
-\:Chytøej¹í: Pro $v\notin V(T)$ si pamatujeme $D(v)=\min\left\{w(uv):u\in T\right\}$. Pøi ka¾dém
-prùchodu hlavním cyklem pak procházíme v¹echna~$D(v)$ (to v¾dy trvá $\O(n)$) a pøi pøidání vrcholu do~$T$ kontrolujeme
-okolní~$D(w)$ pro $vw\in E$ a pøípadnì je sni¾ujeme (za~ka¾dou hranu~$\O(1)$). Èasovou slo¾itost tím celkovì
-zlep¹íme na~$\O(n^2+m)=\O(n^2)$.
-\:Také se dá pou¾ít halda pro uchovávání hran nebo hodnot~$D(v)$.
-\endlist
-
-\h{Borùvkùv algoritmus}
-
-\s{Algoritmus:}
-
-\algo
-\algin Graf~$G$ s~ohodnocením~$w$.
-\:$F\leftarrow(V(G),\emptyset)$
-\:Dokud $F$ má alespoò dvì komponenty:
-\::Pro ka¾dou komponentu $T_i$ grafu $F$ vybereme nejlehèí incidentní hranu $t_i$.
-\::V¹echny hrany $t_i$ pøidáme do $F$.
-\algout Minimální kostra~$F$.
-\endalgo
-
-\s{Vìta:} Borùvkùv algoritmus se zastaví po $\left\lceil \log_2 n\right\rceil$ iteracích a vydá minimální kostru grafu $G$.
-
-\proof
-V¹imnìme si nejprve, ¾e po~$k$ iteracích mají v¹echny komponenty grafu~$F$ minimálnì $2^k$ vrcholù
-(indukcí -- na~poèátku jsou v¹echny komponenty jednovrcholové, v~ka¾dé dal¹í
-iteraci se komponenty sluèují do~vìt¹ích,
-ka¾dá s~alespoò jednou sousední, tak¾e se velikosti komponent minimálnì zdvojnásobí).
-Proto nejpozdìji po~$\left\lceil \log_2 n\right\rceil$ iteracích u¾~velikost komponenty dosáhne poètu v¹ech vrcholù a algoritmus
-se zastaví.
-
-Hrany mezi ka¾dou komponentou a~zbytkem grafu tvoøí øez, tak¾e podle øezového
-lemmatu v¹echny hrany pøidané do~$F$ musí být souèástí (jednoznaènì
-urèené) minimální kostry. Graf $F\subseteq G$ je tedy v¾dy les a a¾ se
-algoritmus zastaví, bude roven minimální kostøe.
-\qed
-
-\s{Implementace:}
-\itemize\ibull
-\:Inicializace pøímoèará.
-\:Pomocí DFS rozlo¾íme les na komponenty. Pro ka¾dý vrchol si pamatujeme èíslo komponenty.
-\:Pro ka¾dou hranu zjistíme, do které komponenty patøí, a pro ka¾dou komponentu
-si uchováme nejlehèí hranu.
-\endlist
-
-\>Takto doká¾eme ka¾dou iteraci provést v~èase $\O(m)$ a celý algoritmus dobìhne v~$\O(m\log n)$.
-
-\h{Kruskalùv neboli hladový algoritmus}
-
-\s{Algoritmus:}
-
-\algo
-\algin Graf~$G$ s~ohodnocením~$w$.
-\:Setøídíme v¹echny hrany z $E(G)$ tak, aby: $w(e_1)<...<w(e_m)$.
-\:$F\leftarrow (V(G),\emptyset)$.
-\:Pro $i=1$ do $m$:
-\::Pokud $F+e_i$ je acyklický, provedeme $F\leftarrow F+e_i$.
-\algout Minimální kostra~$F$.
-\endalgo
-
-\s{Vìta:} Kruskalùv algoritmus se zastaví po~$m$ iteracích a vydá minimální kostru.
-
-\proof
-Ka¾dá iterace algoritmu zpracovává jednu hranu, tak¾e iterací je~$m$. Indukcí doká¾eme,
-¾e~$F$ je v¾dy podgrafem minimální kostry: prázdné poèáteèní~$F$ je podgrafem èehokoliv,
-ka¾dá hrana, kterou pak pøidáme, je minimální v~øezu oddìlujícím nìjakou
-komponentu~$F$ od~zbytku grafu (ostatní hrany tohoto øezu je¹tì nebyly
-zpracovány, a~tudí¾ jsou tì¾¹í). Naopak ¾ádná hrana, kterou jsme se rozhodli
-do~$F$ nepøidat, nemù¾e být souèástí minimální kostry, jeliko¾ s~hranami,
-o~kterých ji¾ víme, ¾e v~minimální kostøe le¾í, tvoøí kru¾nici. \qed
-
-% Toto jsme na pøedná¹ce dìlali sporem a mì se to líbí více, ale oba dùkazy
-% urèitì fungují.
-
-\s{Implementace:}
-\itemize\ibull
-\:Setøídìní v èase $\O(m\log m)=\O(m\log n)$.
-\:Pak potøebujeme udr¾ovat komponenty souvislosti grafu~$F$, abychom umìli rychle
-  urèit, jestli právì zpracovávaná hrana vytvoøí kru¾nici. Potøebujeme tedy strukturu
-  pro udr¾ování komponent souvislosti, které se $m$-krát zeptáme, zda dva vrcholy
-  le¾í v~té¾e komponentì (tomu budeme øíkat operace \<Find>), a~$(n-1)$-krát spojíme
-  dvì komponenty do jedné (\<Union>).
-\endlist
-
-\>Kruskalùv algoritmus tedy pobì¾í v~èase $\O(m\log n + mT_f + nT_u)$, kde~$T_u$ je
-èas na~provedení jedné operace \<Union> a $T_f$ na~operaci \<Find>.
-
-\s{Jednoduchá struktura pro komponenty:}
-Budeme si pamatovat v~poli èísla komponent, ve~kterých le¾í jednotlivé
-vrcholy. \<Find> zvládneme v~èase $\O(1)$, ale \<Union> bude stát $\O(n)$. Celý
-algoritmus pak pobì¾í v~èase $\O(m\log n+ m + n^2) = \O(m\log n+n^2)$.
-
-\s{Chytøej¹í struktura:} Ka¾dou komponentou si ulo¾íme jako strom orientovaný smìrem ke koøeni
--- ka¾dý vrchol si pamatuje svého otce, navíc ka¾dý koøen si pamatuje velikost
-komponenty. 
-%Hloubku podstromu? Zaøazujeme mìlèí pod hlub¹í, ne nutnì men¹í pod vìt¹í.
-%Myslím, ¾e s hloubkami to funguje lépe, ovìøit.
-Operace \<Find> vystoupá z~obou vrcholù ke~koøeni a koøeny porovná. \<Union>
-rovnì¾ najde koøeny a pøipojí koøen men¹í komponenty pod koøen té vìt¹í. Obojí
-zvládneme v~èase lineárním v~hloubce stromu a jak si uká¾eme, tato hloubka je
-v¾dy nejvý¹e logaritmická, a proto celý Kruskalùv algoritmus pobì¾í v~èase
-$\O(m\log n + m\log n + n\log n) = \O(m\log n)$.\foot{%
-Drobnou úpravou bychom mohli dosáhnout daleko efektivnìj¹í struktury, ale tu
-bychom neupotøebili,
-jeliko¾ by nás stejnì brzdilo tøídìní, a analýza slo¾itosti by byla \dots\ inu, slo¾itìj¹í.}
-
-% V originále je jen "strom hloubky..." a to pøece pro obecný (napø.
-% zdegenerovaný) strom neplatí.
-
-% Mimochodem, zde se u¾ mluví o spojování podle hloubky, tak¾e to nahoøe je asi
-% vá¾nì chyba
-\s{Lemma:} \<Union-Find> strom hloubky $h$ má alespoò $2^h$ prvkù.
-
-\proof
-Indukcí: Pokud \<Union> spojí strom s~hloubkou $h$ s jiným s~hloubkou men¹í ne¾ $h$, pak hloubka výsledného
-stromu zùstává~$h$. Pokud spojuje dva stromy stejné hloubky~$h$, pak má výsledný strom hloubku $h+1$.
-Z~indukèního pøedpokladu víme, ¾e strom hloubky $h$ má minimálnì $2^h$ vrcholù,
-a~tedy výsledný strom hloubky $h+1$ má alespoò $2^{h+1}$ vrcholù. \qed
-
-% Union-find je jsme mìli na pøedná¹ce detailnìji rozebraný (i s pseudokódem),
-% není to nutnost, ale mù¾u ho klidnì rozepsat trochu výøeènìji, aby mìl ètenáø
-% vìt¹í jistotu. Zále¾í na tobì.
