\prednaska{2}{Rozdìl a panuj}{(zapsali J. Záloha a P. Ba¹ista)}
-\s{O èem bude dne¹ní pøedná¹ka?} Pøevá¾nì o metodì Rozdìl a panuj (\>{\sl Divide et Impera})
+\s{O èem bude dne¹ní pøedná¹ka?} Pøevá¾nì o metodì Rozdìl a panuj (\>{\sl Divide et Impera}).
\noindent
-Pro porovnávání algoritmù si musíme zavést nìjaké kritérium. Vìt¹inou se zajímáme o èas a pamìt, které spotøebují pro svùj bìh. Proto, abychom v¹ak mohli takto algoritmy porovnávat bez ohledu na prostøedí, stroj a podobné vìci, zavádíme takzvanou $O$ notaci.
+Pro porovnávání algoritmù si musíme zavést nìjaké kritérium. Vìt¹inou se zajímáme o èas a pamì», které spotøebují pro svùj bìh. Proto, abychom mohli takto algoritmy porovnávat bez ohledu na prostøedí, poèítaè a podobné vìci, zavádíme takzvanou $\O${ }notaci.
\s{Definice:} Pøedpokládejme, ¾e funkce, které porovnávame jsou dle následujícího pøedpisu:
-$f:{N} \rightarrow {R}^{+}$
+$f:M \rightarrow \bb{R}^{+}$, kde $M \subset \bb{N}$.
\noindent
-Potom øekneme, ¾e$ f(n) $ je $ O(g(n))$ právì tehdy kdy¾ existuje nìjaké $c>0: \forall ^{*} n \in {N}: f(n) \leq c.g(n)$
+Potom øekneme, ¾e $f(n)$ je $\O(g(n))$ právì tehdy kdy¾: $\exists$ $c>0, c \in \bb{R}:$ $\forall ^{*} n \in \bb{N}:$ $f(n) \leq c \cdot g(n)$.
-\>{\sl Poznámka:} $ \forall ^{*} n$ znamená $\exists n_{0}: \forall n \geq n_{0} :$. Nebo je¹tì jinak: výrok platí pro v¹echna $n$ a¾ na koneèný poèet vyjímek. Tato definice tedy øíká, ¾e funkci $f(n)$ je mo¾né ohranièit shora nìjakým reálným násobkem funkce $g(n)$ pro $\forall n \in N$. Èasto zapisujeme $f(n)=O(g(n))$ - jedná se o za¾itou vìc, ale uvìdomme si v¹ak, ¾e tento zápis neoznaèuje rovnost! (proto¾e platí napøíklad $\log{n}=O(n)$ ale neplatí $n=O(\log{n})$ To znamená, ¾e neplatí symetrie, odtud vyplývá, ¾e nemù¾e jít o rovnost). Ale: "Je to èuòaèina." Formálnì jde o nìjakou mno¾inu nebo tøídu funkcí $f(n)$, pro které platí, ¾e se dají shora ohranièit kladným reálným násobkem funkce $g(n)$. A potom pí¹eme $f \in O(g)$.
+\>{\sl Poznámka:} $ \forall ^{*} n \in \bb{N}$ $\Longleftrightarrow \exists$ $n_{0} \in \bb{N}:$ $\forall n \geq n_{0}, n \in \bb{N}$. Tedy $ \forall ^{*} n \in \bb{N}$ znamená, ¾e výrok platí pro v¹echna $n \in \bb{N}$ a¾ na koneèný poèet vyjímek. $\O$ notace tedy vyjadøuje, ¾e funkce $f(n)$ je men¹í nejvý¹e rovná nìjakému reálnému násobku funkce $g(n)$ pro $\forall ^{*} n \in \bb{N}$. Tento fakt se zapisuje takto: $f(n)=\O(g(n))$. Zde se jedná o za¾itou vìc, ale~je nutné si uvìdomit, ¾e tento zápis neoznaèuje rovnost! Je to proto, ¾e napøíklad platí: $\log{n}=\O(n)$, ale neplatí $n=\O(\log{n})$. To znamená, ¾e neplatí symetrie, a tudí¾ nemù¾e jít o rovnost ve smyslu ekvivalnece. \uv{Je to èuòaèina.} Formálnì tuto skuteènost poipsujeme, ¾e jde o nìjakou mno¾inu nebo tøídu funkcí $f(n)$, pro které platí, ¾e se dají shora ohranièit kladným reálným násobkem funkce $g(n)$. A potom zapisujeme $f \in \O(g)$. Napøíklad:
-napøíklad:
+$2{,}5n^{2} \in \O(n^{2})$
-$$
-2,5n^{2} \dots O(n^{2})
-$$
-$$
-2,5n^{2}+30n \dots O(n^{2})
-$$
+$2{,}5n^{2}+30n \in \O(n^{2})$.
-Platí:
+\noindent
+Pro dvì tøídy funkcí $\O(f)$ a $\O(g)$ platí:
$$
-O(f)+O(g)=O(f+g)
+\O(f)+\O(g) \in \O(f+g)
$$
-proto¾e:
-
-$$
-f^{\prime} \leq c.f
-$$
-$$
-g^{\prime} \leq d.g
-$$
+\noindent
+proto¾e pro v¹echny funkce $f^{\prime} \in \O(f)$ a $g^{\prime} \in \O(g)$ platí:
$$
-f^{\prime} + g^{\prime} \leq c.f+d.g \leq (c+d).(f+g)
+\eqalign{
+f^{\prime} &\leq c\cdot f \cr
+g^{\prime} &\leq d\cdot g \cr
+f^{\prime}+g^{\prime} \leq c\cdot f+d\cdot g &\leq (c+d)\cdot (f+g)
+}
$$
+
\noindent
-A zde vidíme, ¾e $(c+d)$ se schová do $O$. Naprosto stejnì se doká¾e odobný vztah také pro násobení:
+A zde vidíme, ¾e $(c+d)$ se schová do $\O$. Naprosto stejnì se uká¾e obdobný vztah pro násobení:
$$
-O(f).O(g)=O(f.g)
+\O(f).\O(g)=\O(f.g)
$$
+\noindent
Rovnì¾ platí:
$$
-O(f+g)=O(max(f,g))
+\O(f+g)=\O(\max(f,g))
$$
\noindent
-nebot libovolný $n$-násobek funkce s exponentem $k$ je v¾dy asymmptoticky men¹í ne¾ $n$ násobek funkce s exponentem $l>k$.
