\s{Definice:} 3-SAT je takový SAT, kde ka¾dá klauzule obsahuje nejvý¹e tøi literály.
\s{Pøevod 3-SAT na SAT:}
-Platí identita, 3-SAT splòuje vlastnosti SATu, proto 3-SAT = SAT (3-SAT je alespoò tak tì¾ký, jako SAT)
+Platí identita, 3-SAT splòuje vlastnosti SATu, proto 3-SAT = SAT (3-SAT je alespoò tak tì¾ký jako SAT)
\s {Pøevod SAT na 3-SAT:}
Musíme formuli pøevést tak, abychom neporu¹ili splnitelnost.
--- /dev/null
+\input lecnotes.tex
+
+\prednaska{11}{NP-úplné problémy}{(zapsali F. Kaèmarik, R. Krivák, D. Remi¹)}
+
+Dosud jsme zkoumali problémy, které se nás ptali na to, jestli nìco existuje. Napøíklad: dostali jsme formuli a problém splnitelnosti se nás ptal, zda existuje ohodnocení promìnných takové, ¾e formule platí? Nebo v~pøípade nezávislých mno¾in jsme dostali graf a èíslo $k$ a ptali jsme se, jestli v~grafu existuje nezávislá mno¾ina, která je veliká alespoò $k$? Tyto otázky mìli spoleèné to, ¾e kdy¾ nám nìkdo zadal nìjaký objekt, pak jsme umìli efektivnì øíci, zda je to objekt, který hledáme. Napøíklad: pokud dostaneme ohodnocení promìnných logické formule, staèí jen dosadit a spoèítat, ¾e formule je $true$ nebo $false$. Zjistit, ¾e nìjaký objekt je ten, který hledáme, umíme efektivnì. Tì¾ké na tom je takový objekt najít. Co¾ vede k~definici obecných vyhledávacích problémù, kterým se øíká tøída problémù NP. Definujme si ji poøádnì, ale nejdøíve zaèneme tro¹ièku jednodu¹¹í tøídou.
+
+\s{Definice:} P je {\I tøída rozhodovacích problémù}, které jsou øe¹itelné v~polynomiálním èase. Jinak øeèeno, problém
+$L \in \rm{P} \Leftrightarrow \exists $ polynom $f$ a~$\exists$ algoritmus $A$ takový, ¾e $\forall x: L(x)=A(x)$ a $A(x)$ dobìhne v~èase $\leq f(x)$
+
+Tøída P odpovídá tomu, o èem jsme se shodli, ¾e umíme efektivnì øe¹it. Nadefinujme tedy tøídu NP:
+
+\s{Definice:} NP je {\I tøída rozhodovacích problémù} takových, ¾e $L \in \rm{NP}$ právì tehdy, kdy¾ $\exists $ problém
+$K\in\rm{P}$ a $\exists$ polynom $g$ takový, ¾e pro
+$\forall x$ platí $L(x)=1 \Leftrightarrow \exists $ nápovìda $ y: \vert y \vert \leq g(\vert x \vert)$ a souèasnì $K(x,y)=1$.
+
+\s{Pozorování:} Splnitelnost logických formulí je v~NP.
+
+Máme-li lineárnì velké nápovìdy, co¾ jsou ohodnocení promìnných zadané formule, odpovíme ano, formule je splnitelná, pokud nám nìkdo mù¾e odpovìdìt na splòující ohodnocení. Tak¾e $K$ nám ovìøí, èi dosazením ohodnocení je formule splnìna a ptáme se tedy, èi existuje nápovìda taková, ¾e existuje ohodnocení takové, ¾e formule je splnìna. Splnitelnost logických formulí je urèitì v~tøídì NP.
+V¹imneme si, ¾e celá tøída P le¾í uvnitø NP. v~podstatì øíkáme, ¾e kdy¾ máme problém, který umíme øe¹it v~polynomiálním èase bez nápovìdy, tak to zvládneme v~polynomiálním èase i s~nápovìdou.
+
+Nasbírali jsme problémy, které jsou v~NP, ale nevíme, jestli jsou v~P. Dokonce by se dalo øíct, ¾e jsou to nejtì¾¹í problémy v~NP.
+Nadefinujme si:
+
+\s{Definice:} Problém $L$ je NP-tì¾ký právì tehdy, kdy¾ je pøevoditelný:
+$\forall M \in \rm{NP}: M~\rightarrow~L$.
+Také platí, ¾e pokud umíme øe¹it nìjaký NP problém v~polynomiálním èase, pak umíme vyøe¹it v~polynomiálním èase v¹e v~NP,
+pokud nìjaké takové~$\in~\rm{P} \Rightarrow \rm{P}=\rm{NP} $.
+
+My se budeme zabývat problémy, které jsou NP tì¾ké a samotné jsou v~NP. Takovým problémùm se øíká NP-úplné.
+
+\s{Definice:} Problém $L$ je NP úplný právì tehdy, kdy¾ $L$ je NP-tì¾ký a $L \in \rm{NP}$.
+
+NP-úplné problémy jsou ve své podstatì nejtì¾¹í problémy, které le¾í v~NP. Proto¾e le¾í v~NP platí, ¾e libovolný problém z NP umíme pøevést na nì. Kdybychom umìli pøevést nìjaký NP-úplný problém, který je øe¹itelný v~polynomiálním èase, pak v¹echno v~NP je øe¹itelné v~polynomiálním èase. Bohu¾el, to se neví. Otázka jestli $\rm{P}=\rm{NP}$ je asi nejznámìj¹í otevøený problém v~celé teoretické informatice.
+
+Uká¾eme si nìjaký NP-úplný problém. Velmi se nám bude hodit následující vìta:
+
+\s{Vìta (Cookova):} SAT je NP-úplný.
+
+\>Dùkaz je pøíli¹ technický, jenom ho pøibli¾nì naznaèíme pozdìji. Pøímím dùsledkem je, ¾e cokoli v~NP je pøevoditelné na SAT.
+Dá se dokázat vìtièka:
+
+\s{Vìtièka:} $L$ je NP-úplný a $L$ se dá pøevést na $M$ ($L \rightarrow M$), $M \in \rm{NP} \Rightarrow M$ je také NP-úplný.
+
+\proof
+Tuto vìtièku staèí dokázat pro NP tì¾kost, NP-úplnost plyne okam¾itì z toho, ¾e problémy jsou NP tì¾ké a le¾í v~NP, podle pøedpokladu.
+Víme ¾e $L$ se dá pøevést na $M$ a také z toho, ¾e $L$ je NP-úplný plyne, ¾e ka¾dý problém $Q$ z $NP$ se $Q$ dá pøevést na $L$. $Q$ se dá pøevést na $L$, $L$ se dá pøevést na $M$. Z toho plyne, ¾e $Q$ se dá pøevést na $M$. Staèí tedy slo¾it funkci $f(Q \rightarrow L)$ s~funkcí $g(L \rightarrow M)$. To urèitì také spoèteme v~polynomiálním èase. Tak nahlédneme, ¾e v¹echny problémy z $NP$ se dají pøevést na problém $M$.
