From f218156e6dbe86b7d1cb0d9ef3796274c84bd70d Mon Sep 17 00:00:00 2001
From: Martin Mares <mj@ucw.cz>
Date: Wed, 29 Apr 2009 14:14:50 +0200
Subject: [PATCH] Nove zapisky

---
 3-dfs/3-dfs.tex        |  189 ++++++
 3-dfs/Makefile         |    3 +
 3-dfs/img1_stvorec.eps |  121 ++++
 3-dfs/img2_dfs.eps     |  211 +++++++
 3-dfs/img3_dfs.eps     |  264 +++++++++
 3-dfs/img4_susedia.eps |  245 ++++++++
 3-dfs/img5_dfso.eps    |  259 ++++++++
 3-dfs/img7_hrany.eps   |  179 ++++++
 3-dfs/praseci-graf.eps |  Bin 0 -> 40201 bytes
 8-rozdel/8-rozdel.tex  |  324 ++++++++++
 8-rozdel/Makefile      |    3 +
 8-rozdel/figure.eps    | 1189 +++++++++++++++++++++++++++++++++++++
 8-rozdel/petice.eps    | 1283 ++++++++++++++++++++++++++++++++++++++++
 13 files changed, 4270 insertions(+)
 create mode 100644 3-dfs/3-dfs.tex
 create mode 100644 3-dfs/Makefile
 create mode 100644 3-dfs/img1_stvorec.eps
 create mode 100644 3-dfs/img2_dfs.eps
 create mode 100644 3-dfs/img3_dfs.eps
 create mode 100644 3-dfs/img4_susedia.eps
 create mode 100644 3-dfs/img5_dfso.eps
 create mode 100644 3-dfs/img7_hrany.eps
 create mode 100644 3-dfs/praseci-graf.eps
 create mode 100644 8-rozdel/8-rozdel.tex
 create mode 100644 8-rozdel/Makefile
 create mode 100644 8-rozdel/figure.eps
 create mode 100644 8-rozdel/petice.eps

diff --git a/3-dfs/3-dfs.tex b/3-dfs/3-dfs.tex
new file mode 100644
index 0000000..c25d20b
--- /dev/null
+++ b/3-dfs/3-dfs.tex
@@ -0,0 +1,189 @@
+\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...
+Díky tomuto zpùsobu procházení se nìkdy té¾ nazýva {\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{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
+
+\s{Popis algoritmu:}
+Ná¹ algoritmus na zaèátku do fronty $Q$ vlo¾í poèáteèní vrchol $v_0$. Dále si v poli $Z$ budeme pro ka¾dý vrchol pamatovat znaèku, zda ho je¹tì máme nav¹tívit, 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 ($Z[w]=0$), 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.
+
+\>{\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 - indukcí dle doby bìhu algoritmu:
+
+$v_0 \rightarrow v_0$ ... triviální
+
+Oznaèujeme $v$ pøes hranu $uv \Rightarrow u$ oznaèené $\Rightarrow$ (IP) $\exists$ cesta z $v_0$ do $u \Rightarrow$
+
+$\Rightarrow \exists$ cesta z $v_0$ do $v$
+
+\noindent
+\uv{$\Longleftarrow$} Sporem: Nech» existuje neoznaèený vrchol $v$, dosa¾itelný po nìjaké cestì z $v_0$. Uva¾me nejkrat¹í cestu $(v_0, v)$: $v_0, \dots, u, v$. Pøedchozí vrchol na této cestì - $u$ - musí být oznaèený. Vrchol $u$ se dostane do fronty, pak je z ní vybrán a tím se zpracuje i vrchol $v$ $\Rightarrow$ SPOR \qed
+
+\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{img1_stvorec.eps}{}{\epsfxsize}
+
+\s{1. matice sousednosti}