+nebo» libovolný kladný $c_1$-násobek funkce s exponentem $k$ je v¾dy asymptoticky men¹í ne¾ $c_2$-násobek funkce s~exponentem $l>k$.
To nám umo¾nuje zanedbávat pomaleji rostoucí èleny:
$$
-O(n^{2})+O(n)=O(n^{2}+n)=O(n^{2})
+\O(n^{2})+\O(n)=\O(n^{2}+n)=\O(n^{2})
$$
\noindent
-$O$ notace v¹ak popisuje nejhor¹í pøípad. U nìkterých algoritmù nejhor¹í, nejlep¹í a tudí¾ i prùmìrný pøípad splývají. Je ale mnoho algoritmù, které se v tìchto parametrech diametrálnì li¹í, proto zavádíme dal¹í notace.
-
+$\O$ notace popisuje horní odhad asymptotického chování algoritmù. Mnohdy v¹ak potøebujeme také urèit jeho spodní hranici, popøípadì je odhadnout obì. U nìkterých algoritmù sice splývají, ale u nìkterých ne, tak¾e zavádíme dal¹í notace:
\s{Definice:}
\itemize\ibull
-\:$f(n)$ je $\Omega(g(n)) \Longleftrightarrow \exists c>0: \forall ^{*} n \in {N}: f(n) \geq c.g(n)$
-$\Omega$ notace øíká, ¾e algoritmus se v¾dy chová stejnì èi lépe ne¾ nìjaký $c$-násobek funkce $g$.
-\:$f(n)$ je $\Theta(g(n)) \Longleftrightarrow f(n)$ je $O(g(n)) \wedge f(n) $ je $ \Omega(g(n))$
+\:$f(n) \in \Omega(g(n)) \Longleftrightarrow \exists$ $c>0:$ $\exists$ $g(n): \forall ^{*} n \in {\bb N}: f(n) \geq c\cdot g(n)$
+
+$\Omega$ notace øíká, ¾e hodnota funkce $f$ je v¾dy stejná nebo vy¹¹í ne¾ nìjaký $c$-násobek funkce $g$, a tedy $g \in \O(f)$.
+\:$f(n) \in \Theta(g(n)) \Longleftrightarrow f(n) \in O(g(n)) \wedge f(n) \in \Omega(g(n))$
nebo:
-$f(n)$ je $\Theta(g(n)) \Longleftrightarrow \exists c_{1},c_{2}: c_{1}.g(n) \leq f(n) \leq c_{2}.g(n)$
-existují-li nezáporné konstanty $c_{1},c_{2}$ takové, ¾e se funkce $f(n)$ dá ohranicit $c_{1}$ a $c_{2}$ násobky funkce $g(n)$
+$f(n) \in \Theta(g(n)) \Longleftrightarrow \exists$ $c_{1},c_{2} > 0:\exists$ $g(n) : c_{1}\cdot g(n) \leq f(n) \leq c_{2}\cdot g(n)$ To znamená, ¾e existují nezáporné reálne konstanty $c_{1},c_{2}$ takové, ¾e se funkce $f(n)$ dá ohranièit $c_{1}$ a $c_{2}$ násobky funkce $g(n)$.
\endlist
\noindent
-$\Theta$ notace tedy vyjadøuje, ¾e nejlep¹í a nejhor¹í pøípad chování algoritmu jsou stejné tøídy, li¹í se nanejvý¹ multiplikativní konstantou.
+$\Theta$ notace tedy vyjadøuje, ¾e chování algoritmu je shora i zespoda odhadnuto nìjakými kladnými rálnymi násobky funkce $g$. Proto je zøejmé, ¾e se v¾dy bude asymptoticky chovat stejnì.
\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
\: $\Theta(\log{( \log{n} )})$
-\: $\Theta(\log{n})$
-\: $\Theta(n^{\varepsilon}), \varepsilon \in (0,1)$
-\: $\Theta(n)$
-\: $\Theta(n^{2})$
+\: $\Theta(\log{n}) \ldots$logaritmická
+\: $\Theta(n^{\varepsilon}), \varepsilon \in (0,1) \ldots$ sublineární
+\: $\Theta(n) \ldots$ lineární
+\: $\Theta(n^{2}) \ldots$ kvadratická
$\vdots$
-\: $\Theta(n^{k}), k \in {N}$
+\: $\Theta(n^{k}), k \in {\bb N} \ldots$ polynomiální
$\vdots$
-\: $\Theta(2^{n})$
-\: $\Theta(3^{n})$
+\: $\Theta(2^{n}) \ldots$ exponenciální pøi základu $2$
+\: $\Theta(3^{n}) \ldots$ exponenciální pøi základu $3$
$\vdots$
-\: $\Theta(k^{n}), k \in {R}^{+}, k > 1$
+\: $\Theta(k^{n}), k \in \bb{R}^{+},$ $k > 1 \ldots$ exponenciální pøi základu $k$
$\vdots$
-\: $\Theta(n!)$
+\: $\Theta(n!) \ldots$ faktoriálová
$\vdots$
\: $\Theta(n^{n})$
+
+$\vdots$
\endlist
-\>{\sl Poznámka:} Pøi logaritmech a odhadech slo¾itosti se dá v¾dy hovoøit o logaritmu s libovolným základem, proto¾e platí:
+\>{\sl Poznámka:} Pøi logaritmech a odhadech slo¾itosti se dá v¾dy hovoøit o logaritmu s~libovolným základem, proto¾e~platí:
$$
-\log_k{n}={{\log_c{n}}\over{\log_c{k}}}={{1}\over{\log_c{k}}}.\log_c{n}
+\log_k{n}={{\log_c{n}}\over{\log_c{k}}}={{1}\over{\log_c{k}}}\cdot \log_c{n}
$$
kde ${1}\over{\log_c{k}}$ je jen konstanta, tak¾e ji mù¾eme zanedbat.