+\qed
+
+\s{Dùsledek:} Cokoliv, na co jsme umìli pøevést SAT je také NP-úplné. Napøíklad nezávislá mno¾ina, v¹echny varianty SATu, klika v~grafu...
+
+\figure{p-np.eps}{Obrázek 11.1}{2.5cm}
+
+\s{Katalog NP-úplných problémù:}
+\itemize\ibull
+\:{\I logické:}
+SAT, 3-SAT, 3,3-SAT, obvodový SAT, SAT pro obecné formule
+\:{\I grafové:}
+Nezávislá mno¾ina, klika v~grafu, 3D párování, 3-barvení, Hamiltonova cesta/kru¾nice
+\:{\I èíselné:}
+problém batohu, loupe¾níci, $Ax=b$, celoèíselné lineární programování
+\endlist
+
+\h { Pøevoditelnost 3,3-SAT na 3D-párování }
+
+Kdy¾ chceme ukázat, ¾e na nìco se dá pøevést SAT, potøebujeme obvykle dvì vìci. Konstrukci, která bude simulovat promìnné, nìco co nabývá dvou stavù $true$/$false$ a nìco co bude reprezentovat klauzule a umí zaøídit, aby ka¾dá klauzule byla splnìna alespoò jednou promìnnou.
+Jenom pro pøipomenutí, máme mno¾inu klukù, dìvèat, zvíøátek a nìjaké trojice, kdo se s~kým snese a chceme vybrat trojice tak, aby se v~nich ka¾dý kluk, holka, zvíøátko vyskytovalo právì jednou.
+Najdeme si takovouto konfiguraci:
+
+\figure{3d.eps}{Obrázek 11.2}{4cm}
+
+\>4 zvíøátka, 2 kluci, 2 dívky a~takové 4 trojice, které oznaèíme $A, B, C, D$. Je¹tì pøedpokládáme, ¾e zvíøátka se mohou úèastnit nìjakých jiných trojic, ale tito ètyøi lidé se vyskytují pouze v~tìchto ètyøech trojicích a~nikde jinde.
+V¹imneme si, ¾e existují právì dvì mo¾nosti, jak tento obrázek spárovat. Abychom spárovali kluka $k_1$, tak si musíme vybrat $A$ nebo $B$. Kdy¾ si vybereme $A$, $k_1$ i $d_2$ u¾ jsou spárovaní tak¾e si nesmíme vybrat $B$ ani $D$. Pak jediná mo¾nost, jak spárovat $d_1$ a~$k_2$ je $C$. Jedna mo¾nost je tedy vybrat si $A$ a $C$ a jeliko¾ je obrázek symetrický, tak kdy¾ vybereme místo $A$ trojici $B$, tak dostaneme $B$ a~$D$. V¾dy si vybereme dvì protìj¹í trojice v~obrázku. Takovýto obrázek budeme pou¾ívat k~reprezentaci promìnných. Pro ka¾dou promìnnou si nakreslíme takový obrázek a~to, ¾e $A$ bude spárované s~$C$ bude odpovídat tomu, ¾e $x=1$
+a~spárování $B$ a~$D$ odpovídá $x=0$. Pokud jsme pou¾ili $A$ a~$C$, zvíøata se sudými èísly ,tj. $z_2$ a~$z_4$, horní a~dolní jsou nespárovaná a~pokud jsme pou¾ili $B$ a~$D$, zvíøátka $z_1$ a~$z_3$ zùstala nespárovaná. Pøes tyto nespárovaná zvíøátka mù¾eme pøedávat informaci, jestli promìnná $x$ má hodnotu $true$ nebo $false$ do dal¹ích èástí grafu.
+Zbývá vymyslet, jak reprezentovat klauzule. Klauzule budou vypadat jako trojice literálù:
+$\kappa = (x \lor y \lor \lnot r) $
+Potøebujeme zajistit, aby $x$ bylo nastavené na $1$ nebo $y$ bylo nastavené na $1$ nebo $r$ na $0$.
+
+\figure{klauzule.eps}{Obrázek 11.3}{4cm}
+
+\>Pro takovouto klauzuli si poøídíme dvojici kluk-dívka, kteøí budou figurovat ve tøech trojicích se tøemi rùznými zvíøátky, co¾ mají být volná zvíøátka z~obrázku pro pøíslu¹né promìnné. A~zaøídíme to tak, aby ka¾dé zvíøátko bylo pou¾ité maximálnì v~jedné takové trojici, co¾ jde proto, ¾e ka¾dý literál se vyskytuje maximálnì dvakrát a~pro ka¾dý literál máme dvì volná zvíøátka,
+z~èeho¾ plyne, ¾e zvíøátek je dost pro v¹echny klauzule. Pak nám ale urèitì zbude $2p-k$ zvíøátek, kde $p$ je poèet promìnných, tak pøidáme je¹tì $2p-k$ párù kluk-dìvèe, kteøí milují v¹echna zvíøátka a~ti vytvoøí zbývající páry. Pokud formule byla splnitelná, pak ze splòujícího ohodnocení mù¾eme vyrobit párování s~na¹í konstrukcí. Obrázek pro ka¾dou promìnnou spárujeme podle ohodnocení, tj. promìnná je $0$ nebo $1$ a~pro ka¾dou klauzuli si vybereme nìkterou z~promìnných, kterými je ta klauzule splnìna. Funguje to také ale i~opaènì. Kdy¾ nám nìkdo dá párovaní v~na¹í konstrukci, pak z nìho doká¾eme vyrobit splòující ohodnocení dané formule. Podíváme se, v~jakém stavu je promìnná a~to je v¹echno. Z~toho, ¾e jsou správnì spárované klauzule u¾ okam¾itì víme, ¾e jsou v¹echny splnìné. Zbývá ovìøit, ¾e na¹e redukce funguje v~polynomiálním èase. Pro ka¾dou klauzuli spotøebujeme konstantnì mnoho èasu, $2p-k$ je také polynomiálnì mnoho a~kdy¾ to seèteme, pak máme polynomiální èas vzhledem k~velikosti vstupní formule. Tím je pøevod hotový a~mù¾eme 3D-párování zaøadit mezi NP-úplné problémy.
+
+
+%RK
+
+
+\h{Náznak dùkazu Cookovy vety}
+
+Abychom mohli budovat teorii NP-úplnosti, tak potøebujeme alespoò jeden problém, o kterém doká¾eme, ¾e je NP-úplný z definice. Cookova vìta øíká o NP-úplnosti SAT-u, ale nám se to hodí dokázat o tro¹ku jiném problému -- {\I obvodovém SAT-u}.
+
+\>{\I Obvodový SAT} je splnitelnost, která nepracuje s~formulemi, ale s~boolovskými obvody. Ka¾dá formule se dá pøepsat do boolovského obvodu, který ji poèítá, tak ¾e dává smysl zavést splnitelnost pro obvody. Na¹e obvody budou mít nìjaké vstupy a~jenom jeden výstup. Budeme se ptát, jestli se vstupy tohoto obvodu dají nastavit tak, abychom na výstupu dostali $1$.