+
+Matice sousednosti je pole $A$ o velikosti $n \times n$, jeho¾ prvky na
+souøadnicích $i, j$ jsou dány následujícím pøedpisem:
+
+$$ A_{i,j} = \left\{ \matrix {1 \Leftrightarrow \{i,j\} \in E  \cr
+				0 \Leftrightarrow \{i,j\} \notin E \cr            			
+				}
+\right.$$
+
+Ná¹ graf z obrázku vý¹e by tedy v maticové reprezentaci vypadal takto:
+
+$$\bordermatrix{
+  & A & B & C & D\cr
+A & 0 & 1 & 1 & 0\cr
+B & 1 & 0 & 1 & 1\cr
+C & 1 & 1 & 0 & 1\cr
+D & 0 & 1 & 1 & 0\cr
+}$$
+
+S touto maticí se pracuje velmi snadno (napø. v¹echny sousedy i-tého vrcholu
+zjistíme jednodu¹e tak, ¾e projdeme i-tý øádek matice, co¾ pøedstavuje èasovou
+slo¾itost $O(n)$), ale má i jednu zøejmou nevýhodu: její velikost je v¾dy
+kvadratická bez ohledu na to, jak "øídký" je graf. U grafu s mnoha vrcholy, ale
+s malým poètem hran, tedy budeme zbyteènì plýtvat místem v pamìti (napø.: strom má $n-1$ hran; rovinný graf nejvý¹e $3n-6$ hran...).
+
+\noindent
+BFS bì¾í v èase: $\Theta(n^2)$
+
+\s{2. seznam sousedù}
+
+V pamìti poèítaèe mù¾eme seznam sousedù uchovávat dvìma poli: polem vrcholù
+$V$ grafu, jeho¾ prvky postupnì pro ka¾dý vrchol udávají index na zaèátek
+odpovídajícího úseku v poli $E$, ve kterém by byli ulo¾eni jeho sousedé.
+\figure{img4_susedia.eps}{Znázornìní polí seznamu sousedù.}{\epsfxsize}
+
+Pøípadnì si mù¾eme vystaèit s jedním polem indexovaným vrcholy, pøièem¾
+v ka¾dém prvku pole bude spojový seznam $L(v)={w: vw \in E(G)}$.
+
+Na tuto reprezentaci u¾ staèí prostor $O(n + m)$, co¾ u¾ je, na rozdíl od
+pøedchozího kvadratického prostoru, docela pøíjemné.
+
+\noindent
+BFS bì¾í v èase: $$\Theta(n+\sum_{v\in V(G)} deg(v)) = \Theta(n+m)$$
+
+\s{3. orákulum}
+
+Dal¹í mo¾ností je pak jakési orákulum, které nám øekne (spoèítá), kam hrany z daného vrcholu vlastnì vedou...
+
+\>
+
+\s{Roz¹íøení algoritmu:}
+\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 vrcholy $u$ z $Q$
+\:: $\forall w: \{u,w\} \in E$:
+\::: 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
+
+\noindent
+Pøidali jsme pole D, ve kterém bude ulo¾ena vzdálenost od poèáteèního vrcholu.
+Dále je pøidáno pole P, které je indexováno vrcholy a které si pamatuje pøedchùdce vrcholu.
+
+\>
+
+\s{Definice:} {\I Fáze bìhu algoritmu}
+\>$F_0$...zpracovává vrchol $v_0$
+
+\>$F_{i+1}$...zpracovává vrcholy ulo¾ené do fronty $Q$ bìhem fáze $F_i$
+
+
+\s{Lemma:} Na konci BFS $\forall v \in V(G)$ dosa¾itelný z $v_0$ platí: $v \in F_i \Leftrightarrow d(v_0,v) = i$
+
+\proof
+\uv{$\Longrightarrow$}: indukcí podle $i$
+
+\> Pro $F_0$ triviální pravda $d(v_0, v_0) = 0$
+
+\> Pro $i>0$: Ve fázi $F_{i-1}$ musí být vrchol $v$ takový, ¾e $d(v_0,v)=i$, musí být oznaèen
+a zároveò nemohl být oznaèen døíve a tudí¾ patøí do $F_i$.