\>{\sl Pøíklady:}
-\s{Eukleidùv algoritmus:} Pokd jej pustím na 2 èísla o $n$ bitech, potom poèet iterací bude $O(n)$, ka¾dá iterace trvá $O(n^{2})$ krokù. Tak¾e celkovì má tento algoritmus èasovou slo¾itost: $O(n^{3})$
+\s{Eukleidùv algoritmus:} Pokud jej pustíme na $2$ èísla o $n$ bitech, poèet iterací bude $\O(n)$, ka¾dá iterace trvá $\O(n^{2})$ krokù. Tak¾e celkovì má tento algoritmus èasovou slo¾itost $\O(n^{3})$.
-\s{Rozdìl a panuj:} A nyní pøestaòme chodit okolo horké ka¹e a øekòeme si, co to ono vý¹e zmiòované ``rozdìl a panuj'' znamená. Mìjme nìjaký problém, který má tu vlastnost, ¾e kdy¾ jej rozdìlím na nìjaké podproblémy, které mají stejný charakter a ty vyøe¹ím, tak slo¾ením jejich øe¹ení získám øe¹ení pùvodního problému. Ten po¾adavek na stejný charakter je podstatný, nebot nám umo¾ní se podívat na tyto podproblémy pod stejným úhlem a opìt je rozdìlit na ``podpodproblémy'' a tak dále, a¾ se dostanu na úroveò, kterou je mo¾né vyøe¹it triviálnì, popøípadì jiným ménì nároèným zpùsobem, teï je provedeno rozdìlení. Po jejich vyøe¹ení se zaènu vynoøovat z logické rekurze a na jednotlivých hladinách skládám øe¹ení, a¾ se octnu na hladinì pùvodního problému, tu slo¾ím a u¾ mohu panovat.
+\s{Rozdìl a panuj:} A nyní pøestaòme chodit kolem horké ka¹e a øekòeme si, co to ono vý¹e zmiòované \uv {rozdìl a panuj} znamená. Mìjme nìjaký problém, který má tu vlastnost, ¾e kdy¾ jej rozdìlíme na nìjaké podproblémy, které mají stejný charakter a ty vyøe¹íme, tak slo¾ením jejich øe¹ení získáme øe¹ení pùvodního problému. Po¾adavek na stejný charakter podproblémù je podstatný, nebo» nám umo¾ní se podívat na tyto podproblémy pod stejným úhlem a opìt je rozdìlit na \uv {podpodproblémy} a tak dále, a¾ se dostaneme na úroveò, kterou je mo¾né vyøe¹it triviálnì, popøípadì jiným, ménì nároèným, zpùsobem. (V této chvíli je dokonèena èást rozdìlování.) Po jejich vyøe¹ení se zaèneme vynoøovat z rekurze a na jednotlivých hladinách skládat øe¹ení, a¾ se ocitneme na hladinì pùvodního problému. Po~posledním slo¾ení je pùvodní úloha vyøe¹ena.
-\s{Odboèka:} \>{\sl (Mergesort)} Pøi prùbìhu algoritmu mergesort nejprve rozdìlujeme vstup na dvì ``stejnì'' (v hor¹ím pøípadì a¾ na jednotku) velká pole. To nám zabere na ka¾dé hladinì konstantní práci:
-$T(1) \dots O(1)$.
+\s{Odboèka:} \>{\sl (Mergesort)} Pøi prùbìhu algoritmu mergesort nejprve rozdìlujeme vstup na dvì \uv{stejnì} (v hor¹ím pøípadì a¾ na jednotku) velké èásti. To nám zabere na ka¾dé hladinì konstantní práci $\O(1)$.
\noindent
-Kdy¾ se v¹ak vynoøujeme z logické rekurze, musíme na ka¾dé hladinì strávit linárnì èasu sluèováním:
-$T(n)=2.T({{n}\over{2}})+O(n)$
-
-\s{Strom volání: }
+Kdy¾ se v¹ak vynoøujeme z rekurze, musíme na ka¾dé hladinì strávit linárnì èasu sluèováním:
-\figure{figure1.eps}{Strom volání}{2in}
+$T(n)=2 \cdot T({{n}/{2}})+\O(n)$.
\noindent
-Souèet práce pøes jednu hladinu stromu je $O(n)$. A jedodu¹e odvodíme, ¾e celkový poèet hladin je $O(\log{n})$, tudí¾ jsme si ukázali slo¾itost $O(n\log(n))$.
+Souèet práce pøes jednu hladinu stromu je $\O(n)$. Víme, ¾e celkový poèet hladin je $\O(\log{n})$, a tudí¾ jsme ukázali, ¾e~èasová slo¾itost algoritmu je $\O(n \cdot \log n)$.