+
+\>Nejprve doká¾eme NP-úplnost {\I obvodového SAT-u} a~pak uká¾eme, ¾e se dá pøevést na formulový SAT. Tím bude dùkaz Cookovy vìty hotový.
+
+\s{Vìta:} Obvodový SAT je NP-úplný.
+
+\proof
+Náznak. Na základì zku¹eností principu poèítaèù intuitivnì chápeme, ¾e pokud nìjaký problém $L \in \rm{P}$, pak existuje polynomiálnì velký boolovský obvod, který poèítá $L$.
+
+\>Dovolíme si drobnou úpravu v~definici tøídy NP. Budeme chtít, aby nápovìda byla pravì tak velká jako velikost vstupu (tedy: $\vert y \vert = g(\vert x \vert)$). Proè je taková úprava BÚNO? Jistì si dovedete pøedstavit, ¾e pùvodní nápovìdu doplníme
+na po¾adovanou délku nìjakými \uv{mezerami}, které program ignoruje. (Tedy upravíme program tak, aby mu nevadilo, ¾e dostane na konci nápovìdy nìjak kódované mezery.)
+
+\>Máme nìjaký problém $L$ z~NP a~chceme dokázat, ¾e $L$ se dá pøevést na obvodový SAT. Kdy¾ nám nìkdo pøedlo¾í nìjaký vstup $x$ (chápeme jako vektor $(x_1, x_2, \dots)$), spoèítáme velikost nápovìdy $g(\vert x\vert)$. Víme, ¾e kontrolní algoritmus $K$ (který kontroluje, zda nápovìda je správnì) je v~P. Vyu¾ijeme intuice v~obvodech, abychom získali obvod, který pro konkrétní velikost vstupu $x$ poèítá to, co kontrolní algoritmus $K$. Na vstupu tohoto obvodu bude $x$ (vstup problému $L$) a~nápovìda $y$.
+Na výstupu nám øekne, jestli je to nápovìda, která k~tomu patøí nebo nepatøí. Velikost vstupu tohoto obvodu bude tedy $\vert x\vert + g(\vert x\vert)$, co¾ je polynom.
+
+\figure{kobvod.eps}{Obrázek 11.4 - Obvod pro kontrolní algoritmus $K$}{2.3cm}
+
+\>V tomto obvodu zafixujeme vstup $x$ (na místa vstupu dosadíme konkrétní hodnoty z $x$). Tím získáme obvod, jeho¾ vstup je jen $y$ a~ptáme se, zda za $y$ mù¾eme dosadit nìjaké hodnoty tak, aby na výstupu byla 1. Jinými slovy, ptáme se, zda je tento obvod splnitelný.
+
+\>Pro libovolný problém z~NP tak doká¾eme vyrobit funkci, která pro ka¾dý vstup $x$ v~polynomiálním èase vytvoøí obvod, který je splnitelný pravì tehdy, kdy¾ odpovìï tohoto problému na vstup $x$ má být jedna. Tedy libovolný problém z~NP se dá
+v~polnomiálním èase pøevést na obvodový SAT.
+\qed
+
+\s{Lemma:} Obvodový SAT se dá pøevést na 3-SAT.
+
+\proof
+Budeme postupnì budovat formuli v~konjunktivní normální formì. Pro ka¾dé hradlo v~obvodu zavedeme novou promìnnou popisující jeho výstup. Pøidáme klauzule, které nám kontrolují, ¾e toto hradlo máme ohodnocené konzistentnì. Ka¾dý boolovský obvod se dá pøevést na ekvivalentní obvod, v~kterém se vyskytují jen hradla {\sc and} a {\sc not}, tak ¾e staèí najít klauzule odpovídající tìmto hradlùm.
+
+\>{\I Pøevod hradla \sc not}: na vstupu hradla budeme mít nìjakou promìnnou $x$ (která pøi¹la buïto pøímo ze~vstupu toho celého obvodu nebo je to promìnná, která vznikla na výstupu nìjakého hradla) a na výstupu promìnnou $y$. Pøidáme klauzule, které nám zaruèí, ¾e jedna promìnná bude negací té druhé:
+$$\matrix{ (x \lor y), \cr
+ (\neg{x} \lor \neg{y}). \cr }$$
+\figure{not.eps}{Obrázek 11.5 - Hradlo \sc not}{0.8cm}
+
+\>{\I Pøevod hradla \sc and}: Hradlo má vstupy $x, y$ a~výstup $z$. Potøebujeme pøidat klauzule, které nám popisují, jak se má hradlo {\sc and} chovat. Tyto vztahy pøepí¹eme do~konjunktivní normální formy:
+$$
+\left. \matrix{
+ x\ \&\ y \Rightarrow z \cr
+ \neg{x} \Rightarrow \neg{z} \cr
+ \neg{y} \Rightarrow \neg{z} \cr
+}
+\ \quad
+ \right\}
+\quad
+\matrix{
+ (z \lor \neg{x} \lor \neg{y}) \cr
+ (\neg{z} \lor x) \cr
+ (\neg{z} \lor y) \cr
+ } $$
+\figure{and.eps}{Obrázek 11.6 - Hradlo \sc and}{0.9cm}
+
+\>Kdy¾ chceme pøevádìt obvodový SAT na 3-SAT, tak ten obvod nejdøíve pøelo¾íme na obvod, v~kterém jsou jen hradla {\sc and} a~{\sc not} a~pak hradla tohoto obvodu pøelo¾íme na klauzule. Formule vzniklá z~takovýchto klauzulí je splnitelná pravì tehdy, kdy¾ je splnitelný daný obvod.
+\qed
+
+\s{Poznámka:}
+Kdy¾ jsme zavádìli SAT, omezili jsme se jen na formule, které jsou
+v~konjunktivní normální formì (CNF). Teï u¾ víme, ¾e splnitelnost obecné
+boolovské formule doká¾eme pøevést na obvodovou splnitelnost a tu pak
+pøevést na 3-SAT. SAT bychom si tedy mohli definovat i jako problém
+splnitelnosti obecných boolovských formulí.
+
+
+\bye
+
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\input lecnotes.tex
-\prednaska{12}{Aproximaèní algoritmy}{(???)}
+\prednaska{12}{Aproximaèné algoritmy}{(F. Ha¹ko, J. Menda)}
-\h{Pseudopolynomiální algoritmus pro problém batohu}
+\>Na~minulých predná¹kach sme sa zaoberali rôzne »a¾kými rozhodovacími problémami. Táto sa zaoberá postupmi ako sa v~praxi vysporiada» s~rie¹ením týchto problémov.
-\s{POZOR:} Na~pøedná¹ce byla jen verze bez cen, nauète se, prosím, obì. --M.M.