-
-\bye
diff --git a/6-kostry/Makefile b/6-kostry/Makefile
deleted file mode 100644
index b86ea6f..0000000
--- a/6-kostry/Makefile
+++ /dev/null
@@ -1,3 +0,0 @@
-P=6-kostry
-
-include ../Makerules
diff --git a/8-rozdel/8-rozdel.tex b/8-rozdel/8-rozdel.tex
deleted file mode 100644
index 6103891..0000000
--- a/8-rozdel/8-rozdel.tex
+++ /dev/null
@@ -1,408 +0,0 @@
-\input ../lecnotes.tex
-
-\prednaska{8}{Rozdìl a~panuj}{}
-
-Známá strategie {\sl Divide et Impera} (nìkdy pøekládáno spí¹e jako \uv {Roze¹tvi a~panuj}) pochází z dob antického Øíma\foot{Aè se touto strategií øím¹tí panovníci øídili, není nalezen ¾ádný antický zdroj výroku; ten je pøipisován a¾ renesanènímu N. Machiavellimu.
-}, kdy panovník zasel nesvár mezi místní kmeny (èím¾ je rozdìlil a~roze¹tval) a~následnì pøi¹el, urovnal napìtí, aby je mohl opìt ovládnout a~panovat a~aby byli v¹ichni spokojení.
-
-Tato strategie ov¹em pøetvrává (aè tøeba V~jiných oblastech) a~pomáhá øe¹it ty problémy, které se dají rozdìlit na~men¹í jednodu¹¹í podproblémky.
-
-Jak tedy algoritmus typu {\it rozdìl a~panuj} pracuje? Mìjme problém, který má tu vlastnost, ¾e kdy¾ jej rozdìlíme na~vhodné podproblémy, které mají stejný charakter, a~ty vyøe¹íme, slo¾ením jejich øe¹ení mù¾eme získat øe¹ení pùvodního problému. Algoritmus tedy bude rekurzivnì volat sám sebe, ne¾ se dostane k~podproblému
-nìjaké konstantní velikosti, který u¾ umí vyøe¹it triviálnì, a~pak se zaène z~rekurze vracet a~skládat jednotlivá dílèí øe¹ení.
-
-\h{Pøíklad 1 -- MergeSort:}
-
-Tento tøídící algoritmus pracuje na~principu, ¾e vstup rozdìlíme na~dvì (skoro) stejnì velké èásti, které rekurzivním voláním setøídíme, a~nakonec výsledné dvì posloupnosti slijeme do~jedné. 
-
-\s{Algoritmus:}
-
-\algo
-\algin posloupnost $x_1,\dots, x_n$.
-\:Pokud $n \leq 1 \Rightarrow$ vrátíme vstup.
-\:$y_1,\dots,y_{\lfloor n/2 \rfloor} \leftarrow$ \<MergeSort> $(x_1,\dots,x_{\lfloor n/2 \rfloor})$.
-\:$z_1,\dots,z_{\lceil n/2 \rceil} \leftarrow$ \<MergeSort> $(x_{\lfloor n/2 \rfloor + 1},\dots,x_n)$.
-\:Vrátíme $\<Merge>(y_1,\dots,y_{\lfloor n/2 \rfloor}; z_1,\dots,z_{\lceil n/2 \rceil})$.
-\endalgo
-
-\noindent
-Na slití dvou setøídìných posloupností do~jedné pou¾íváme funkci Merge:
-
-{\bo $\<Merge>(y_1, \dots,y_a;z_1, \dots,z_b)$:}
-\algo
-\:$i \leftarrow 1, j \leftarrow 1, k \leftarrow 1$.
-\:Dokud $k \leq a+b$:
-\::Je-li $(j>b)$ nebo $(i \leq a) \& (y_i < z_j) \Rightarrow x_k \leftarrow y_i, k|++|, i|++|$.
-\::Jinak $\Rightarrow x_k \leftarrow z_j, k|++|, j|++|$.
-\:Vrátíme $x_1, \dots,x_n$.
-\endalgo
-
-\s{Pozorování:}
-\<Merge> trvá $\Theta (n)$, nebo» ka¾dou ze~slévaných posloupností projdeme právì jednou.
-
-\s{Èasová slo¾itost MergeSortu:}
-
-Rozdìlování a~slévání nám trvá lineárnì dlouho, tak¾e pro èasovou slo¾itost MergeSortu platí tato rekurentní rovnice: $$T(n)= 2 \cdot T(n/2) + c \cdot n.$$ 
-Pøitom $T(n)$ zde znaèí èas strávený na~vstupu délky $n$ a~$c$ je nìjaká vhodná konstanta. Pro jednoduchost BÚNO pøedpokládejme, ¾e $n$ je mocnina dvojky. Zároveò si jednotky zvolme tak, ¾e bude platit $T(1)=1$. Mù¾eme si tedy rekurentní vztah rozepsat následovnì:
-
-$$\eqalign{
-T(n) &= 2 \cdot (2T(n/4) + c \cdot n/2) + c \cdot n = 4T(n/4) + 2cn = \cr
-		 &= 4(2T(n/8) + c(n/4)) + 2cn = 8T(n/8) + 3cn = \dots\cr
-		 &= 2^k T(n/2^k) + kcn.
-}$$
-
-Pokud zvolíme $k = \log_2{n}$, získáme:
-$$T(n) = 2^{\log_2{n}} \cdot T(n/2^{\log_2{n}}) + \log_2{n} \cdot c \cdot n = n \cdot T(1) + c \cdot n \cdot \log_2{n} = \Theta(n \log{n}). $$
-
-Ke stejnému výsledku mù¾eme ale dojít také úplnì jinou cestou. Pøedstavme si strom rekurzivních volání. Ka¾dý vrchol má dva syny (dìlíme vstup na~dvì èásti), v~nich¾ jsou vstupy polovièní velikosti. V~ka¾dém vrcholu trávíme èas lineární s~velikostí jeho vstupu, souèet velikostí vstupù pøes ka¾dou hladinu je~$n$ a~hloubka stromu musí být $\Theta(\log n)$. Vyjde nám tedy, ¾e $T(n)=\Theta(n\log n)$.\foot{Po pozornìj¹ím zamy¹lení si ètenáø mù¾e uvìdomit, ¾e se jedná vlastnì pouze o~jiný pohled na~stejný dùkaz jako rozepisování rekurentního vzorce.}
-
-\s{Pamì»ová slo¾itost MergeSortu:}
-
-$M(n) = d \cdot n + M(n/2) = d \cdot n + d \cdot n/2 + d \cdot n/4 + \dots \leq 2d n = \Theta(n).$
-
-Tento vztah platí pro nìjakou vhodnou konstantu $d$. Mù¾eme ho opìt nahlédnout napøíklad ze stromu rekurzivních volání. Podívejme se na~libovolnou cestu od~koøene do~listu. V~jednotlivých vrcholech potøebujeme pamìti pøesnì $d \cdot n/2^k$, kde $k$ je èíslo hladiny. Kdy¾ tyto hodnoty seèteme pøes v¹echny vrcholy na~této cestì, výsledek bude konvergovat k~$2dn$, co¾ dává pamì»ovou slo¾itost $\Theta(n)$.
-
-\s{Závìr:}
-Mergesort bì¾í V~èase $\Theta(n \log{n})$ a~pamìti $\Theta(n)$. Lineární pamì»ová slo¾itost není výhodná, ale na~druhou stranu se tento algoritmus velmi hodí napøíklad na~tøídìní lineárních spojových seznamù.
-
-\h{Pøiklad 2 -- Násobení èísel:} 
-Pokud násobíme dvì èísla $X$ a~$Y$ (obì délky $n$; pokud bylo jedno krat¹í, tak ho doplníme nulami zleva tak, aby byla obì èísla stejnì dlouhá) zpùsobem, který nás uèili na~základní ¹kole, výsledný algoritmus má èasovou slo¾itost $\Theta(n^2)$. Proto¾e se jedná o~dost èastou operaci, zamysleme se, zda by ne¹la zrychlit. Nasmìrujme na¹e úvahy na~postup {\it rozdìl a~panuj}. Rozdìlíme ka¾dého èinitele na~dvì stejnì dlouhé èásti. Pro jednoduchost pøedpokládejme, ¾e toto roz¹tìpení èinitele probìhne v¾dy bez zbytku:
-$$
-X=A \cdot 10^{{n}/2}+B, \qquad Y=C \cdot10^{{n}/{2}}+D.
-$$
-Zde $A, B, C, D$ jsou u¾ jen $(n/2)$-ciferná èísla. Pùvodní souèin získáme jako:
-$$
-XY=(A\cdot 10^{{n}/{2}}+B) (C\cdot 10^{{n}/{2}}+D)=AC \cdot 10^{n}+(AD+BC)\cdot 10^{{n}/{2}}+BD.