-\s{Rychlej¹í algoritmus pro násobení:} \>{\sl (rychlej¹í ne¾ $O(n^{2})$)} Pokud násobíme dvì èísla zpùsobem, který nás uèili na základní ¹kole, dostaneme se na èasovou slo¾itost $O(n^2)$. Proto¾e se jedná o dost èastou operaci, zamyslíme se, jestli by ne¹la zjednodu¹it. Nasmìrujme na¹e úvahy na postup ``rozdìl a panuj''. Rozdìlíme ka¾dého èinitele na dvì stejnì dlouhé èásti a pro jednoduchost pøedpokládejme, ¾e dìlení probìhne v¾dy bez zbytku:
+\s{Rychlej¹í algoritmus pro násobení:} \>{\sl (rychlej¹í ne¾ $\O(n^{2})$)} Pokud násobíme dvì èísla $X$ a $Y$ zpùsobem, který nás uèili na základní ¹kole, dostaneme se na èasovou slo¾itost $\O(n^2)$. Proto¾e se jedná o dost èastou operaci, zamysleme se, jestli by ne¹la zrychlit. Nasmìrujme na¹e úvahy na postup \uv{rozdìl a panuj}. Rozdìlíme ka¾dého èinitele na dvì stejnì dlouhé èásti a pro jednoduchost pøedpokládejme, ¾e toto roz¹tìpení èinitele probìhne v¾dy bez zbytku:
$$
-X=A.10^{{n}\over{2}}+B
+X=A \cdot 10^{{n}/2}+B
$$
$$
-Y=C.10^{{n}\over{2}}+D
+Y=C \cdot10^{{n}/{2}}+D
$$
\noindent
-Potom získáme výsledek pùvodního výrazu jako:
+Zde $A, B, C, D$ jsou u¾ jen $n/2$ ciferná èísla. Potom získáme pùvodní souèin $X \cdot Y$ jako:
\noindent
-$X.Y=(A.10^{{n}\over{2}}+B).(C.10^{{n}\over{2}}+D)=A.C.10^{n}+A.D.10^{{n}\over{2}}+B.C.10^{{n}\over{2}}+B.D=A.C.10^{n}+(A.D+B.C).10^{{n}\over{2}}+B.D$
+$X \cdot Y=(A\cdot 10^{{n}/{2}}+B)\cdot (C\cdot 10^{{n}/{2}}+D)=A\cdot C\cdot 10^{n}+A\cdot D\cdot 10^{{n}/{2}}+B\cdot C\cdot 10^{{n}/{2}}+B\cdot D=A\cdot C\cdot 10^{n}+(A\cdot D+B\cdot C)\cdot 10^{{n}/{2}}+B\cdot D$
\noindent
-Nyní, jak vidíme, staèí spoèítat souèin ètyø ${n}\over{2}$ ciferných císel. Spoèítejme, jak se tím zmìní celková èasová slo¾itost:
+Nyní, jak vidíme, staèí spoèítat souèin ètyø ${n}/{2}$ ciferných císel. Uva¾me, jak se tím zmìní celková èasová slo¾itost:
\noindent
-$T(n)=4.T({{n}\over{2}})+O(n)=4.T({{n}\over{2}})+c.n=4.(4.T({{n}\over{4}})+c.{{n}\over{2}})+c.n=4^{2}.T({{n}\over{4}})+2.c.n+c.n=4^{2}.T({{n}\over{4}})+3.c.n=4^{2}.(4.T({{n}\over{8}})+c.{{n}\over{4}})+3.c.n=4^{3}.T({{n}\over{8}})+4.c.n+3.c.n=4^{3}.T({{n}\over{8}})+7.c.n=\ldots$
+$T(n)=4\cdot T({{n}/{2}})+O(n)=4\cdot T({{n}/{2}})+c\cdot n=4\cdot (4\cdot T({{n}/{4}})+c\cdot {{n}/{2}})+c\cdot n=4^{2}\cdot T({{n}\over{4}})+2\cdot c\cdot n+c\cdot n=4^{2}\cdot T({{n}/{4}})+3\cdot c\cdot n=4^{2}\cdot (4\cdot T({{n}/{8}})+c\cdot {{n}/{4}})+3\cdot c\cdot n=4^{3}\cdot T({{n}/{8}})+4\cdot c\cdot n+3\cdot c\cdot n=4^{3}\cdot T({{n}/{8}})+7\cdot c\cdot n=\ldots$
\noindent
takto bychom mohli pokraèovat dále, a¾ bychom se dostali na:
$$
-T(n)=4^{4}.T({{n}\over{16}})+15.c.n
+T(n)=4^{4}\cdot T\left({{n}\over{16}}\right)+15\cdot c\cdot n
$$
$$
-T(n)=4^{5}.T({{n}\over{32}})+31.c.n
+T(n)=4^{5}\cdot T\left({{n}\over{32}}\right)+31\cdot c\cdot n
$$
$$
\vdots
$$
Odtud mù¾eme vypozorovat, ¾e se vztah pro $T(n)$ vyvíjí zøejmì podle vzorce:
$$
-T(n)=4^{k}.T({{n}\over{2^{k}}})+2^{k-1}.c.n+2^{k-2}.c.n+2^{k-3}.c.n+2^{k-4}.c.n+\ldots+2^{0}.c.n
-$$
-$$
-T(n)=4^{k}.T({{n}\over{2^{k}}})+(2^{k}-1).c.n