+\h{Prvý spôsob: ©peciálny prípad}
-\s{Problém batohu:} Je dána mno¾ina $n$~pøedmìtù s~hmotnostmi $h_1,\ldots,h_n$
-a cenami $c_1,\ldots,c_n$ a batoh, který unese hmotnost~$H$. Naleznìte takovou
-podmno¾inu pøedmìtù, jejich¾ celková hmotnost je nejvý¹e~$H$ a celková cena je
-maximální mo¾ná.
+\>Èasto si vystaèíme s~vyrie¹ením ¹peciálneho prípadu NP~problému, ktorý le¾í v~P. Napríklad, ak rie¹ime grafovú úlohu, tak nám mô¾e staèi» rie¹enie pre~¹peciálny graf (strom, bipartitný graf,$\ldots$). Farbenie grafu je µahké pre~nejaký malý poèet farieb. 2SAT, ako ¹peciálny prípad SAT-u sa dá rie¹i» v~lineárnom èase.
-Tento problém je zobecnìním problému batohu z~minulé pøedná¹ky dvìma smìry:
+\s{Problém: Maximálna nezávislá mno¾ina v strome (nie rozhodovacia)}
+
+\>{\I Vstup:} zakorenený strom~$T$
+\>{\I Vstup:} nezávislá mno¾ina vrcholov~$M$
+
+\>BUNV mô¾eme predpoklada», ¾e v~$M$ sú v¹etky listy~$T$. Ak by nejaký list $l$ nebol v~$M$, tak sa pozrieme na jeho otca:
+\itemize\ibull
+\:ak otec nie je v~$M$, tak vytvoríme novú nezávislú mno¾inu~$M'$ obsahujúcu aj~$l$ (veµkos» nezávislej mno¾iny stúpla o~1).
+\:ak tam otec je, tak ho z~$M$ vyjmeme a~namiesto neho vlo¾íme~$l$ (veµkos» nezávislej mno¾iny sa nezmen¹ila).
+\endlist
+\>Tieto listy aj ich otcov z~$T$ odstránime a~postup opakujeme. $T$ sa mô¾e rozpadnú» na~les; potom tento postup aplikujeme na~v¹etky stromy v~lese.
+
+\s{Algoritmus:}
+\algo
+\:Polo¾íme $M1:={listy stromu T}$.
+\:Polo¾íme $M2:={otcovia vrcholov z~M1}$.
+\:Vrátime $M1 \cup MaxNz(T\setminus(M1 \cup M2)$
+\endalgo
+\>{\I Poznámka:} toto doká¾eme naprogramova» v \O(n) (vrcholy máme vo fronte a prechádzame).
+
+\s{Problém: Batoh}
+
+\>Je daná mno¾ina $n$~predmetov s~hmotnos»ami $h_1,\ldots,h_n$
+a cenami $c_1,\ldots,c_n$ a~batoh, ktorý unesie hmotnos»~$H$. Nájdite takú
+podmno¾inu predmetov, ktorých celková hmotnos» je najviac~$H$ a~celková cena je
+maximálna mo¾ná.
+
+\>Tento problém je zobecnìním problému batohu z~minulé pøedná¹ky dvìma smìry:
Jednak místo rozhodovacího problému øe¹íme optimalizaèní, jednak pøedmìty
mají ceny (pøedchozí verze odpovídala tomu, ¾e ceny jsou rovny hmotnostem).
Uká¾eme si algoritmus pro øe¹ení tohoto obecného problému, jeho¾ èasová
slo¾itost bude polynomiální v~poètu pøedmìtù~$n$ a souètu v¹ech cen~$C=\sum_i c_i$.
-Pou¾ijeme dynamické programování. Pøedstavme si problém omezený na~prvních~$k$
+\>Pou¾ijeme dynamické programování. Pøedstavme si problém omezený na~prvních~$k$
pøedmìtù. Oznaème si $A_k(c)$ (kde $0\le c\le C$) minimální hmotnost
podmno¾iny, její¾ cena je právì~$c$. Tato $A_k$ spoèteme indukcí podle~$k$:
Pro $k=0$ je urèitì $A_0(0)=0$, $A_0(c)=\infty$ pro $c>0$. Pokud ji¾ známe
$$
Tímto zpùsobem v~èase $\O(C)$ spoèteme jednu mno¾inu, v~èase $\O(nC)$ pak v¹echny.
-Podle $A_n$ snadno nalezneme maximální cenu mno¾iny, která se vejde do batohu. To bude
+\>Podle $A_n$ snadno nalezneme maximální cenu mno¾iny, která se vejde do batohu. To bude
nejvìt¹í~$c^*$, pro nì¾ je $A_n(c^*) < \infty$. Jeho nalezení nás stojí èas $\O(C)$.
-A~jak zjistit, které pøedmìty do~nalezené mno¾iny patøí? Upravíme algoritmus,
+\>A~jak zjistit, které pøedmìty do~nalezené mno¾iny patøí? Upravíme algoritmus,
aby si pro ka¾dé $A_k(c)$ pamatoval $B_k(c)$, co¾ bude index posledního pøedmìtu,
který jsme do~pøíslu¹né mno¾iny pøidali. Pro nalezené $c^*$ tedy bude $i=B_n(c^*)$
poslední pøedmìt v~nalezené mno¾inì, $i'=B_{i-1}(c^*-c_i)$ ten pøedposlední
a tak dále. Takto v~èase $\O(n)$ rekonstruujeme celou mno¾inu od~posledního
prvku k~prvnímu.
-Ukázali jsme tedy algoritmus s~èasovou slo¾itostí $\O(nC)$, který vyøe¹í
-problém batohu. Jeho slo¾itost není polynomem ve~velikosti vstupu ($C$~mu¾e být a¾ exponenciálnì
+\>Ukázali jsme tedy algoritmus s~èasovou slo¾itostí $\O(nC)$, který vyøe¹í
+problém batohu. Jeho slo¾itost není polynomem ve~velikosti vstupu ( $C$~mu¾e být a¾ exponenciálnì
velké vzhledem k~velikosti vstupu), ale pouze ve~velikosti èísel na~vstupu.
Takovým algoritmùm se øíká {\I pseudopolynomiální.}
spoèteme ze~$Z_{k-1}$ a ze~$Z_n$ vyèteme výsledek. V¹echny tyto mno¾iny
mají nejvý¹e $H$ prvkù, tak¾e celková èasová slo¾itost algoritmu je~$\O(nH)$.
+\h{Druhý spôsob: Aproximácia}
+
+\>V predchádzajúcich problémoch sme sa zamerali na ¹peciálne prípady. Obèas v¹ak také ¹tastie nemáme a~musíme vyrie¹i» NP-úplný problém. Mo¾eme ustúpi» tak, ¾e nebudeme rie¹i» nieèo, èo je úplne optimálne, ale je to nejaky pomer optimalnosti ({\I aproximácia}), t.j. vieme o~koµko maximálne je na¹e rie¹enie hor¹ie ako optimálne.