+
+\> \uv{$\Longleftarrow$}: Ka¾dý vrchol padne do nìjaké fáze (viz. minulé lemma)
+\qed
+\>
+
+\>{\I Pozorování:} $v_0v_1,...,v_{k-1}$ je nejkrat¹í cesta z $v_0$ do $v_{k-1}$
+
+\> rekonstrukce nejkrat¹í cesty: P[v], P[P[v]], P[P[P[v]]], ...
+
+\>{\I Pozorování:} BFS u neorientovaného grafu projde celou komponentu souvislosti.
+
+\>{\I Pozorování:} Pokud BFS postupnì spou¹tíme na dosud neobarvené vrcholy v neorientovaném grafu, nalezneme nakonec v èase $O(n+m)$ v¹echny komponenty souvislosti.
+
+\>
+
+\s{Vìta:} $BFS(v_0)$ v èase $\Theta(m+n)$ spoète:
+\itemize\ibull
+\:vrcholy dosa¾itelné z $v_0$
+\:vzdálenosti tìchto vrcholù od $v_0$
+\:strom nejkrat¹ích cest z $v_0$
+\endlist
+
+\>
+
+\h{Prohledávání do hloubky (DFS) {\I Depth First Search} }
+
+\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 vrcholy dosa¾itelné z $v_0$.
+
+\figure{img5_dfso.eps}{Graf a znázornìní prùbìhu DFS s jednotlivými hranami:}{\epsfxsize}
+
+\s{Typy hran ($v \rightarrow w$):}
+
+\itemize\ibull
+\:Stromové hrany ... po nich DFS pro¹lo $\{(A \rightarrow B), (B \rightarrow C), (B \rightarrow D)\}$
+\:Zpìtné hrany $<<>_v>_w$... vedou do pøedchùdce $v$ ve stromu $\{(C \rightarrow A)\}$
+\:Dopøedné hrany $<<>_w>_v$... vedou do potomka $v$ $\{(A \rightarrow D)\}$
+\:Pøíèné hrany $<>_w<>_v$... vedou do vrcholu $v$ v sousedním podstromì, v¾dy zprava doleva $\{(D \rightarrow A)\}$
+\endlist
+
+\>$<>_v = <in(v), out(v)>$
+
+\>$<>_w = <in(w), out(w)>$
+
+\>
+
+\>{\I Pozorování:} Hrany, po kterých DFS pro¹lo, tvoøí strom.
+
+\>{\I Pozorování:} Intervaly (in[v], out[v]) $\forall v \in V(G) $ tvoøí dobré uzávorkování. (intervaly synù disjunktnì vyplòují otce $\Rightarrow$ intervaly se nemohou køí¾it).
+
+\bye
diff --git a/3-dfs/Makefile b/3-dfs/Makefile
new file mode 100644
index 0000000..62ff1bc
--- /dev/null
+++ b/3-dfs/Makefile
@@ -0,0 +1,3 @@
+P=3-dfs
+
+include ../Makerules
diff --git a/3-dfs/img1_stvorec.eps b/3-dfs/img1_stvorec.eps
new file mode 100644
index 0000000..13e4481
--- /dev/null
+++ b/3-dfs/img1_stvorec.eps
@@ -0,0 +1,121 @@
+%!PS-Adobe-2.0 EPSF-2.0
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diff --git a/8-rozdel/8-rozdel.tex b/8-rozdel/8-rozdel.tex
new file mode 100644
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+++ b/8-rozdel/8-rozdel.tex
@@ -0,0 +1,324 @@
+\input ../lecnotes.tex
+
+\prednaska{8}{Rozdìl a panuj}{(zapsali)}
+
+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, tak není nalezen ¾ádný antický zdroj výroku a 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 podproblémky, které se øe¹í snáze.