+T(n)=4^{k}\cdot T\left({{n}\over{2^{k}}}\right)+2^{k-1}\cdot c\cdot n+2^{k-2}\cdot c\cdot n+2^{k-3}\cdot c\cdot n+2^{k-4}\cdot c\cdot n+\ldots+2^{0}\cdot c\cdot n,
$$
-kde k je poèet vìtvení a n je velikost úlohy. Kdy¾ uvá¾íme, ¾e se strom volaní v¾dy vìtví pravidelnì na dva podstromy, tak platí: $k \approx \log{n}$. Kdy¾ dosadíme:
$$
-T(n)=4^{\log{n}}.T({{n}\over{2^{\log{n}}}})+(2^{\log{n}}-1).c.n
+T(n)=4^{k}\cdot T\left({{n}\over{2^{k}}}\right)+(2^{k}-1)\cdot c\cdot n
$$
+kde $k$ je poèet vìtvení a $n$ je velikost úlohy. Kdy¾ uvá¾íme, ¾e se strom volaní v¾dy vìtví pravidelnì na dva podstromy, tak platí: $k =\left\lceil \log{n} \right\rceil$. Dosadíme:
$$
-T(n)=2^{\log{n}}.2^{\log{n}}.T({{n}\over{2^{\log{n}}}})+(2^{\log{n}}-1).c.n
+\eqalign{
+T(n)&=4^{\log{n}}\cdot T\left({{n}\over{2^{\log{n}}}}\right)+\left(2^{\log{n}}-1\right)\cdot c\cdot n\cr
+T(n)&=2^{\log{n}}\cdot 2^{\log{n}}\cdot T\left({{n}\over{2^{\log{n}}}}\right)+\left(2^{\log{n}}-1\right)\cdot c\cdot n\cr
+T(n)&=n\cdot n\cdot T\left({{n}\over{n}}\right)+(n-1)\cdot c\cdot n\cr
+T(n)&=n^{2}\cdot T(1)+(n-1)\cdot n\cdot c\cr
+T(n)&=n^{2}\cdot T(1)+(n^{2}-n) \cdot c\cr
+T(n)&=n^{2}\cdot T(1)+n^{2} \cdot c-n \cdot c\cr
+T(n)&=n^{2}\cdot (T(1)+c)-c \cdot n\cr
+}
$$
-$$
-T(n)=n.n.T({{n}\over{n}})+(n-1).c.n
-$$
-$$
-T(n)=n^{2}.T(1)+(n-1).n.c
-$$
-$$
-T(n) \doteq n^{2}.(T(1)+c)
-$$
-Pokud $T(1)$ a $c$ jsou konstanty, mù¾eme psát: $T(n)$ je $O(n^{2})$. Tak¾e jsme si pøíli¹ nepomohli, proto¾e i klasický algoritmus na násobení má kvadratickou èasovou slo¾itost. Podívejme se v¹ak, jako vypadá tabulka vìtvení pro daný algoritmus:
+Pokud $T(1)$ a $c$ jsou konstanty, mù¾eme psát: $T(n) \in \O(n^{2})$. Tak¾e jsme si pøíli¹ nepomohli, proto¾e i klasický algoritmus na násobení má kvadratickou èasovou slo¾itost. Podívejme se v¹ak, jak vypadá tabulka vìtvení pro daný algoritmus:
\medskip
\vbox{\halign{# \quad \vrule \quad & # \quad \vrule \quad & #\cr
poèet vìtvení & poèet úloh & velikost podúlohy\cr
\noalign{\medskip\hrule\bigskip}
-0 & $4^{0}$ & ${n}\over{2^{0}}$\cr
-1 & $4^{1}$ & ${n}\over{2^{1}}$\cr
-2& $4^{2}$ & ${n}\over{2^{2}}$\cr
-3 & $4^{3}$ & ${n}\over{2^{3}}$\cr
+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 & $4^{k}$ & ${n}\over{2^{k}}$\cr}}
+k & $4^{k}$ & ${n}/{2^{k}}$\cr}}
\medskip
\noindent
-Naskýtá se otázka, jestli bychom nemohli, kdy¾ se pozornì zamyslíme, èasovou nároènost zlep¹it. Existuje nìkolik mo¾ností:
-- zlep¹it èlen $O(n) \ldots c.n$, to znamená zlep¹it èas spojování podúloh $\rightarrow$ to v¹ak rychleji nejde (pokud ètenáø nevìøí, mù¾e si dokázat)
-- sní¾it poèet vìtvení $\rightarrow$ nech» se algoritmus nevìtví na $4$ vìtve, ale na ménì. To, ale jak dále uvidíme u¾ mo¾né je\vdots Staèí si uvìdomit, ¾e vlastnì potøebujeme spoèítat:
+Naskýtá se otázka, jestli bychom nemohli, èasovou slo¾itost zlep¹it. Existuje nìkolik mo¾ností:
+
+\itemize\ibull
+\: zlep¹it èlen $c \cdot n$, to znamená zlep¹it èas spojování podúloh. Toto v¹ak rychleji nejde (pokud ètenáø nevìøí, mù¾e si to dokázat).
+\: sní¾it poèet vìtvení. Nech» se tedy algoritmus nevìtví na $4$ vìtve, ale na ménì. To, ale jak dále uvidíme u¾ mo¾né je.