+
+\s{Problém: Obchodný cestujuci}
+
+\>{\I Vstup:} neorientovaný graf $G$, popisujúci nejaku krajinu a~ka¾dá hrana je ohodnotená funkciou $w: E(G)\rightarrow R^+_0$
+\>{\I Vystup:} Hamiltonovská kru¾nica (v¹etky vrcholy grafu), a~to tá najkrat¹ia (podµa ohodnotenia).
+
+\>Tento problém je hneï na~prvý pohµad nároèný - u¾ problém, èi existuje Hamiltonovská kru¾nica je NP-úplný. BUNV nech graf~$G$ je úplný (doplnime zvy¹né hrany ohodnotené $max(w)+1$ alebo viac, nie v¹ak nekoneènom, lebo by neplatila trojuholníková nerovnos»). Vyrie¹me tento problém najprv za~predpokladu, ¾e vrcholy grafu spåòajú trojuholníkovú nerovnos», potom bez nej.
+
+\>{\I a) trojuholníková nerovnos»:} $\forall x,y,z \in V: w(xz)\le w(xy)+w(yz)$
+
+\>Existuje pekný algoritmus, ktory nájde Hamiltonovsku kru¾nicu, èo je
+maximálne dvakrát tak veµká ako najoptimálnej¹ia.
+
+\>Nájdeme najmen¹iu kostru a~obchodnému cestujúcemu poradíme, nech ide po~nej (staèí zakoreni» a~prejs» do~håbky). Problémom v¹ak je, ¾e daný sled obsahuje ka¾dý vrchol viackrát a~preto musíme nahradi» nepovolené vracania sa, t.j.~pre ka¾dý vrchol nájs» e¹te nenav¹tívený vrchol v~na¹om slede a~ís» priamo naò. Keï¾e platí trojuholníková nerovnos», tak si týmito skratkami neu¹kodíme. Nech minimálna kostra má váhu~$T$. Váha obídeného sledu tak bude~$2T$. Skrátenia urèite nezväè¹ujú, tak¾e váha nájdene Hamiltonovskej kru¾nice bude nanajvý¹~$2T$.
+
+\>Ak máme Hamiltonovskú kru¾nicu~$C$ a~z~nej vy¹krtneme hranu, tak máme kostru grafu~$G$ s~váhou najviac~$w(C)$, teda to aspoò takú, aká je váha minimálnej kostry - $T$. To je optimálny prípad Hamiltonovskej kru¾nice. Ak to teda zlo¾íme dohromady, algoritmus nám vráti Hamiltonovskú kru¾nicu s~váhou najviac dvojnásobnou od~optimálnej Hamiltonovskej kru¾nice. Takéto algoritmy sa nazývajú {\I 2-aproximaèné}, keï¾e rie¹enie je maximálne dvojnásobné od~optimálneho.
+
+\>{\I b) bez~trojuholníkovej nerovnosti}
+\>Tu sa budeme naopak sna¾i» ukáza», ¾e ¾iaden polynomiálny aproximaèný algoritmus neexistuje.
+
+\s{Veta:} Ak existuje polynomiálny $(1+\varepsilon)$-aproximaèný algoritmus pre~algoritmus obchodného cestujúceho bez~trojuholníkovej nerovnosti pre~µubovoµné $\varepsilon>0$, potom $P~=~NP$.
+\>Dôkaz: Uká¾eme, ¾e v~tom prípade doká¾eme v~polynomiálnom èase nájs» Hamiltonovskú kru¾nicu.
+
+\>Dostali sme graf $G$, v~ktorom hµadáme Hamiltonovskú kru¾nicu. Doplníme $G$ na~uplný graf~$G'$ a~váhy hrán~$G'$
+\itemize\ibull
+\: $w(e) = 1$, ak $e \in E(G)$
+\: $w(e) = c \ll 1$, ak $e \in E(G)$
+\endlist
+\>Ak existuje Hamiltonovská kru¾nica v~$G'$ zlo¾ená iba z~hrán, ktoré boli pôvodne v~$G$, tak optimálné rie¹enie bude ma» váhu $n$, inak bude urèite minimálne $n-1+c$. Ak máme aproximaèný algoritmus s~pomerom $1+\varepsilon$, musí by»
+$$
+\eqalign{
+(1+\varepsilon)\cdot n < n-1+c
+c > \varepsilon\cdot n+1
+}
+$$
+\>Ak by taký algoritmus existoval, tak na~neho máme polynomiálny algoritmus
+na~Hamiltonovsku kru¾nicu. Inak neexistuje ani pseudo-polynomialny algoritmus.
+
\h{Aproximaèní schéma pro problém batohu}
\s{POZOR:} Verze algoritmu, kterou jsem øíkal na~pøedná¹ce, obsahovala jednu
Nyní si v¹imneme, ¾e velikost toku mù¾eme mìøit pøes libovolný separátor: je to mno¾ství
tekutiny, které teèe pøes separátor z~$A$ do~$B$ minus to, které se vrací zpátky
-(zatím jsme velikost mìøili u~zdroje, vlastnì na~triviálním separátoru $A=\{v\}$, $B=\overline A).$
+(zatím jsme velikost mìøili u~zdroje, vlastnì na~triviálním separátoru $A=\{z\}$, $B=\overline A).$
\s{Lemma:} Nech» $A\subseteq V(G),z\in A,s\not\in A$ a $f$ je libovolný tok. Potom platí,
¾e: $$\vert f\vert=f(A,\overline A)-f(\overline A,A).$$
\figure{cesta.eps}{Pøíklad zlep¹ující cesty.}{3in}
\s{Definice:} {\I Zlep¹ující cesta} $P$ ze $z$ do $s$ je {\I nasycená}, pokud:
-$$\forall e \in P:\cases{
+$$\exists e \in P:\cases{
f(e)=c(e) & \hbox{je-li $e$ orientovaná po smìru} \cr
f(e)=0 & \hbox{je-li $e$ orientovaná proti smìru} \cr
}$$
-%version 1.6
+%version 1.8
\input lecnotes.tex
Pokud platí, ¾e:
\numlist \ndotted
\:ve~vrcholu~$u$ je nenulový pøebytek, tj. $f^{\Delta}(u) > 0$,
- \:vrchol~$u$ je vý¹ ne¾ vrchol~$v$, tj. $h(u) > h(v)$ a
+ \:vrchol~$u$ je vý¹ ne¾ vrchol~$v$, tj. $h(u) > h(v)$, a
\:hrana $uv$ má nenulovou rezervu, tj. $r(uv)>0$,
\endlist
\noindent pøevedeme tok o~velikosti $\delta:=\min(f^{\Delta}(u),r(uv))$ z~$u$ do~$v$ tímto zpùsobem:
\:$\forall e \in E: f(e)\leftarrow 0$ (po~hranách na~poèátku nenecháme protékat nic).
\:$\forall zu \in E : f(zu)\leftarrow c(zu)$ (ze~zdroje pustíme maximální mo¾nou vlnu).