-$$
-Nyní, jak vidíme, staèí spoèítat souèin ètyø $(n/2)$-ciferných èísel a~pak výsledky spolu seèíst. Uva¾me,
-jakou bude mít tento algoritmus èasovou slo¾itost:
-$$T(n) = 4T(n/2) + cn.$$
-Toto platí pro nìjakou vhodnou konstantu $c$ (výraz $cn$ je re¾ie na~sèítání).  Jednotky si zvolme tak, aby platilo: $$T(1)=1.$$
-Jak takovou rekurenci vyøe¹íme? Máme opìt dvì mo¾nosti:
-
-\>{\sl 1. zpùsob: Øe¹ení rozepsáním rekurentního vztahu:}
-$$\eqalign{
-T(n)&= 4T(n/2)+cn = \cr
-    &= 4\cdot (4T(n/4)+cn/2)+cn = 4^2T(n/4)+2cn+cn = 4^2T(n/4)+3cn = \cr
-    &= 4^2\cdot (2T(n/8)+cn/4)+3cn = 4^3T(n/8)+4cn+3cn = 4^3T(n/8)+7cn = \cr
-    &\dots\cr
-}$$
-Odtud snadno vypozorujeme, ¾e jednotlivé vztahy se vyvíjejí podle vzorce
-$T(n)=4^kT(n/2^k) + (2^k-1)cn.$ Pro $k=\lceil\log_2 n\rceil$ je ov¹em
-$2^k\le 1$, tak¾e $T(n/2^k)=\Theta(1)$ a~dostaneme (horní celou èást zanedbáme,
-ta ovlivní jen konstanty):
-$$
-T(n) = 4^{\log_2 n}\Theta(1) + (2^{\log_2 n}-1)cn = n^2\Theta(1) + (n-1)cn = \Theta(n^2).
-$$
-
-\>{\sl 2. zpùsob: Úvaha o~stromu:} Nakreslíme si strom rekurzivních volání
-na¹eho algoritmu:
-\fig{figure.eps}{4in}
-Na~$i$-té hladinì stromu le¾í $4^i$ vrcholù, v~nich jsou vstupy velikosti
-$n/2^i$, tak¾e na~celé hladinì trávíme èas celkem $\Theta(4^i\cdot n/2^i)
-= \Theta(2^in)$. Velikosti vstupù klesají exponenciálnì, tak¾e celý strom
-je hluboký $k=\log_2 n$ (opìt si dovolíme zapomenout na~horní celou èást).
-Celkem tedy spotøebujeme èas $\sum_{i=0}^{k}\Theta(2^in) = \Theta(n\cdot\sum_{i=0}^k 2^i) = \Theta(n^2)$.
-
-Oba zpùsoby analýzy se tedy shodují na~tom, ¾e ná¹ algoritmus má kvadratickou èasovou
-slo¾itost a~¾e jsme si oproti klasickému algoritmu nikterak nepomohli.
-Podívejme se je¹tì jednou na~to, jak se ná¹ algoritmus vìtví:
-$$\vbox{\halign{\hfil#\hfil \quad & \hfil#\hfil \quad &\hfil#\hfil\cr
-hloubka & poèet úloh & velikost podúlohy\cr
-\noalign{\smallskip\hrule\medskip}
-0 & $4^{0}$ & ${n}/{2^{0}}$\cr
-1 & $4^{1}$ & ${n}/{2^{1}}$\cr
-2 & $4^{2}$ & ${n}/{2^{2}}$\cr
-3 & $4^{3}$ & ${n}/{2^{3}}$\cr
-\vdots & \vdots & \vdots\cr
-$k$ & $4^{k}$ & ${n}/{2^{k}}$\cr}}$$
-Naskýtá se otázka, jestli bychom nemohli èasovou slo¾itost zlep¹it. Toho bychom
-mohli dosáhnout napøíklad zlep¹ením èlenu $cn$ v~na¹í rekurenci, èili zefektivnìním
-spojování podúloh. To ov¹em není pøíli¹ nadìjné (pokud ètenáø nevìøí, mù¾e si to dokázat),
-tak¾e místo toho vyu¾ijeme druhou ¹anci a~to omezení vìtvení ze~ètyø vìtví na~tøi.
-Pøipomeòme si, ¾e potøebujeme spoèítat:
-$$
-XY=AC\cdot 10^{n}+(AD+BC)\cdot 10^{n/2}+BD.
-$$
-Pøitom ale nepotøebujeme znát souèiny $AD$ ani $BC$ samostatnì, nám staèí zjistit hodnotu celého výrazu $AD+BC$. Kdy¾ budeme znát hodnotu výrazù: $AC$, $BD$ a~$(A + B)(C + D)$ (k tomu nám staèí 3 násobení a~2 sèítání), tak mù¾eme výraz $AD + BC$ získat následovnì:
-$$(A + B)(C + D) - AC - BD = AC + AD + BC + BD - AC - BD = AD + BC$$
-Nyní nám ji¾ staèí jen tøi násobení, ale potøebujeme dvì sèítání a~jedno odèítání navíc. Nicménì tato komplikace je zanedbatelná oproti práci u¹etøené men¹ím vìtvením. (Nová dvì sèítání a~jedno odèítání se V~èasové slo¾itosti schová do~$cn$.) Podívejme se opìt na~tabulku:
-$$\vbox{\halign{\hfil#\hfil \quad & \hfil#\hfil \quad &\hfil#\hfil\cr
-hloubka & poèet úloh & velikost podúlohy\cr
-\noalign{\smallskip\hrule\medskip}
-0 & $3^{0}$ & ${n}/{2^{0}}$\cr
-1 & $3^{1}$ & ${n}/{2^{1}}$\cr
-2& $3^{2}$ & ${n}/{2^{2}}$\cr
-3 & $3^{3}$ & ${n}/{2^{3}}$\cr
-\vdots & \vdots & \vdots\cr
-$k$ & $3^{k}$ & ${n}/{2^{k}}$\cr}}$$
-Ná¹ rekurentní vztah po~zbavení se jednoho násobení vypadá následovnì:
-$$T(n) = 3T(n/2)+ cn.$$
-
-Opìt uva¾me, kolik práce spotøebujeme v~souètu pøes v¹echny hladiny (hloubka stromu
-$k$ je opìt $\lceil\log_2 n\rceil$ a~horní celou èást zanedbáme):
-$$\sum_{i=0}^{k}3^{i}\cdot {{n}\over{2^{i}}}=\sum_{i=0}^{k} \left( {{3}\over{2}} \right) ^{i}\cdot n=n\cdot \sum_{i=0}^{k} \left( {{3}\over{2}} \right) ^{i}=n\cdot {{ \left( {{3}\over{2}} \right) ^{k+1}-1}\over{{{3}\over{2}}-1}}=
-$$
-$$
-=n\cdot {{ \left( {3}\over{2} \right) ^{k+1}-1}\over{{{1}\over{2}}}}=2\cdot n\cdot  \left[ \left( {{3}\over{2}} \right) ^{k+1}-1 \right] = \Theta \left( n\cdot  \left( {{3}\over{2}} \right) ^{\log_2{n}} \right) =
-$$
-$$
-=\Theta \left( n\cdot {{3^{\log_2{n}}}\over{2^{\log_2{n}}}} \right)=\Theta \left( n\cdot {{3^{\log_2{n}}}\over{n}} \right)=\Theta \left( 3^{\log_2{n}} \right)=\Theta \left( (2^{\log_2{3}})^{\log_2{n}} \right)=
-$$
-$$
-=\Theta \left( 2^{(\log_2{n}) \cdot \log_2{3}} \right)=\Theta \left( (2^{\log_2{n}})^{\log_2{3}} \right)=\Theta \left( n^{\log_2{3}} \right) =\Theta \left( n^{1.585} \right).
-$$
-Upravený algoritmus má u¾ tedy lep¹í èasovou slo¾itost, konkrétnì $\Theta(n^{1.585})$.
-V~praxi bychom samozøejmì èinitele ne¹tìpili a¾ na~jednociferná èísla,
-ale zastavili se u~nìjaké dostateènì malé délky (øeknìme 50~cifer) a~tam
-pøepnuli na~kvadratický algoritmus, který má men¹í re¾ii.
-
-(Mimochodem, asymptoticky tato slo¾itost není nejlep¹í známá, pro násobení èísel existují efektívnìj¹í algoritmy, které
-dosahují èasové slo¾itosti $\Theta(n \log{n})$, ale jednak mají vysoké multiplikativní konstanty a~druhak jsou v~nich u¾ potøeba trochu
-pokroèilej¹í techniky, jako je diskrétní Fourierova transformace, tak¾e si je necháme na~pøí¹tí semestr.)