+\endlist
+
+\noindent
+Staèí si uvìdomit, ¾e vlastnì potøebujeme spoèítat:
$$
-X.Y=A.C.10^{n}+(A.D+B.C).10^{{n}\over{2}}+B.D
+X\cdot Y=A\cdot C\cdot 10^{n}+(A\cdot D+B\cdot C)\cdot 10^{{n}\over{2}}+B\cdot D
$$
-Pøièem¾ ale nepotøebujeme znát souèiny $A.D$ ani $B.C$ samostatnì, nebo» nám staèí zjistit èlen $A.D+B.C$. Jako kdybychom poèítali $A.C$, $B.D$ a potom $(A+B).(C+D)=A.C+A.D+B.C+B.D$, tak odèítáním $(A.D+B.C)$ od $A.C+A.D+B.C+B.D$ dostaneme hledaný prostøední èlen: $A.D+B.C$, který potøebujeme, abychom spoèetli $X.Y$. Nyní nám ji¾ staèí jen tøi násobení, jedno sèítání a jedno odèítání navíc.Otázka je, zda-li to bude výhodné. Sèítání i odèítání nám zaberou nanejvý¹e lineární èas, tak¾e to skuteènì je výhodná úprava. Jak se tím zmìní výsledný èas? Podívejme se opìt na tabulku vìtvení:
+Pøitom ale nepotøebujeme znát souèiny $A\cdot D$ ani $B\cdot C$ samostatnì, nebo» nám staèí zjistit èlen $A\cdot D+B\cdot C$. Kdybychom poèítali $A\cdot C$, $B\cdot D$ a potom $(A+B)\cdot (C+D)=A\cdot C+A\cdot D+B\cdot C+B\cdot D$, tak odèítáním $(A\cdot C+B\cdot D)$ od $A\cdot C+A\cdot D+B\cdot C+B\cdot D$ dostaneme hledaný prostøední èlen: $A\cdot D+B\cdot C$. Nyní nám ji¾ staèí jen tøi násobení, ale potøebujeme tøi sèítání a jedno odèítání navíc. Otázka je, zda-li to bude výhodné. Sèítání i~odèítání nám zaberou nanejvý¹e lineární èas, tak¾e to skuteènì je výhodná úprava. Jak se tím zmìní výsledný èas? Podívejme se opìt na tabulku vìtvení:
\medskip
\vbox{\halign{# \quad \vrule \quad & # \quad \vrule \quad & #\cr
poèet vìtvení & poèet úloh & velikost podúlohy\cr
\noalign{\medskip\hrule\bigskip}
-0 & $3^{0}$ & ${n}\over{2^{0}}$\cr
-1 & $3^{1}$ & ${n}\over{2^{1}}$\cr
-2& $3^{2}$ & ${n}\over{2^{2}}$\cr
-3 & $3^{3}$ & ${n}\over{2^{3}}$\cr
+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 & $3^{k}$ & ${n}\over{2^{k}}$\cr}}
+k & $3^{k}$ & ${n}/{2^{k}}$\cr}}
\medskip
\noindent
-Spoèítejme si práci,která se musí udìlat na jedné hladinì. Pøedpokládejme, ¾e $k \approx \log{n}$. Dostávame:
-$$\sum_{k=0}^{n}3^{k}.{{n}\over{2^{k}}}=\sum_{k=0}^{n} \left( {{3}\over{2}} \right) ^{k}.n=n.\sum_{k=0}^{n} \left( {{3}\over{2}} \right) ^{k}=n.{{ \left( {{3}\over{2}} \right) ^{k+1}-1}\over{{{3}\over{2}}-1}}=
+Spoèítejme si práci, která se musí udìlat na jedné hladinì. Pøedpokládejme, ¾e $k= \lceil \log_2{n} \rceil$. Dostávame:
+$$\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{{ \left( {3}\over{2} \right) ^{\log{n}+1}-1}\over{{{1}\over{2}}}}=2.n. \left[ \left( {{3}\over{2}} \right) ^{\log{n}+1}-1 \right] \approx O \left( n. \left( {{3}\over{2}} \right) ^{\log{n}} \right) =
+=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.{{3^{\log{n}}}\over{2^{\log{n}}}} \right)=O \left( n.{{3^{\log{n}}}\over{n}} \right)=O \left( 3^{\log_2{n}} \right)=O \left( (2^{\log_2{3}})^{\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}.\log_2{3}} \right)=O \left( (2^{\log_2{n}})^{\log_2{3}} \right)=O \left( n^{\log_2{3}} \right) \doteq O \left( n^{1,585} \right)
+=\O \left( 2^{\log_2{n}.\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)
$$
-Z toho vyplývá, ¾e jsme na¹li algoritmus s èasovou slo¾itostí men¹í ne¾ $O(n^{2})$. "Rozumné" implementace tohoto algoritmu jsou v¹ak trochu modifikované, a to tak, ¾e se rekuzivnì nevolají a¾ na jednociferná èísla, ale asi na 50 ciferná, a ty u¾ násobí standardním zpùsobem, nebo» re¾ie algoritmu není nulová a takto se dosahuje nejlep¹ích výsledkù.
+Z toho vyplývá, ¾e jsme na¹li algoritmus s èasovou slo¾itostí men¹í ne¾ $\O(n^{2})$. \uv{Rozumné} implementace tohoto algoritmu jsou v¹ak trochu modifikované. A to tak, ¾e rekuzivnì ne¹tìpí èinitele a¾ na jednociferná èísla, ale konèí asi na 50 ciferných, a ty se u¾ vynásobí standardním zpùsobem, nebo» re¾ie rekurzivního algoritmu není nulová a~takto se dosahuje nejlep¹ích výsledkù.
\noindent
-Pro násobení èísel existuje je¹tì efektívnìj¹í algoritmus, který má èasovou slo¾itost $O(n.\log{n})$, av¹ak tento u¾ vyu¾ívá rùzné pokroèilé techniky jako diskrétní Fourierova transformáce a podobnì, tak¾e jej zde nebudeme rozebírat.
+Pro násobení èísel existuje je¹tì efektívnìj¹í algoritmus, který má èasovou slo¾itost $\O(n.\log{n})$, av¹ak tento u¾ vyu¾ívá rùzné pokroèilé techniky jako diskrétní Fourierova transformace a podobnì, tak¾e jej zde nebudeme rozebírat.