\:Dokud $\exists u \in V \setminus \{z,s\}, f^{\Delta}(u)>0$:
-\::Pokud $\exists uv \in E, r(uv)>0$ a $h(u)>h(v)$, pøevedeme pøebytek po~hranì $uv$.
+\::Pokud $\exists uv \in E, r(uv)>0$ a $h(u)>h(v)$: pøevedeme pøebytek po~hranì $uv$.
\::V~opaèném pøípadì zvedneme $u$.
\:Vrátíme tok $f$ jako výsledek.
\endalgo
\proof
Nejprve uká¾eme, ¾e $f$ je tok, a pak jeho maximalitu. Vyjdìme z~toho, ¾e $f$ je vlna a algoritmus se mù¾e zastavit, jen pokud nastanou oba následující pøípady souèasnì:
\itemize\ibull
-\:Ve~vrcholech grafu nejsou ¾ádné pøebytky (mimo $z$ a $s$), proto¾e jinak by se algoritmus nezastavil a pokraèoval dále ve~výpoètu. Tudí¾ $f$ je tok. %todo check ? mimo~ ?
+\:Ve~vrcholech grafu nejsou ¾ádné pøebytky (mimo $z$ a~$s$), proto¾e jinak by se algoritmus nezastavil a pokraèoval dále ve~výpoètu. Tudí¾ $f$ je tok.
\:Neexistuje nenasycená cesta ze~zdroje do~stoku, èím¾ z~{\I Ford-Fulkersonovy vìty} okam¾itì vyplývá, ¾e $f$ je tok maximální. A jak tuto neexistenci nahlédneme? Pro~spor pøedpokládejme, ¾e nìjaká nenasycená cesta~$P$ ze~$z$ do~$s$ existuje. Tato cesta mù¾e mít maximálnì $N-1$ hran. O~nich víme, ¾e v¹echny mají kladnou rezervu, a dále víme, ¾e po~celou dobu výpoètu je vý¹ka zdroje $N$ a vý¹ka stoku $0$. Tak¾e celkový spád cesty $P$ je $N$, co¾ ale znamená, ¾e na cestì $P$ existuje hrana s~kladnou rezervou, která má spád alespoò $2$. To je v¹ak v~rozporu s~invariantem~S.
\qeditem
\endlist
\s{Definice:} {\I Separátor $S_n$}
je sí», ve které jsou v¾dy~$i$-tý a~$(i+{n/2})$-tý prvek vstupu
(pro $i=0,\dots, {n/2}-1$) propojeny komparátorem, minimum bude~$i$-tým,
-maximum $(i+1)$-ním prvkem výstupu.
+maximum $(i+{n/2})$-ním prvkem výstupu.
\figure{sortnet.3}
{$(y_i, y_{i+{n/2}}) = CMP(x_i, x_{i+{n/2}})$} {300pt}
se prohodí. Pokud ho nenajdeme, separátor neudìlá vùbec nic a~obì
tvrzení lemmatu zøejmì tedy platí. Øeknìme, ¾e $x_m$ je maximum
vstupní posloupnosti. Pak~$k$ bude jistì men¹í ne¾~$m$
-a~$x_{k+{n/2}}$ bude vìt¹í ne¾~$m$, mezi~$k$ a~$m$ je tedy vstupní
+a~$k+{n/2}$ bude vìt¹í ne¾~$m$, mezi~$k$ a~$m$ je tedy vstupní
posloupnost neklesající, mezi $k+{n/2}$ a~$n-1$ nerostoucí.
Uvìdomíme si, ¾e pro ka¾dé~$i$, $k\leq i\leq {n/2}-1$ se prvky
$x_i$ a~$x_{i+{n/2}}$ prohodí. Úsek mezi~$k$ a~${n/2}-1$ tedy
\itemize\ibull
\s{Definice:}
\itemize\ibull
-\:{\I Abeceda $\Sigma$} je koneèná mno¾ina znakù, ze kterých tvoøíme text, øetìzece, slova jako koneèné posloupnosti znakù ze $\Sigma$. Pøíkladem extrémních abeced je lineární abeceda slo¾ená z~nul a jednièek. Pøíklad s~druhého konce je abeceda, která má jako znaky slova èeského jazyka. V algoritmech nebudeme uva¾ovat velikost abecedy (poèet znakù).
+\:{\I Abeceda $\Sigma$} je koneèná mno¾ina znakù, ze kterých tvoøíme text, øetìzece, slova jako koneèné posloupnosti znakù ze $\Sigma$. Pøíkladem extrémních abeced je binární abeceda slo¾ená z~nul a jednièek. Pøíklad z~druhého konce je abeceda, která má jako znaky slova èeského jazyka. V algoritmech nebudeme uva¾ovat velikost abecedy (poèet znakù).
\:{\I $\Sigma$*} je mno¾ina v¹ech slov nad abecedou $\Sigma$.
\endlist
\s{Znaèení:}
\s{Vyhledávaní:}
\algo
\:$\alpha \leftarrow \varepsilon$.
-\:Pro $C\in\Sigma$ postupnì:
-\:$\indent$Dokud $\neg \exists d(\alpha , C) \wedge \alpha\neq\varepsilon : \alpha \leftarrow z(\alpha)$.
-\:$\indent$Jestli¾e $\exists d(\alpha , C) \Rightarrow \alpha \leftarrow d(\alpha , C)$.
-\:$\indent$Jestli¾e $\alpha = \iota \Rightarrow$ hledané slovo je v~textu.
+\:Pro $C\in\sigma$ postupnì:
+\::Dokud $\neg \exists d(\alpha , C) \wedge \alpha\neq\varepsilon$: $\alpha \leftarrow z(\alpha)$.
+\::Pokud $\exists d(\alpha , C)$: $\alpha \leftarrow d(\alpha , C)$.
+\::Pokud $\alpha = \iota$: vrátíme, ¾e hledané slovo je v~textu.
\endalgo
\s{Alternativa:}
\algo
\:$k \leftarrow 0$.
-\:Pro $C\in\Sigma$ postupnì:
-\:$\indent$Dokud $C\neq \iota[k] \wedge k>0: k \leftarrow z(k)$.
-\:$\indent$Jestli¾e $C=\iota[k] \Rightarrow k++$.
-\:$\indent$Jestli¾e $k = J \Rightarrow$ hledané slovo je v~textu.
+\:Pro $C\in\sigma$ postupnì:
+\::Dokud $C\neq \iota[k] \wedge k>0$: $k \leftarrow z(k)$.
+\::Pokud $C=\iota[k]$: $k \leftarrow k+1$.
+\::Pokud $k = J$: vrátíme, ¾e hledané slovo je v~textu.
\endalgo
-\s{Invariant:} Stav po pøeètení vstupu $\beta$: $\alpha(\beta)$ $=$ nejdel¹í suffix $\beta$, který je prefixem $\iota$.
+\s{Invariant:} Stav po pøeètení vstupu $\beta$: $\alpha(\beta)$ je nejdel¹í suffix $\beta$, který je prefixem $\iota$.