-
-\h{Kuchaøková vìta {\it (Master Theorem)}}
-
-Metody øe¹ení rekurentních rovnic z pøedchozích dvou pøíkladù
-by jistì fungovaly i na~jiné algoritmy, ale proè poka¾dé zbyteènì upravovat tolik  výrazù? Radìji si doká¾eme obecnou vìtu, která pùjde pou¾ít na~vìt¹inu
-takovýchto rekurencí. Øíká se jí {\it Master Theorem} nebo také (vzhledem k~tomu,
-jak se pou¾ívá) {\it Kuchaøková vìta}.
-
-\s{Vìta:} \>{\sl (Master Theorem)} 
-
-Pøedpokládejme, ¾e $T(1)=\Theta(1)$ a~$T(n)=a\cdot T(\lceil {{n}\over{b}} \rceil)+\Theta(n^d)$, kde $a \geq 1$, $b>1$, $d \geq 0$ a~$a,b \in \bb N$. Potom $T(n)$ je:
-
-\smallskip
-
-\halign{#&#&#\cr
-\indent & $\Theta(n^d)$ & kdy¾ $a<b^d$,\cr
-& $\Theta(n^d\cdot \log{n})$ & kdy¾ $a=b^d$,\cr
-& $\Theta(n^{\log_b{a}})$ & kdy¾ $a>b^d$.\cr}
-
-\proof Pøedpokládejme, ¾e $n=b^k, k \in \bb{N}$, aby platilo $\lceil
-{{n}\over{b}} \rceil = {{n}\over{b}}$. Pou¾ijeme opìt \uv{dùkaz stromem}.
-Strom rekurzivních volání se v¾dy vìtví na~stejný poèet vìtví, konkrétnì~$a$,
-a~velikosti vstupù klesají $b$-krát. Podívejme se na~tabulku:
-$$\vbox{\halign{\hfil#\hfil \quad & \hfil#\hfil \quad & \hfil#\hfil \quad & \hfil#\hfil \cr
-poèet vrcholù na~hladinì & velikost vstupu & èas ve vrcholu & èas na~hladinì \cr
-\noalign{\medskip\hrule\medskip}
-$1$ & $n$ & $\Theta(n^d)$ & $\Theta(n^d)$\cr
-$a$ & $n/{b^1}$ & $\Theta((n/b^1)^d)$ & ${\Theta(a^1 \cdot ({n/{b^1}})^d)}$\cr
-$a^2$ & $n/{b^2}$ & $\Theta((n/b^2)^d)$ & ${\Theta(a^2 \cdot ({n/{b^2}})^d)}$\cr
-$a^3$ & $n/{b^3}$ & $\Theta((n/b^3)^d)$ & ${\Theta(a^3 \cdot ({n/{b^3}})^d)}$\cr
-\vdots & \vdots  & \vdots & \vdots\cr
-$a^k$ & $n/{b^k}$  & $\Theta((n/b^k)^d)$ & ${\Theta(a^k \cdot ({n/{b^k}})^d)}$\cr}}$$
-
-\noindent
-Celkový èas potøebný na~vyøe¹ení v¹ech dílèích podúloh je následovný:
-$$
-T(n)=\sum_{i=0}^k\Theta \left( a^i \left( {n\over{b^i}} \right) ^d \right)=\sum_{i=0}^k\Theta \left( n^d \left( {a\over{b^d}} \right) ^i \right)=\Theta \left( n^d \cdot \sum_{i=0}^k \left( {a\over{b^d}} \right) ^i \right).
-$$
-V¹imnìme si sumy $\sum_{i=0}^k \left( {a\over{b^d}} \right) ^i$. Jedná se vlastnì o~geometrickou øadu s kvocientem $q={a\over{b^d}}$. Rozli¹me následující pøípady:
-
-\>{\I 1.} $q<1$: Práce na~jednotlivých hladinách exponenciálnì ubývá a~souèet øady (i~kdyby byla nekoneèná) se dá omezit nìjakou konstantou, tedy $T(n)=\Theta(n^d)$.
-
-\>{\I 2.} $q=1$: Práce na~jednotlivých hladinách je stejnì, to znamená, ¾e suma je právì $1 + \log_b n$, a~tedy $T(n) = \Theta(n^d \cdot \log{n})$.
-
-\>{\I 3.} $q>1$: Práce na~jednotlivých hladinách pøibývá, tak¾e musíme geometrickou øadu seèíst poctivì. Víme, ¾e souèet geometrické øady od~$0$ do~$k$ s kvocientem $q$ a~prvním èlenem $1$ je: ${q^{k+1} - 1}\over{q - 1}$, co¾ pøibli¾ne odpovídá $q^k$. Pak tedy platí: $$T(n) = \Theta(n^d \cdot q^{\log_b{n}}).$$ Tento výraz vypadá ponìkud o¹klivì, ale je¹tì ho trochu (alespoò kosmeticky) upravíme:
-$$
-\Theta\left(n^d \cdot q^{\log_b{n}}\right)=\Theta\left({ a^{\log_b{n}} \cdot n^d \over (b^d)^{\log_b{n}}}\right)=\Theta\left({\left(b^{\log_b{a}}\right)^{\log_b{n}} \cdot n^d \over{\left(b^d\right)^{\log_b{n}}}}\right)=
-$$
-$$
-=\Theta\left({\left(b^{\log_b{n}}\right)^{\log_b{a}} \cdot n^d \over{\left(b^{\log_b{n}}\right)^d}}\right)
-=\Theta\left({n^{\log_b{a}} \cdot n^d \over{n^d}}\right)
-=\Theta\left(n^{\log_b{a}}\right).
-$$
-Tyto tøi pøípady pøesnì odpovídají rozdìlení pøípadù v~tvrzení vìty.
-
-\noindent
-Vra»me se nyní k~mo¾nosti, kdy $n$ není mocnina~$b$. Pro obecné $n \geq 1$ zvolme: 
-
-\itemize\ibull
-\:$n^+ \dots$ nejbli¾¹í vy¹¹í mocnina $b$
-\:$n^-$ \dots nejbli¾¹í ni¾¹í mocnina $b$
-\endlist
-
-Platí tedy, ¾e ${n \over b} \leq n^- \leq n \leq n^+ \leq n \cdot b$. Zajisté i platí $T(n^-) \le T(n) \le T(n^+)$. Ale $T(n^-) = \Theta((n^-)^c) = \Theta(n^c)$ a~$T(n^+) = \Theta((n^+)^c) = \Theta(n^c)$. Pak tedy i $T(n) = \Theta(n^c)$. Vìta tedy platí i v~pøípadì, ¾e $n$ není mocnina~$b$.
-\qed
-
-\s{Pøíklad:}
-Porovnejme si nìkteré známé algoritmy a~jejich èasovou slo¾itost pomocí \>{\sl Master Theoremu}:
-$$\vbox{\halign{# \quad  \quad & # \quad  \quad & # \quad  \quad & # \quad  \quad & #\cr
-algoritmus & $a$ & $b$ & $d$ & èasová slo¾itost\cr
-\noalign{\smallskip\hrule\medskip}
-Mergesort & 2 & 2 & 1 & $\Theta({n \cdot \log{n}})$\cr
-Násobení I. & 4 & 2 & 1 & $\Theta(n^2)$\cr
-Násobení II. & 3 & 2 & 1 & $\Theta(n^{\log_2{3}})$\cr
-Binární vyhledávání & 1 & 2 & 0 & $\Theta(\log{n})$\cr}}$$
-
-
-\h{Hledání $k$-tého nejmen¹ího prvku (mediánu)}
-
-V tomto oddílu se budeme zabývat tím, jak co nejrychleji najít V~jakékoli posloupnosti $n$ èísel $k$-tý nejmen¹í prvek, popøípadì medián. Pro ty, kdo medián neznají, tu máme definici:
-
-\s{Definice:}
-{\I Medián} posloupnosti $a_1, a_2,\ldots , a_n$ je takové $a_i$, ¾e nejvý¹e $n/2$ prvkù je men¹ích ne¾ $a_i$ a~nejvý¹e $n/2$ prvkù je vìt¹ích ne¾ $a_i$. Platí tedy, ¾e medián je buï $\lfloor n/2\rfloor$-tý, nebo $\lceil n/2\rceil$-tý nejmen¹í prvek posloupnosti.
-
-Nejjednodu¹¹ím øe¹ením by urèitì bylo celou posloupnost nejdøíve setøídit a~pak u¾ jednodu¹e vybrat po¾adovaný prvek. To bychom dokázali V~celkem slu¹ném èase $\Theta(n\log n)$, ale u¾ teï mù¾eme prozradit, ¾e to jde V~èase $\Theta(n)$. Jak?
-
-Pou¾ijme metodu {\it rozdìl a~panuj}. Nìjakým zpùsobem si zvolíme jeden prvek posloupnosti a~nazveme ho {\it pivot}. Poté rozdìlíme zadanou posloupnost na~tøi disjunktní mno¾iny. do~první dáme v¹echny prvky men¹í ne¾ pivot, do~druhé rovné pivotu jako pivot a~do tøetí vìt¹í ne¾ pivot. Tímto máme zaji¹tìno, ¾e prvky z první mno¾iny jsou urèitì men¹í ne¾ prvky z druhé a~ty ne¾ prvky z tøetí.