+
+Jak tedy algoritmus \uv {rozdìl a panuj} pracuje? Mìjme nìjaký problém, který má tu vlastnost, ¾e kdy¾ jej rozdìlíme na nìjaké 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) stejné 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 $k < a+b$:
+\::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.
+
+\h{Èasová slo¾itost MergeSort:}
+
+Rozdìlování a slévání nám trvá lineárnì dlouho, tak¾e pro èasovou slo¾itost MergeSortu platí tato rekurentní rovnice: $$T(n)= 2 \cdot T(n/2) + c \cdot n.$$ 
+Pøièem¾ $T(n)$ je èas strávený na vstup délky $n$ a $c$ je nìjaká vhodná konstanta. Pro jednoduchost BÚNO pøedpolkládejme, ¾e $n$ je mocnina dvojky. Zároveò víme, ¾e $T(1)=1$. Mù¾eme si tedy rekurentní vztah rozepsat:
+
+$$\eqalign{
+T(n) &= 2 \cdot (2T(n/4) + c \cdot n/2) + c \cdot n = 4T(n/4) + 2cn = \cr
+		 &= 4(2T(n/8) + c(n/4)) + 2cn = \cr
+		 &= 8T(n/8) + 3cn = \dots\cr
+		 &= 2^k T(n/2^k) + kcn
+}$$
+
+Pokud zvolíme $k = \log_2{n}$, tak 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 $\O(\log n)$. Vyjde nám tedy, ¾e $T(n)=\O(n\log n)$.\foot{Po pozornìj¹ím zamy¹lení si ètenáø mù¾e uvìdomit, ¾e se jedná vlastnì pouze o jiný pohled na stejný dùkaz jako rozepisování rekurentního vzorce.}
+
+\h{Pamì»ová slo¾itost MergeSort:}
+
+$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) $
+
+\s{Závìr:}
+Mergesort bì¾ v èase $\Theta(n \log{n})$ a pamìti $\Theta(n)$. Velmi se hodí na tøídìní lineárních spojových seznamù.
+
+
+\h{Pøiklad 2 -- Násobení èísel:} 
+Pokud násobíme dvì èísla $X$ a $Y$ (obì délky $n$, tedy pokud bylo jedno krat¹í, tak ho prodlou¾íme tak, aby byla stejnì dlouhá) zpùsobem, který nás uèili na základní ¹kole, dostaneme se na èasovou slo¾itost $\Theta(n^2)$. Proto¾e se jedná o~dost èastou operaci, zamysleme se, zda by ne¹la zrychlit. Nasmìrujme na¹e úvahy na postup \uv{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. Uva¾me,
+jakou bude mít tento algoritmus èasovou slo¾itost:
+$$T(n) = 4T(n/2)+ cn.$$
+Toto platí pro nìjakou vhodnou konstantu $c$. A zároveò pro násobení jednociferných èísel platí: $$T(1)=1.$$
+Jak takovou rekurenci vyøe¹íme? Máme opìt dvì mo¾nosti:
+
+\>{\sl 1. zpùsob: Øe¹ení pomocí rozepsání 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 vyvíjí 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 máme $4^i$ 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 trávíme 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, 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 buïto 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, nebo» nám staèí
+zjistit celý èlen $AD+BC$. Kdybychom poèítali $AC$, $BD$
+a potom $(A+B)(C+D)=AC+AD+BC+BD$, tak odèítáním $(AC+BD)$ od $AC+AD+BC+BD$ dostaneme
+hledaný prostøední èlen $AD+BC$. Nyní nám ji¾ staèí jen tøi
+násobení, ale potøebujeme tøi sèítání a jedno odèítání navíc.