Z~invariantu vyplývá korektnost vyhledávací èásti algoritmu KMP.
\proof
\prednaska{7}{Vyhledávání v textu}{(zapsali J. Kunèar, M. Demin a J. Chludil)}
-Na minulých predná¹kách jsme si ukázali, jak se v textu vyhledává slovo. Teï si ov¹em úlohu zobecníme a uká¾eme si, jak v kupce sena vyhledat více ne¾ jednu jehlu.
+Na minulých predná¹kách jsme si ukázali, jak se v~textu (senì) vyhledává slovo (jehlu). Teï si ov¹em úlohu zobecníme a uká¾eme si, jak v kupce sena hledat souèasnì více ne¾ jednu jehlu.
+
+\h{Hledání výskytu v¹ech slov}
-\h{Zopakujeme si základní znaèení}
\itemize\ibull
-\:$\iota_1, \ldots, \iota_k$ -- vyhledávaná slova (jehly)
-\:$\sigma$ -- text, kde se hledá (seno)
+\:$\iota_1, \ldots, \iota_k$ -- vyhledávaná slova (jehly) délek $ J_i= \vert\iota_i\vert $
+\:$\sigma$ -- text (seno) délky $ S= \vert\sigma\vert $
\endlist
+Nejprve si øekneme, jak chceme, aby vypadal výstup. Výstupem pro nás budou v¹echny uspoøádané dvojice $(i,j)$ ($i$ je index jehly, kterou jsme nalezli, a $j$ je poèáteèní pozice v senì, kde se jehla nachází) takové, ¾e $$\iota_i=\sigma[j:j+J_i].$$ Postavme si proto vyhledávací automat podobný tomu, který jsme vidìli na minulé pøedná¹ce. Tento automat nám v¹echny takové uspoøádané dvojice najde.
-\h{Hledání výskytu v¹ech slov}
-Nejprve si øekneme, jak chceme, aby vypadal výstup, a poté jak ho dosáhnout. Výstupem pro nás budou v¹echny uspoøádané dvojice $(i,j)$ takové, ¾e $$\iota_i=\sigma[j:j+\vert\iota_i\vert]$$ Postavme si proto vyhledávací automat, který najde v¹echny takové uspoøádané dvojice.
\h{Vyhledávací automat}
-{\I Vyhledávací automat} je vlastnì strom\foot{http://en.wikipedia.org/wiki/Trie}, kde ka¾dý vrchol mù¾e mít stupeò a¾ do velikosti abecedy a kde jednotlivé hrany odpovídají písmenùm této abecedy. Vrcholy, ve kterých konèí slovo, jsou oznaèené (na obrazcích èernì). Dále si èasem do tohto vyhledávacího stromu pøidáme zpìtné hrany a \uv{zkratky}.
+Vyhledávací automat je strom\foot{http://en.wikipedia.org/wiki/Trie}, kde ka¾dý vrchol mù¾e mít stupeò a¾ do velikosti abecedy a kde jednotlivé hrany odpovídají písmenùm této abecedy. Vrcholy, ve kterých konèí slovo, jsou oznaèené (na obrazcích èernì). Dále si èasem do tohoto vyhledávacího stromu pøidáme zpìtné hrany a \uv{zkratky}.
-\s{Definice:} {\I Stav} je pozice ve stromì, která odpovídá nejdel¹ímu prefixu vyhovující jehly v senì (platí rovnì¾ stejný invariant z pøedchozí pøedná¹ky).
-
-\s{Definice:} {\I Zpìtná hrana} $z$($\alpha$) := nejdel¹í vlastní suffix\foot{definováno na 6. pøedná¹ce} slova $\alpha$, který je stavem.
+\s{\I{Stavy}} automatu jsou urèeny vrcholy stromu, pro které platí rovnì¾ stejný {\I{invariant}} z pøedchozí pøedná¹ky.\par
+\s{\I{Zpìtná hrana}} $z$($\alpha$) := nejdel¹í vlastní suffix\foot{definováno na minulé pøedná¹ce} slova $\alpha$, který je stavem.\par
\figure{vyhl_automat_dopr.eps}{Vyhledávací automat}{1in}
\h{Výstup z automatu}
Pøi vypisování výsledkù mu¾eme narazit na urèité problémy, které jsou dobøe vidìt na následujícím obrázku. První problém urèitì nastane, proto¾e v automatu není pøesnì øeèeno, které slovo konèí v jakém vrcholu.
Napøíklad ve stavu, kde konèí slovo BARBARA, konèí také slovo ARA, ale o tom nevíme.
-Druhý problém nastává, kdy¾ v automatu není zaznaèen konec slova. Pøíkladem je seno BARAB (jednoduché k~nahlédnutí, viz obrázek).
+Druhý problém nastává, kdy¾ v automatu není informace o~konci slova. Pøíkladem je seno BARAB (jednoduché k~nahlédnutí, viz obrázek).
Teï nám nezbývá nic jiného, ne¾ najít øe¹ení tìchto záludných problémù. Øe¹e¹í se nám naskýtá hned nìkolik:
\itemize\ibull
-\:Projdeme v¹echy zpìtné hrany a vypí¹eme slova, jen¾ v daných stavech konèí. Toto øe¹ení funguje, ale je pomalé, proto¾e procházíme v¹echny zpìtné hrany.
-\:Pøedpoèítání mno¾in. Najdeme mno¾inu slov tak, aby celková velikost slov byla vìt¹í ne¾ lineární. Funkèní, ale konstrukce je pomalá.
-\:$\<slovo>(s) =$ index slova $\iota$, které konèí ve stavu $s$, nebo 0, \par
-$\<out>(s) =$ nejbli¾¹í vrchol, do kterého se lze z $s$ dostat po zpìtných hranách a $\<slovo>(v) \ne 0$ (konèí tam slovo).
-\figure{Graphic2.eps}{Vyhledávací automat -- se zpìtnými hranami}{1.3in}
+\:Projdeme v¹echy zpìtné hrany a vypí¹eme slova, je¾ v daných stavech konèí. Toto øe¹ení funguje, ale je pomalé, proto¾e poka¾dé procházíme v¹echny zpìtné hrany.
+%\:Pøedpoèítání mno¾in slov. Najdeme mno¾inu slov tak, aby celková velikost slov byla vìt¹í ne¾ lineární. Funkèní, ale konstrukce je pomalá.
+\:Pro následující øe¹ení, jen¾ spoèívá v nalezení zkratek ve stromì, si zavedeme toto znaèení:\par
+\s{\<slovo>($s$)} = index slova $\iota$, které konèí ve stavu $s$, nebo $\emptyset$ \par
+\s{\<out>($s$)} = nejbli¾¹í vrchol, do kterého se lze z $s$ dostat po zpìtných hranách a \<slovo(v)> $\ne 0$ (konèí tam slovo)
+\figure{Graphic2.eps}{Vyhledávací automat se zpìtnými hranami}{1.3in}
\endlist
-\>Jako vhodné øe¹ení tohoto problému se naskýtá poslední bod. Podle nìho vytvoøíme algoritmus na vyhledávání \uv{jehel v senì}.