-O tom, jak jsou prvky uspoøádány uvnitø tìchto mno¾in, ale nic nevíme.
-
-V posledním kroku na¹eho algoritmu se pak rozhodneme, na~kterou mno¾inu svùj algoritmus rekurzivnì zavoláme. Pokud je $k$ men¹í nebo rovno ne¾ velikost první mno¾iny, pokraèujeme V~první mno¾inì, pokud je $k$ men¹í nebo rovno ne¾ souèet velikostí první a~druhé mno¾iny, pak hledaným prvkem je právì vybraný pivot a~algoritmus skonèí, a~nakonec pokud ani jedna podmínka splnìna nebyla, pustíme se do~hledání ve tøetí mno¾inì, ov¹em u¾ nehledáme $k$-tý nejmen¹í prvek, ale $l$-tý, kde $l$ se rovná $k$ minus velikost prvních dvou mno¾in. Pro vìt¹í názornost zapí¹eme tento algoritmus formálnìji:
-
-\algo
-{\bo Select($k,X$):} (Hledání $k$-tého nejmen¹ího prvku V~mno¾ine $X$)
-\:Jestli¾e $\vert X\vert \le 1$, vyøe¹íme triviálnì.
-\:Zvolíme pivota $p \in X$.
-\:Rozdìlíme mno¾inu X na~tøi podmno¾iny: $L = \{x \in X; x < p\},$ $ S = \{x \in X; x = p\}, P = \{x \in X; x > p\}$.
-\:Jestli¾e $k \le \vert L\vert$, vrátíme výsledek funkce \<Select>($k$, $L$).
-\:Jestli¾e $ \vert L\vert < k \le \vert L\vert + \vert S\vert$, vrátíme pivota $p$.
-\:Jestli¾e $ \vert L\vert + \vert S\vert < k$, vrátíme výsledek funkce \<Select>($k - \vert L\vert - \vert S\vert, P$).
-\endalgo
-Na první pohled je vidìt, ¾e se algoritmus zastaví (vstup se v¾dy zmen¹í alespoò o~1) a~¾e vydá v¾dy správný výsledek. Jak je to ov¹em s èasovou slo¾itostí? Rozdìlení do~mno¾in a~podmínky V~druhém a~tøetím kroku mají lineární slo¾itost, èemu¾ se nevyhneme. Pøi ne¹»astné volbì pivota se nám mù¾e stát, ¾e poèet rekurentních volání mù¾e být a¾ $n$, tedy celková slo¾itost V~nejhor¹ím pøípadì je $\Theta(n^2)$, èím¾ jsme si oproti prostému setøídìní je¹tì pohor¹ili. Co s tím? Jak je vidìt, velmi dùle¾itá je volba pivota. Tu mù¾eme provést nìkolika zpùsoby:
-
-a) Pivot by se V~setøídìné posloupnosti vyskytoval uprostøed, vstup by se tedy stále pùlil. Èasovou slo¾itost vypoèteme z rekurentního zápisu:
-$$ T(n) = T\left({n \over 2}\right) + \Theta(n) = \Theta\left(n + {n \over 2} + {n \over 4} + \dots\right) = \Theta(n). $$
-To by bylo sice skvìlé, ale nalezení takového pivota je vlastnì vyøe¹ení úlohy hledání mediánu, o~co¾ se sna¾íme. Tedy jsme si vùbec nepomohli.
-
-b) Pivot by se V~setøídìné posloupnosti náchazel V~prostøedních dvou ètvrtinách (nazvìme tento prvek \uv{\I l¾imedián}). Tím bychom V~ka¾dém kroku urèitì odstranili mno¾inu velikosti alespoò ètvrtiny vstupu. Èasová slo¾itost tohoto øe¹ení by byla:
-
-$$ T(n) = T\left({3 \over 4}n\right) + \Theta(n) = \Theta\left(n + {3 \over 4}n + {9 \over 16}n + \dots\right) = \Theta(n). $$
-
-Tímto bychom tedy také dosáhli lineární èasové slo¾itosti. Ale jak vybrat pivota tak, aby se nacházel V~prostøedních dvou ètvrtinách a~aby nám nám to nepokazilo lineární slo¾itost?
-
-Zkusme vybrat pivota náhodnì. Pravdìpodobnost, ¾e vytáhneme zrovna l¾imedián, je alespoò $1/2$. (Pokud by se prvky nemohly opakovat a~byl jich sudý poèet, byla by to pøesnì $1/2$.) Tuto pravdìpodobnost si oznaème $p$. 
-
-\s{Volba l¾imediánu}
-\algo
-\:Vybereme rovnomìrnì náhodnì pivota z mno¾iny $X$.
-\:Otestujeme, zda je pivot l¾imedián.
-\:Pokud není $\Rightarrow$ pokraèujeme znovu od~zaèátku, jinak konec.
-\endalgo
-
-Pivota vybíráme rovnomìrnì náhodnì, tedy ka¾dý prvek posloupnosti má stejnou pravdìpodobnost, ¾e bude vybrán. 
-
-Oznaème si $T$ náhodnou velièinu znaèící dobu bìhu algoritmu. Potom støední hodnota této náhodné velièiny ${\bb E}[T] = \Theta(n) \cdot {\bb E}$[{\I poèet~prùchodù~cyklem}].
-
-Teï si doka¾me, ¾e budeme-li náhodnì vybírat pivota tak dlouho, a¾ se strefíme do~l¾imediánu, tak \uv{v prùmìru} budeme muset tahat jen $1/p$-krát.
-
-\s{Lemma: } {\I (O d¾bánu a~vodì)}
-Èekání na~náhodnou událost, která nastává s pravdìpodobností $p$, trvá V~prùmìru $1/p$.
-
-\proof
-Oznaème $N$ poèet pokusù. Potom støední hodnota poètu pokusù je:
-$$\eqalign{
-		{\bb E}[N] & = p \cdot 1 +  (1-p) \cdot (1 + {\bb E}[N])\cr
-		{\bb E}[N] & = p + 1 +  {\bb E}[N] - p -  p \cdot {\bb E}[N]\cr
-    {\bb E}[N] \cdot (1 - 1 + p) & = 1\cr
-    {\bb E}[N] & = 1/p\cr
-}$$
-S pravdìpodobností $p$ událost nastane a~s odvrácenou pravdìpodobností ($1-p$) jsme jeden pokus promarnili a~musíme celý proces opakovat. Jednoduchými úpravami nám vyjde, ¾e støední hodnota poètu pokusù je $1/p$.
-
-\noindent
-\uv{V prùmìru se tedy chodí $1/p$-krát se d¾bánem pro vodu, ne¾ se ucho utrhne...}
-\qed
-
-Z lemmatu tedy plyne, ¾e V~na¹em pøípadì ${\bb E}$[{\I poèet prùchodù cyklem}]$ \le 2 $. V~prùmìru tedy na~druhý pokus vytáhneme l¾imedián. Ten pak pou¾ijeme jako pivot. Tímto tudí¾ dosáhneme prùmìrné èasové slo¾itosti $\Theta(n)$.
-
-\s{Vìta: } Pravdìpodobnostním algoritmem lze najít $k$-tý nejmen¹í prvek z $n$ prvkù V~prùmìrném èase $\Theta(n)$.
-
-\s{Poznámka: } Kdy¾ budeme volit pivota náhodnì a~nebudeme se starat o~to, zda se jedná o~l¾imedián, èi ne, tak dostaneme také prùmìrný èas $\Theta(n)$.
-
-K volbì l¾imediánu jsme volili v¾dy náhodného pivota, jednalo se tedy o~randomizovaný algoritmus. Stejný výsledek nám ale vyjde i pøi deterministickém algoritmu, kdy budeme volit pivota v¾dy stejnì (napø. na~první pozici ve vstupní posloupnosti), ale budeme mít nìjak zaruèeno, ¾e vstupy budou dostateènì náhodné. U randomizovaného algoritmu se jednalo o~prùmìr pøes náhodná èísla, u deterministického algoritmu o~prùmìr pøes náhodné vstupy.
-
-U¾ tedy víme, ¾e kdy¾ budeme volit pivota hodnì ¹patnì, tak se mù¾eme dostat a¾ na~èasovou slo¾itost $\Theta(n^2)$. Kdy¾ budeme volit o~nìco lépe, tak se mù¾eme dosáhnout prùmìrné èasové slo¾itosti $\Theta(n)$. Existuje ale i algoritmus, který pracuje v¾dy (nejen V~prùmìrném pøípadì) V~èase $\Theta(n)$. Podívejme se, jak bude vypadat \dots
-
-\s{Volba pivota:}
-\algo
-\:Rozdìlíme vstup na~pìtice \dots $\Theta(n)$
-\:Spoèteme medián ka¾dé pìtice \dots $\Theta(n)$
-\:Spoèteme medián mediánù pìtic tak, ¾e rekurzivnì zavoláme tentý¾ algoritmus \<Select>($n / 10$, {mediány pìtic}). Tím získáme pivota.