+Uká¾eme, ¾e tato komplikace je zanedbatelná oproti práci u¹etøené
+men¹ím vìtvením. 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 bude tedy vypadat:
+$$T(n) = 3T(n/2)+ cn.$$
+
+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):
+$$\sum_{i=0}^{k}3^{i}\cdot {{n}\over{2^{i}}}=\sum_{i=0}^{k} \left( {{3}\over{2}} \right) ^{i}\cdot n=n\cdot \sum_{i=0}^{k} \left( {{3}\over{2}} \right) ^{i}=n\cdot {{ \left( {{3}\over{2}} \right) ^{k+1}-1}\over{{{3}\over{2}}-1}}=
+$$
+$$
+=n\cdot {{ \left( {3}\over{2} \right) ^{k+1}-1}\over{{{1}\over{2}}}}=2\cdot n\cdot  \left[ \left( {{3}\over{2}} \right) ^{k+1}-1 \right] = \O \left( n\cdot  \left( {{3}\over{2}} \right) ^{\log_2{n}} \right) =
+$$
+$$
+=\O \left( n\cdot {{3^{\log_2{n}}}\over{2^{\log_2{n}}}} \right)=\O \left( n\cdot {{3^{\log_2{n}}}\over{n}} \right)=\O \left( 3^{\log_2{n}} \right)=\O \left( (2^{\log_2{3}})^{\log_2{n}} \right)=
+$$
+$$
+=\O \left( 2^{(\log_2{n}) \cdot \log_2{3}} \right)=\O \left( (2^{\log_2{n}})^{\log_2{3}} \right)=\O \left( n^{\log_2{3}} \right) =\O \left( n^{1.585} \right).
+$$
+Upravený algoritmus u¾ tedy má lep¹í èasovou slo¾itost, konkrétnì $\O(n^{1.585})$.
+V~praxi bychom samozøejmì pro 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 zrovna nejlep¹í, pro násobení èísel existují efektívnìj¹í algoritmy, které
+dosahují èasové slo¾itosti $\O(n \log{n})$, ale jednak mají vysoké multiplikativní konstanty a druhak jsou na~to u¾ potøeba trochu
+pokroèilej¹í techniky, jako je diskrétní Fourierova transformace, tak¾e
+si jej necháme na~pøí¹tí semestr.)
+
+
+
+
+
+
+
+
+
+\h{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 Master Theorem nebo také (vzhledem k~tomu,
+jak se pou¾ívá) Kuchaøková vìta.
+
+\s{Vìta:} \>{\sl (Master Theorem)} 
+
+Pøedpokládejme, ¾e $T(1)=\O(1)$ 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
+Celkem je tedy èas potøebný na vyøe¹ení v¹ech dílèích podúloh na v¹ech hladinách:
+$$
+T(n)=\sum_{i=0}^k\Theta \left( a^i \left( {n\over{b^i}} \right) ^d \right)=\sum_{i=0}^k\Theta \left( n^d \left( {a\over{b^d}} \right) ^i \right)=\Theta \left( n^d \sum_{i=0}^k \left( {a\over{b^d}} \right) ^i \right)
+$$
+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 sumy (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 souèet sumy je právì $\log_b n (+1)$, a tedy $T(n) = \Theta(n^d \cdot \log_b(n))$.
+
+\>{\I 3.} $q>1$: Práce na jednotlivých hladinách pøibývá, tak¾e musíme geometrickou øadu seèíst poctivì: $T(n) = \Theta(n^d \cdot q^{\log_b{n}})$. Tento výraz vypadá ponìkud o¹klivì,
+ale je¹tì ho trochu (alespoò kosmeticky) upravíme:
+$$
+\Theta\left(n^d \cdot \left(q\right)^{\log_b{n}}\right)=\Theta\left({ a^{\log_b{n}} \cdot n^d \over (b^d)^{\log_b{n}}}\right)=\Theta\left({\left(b^{\log_b{a}}\right)^{\log_b{n}} \cdot n^d \over{\left(b^d\right)^{\log_b{n}}}}\right)=
+$$
+$$
+=\Theta\left({\left(b^{\log_b{n}}\right)^{\log_b{a}} \cdot n^d \over{\left(b^{\log_b{n}}\right)^d}}\right)
+=\Theta\left({n^{\log_b{a}} \cdot n^d \over{n^d}}\right)
+=\Theta\left(n^{\log_b{a}}\right).