-\s{Vìta:} \>{\sl (Master Theorem)}
+Na¹e pozorování se nyní pokusíme shrnout ve vìtì Master Theorem, èesky tì¾ nazývané (ne náhodou) \uv{Kuchaøková vìta}:
-\noindent
-Pokud $T(1)=O(1)$ a $T(n)=a.T(\lceil {{n}\over{b}} \rceil)+O(n^d)$, kde $a \geq 1$, $b>1$ a $d \geq 0$, potom $T(n)$ je:
-\halign{#&#\cr
-$O(n^d)$ & kdy¾ $a<b^d$\cr
-$O(n^d.\log{n})$ & kdy¾ $a=b^d$\cr
-$O(n^{\log_a{b}})$ & kdy¾ $a>b^d$\cr}
+\s{Vìta:} \>{\sl (Master Theorem)} Pøedpokládejme, ¾e $T(1) \in \O(1)$ a $T(n)=a\cdot T(\lceil {{n}\over{b}} \rceil)+\O(n^d)$, kde $a \geq 1$, $b>1$ a $d \geq 0$. Potom $T(n)$ je:
+
+\smallskip
+\halign{#&#&#\cr
+\indent & $\O(n^d)$ & kdy¾ $a<b^d$\cr
+& $\O(n^d\cdot \log{n})$ & kdy¾ $a=b^d$\cr
+& $\O(n^{\log_a{b}})$ & kdy¾ $a>b^d$\cr}
-\proof \>{\sl 1. pøípad: }Pøedpokládejme nejdøíve, ¾e $n=b^x, x \in {N}$, aby platilo $\lceil {{n}\over{b}} \rceil = {n}\over{b}$. Uká¾eme si "dùkaz stromem":
+\proof \>{\sl 1. pøípad: }Pøedpokládejme nejdøíve, ¾e $n=b^m, m \in \bb{N}$, aby platilo $\lceil {{n}\over{b}} \rceil = {{n}\over{b}}$. Uká¾eme si \uv{dùkaz stromem}:
-\figure{figure2.eps}{Dùkaz stromem}{4in}
+\figure{figure.eps}{}{4in}
\noindent
Jak vidíme, strom sa v¾dy vìtví na stejný poèet vìtví - oznaème si jejich poèet $a$ a sestavme si tabulku vìtvení:
\vbox{\halign{# \quad \vrule \quad & # \quad \vrule \quad & #\cr
poèet vìtvení & velikost podúlohy & èas potøebný na vyøe¹ení v¹ech podúloh\cr
\noalign{\medskip\hrule\bigskip}
-$1$ & $n$ & $O(n^d)$\cr
-$a$ & ${n}\over{b^1}$ & $O(({{n}\over{b^1}})^d).a^1$\cr
-$a^2$ & ${n}\over{b^2}$ & $O(({{n}\over{b^2}})^d).a^2$\cr
-$a^3$ & ${n}\over{b^3}$ & $O(({{n}\over{b^3}})^d).a^3$\cr
+$1$ & $n$ & $\O(n^d)$\cr
+$a$ & ${n}\over{b^1}$ & $\O(({{n}\over{b^1}})^d).a^1$\cr
+$a^2$ & ${n}\over{b^2}$ & $\O(({{n}\over{b^2}})^d).a^2$\cr
+$a^3$ & ${n}\over{b^3}$ & $\O(({{n}\over{b^3}})^d).a^3$\cr
\vdots & \vdots & \vdots\cr
-$a^k$ & ${n}\over{b^k}$ & $O(({{n}\over{b^k}})^d).a^k$\cr}}
+$a^k$ & ${n}\over{b^k}$ & $\O(({{n}\over{b^k}})^d).a^k$\cr}}
\medskip
\noindent
Zkoumejme èas potøebný na vyøe¹ení v¹ech podúloh na jedné hladinì:
$$
-O \left( \left( {{n}\over{b^k}} \right) ^d \right) .a^k=O \left( a^k.n^d \left( {{1}\over{b^k}} \right) ^d \right)=O \left( a^k.n^d \left( {{1}\over{b^d}} \right) ^k \right)=O \left( n^d \left( {{a}\over{b^d}} \right) ^k \right)
+\O \left( \left( {{n}\over{b^k}} \right) ^d \right) \cdot a^k=\O \left( a^k\cdot n^d \left( {{1}\over{b^k}} \right) ^d \right)=\O \left( a^k.n^d \left( {{1}\over{b^d}} \right) ^k \right)=\O \left( n^d \left( {{a}\over{b^d}} \right) ^k \right)
$$
Celkem je tedy èas potøebný na vyøe¹ení v¹ech podúloh na v¹ech hladinách (to znamená celé úlohy):
$$
-T(n)=\sum_{k=0}^{\log_b{n}}O \left( n^d \left( {{a}\over{b^d}} \right) ^k \right)=O \left( n^d \sum_{k=0}^{\log_b{n}} \left( {{a}\over{b^d}} \right) ^k \right)
+T(n)=\sum_{k=0}^{\log_b{n}}\O \left( n^d \left( {{a}\over{b^d}} \right) ^k \right)=\O \left( n^d \sum_{k=0}^{\log_b{n}} \left( {{a}\over{b^d}} \right) ^k \right)
$$
-V¹imnìme si výrazu ${a}\over{b^d}$. Na jeho hodnotì závisí hodnota výsledné sumy, proto¾e se vlastnì jedná o kvocient geometrické posloupnosti. Proto rozli¹me následující pøípady:
+V¹imnìme si èlenu ${a}\over{b^d}$. Na jeho hodnotì závisí hodnota výsledné sumy, proto¾e se vlastnì jedná o kvocient geometrické øady. Proto rozli¹me následující pøípady:
+
+\>{\I 1.} ${{a}\over{b^d}}<1$ --- jinými slovy, práce na jednotlivých hladinách exponenciálnì ubývá a souèet sumy se dá omezit nìjakou konstantou, která se schová do $\O$, a tak mù¾eme psát $T(n) \in \O(n^d)$.
+
+\>{\I 2.} ${{a}\over{b^d}}=1$ --- práce na jednotlivých hladinách je stejnì, to znamená, ¾e ¾e souèet sumy je právì $\log_b(n)$ a tedy: $T(n) \in \O(n^d \cdot \log_b(n))$.
-\>{\I 1} ${a}\over{b^d}<1$ jinými slovy, práce na jednotlivých hladinách exponenciálnì ubývá a souèet sumy se dá omezit nìjakou konstantou, která se schvá do $O$, a tak mù¾eme psát $T(n)=O(n^d)$.
+\>{\I 3.} ${{a}\over{b^d}}>1$ --- to znamená, ¾e práce na jednotlivých hladinách pøibývá, a potom musíme psát: $T(n) \in \O(n^d.({{a}\over{b^d}})^{\log_b{n}})$.