-\endalgo
-
-Pro úplnost se podívejme, jak bude vypadat celý algoritmus na~hledání $k$-tého nejmen¹ího prvku posloupnosti za pou¾ití na¹í nové volby pivota:
-
-\s{Deterministický lineární algoritmus na~výbìr mediánu lineárnì}
-\algo
-\algin $X = x_1 \dots x_n$, $k$ ($1 \le k \le n$)
-\:Pokud $n \le 5 \Rightarrow$ vyøe¹íme hrubou silou.
-\:Vstup rozdìlíme na~pìtice $P_1 \dots P_{\lceil n/5 \rceil}$.
-\:$\forall i: m_i \leftarrow$ medián($P_i$).
-\:$p \leftarrow$ \<Select>($m_1 \dots m{\lceil n/5 \rceil}, \lceil n/10 \rceil$).
-\:Rozdìlíme $X$ na~$L,S,P$
-\:Rekurzivnì se zavoláme na~jednu z $L,S,P$ (tu, ve které se má vyskytovat hledaný prvek)
-\endalgo
-
-Ná¹ deterministický algoritmus tedy roz¹iøuje pùvodní algoritmus tak, ¾e v¾dy, kdy¾ je potøeba zvolit pivota z posloupnosti X, tak si posloupnost rozdìlí na~pìtice a~v ka¾dé spoète medián. Následnì potøebuje medián z tìchto mediánù, co¾ bude výsledný pivot -- medián V~pùvodní posloupnosti. Ten získá tak, ¾e se znovu zavolá na~\<Select> s parametry posloupnost mediánù a~èíslo $\lceil n/10 \rceil$, nebo» potøebuje právì prostøední prvek této poslounosti, která má délku $\lceil n/5 \rceil$. Kdy¾ máme koneènì výhodného pivota nalezeného, tak mù¾eme pokraèovat, ¾e si posloupnost opìt rozdìlíme na~tøi hromádky -- prvky men¹í ($L$), stejné ($S$) a~vìt¹í ($P$) ne¾ pivot. Následnì jen vybereme hromádku, která odpovídá umístìní $k$-tého nejmen¹ího prvku, a~na tu se rekurzivnì zavoláme.
-
-Abychom dokázali, ¾e tento algoritmus bude mít opravdu lineární èasovou slo¾itost, musíme si nejdøíve dokázat následující lemma:
-
-\s{Lemma:} V~ka¾dém kroku vypadne alespoò ${3/10}\cdot n$ prvkù.
-
-\proof 
-V dùkazu budeme trochu ètenáøe ¹idit tím, ¾e budeme pøedpokládat celoèíselné výsledky (napø. ¾e poèet prvkù posloupnosti, na~kterou se zavoláme, je v¾dy dìlitelný pìti). Ètenáø si mù¾e jako cvièení dokázat, ¾e kdyby výsledky nevycházely celoèíselnì, tak to nevadí.
-
-Pøedstavme si vybrané pìtice seøazené podle velikosti od~nejvìt¹ího prvku a~zakresleme je do~sloupcù. Jejich mediány tedy vyplòují prostøední øadu. Tyto pìtice pak seøaïme podle velikosti jejich mediánù (nejmen¹í vlevo)\foot{Algoritmus ov¹em nikde takto pìtice neøadí! Jen nám to pomù¾e v~dùkazu správnosti.}. Hledaný pivot se tedy nachází (pokud pøedpokládáme pro jednoduchost lichý poèet pìtic) pøesnì uprostøed. o~prvcích nad pivotem a~napravo od~nìj mù¾eme urèitì øíct, ¾e jsou vìt¹í nebo rovny pivotu, prvky pod ním a~nalevo od~nìj jsou zase urèitì men¹í nebo rovny pivotu.
-
-Podle konstrukce algoritmu tedy zaruèenì vypadne jedna nebo druhá skupina prvkù. Obì tyto skupiny pøitom obsahují, jak je vidìt z obrázku, alespoò $3/10 \cdot n$ prvkù.
-\qed
-
-\figure{petice.eps}{Pìtice}{125mm}
-
-Nyní u¾ se tedy mù¾eme pustit do~výpoètu èasové slo¾itosti. V~ka¾dém kroku funkce zavolá sama sebe nejdøíve na~vstup velikosti ${n \over 5}$ a~poté na~vstup velikosti nejvý¹e ${7 \over 10}n$. Ostatní operace zùstávají lineární. Èasovou slo¾itost V~nejhor¹ím pøípadì tedy mù¾eme zapsat rekurentním vzorcem:
-
-$$ T(n) = \Theta(n) + T\left({n \over 5}\right) + T\left({7 \over 10}n\right). $$
-
-Tento rekurentní vzorec ale zatím neumíme obecnì øe¹it. Mohli bychom ho postupnì rozepsat, ale trvalo by dlouho, ne¾ bychom z toho nìco vykoukali. Lep¹í bude rovnou dokázat, ¾e tento rekurentní vzorec odpovídá lineární slo¾itosti, zkusíme tedy dosadit $T(n) = cn$:
-
-$$ cn = n + {1 \over 5}cn + {7 \over 10}cn \quad\Longleftrightarrow\quad c = 10. $$
-
-Tímto jsme tedy dokázali, ¾e èasová slo¾itost tohoto algoritmu V~nejhor¹ím pøípadì je $\Theta(n)$.
-Jeliko¾ to rychleji urèitì není mo¾né, mù¾eme na~závìr této kapitoly zformulovat tuto vìtu:
-
-\s{Vìta:} Hledání $k$-tého nejmen¹ího prvku V~posloupnosti délky $n$ má èasovou slo¾itost V~nejhor¹ím pøípadì $\Theta(n)$. \qed
-
-\s{K zamy¹lení:} Proè jsme zvolili zrovna pìtice? Jak by to dopadlo pro trojice? a~jak pro sedmice? Fungoval by takový algoritmus? Byl by také lineární?
-
-\h{Quicksort}
-
-Ji¾ jsme se seznámili s tøídícím algoritmem {\it Mergesort}. Podívejme se teï na~algoritmus, který je také zalo¾en na~metodì {\it rozdìl a~panuj}, ale chová se V~podstatì opaènì. Namísto slévání men¹ích setøídìných posloupností posloupnost v¾dy rozdìlí na~dvì pokud mo¾no podobnì velké èásti $L$ a~$P$, kdy V~$L$ jsou èísla men¹í ne¾ nìjaký pivot a~v $P$ èísla vìt¹í. Pivot tvoøí samostatnou èást $S$. Obì èásti $L$ i $P$ rekurzivnì setøídí a~následnì vrátí posloupnost, kde budou setøídìné prvky z $L$, pak pivot z $S$ a~poté setøídìné prvky z $P$. Vyu¾ívá tedy podobnou my¹lenku jako hledání $k$-tého nejmen¹ího prvku.
-
-\s{Algoritmus:}
-{\bo \<QS>$(x = x_1, \dots,x_n)$:}
-\algo
-\:Pokud $n \leq 1 \Rightarrow$ vrátíme $x$.
-\:Vybereme pivota.
-\:Rozdìlíme posloupnost na~podposloupnosti $L,S,P$.
-\:$L \leftarrow$ \<QS($L$)>, $P \leftarrow$ \<QS($P$)>
-\:Vrátíme $L + S + P$.
-\endalgo
-
-\s{Pozorování:}
-
-Kdy¾ bude pivot v¾dy medián: $T(n) = \Theta(n) + 2T(n/2) \Rightarrow T(n) = \Theta(n \cdot\log{n})$
-
-Kdy¾ bude pivot v¾dy minimum: $T(n) = \Theta(n) + T(n-1) \Rightarrow T(n) = \Theta(n^2)$
-
-\s{Poznámka:} Hledání mediánu lineárnì je sice hezké a~asymptoticky lineární, ale má tak o¹klivé multiplikativní konstanty, ¾e se V~praxi nepou¾ívá. Doka¾me si tedy, ¾e Quicksort s náhodnou volbou pivota má prùmìrnou èasovou slo¾itost také $\Theta(n\cdot\log{n})$.
-
-\s{Vìta:} Quicksort s náhodnou volbou pivota provede ve støední hodnotì $\Theta(n\cdot\log{n})$ porovnání.
-
-\s{Dùsledek:} Quicksort má prùmìrnou èasovou slo¾itost  $\Theta(n\cdot\log{n})$.