+$$
+Tyto tøi pøípady pøesnì odpovídají rozdìlení pøípadu v~tvrzení vìty.
+
+Vra»me se nyní k~mo¾nosti, kdy $n$ není mocnina~$b$.
+Tehdy platí $b^l<n<b^{l+1}$ pro nìjaké $l \in \bb{N}$. V tomto pøípadì
+zaokrouhleme $n$ a~poèítejme s $n^{\prime}=b^{l+1} < bn$. Jeliko¾ funkce
+$T(n)$ je neklesající, dostaneme $T(n)\le T(n')$ a ve~v¹ech tøech pøípadech
+je výsledek pro~$n'$ maximálnì konstanta-krát vìt¹í ne¾ pro~$n$, co¾ se
+\uv{schová do~$\Theta$.} Vìta tedy platí i v~tomto pøípadì.
+\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:}
+Medián posloupnosti $a_1, a_2,\ldots , a_n$ je takové $m=a_i$, kde nejvý¹e $n/2$ prvkù je men¹ích ne¾ $m$ a nejvý¹e $n/2$ prvkù je vìt¹ích ne¾ $m$.
+
+\s{Poznámka:} Prvky v posloupnosti se mohou i opakovat, my potom jako $k$-tý nejmen¹í prvek myslíme prvek s $k$-tou nejmen¹í hodnotou.
+
+Nejjednodu¹¹ím øe¹ením by urèitì bylo celou posloupnost nejdøíve setøídit a pak u¾ jednodu¹e vybrat po¾adovaný prvek. To bychom dokázali v celkem slu¹ném èase $\O(n\log n)$, ale u¾ teï mù¾eme prozradit, ¾e to jde v èase $\O(n)$. Jak?
+
+Pou¾ijme metodu {\it rozdìl a panuj}. Nìjakým zpùsobem si zvolíme jeden prvek posloupnosti, který nazveme {\it pivot}. Poté rozdìlíme zadanou posloupnost na~tøi disjunktní mno¾iny. Do první dáme v¹echny prvky men¹í ne¾ pivot, do druhé stejné 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 ne¾ velikost první mno¾iny, pokraèujeme v první mno¾inì, pokud je $k$ men¹í 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 rekurencí 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.
+
+\break
+
+b) Pivot by se v setøídìné posloupnosti náchazel v prostøedních dvou ètvrtinách (nazvìme tento prvek \uv{l¾imedián}). Tím bychom v ka¾dém kroku urèitì odstranili mno¾inu velikosti 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, aby nám nám to nepokazilo lineární slo¾itost?
+
+Zkusme vybrat pivota náhodnì. Pravdìpodobnost, ¾e vytáhneme zrovna l¾imedián je $\ge 1/2$. (Pokud by se prvky nemohly opakovat, byla by to pøesnì $1/2$.) Tuto pravdìpodobnost si oznaème p. Teï si doka¾me, ¾e budeme-li náhodnì vybírat pivota tak dlouho, a¾ se ztrefíme do l¾imediánu, tak \uv{v prùmìru} budeme muset tahat jen $1/p$-krát.
+
+\s{Volba l¾imediánu}
+\algo
+\:Vybereme rovnomìrnì náhodnì pivota z mno¾iny $X$.
+\:Otestujeme, zda je pivot l¾imedán.
+\:Pokud není $\Rightarrow GOTO$ 1, jinak konec.
+\endalgo
+
+Oznaème si $T$ jako náhodnou velièinu znaèící dobu bìhu algoritmu. Potom støední hodnota této náhodné velièiny $E[T] = \Theta(n) \cdot E$[{\I poèet prùchodù cyklem}].