-\>{\I 2} ${a}\over{b^d}=1$ práce na jednotlivých hladinách je stejnì, to znamená, ¾e ¾e souèet sumy je právì $\log(n)$ a to nás opravòuje psát: $T(n)=O(n^d \cdot \log(n))$.
+\noindent
+To sice vypadá jako slo¾itý výraz, ale mù¾eme jej dále upravit:
-\>{\I 3} ${a}\over{b^d}>1$ to znamená, ¾e práce na jednotlivých hladinách pøibývá a potom musíme psát:$ T(n)=O(n^d.({{a}\over{b^d}})^{\log_b{n}})$ To sice vypadá jako slo¾itý výraz, ale mù¾eme je dále upravit:
-$O(n^d.({{a}\over{b^d}})^{\log_b{n}})=O(n^d.a^{\log_b{n}}({{1}\over{b^d}})^{\log_b{n}})=O((b^{\log_b{a}})^{\log_b{n}}.n^d.{{1}\over{(b^d)^{\log_b{n}}}})=O((b^{\log_b{n}})^{\log_b{a}}.n^d.{{1}\over{(b^{\log_b{n}})^d}})=O(n^{\log_b{a}}.n^d.{{1}\over{n^d}})=O(n^{\log_b{a}})$
+$$
+\O\left(n^d.\left({{a}\over{b^d}}\right)^{\log_b{n}}\right)=\O\left(n^d.a^{\log_b{n}}\left({{1}\over{b^d}}\right)^{\log_b{n}}\right)=\O\left(\left(b^{\log_b{a}}\right)^{\log_b{n}}.n^d.{{1}\over{\left(b^d\right)^{\log_b{n}}}}\right)=
+$$
+$$
+=\O\left(\left(b^{\log_b{n}}\right)^{\log_b{a}}.n^d.{{1}\over{\left(b^{\log_b{n}}\right)^d}}\right)=\O\left(n^{\log_b{a}}.n^d.{{1}\over{n^d}}\right)=\O\left(n^{\log_b{a}}\right)
+$$
\noindent
-A nyní vidíme, ¾e vìta je správná a zároveò rozdìlení pøípadù je naprosto oprávnìné.
+A nyní vidíme, ¾e vìta v tomto pøípadì platí a ¾e rozdìlení pøípadù je naprosto oprávnìné.
-\>{\sl 2. pøípad: }Vratme se k mo¾nosti $n \neq b^x, x \in {N}$. Potom ale platí: $b^k<n<b^{k+1} \ldots$ v tomto pøípadì zaokrouhleme $n$ a poèítejme s $n'=b^{k+1}$. Pokud platí $b^{k+1}=b.b^k<b.n$, je odtud hned vidìt, ¾e platí $n'<b.n$ a odtud vyplývá, ¾e vstup se nám zvìt¹í pøinejhor¹ím konstanta-krát, co je z asymtotického hlediska, které nás zajímá nejvíce nezajímavé, a tak tuto konstantu "schováme" do $O$ a máme algoritmus stejné slo¾itosti.
+\>{\sl 2. pøípad: }Vra»me se k mo¾nosti $n \neq b^m, m \in \bb{N}$. Potom ale platí: $b^l<n<b^{l+1}$ pro nìjaké $l \in \bb{N}$. V tomto pøípadì zaokrouhleme $n$ a~poèítejme s $n^{\prime}=b^{l+1}$. Potom platí $n^{\prime}=b^{l+1}=b\cdot b^l<b\cdot n$, a odtud je hned vidìt, ¾e $n^{\prime}<b\cdot n$, a odtud vyplývá, ¾e vstup se nám zvìt¹í pøinejhor¹ím $konstanta$-krát, co¾ je z asymtotického hlediska, které nás zajímá nejvíce, nedùle¾ité, a tak tuto konstantu \uv{schováme} do $\O$ a máme algoritmus stejné slo¾itosti. Vidíme tedy, ¾e vìta platí i v tomto pøípadì.
\qed
\noindent
-Porovnejme si nìkteré známé algoritmy a jejich èasovou slo¾itost pomocí \>{\sl Master Theorem}:
+Porovnejme si nìkteré známé algoritmy a jejich èasovou slo¾itost pomocí \>{\sl Master Theoremu}:
\medskip
\vbox{\halign{# \quad \vrule \quad & # \quad \vrule \quad & # \quad \vrule \quad & # \quad \vrule \quad & #\cr
-algoritmus & a & b & d & èasová slo¾itost\cr
+algoritmus & $a$ & $b$ & $d$ & èasová slo¾itost\cr
\noalign{\medskip\hrule\bigskip}
-Mergesort & 2 & 2 & 1 & $O(n.\log{n})$\cr
-Násobení I. & 4 & 2 & 1 & $O(n^2)$\cr
-Násobení II. & 3 & 2 & 1 & $O(n^{\log_2{3}})$\cr
-Binární vyhledávání & 1 & 2 & 0 & $O(\log{n})$\cr}}
+Mergesort & 2 & 2 & 1 & $\O(n\log{n})$\cr
+Násobení I. & 4 & 2 & 1 & $\O(n^2)$\cr
+Násobení II. & 3 & 2 & 1 & $\O(n^{\log_2{3}})$\cr
+Binární vyhledávání & 1 & 2 & 0 & $\O(\log{n})$\cr}}
\medskip
-\s{Domácí úkol nakonec: } Vymyslete algoritmus, který by z $n$ zadaných bodù v rovinì (prostoru) na¹el takové dva, které jsou od sebe nejménì vzdálené (zde existuje takový algoritmus s èasovou slo¾itostí $O(n.\log{n})$).
+\s{Domácí úkol nakonec:} Vymyslete algoritmus, který by z $n$ zadaných bodù v rovinì (prostoru) na¹el takové dva, které~jsou od sebe nejménì vzdálené (zde existuje takový algoritmus s èasovou slo¾itostí $\O(n \cdot \log{n})$).
\bye
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projects / ads1.git / commitdiff