-
-\proof
-Nech» $C$ znaèí poèet porovnání, $y_1, \dots, y_n$ je vstup V~setøídìném poøadí a~$C_{ij}$ znaèí poèet porovnání prvkù $y_i$ s $y_j$ pro $1 \leq i < j \leq n$. Uvìdomme si, ¾e dva prvky porovnává algoritmus nejvý¹e jednou, a~to pouze tehdy, kdy¾ je jeden z nich pivot. Proto $C_{ij}$ nabývá hodnot $0$ nebo $1$. $C_{ij}$ je tedy indikátor jevu, ¾e byly prvky $y_i$ a~$y_j$ porovnány. 
-
-Poèet v¹ech porovnání je pak souèet jednotlivých $C_{ij}$, èili $$C = \sum_{i,j} C_{ij}.$$ 
-Proto platí: $${\bb E}[C] = {\bb E}\left[\sum_{i,j} C_{ij}\right].$$ 
-Díky linearitì støední hodnoty mù¾eme tvrdit, ¾e: $${\bb E}[C] = \sum_{i,j} {\bb E}[C_{ij}].$$
-Støední hodnota indikátoru jevu, ¾e prvky  $y_i$ a~$y_j$ byly porovnány, je rovna pravdìpodobnosti, ¾e $C_{ij}$ je 1. $${\bb E}[C_{ij}] = P[C_{ij} = 1].$$
-Aby $C_{ij}$ se rovnalo $1$ pro prvky $y_i$ a~$y_j$, tak se musel pro podposloupnost $y_i, \dots, y_j$ stát pivotem jako první jeden z $y_i$ nebo $y_j$. Kdyby se tak nestalo, tak spolu pøi tomto rozdìlování prvky $y_i$ a~$y_j$ nebudou porovnány a~dostanou se ka¾dý do~jiné skupiny ($L$ a~$P$), tak¾e u¾ spolu porovnány nikdy ani nebudou.
-Pravdìpodobnost, ¾e $y_i$ nebo $y_j$ se staly pivotem pro posloupnost $y_i, \dots, y_j$  je rovna $2 \over {j-i+1}$ (nebo» pivota vybíráme rovnomìrnì náhodnì). Proto mù¾eme koneènì vyjádøit:
-
-$${\bb E}[C] = \sum_{1 \le i < j \le n} {2\over{j-i+1}} = \sum_{d=1}^{n-1} { \sum_{n=1}^{n-d} {2\over{d + 1}} }\le \sum_{d=1}^{n-1} {2n\over{d + 1}} = 2n\sum_{d=2}^n {1\over d} = \Theta(n \cdot\log{n}).$$
-
-\qed
-
-
-\bye
diff --git a/8-rozdel/Makefile b/8-rozdel/Makefile
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-CurrentInkName_5(Black)eq{IMAGE}{dup/DataSource get/TCC xd/Height get abs{TCC
-pop}repeat}ifelse}{IMAGE}ifelse}ifelse}{2 index 1 ne{CurrentInkName_5(Black)eq
-{IMAGE}{/TCC xd pop pop exch pop abs{TCC pop}repeat}ifelse}{IMAGE}ifelse}
-ifelse}ifelse}{IMAGE}ifelse}ifelse}bd}ifelse/WaldoColor_5 true def/$fm 0 def
-/wfill{1 $fm eq{fill}{eofill}ifelse}bd/@Pf{@sv SepMode_5 0 eq $Psc 0 ne or
-$ink_5 3 eq or{0 J 0 j[]0 d $t $c $m $y $k $n $o @scc_5 pop $ctm setmatrix 72
-1000 div dup matrix scale dup concat dup Bburx exch Bbury exch itransform
-ceiling cvi/Bbury xd ceiling cvi/Bburx xd Bbllx exch Bblly exch itransform
-floor cvi/Bblly xd floor cvi/Bbllx xd $Prm aload pop $Psn load exec}{1 SetGry
-wfill}ifelse @rs @np}bd/F{matrix currentmatrix $sdf{$scf $sca $scp @ss}if $fil
-1 eq{CorelPtrnDoFill}{$fil 2 eq{@ff}{$fil 3 eq{@Pf}{$fil 4 eq
-{CorelShfillDoFill}{$t $c $m $y $k $n $o @scc_5{wfill}{@np}ifelse}ifelse}
-ifelse}ifelse}ifelse $sdf{$dsf $dsa $dsp @ss}if setmatrix}bd/f{@cp F}bd/S{
-matrix currentmatrix $ctm setmatrix $SDF{$SCF $SCA $SCP @ss}if $T $C $M $Y $K
-$N $O @scc_5{matrix currentmatrix $ptm concat stroke setmatrix}{@np}ifelse $SDF
-{$dsf $dsa $dsp @ss}if setmatrix}bd/s{@cp S}bd/B{@gs F @gr S}bd/b{@cp B}bd/_E{
-5 array astore exch cvlit xd}bd/@cc{currentfile $dat readhexstring pop}bd/@sm{
-/$ctm $ctm currentmatrix def}bd/@E{/Bbury xd/Bburx xd/Bblly xd/Bbllx xd}bd/@c{
-@cp}bd/@P{/$fil 3 def/$Psn xd/$Psc xd array astore/$Prm xd}bd/tcc{@cc}def/@B{
-@gs S @gr F}bd/@b{@cp @B}bd/@sep{CurrentInkName_5(Composite)eq{/$ink_5 -1 def}
-{CurrentInkName_5(Cyan)eq{/$ink_5 0 def}{CurrentInkName_5(Magenta)eq{/$ink_5 1
-def}{CurrentInkName_5(Yellow)eq{/$ink_5 2 def}{CurrentInkName_5(Black)eq
-{/$ink_5 3 def}{/$ink_5 4 def}ifelse}ifelse}ifelse}ifelse}ifelse}bd/@whi{@gs
--72000 dup m -72000 72000 l 72000 dup l 72000 -72000 l @cp 1 SetGry fill @gr}
-bd/@neg{[{1 exch sub}/exec cvx currenttransfer/exec cvx]cvx settransfer @whi}
-bd/deflevel 0 def/@sax{/deflevel deflevel 1 add def}bd/@eax{/deflevel deflevel
-dup 0 gt{1 sub}if def deflevel 0 gt{/eax load}{eax}ifelse}bd/eax{{exec}forall}
-bd/@rax{deflevel 0 eq{@rs @sv}if}bd systemdict/pdfmark known not{/pdfmark
-/cleartomark ld}if/wclip{1 $fm eq{clip}{eoclip}ifelse}bd
-% Copyright (c)1992-2003 Corel Corporation
-% All rights reserved.     v12 r0.0
-/z{exch findfont exch scalefont setfont}bd/ZB{9 dict dup begin 4 1 roll
-/FontType 3 def/FontMatrix xd/FontBBox xd/Encoding 256 array def 0 1 255{
-Encoding exch/.notdef put}for/CharStrings 256 dict def CharStrings/.notdef{}
-put/Metrics 256 dict def Metrics/.notdef 3 -1 roll put/BuildChar{exch dup
-/$char exch/Encoding get 3 index get def dup/Metrics get $char get aload pop
-setcachedevice begin Encoding exch get CharStrings exch get end exec}def end
-definefont pop}bd/ZBAddChar{findfont begin dup 4 1 roll dup 6 1 roll Encoding 3
-1 roll put CharStrings 3 1 roll put Metrics 3 1 roll put end}bd/Z{findfont dup
-maxlength 2 add dict exch dup{1 index/FID ne{3 index 3 1 roll put}{pop pop}
-ifelse}forall pop dup dup/Encoding get 256 array copy dup/$fe xd/Encoding exch
-put dup/Fontname 3 index put 3 -1 roll dup length 0 ne{0 exch{dup type 0 type
-eq{exch pop}{$fe exch 2 index exch put 1 add}ifelse}forall pop}if dup 256 dict
-dup/$met xd/Metrics exch put dup/FontMatrix get 0 get 1000 mul 1 exch div 3
-index length 256 eq{0 1 255{dup $fe exch get dup/.notdef eq{pop pop}{5 index 3
--1 roll get 2 index mul $met 3 1 roll put}ifelse}for}if pop definefont pop pop
-}bd/CorelIsValidCharpath{pathbbox 3 -1 roll sub abs 0.5 ge 3 1 roll sub abs 0.5
-ge and}bd/@ftx{{currentpoint 3 -1 roll(0)dup 3 -1 roll 0 exch put dup @gs true
-charpath $ctm setmatrix CorelIsValidCharpath{@@txt}if @gr @np stringwidth pop 3
--1 roll add exch m}forall}bd/@ft{matrix currentmatrix exch $sdf{$scf $sca $scp
-@ss}if $fil 1 eq{/@@txt/@pf ld @ftx}{$fil 2 eq{/@@txt/@ff ld @ftx}{$fil 3 eq
-{/@@txt/@Pf ld @ftx}{$fil 4 eq{/@@txt/CorelShfillDoFill ld @ftx}{$t $c $m $y $k