+
+\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 
+$$\eqalign{
+		&E[N] = 1 + p \cdot 0 +  (1-p) \cdot E[N]\cr
+    &E[N] \cdot (1 - 1 + p) = 1\cr
+    &E[N] = 1/p\cr
+}$$
+\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 $E[T] \le 2 $. V prùmìru tedy na druhý pokus vytáhneme l¾imedián, který pou¾ijeme jako pivot. Tímto tudí¾ dosáhneme prùmìrnì po¾adované èasové slo¾itosti $\Theta(n)$.
+
+\s{Vìta: } K-tý nejmen¹í z $n$ prvkù lze najít pravdìpodobnostním algoritmem v prùmìrném èase $\Theta(n)$.
+
+\bye
diff --git a/8-rozdel/Makefile b/8-rozdel/Makefile
new file mode 100644
index 0000000..4d9a469
--- /dev/null
+++ b/8-rozdel/Makefile
@@ -0,0 +1,3 @@
+P=8-rozdel
+
+include ../Makerules
diff --git a/8-rozdel/figure.eps b/8-rozdel/figure.eps
new file mode 100644
index 0000000..9ba02c2
--- /dev/null
+++ b/8-rozdel/figure.eps
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+get def/SetHsb systemdict/sethsbcolor get def/GetHsb systemdict
+/currenthsbcolor get def/rgb2hsb{SetRgb2 GetHsb}bd/hsb2rgb{3 -1 roll dup floor
+sub 3 1 roll SetHsb GetRgb2}bd/setcmykcolor where{pop/LumSepsDict where{pop
+/SetCmyk_5{LumSepsDict/setcmykcolor get exec}def}{/AldusSepsDict where{pop
+/SetCmyk_5{AldusSepsDict/setcmykcolor get exec}def}{/SetCmyk_5/setcmykcolor ld
+}ifelse}ifelse}{/SetCmyk_5{cmyk2rgb SetRgb}bd}ifelse/currentcmykcolor where{
+pop/GetCmyk/currentcmykcolor ld}{/GetCmyk{GetRgb rgb2cmyk}bd}ifelse
+/setoverprint where{pop}{/setoverprint{/$op xd}bd}ifelse/currentoverprint where
+{pop}{/currentoverprint{$op}bd}ifelse/@tc_5{5 -1 roll dup 1 ge{pop}{4{dup 6 -1
+roll mul exch}repeat pop}ifelse}bd/@trp{exch pop 5 1 roll @tc_5}bd
+/setprocesscolor_5{SepMode_5 0 eq{SetCmyk_5}{0 4 $ink_5 sub index exch pop 5 1
+roll pop pop pop pop SepsColor true eq{$ink_5 3 gt{1 sub neg SetGry}{0 0 0 4
+$ink_5 roll SetCmyk_5}ifelse}{1 sub neg SetGry}ifelse}ifelse}bd
+/findcmykcustomcolor where{pop}{/findcmykcustomcolor{5 array astore}bd}ifelse
+/Corelsetcustomcolor_exists false def/setcustomcolor where{pop
+/Corelsetcustomcolor_exists true def}if CorelIsSeps true eq CorelIsInRIPSeps
+false eq and{/Corelsetcustomcolor_exists false def}if
+Corelsetcustomcolor_exists false eq{/setcustomcolor{exch aload pop SepMode_5 0
+eq{pop @tc_5 setprocesscolor_5}{CurrentInkName_5 eq{4 index}{0}ifelse 6 1 roll
+5 rp 1 sub neg SetGry}ifelse}bd}if/@scc_5{dup type/booleantype eq{dup
+currentoverprint ne{setoverprint}{pop}ifelse}{1 eq setoverprint}ifelse dup _ eq
+{pop setprocesscolor_5 pop}{findcmykcustomcolor exch setcustomcolor}ifelse
+SepMode_5 0 eq{true}{GetGry 1 eq currentoverprint and not}ifelse}bd/colorimage
+where{pop/ColorImage{colorimage}def}{/ColorImage{/ncolors xd/$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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2.39.2