-$n $o @scc_5{show}{pop}ifelse}ifelse}ifelse}ifelse}ifelse $sdf{$dsf $dsa $dsp
-@ss}if setmatrix}bd/@st{matrix currentmatrix exch $SDF{$SCF $SCA $SCP @ss}if $T
-$C $M $Y $K $N $O @scc_5{{currentpoint 3 -1 roll(0)dup 3 -1 roll 0 exch put dup
-@gs true charpath $ctm setmatrix $ptm concat stroke @gr @np stringwidth pop 3
--1 roll add exch m}forall}{pop}ifelse $SDF{$dsf $dsa $dsp @ss}if setmatrix}bd
-/@te{@ft}bd/@tr{@st}bd/@ta{dup @gs @ft @gr @st}bd/@t@a{dup @gs @st @gr @ft}bd
-/@tm{@sm concat}bd/e{/t{@te}def}bd/r{/t{@tr}def}bd/o{/t{pop}def}bd/a{/t{@ta}
-def}bd/@a{/t{@t@a}def}bd/t{@te}def/T{@np $ctm setmatrix/$ttm matrix def}bd/ddt
-{t}def/@t{/$stm $stm currentmatrix def 3 1 roll m $ttm concat ddt $stm
-setmatrix}bd/@n{/$ttm exch matrix rotate def}bd/@s{}bd/@l{}bd/_lineorientation
-0 def/_bitfont null def/_bitlobyte 0 def/_bitkey null def/_bithibyte 0 def
-% Copyright (c)1992-2003 Corel Corporation
-% All rights reserved.     v12 r0.0
-/@ii{concat 3 index 3 index m 3 index 1 index l 2 copy l 1 index 3 index l 3
-index 3 index l clip pop pop pop pop}bd/@i{@sm @gs @ii 6 index 1 ne{/$frg true
-def pop pop}{1 eq{s1t s1c s1m s1y s1k s1n $O @scc_5/$frg xd}{/$frg false def}
-ifelse 1 eq{@gs $ctm setmatrix F @gr}if}ifelse @np/$ury xd/$urx xd/$lly xd
-/$llx xd/$bts xd/$hei xd/$wid xd/$dat $wid $bts mul 8 div ceiling cvi string
-def $bkg $frg or{$SDF{$SCF $SCA $SCP @ss}if $llx $lly Tl $urx $llx sub $ury
-$lly sub scale $bkg{$t $c $m $y $k $n $o @scc_5 pop}if $wid $hei abs $bts 1 eq
-{$bkg}{$bts}ifelse[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse]/tcc load $bts
-1 eq{imagemask}{image}ifelse $SDF{$dsf $dsa $dsp @ss}if}{$hei abs{tcc pop}
-repeat}ifelse @gr $ctm setmatrix}bd/@I{@sm @gs @ii @np/$ury xd/$urx xd/$lly xd
-/$llx xd/$ncl xd/$bts xd/$hei xd/$wid xd $ngx $llx $lly Tl $urx $llx sub $ury
-$lly sub scale $wid $hei abs $bts[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse
-]$msimage false eq $ncl 1 eq or{/$dat $wid $bts mul $ncl mul 8 div ceiling cvi
-string def/@cc load false $ncl ColorImage}{$wid $bts mul 8 div ceiling cvi $ncl
-3 eq{dup dup/$dat1 exch string def/$dat2 exch string def/$dat3 exch string def
-/@cc1 load/@cc2 load/@cc3 load}{dup dup dup/$dat1 exch string def/$dat2 exch
-string def/$dat3 exch string def/$dat4 exch string def/@cc1 load/@cc2 load
-/@cc3 load/@cc4 load}ifelse true $ncl ColorImage}ifelse $SDF{$dsf $dsa $dsp
-@ss}if @gr $ctm setmatrix}bd/@cc1{currentfile $dat1 readhexstring pop}bd/@cc2{
-currentfile $dat2 readhexstring pop}bd/@cc3{currentfile $dat3 readhexstring pop
-}bd/@cc4{currentfile $dat4 readhexstring pop}bd/$msimage false def/COMP 0 def
-/MaskedImage false def L2?{/@I_2{@sm @gs @ii @np/$ury xd/$urx xd/$lly xd/$llx
-xd/$ncl xd/$bts xd/$hei xd/$wid xd/$dat $wid $bts mul $ncl mul 8 div ceiling
-cvi string def $ngx $ncl 1 eq{/DeviceGray}{$ncl 3 eq{/DeviceRGB}{/DeviceCMYK}
-ifelse}ifelse setcolorspace $llx $lly Tl $urx $llx sub $ury $lly sub scale 8
-dict begin/ImageType 1 def/Width $wid def/Height $hei abs def/BitsPerComponent
-$bts def/Decode $ncl 1 eq{[0 1]}{$ncl 3 eq{[0 1 0 1 0 1]}{[0 1 0 1 0 1 0 1]}
-ifelse}ifelse def/ImageMatrix[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse]def
-/DataSource currentfile/ASCII85Decode filter COMP 1 eq{/DCTDecode filter}{COMP
-2 eq{/RunLengthDecode filter}if}ifelse def currentdict end image $SDF{$dsf $dsa
-$dsp @ss}if @gr $ctm setmatrix}bd}{/@I_2{}bd}ifelse/@I_3{@sm @gs @ii @np/$ury
-xd/$urx xd/$lly xd/$llx xd/$ncl xd/$bts xd/$hei xd/$wid xd/$dat $wid $bts mul
-$ncl mul 8 div ceiling cvi string def $ngx $ncl 1 eq{/DeviceGray}{$ncl 3 eq
-{/DeviceRGB}{/DeviceCMYK}ifelse}ifelse setcolorspace $llx $lly Tl $urx $llx sub
-$ury $lly sub scale/ImageDataDict 8 dict def ImageDataDict begin/ImageType 1
-def/Width $wid def/Height $hei abs def/BitsPerComponent $bts def/Decode $ncl 1
-eq{[0 1]}{$ncl 3 eq{[0 1 0 1 0 1]}{[0 1 0 1 0 1 0 1]}ifelse}ifelse def
-/ImageMatrix[$wid 0 0 $hei neg 0 $hei 0 gt{$hei}{0}ifelse]def/DataSource
-currentfile/ASCII85Decode filter COMP 1 eq{/DCTDecode filter}{COMP 2 eq{
-/RunLengthDecode filter}if}ifelse def end/MaskedImageDict 7 dict def
-MaskedImageDict begin/ImageType 3 def/InterleaveType 3 def/MaskDict
-ImageMaskDict def/DataDict ImageDataDict def end MaskedImageDict image $SDF
-{$dsf $dsa $dsp @ss}if @gr $ctm setmatrix}bd/@SetMask{/$mbts xd/$mhei xd/$mwid
-xd/ImageMaskDict 8 dict def ImageMaskDict begin/ImageType 1 def/Width $mwid def
-/Height $mhei abs def/BitsPerComponent $mbts def/DataSource maskstream def
-/ImageMatrix[$mwid 0 0 $mhei neg 0 $mhei 0 gt{$mhei}{0}ifelse]def/Decode[1 0]
-def end}bd/@daq{dup type/arraytype eq{{}forall}if}bd/@BMP{/@cc xd UseLevel 3 eq
-MaskedImage true eq and{7 -2 roll pop pop @I_3}{12 index 1 gt UseLevel 2 eq
-UseLevel 3 eq or and{7 -2 roll pop pop @I_2}{11 index 1 eq{12 -1 roll pop @i}{
-7 -2 roll pop pop @I}ifelse}ifelse}ifelse}bd
-end
-%%EndResource
-%%EndProlog
-%%BeginSetup
-wCorel12Dict begin
-@BeginSysCorelDict
-2.6131 setmiterlimit
-1.00 setflat
-/$fst 128 def
-%%BeginResource: font TimesNewRomanPSMT-NormalItalic
-%!FontType1-1.0: TimesNewRomanPSMT-NormalItalic 001.003
%%Creator: Corel PostScript Engine
10 dict begin
/FontName /TimesNewRomanPSMT-NormalItalic def
/PaintType 0 def
/FontType 1 def
/FontMatrix [0.001 0 0 0.001 0 0] readonly def
/Encoding 256 array 0 1 255 {1 index exch /.notdef put} for
dup 77 /M put
dup 101 /e put
dup 100 /d put
dup 105 /i put
dup 225 /aacute put
dup 110 /n put
dup 121 /y put
dup 112 /p put
dup 116 /t put
dup 99 /c put
dup 72 /H put
dup 108 /l put
dup 97 /a put
dup 253 /yacute put
dup 118 /v put
dup 111 /o put
readonly def
/FontBBox {0 0 0 0} readonly def
currentdict end
currentfile eexec

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