+\>Podle posledního bodu vytvoøíme algoritmus na vyhledávání \uv{jehel v senì}.
\algo
-\:$s \leftarrow \<koøen>$ ($s$ je aktuální stav vyhledávacího automatu).
+\:$s \leftarrow$ \<koøen> ($s$ bude aktuální stav vyhledávacího automatu).
\:Procházíme v¹echny písmena $c$ v senì $\sigma$:
\::$s \leftarrow krok(s,c)$.
-\::Je-li $\<slovo>(s) \ne 0 \Rightarrow$ vypí¹eme $\<slovo>(s)$.
-\::$v \leftarrow out(s)$.
-\::Dokud $v \ne 0 $:
+\::Je-li $\<slovo>(s) \ne 0$, vypí¹eme $\<slovo>(s)$.
+\::$v \leftarrow \<out>(s)$.
+\::Dokud $v \ne 0$:
\:::Vypí¹eme $\<slovo>(v)$.
\:::$v \leftarrow \<out>(v)$.
\endalgo
-\>$\<krok>(s,c) :=$ jeden krok ve vyhledávacím automatu:
+\s{\<krok(s,c)>}:= jeden \<krok> vyhledávacího automatu:
\algo
-\:Dokud $\neg \exists f(s,c) \wedge s \ne$ koøen: $s \leftarrow \<z>(s)$.
-\:Pokud $\exists f(s,c)$: $s \leftarrow \<f>(s,c)$.
+\:Dokud $\not\exists f(s,c) \wedge s \ne$ \<koøen>: $s \leftarrow z(s)$.
+\:Pokud $\exists f(s,c)$: $s \leftarrow f(s,c)$.
\:Vrátíme $s$.
\endalgo
\h{Reprezentace v pamìti}
-První mo¾nost jak reprezentovat vyhledávací automat je pole se seznamem synù. Je to jednoduchá varianta, ale má nevýhodu pro velké abecedy, proto¾e procházení seznamu synù mù¾e trvat neúmìrnì dlouho. Proto se nabízí druhá mo¾nost a to hashovací tabulka $(\<stav>,\<znak>) \rightarrow \<f>(\<stav>,\<znak>)$, kde se \uv{ztratí} pou¾ívání hashovací funkce.
+První mo¾nost, jak reprezentovat vyhledávací automat, je jednorozmìrné pole vrcholù stromu, v nìm¾ ukládáme seznam synù pro ka¾dý vrchol. Je to jednoduchá varianta, ale má nevýhodu pro velké abecedy, proto¾e procházení seznamu synù mù¾e trvat neúmìrnì dlouho. Proto se nabízí druhá mo¾nost a to hashovací tabulka $(\<stav>,\<znak>) \rightarrow f(\<stav>,\<znak>)$, kde se \uv{ztratí} pou¾ívání hashovací funkce.
\h{Slo¾itost}
\itemize\ibull
-\:Kroky 2--5 mají èasovou slo¾itost $\O(\vert \sigma \vert)$, kterou jednodu¹e doká¾eme pomocí potenciálu -- poèet krokù nahoru $ \leq $ poèet krokù dolù $ \leq \vert \sigma \vert$, kde $\vert \sigma \vert$ je délka sena.
-\:Kroky 6--8 mají èasovou slo¾itost $\O(\<poèet výskytù>)$, proto¾e rychleji opravdu nelze v¹echny výskyty vypsat.
+\:Kroky 2.--5. mají èasovou slo¾itost $\O(\vert \sigma \vert)$, kterou jednodu¹e doká¾eme pomocí potenciálu -- poèet krokù nahoru je men¹í nebo roven poètu krokù dolù. A to je maximálnì $S$.
+\:Kroky 6.--8. mají èasovou slo¾itost $\O(\<poèet výskytù>)$, proto¾e rychleji doopravdy nelze v¹echny výskyty vypsat.
\endlist
-\s{Konstrukce automatu (Aho, Corasicková)}
+\s{Konstrukce automatu} (Aho, Corasicková)
\algo
\:Postavíme strom dopøedných hran, $r \leftarrow$ koøen stromu.
-\:Spoèteme $\<slovo>(\ast)$ (spoèteme funkci \<slovo> pro v¹chny stavy).
-\:Spoèteme $z(\ast)$: $z(\beta)=\alpha(\beta[1:])$.
-\itemize\ibull
-\:$z(\beta) = \alpha(\beta[1:])$ -- v¹echny zpìtné hrany vedou do vy¹¹ích hladin
-\:$z(v) = \<krok>(z(u),c)$
-\endlist
+\:Spoèteme $\<slovo>(\ast)$ -- oznaèíme si stavy, kde konèí slova.
+\:Spoèteme $z(\ast)$: $z(\beta)=\alpha(\beta[1:])$:
+ {\parindent=6em \itemize\ibull
+ \:\>\>\>$z(\beta) = \alpha(\beta[1:])$ -- v¹echny zpìtné hrany vedou do vy¹¹ích hladin
+ \:$z(v) = \<krok>(z(u),c)$
+ \endlist}
\figure{Graphic100.eps}{$\<z>(v) = \<krok>(z(u),c)$}{0.7in}
\:$z(r) \leftarrow 0$, do fronty $Q$ pøiøadíme v¹echny syny $r$, pro v¹echny $v$ prvky $Q: z(v) \leftarrow r$.
\:Dokud fronta $Q$ není prázdná:
-\::$u\leftarrow$ vybereme z $Q$.
+\::$u\leftarrow$ vybereme z~$Q$.
\::Pro syny $v$ vrcholu $u$:
-\:::$R \leftarrow \<krok>(z(u)$, znak na hranì \<uv>).
+\:::$R \leftarrow \<krok>(z(u))$ [znak na hranì \<uv>].
\:::$z(v)\leftarrow R$.
\figure{Graphic101.eps}{}{0.7in}
-\:::Je-li $slovo(R) \not= 0 \Rightarrow out(v) \leftarrow R$, jinak $out(v) \leftarrow out(R)$.
+\:::Je-li $slovo(R) \not= 0 \Rightarrow \<out>(v) \leftarrow R$, jinak $\<out>(v) \leftarrow \<out>(R)$.
\figure{Graphic102.eps}{}{0.7in}
\endalgo
\figure{vyhl_automat_full.eps}{Vyhledávací automat -- kompletní}{1in}
\s{Vìta:}
-Algoritmus A-C najde v¹echny výskyty slov $\iota_1, \ldots, \iota_k$ ve slove $\sigma$ v èase: $$\O(\Sigma_i \vert \iota_i \vert + \vert \sigma \vert + \<poèet výskytù>).$$
+Algoritmus Aho-Corasicková najde v¹echny výskyty slov $\iota_1, \ldots, \iota_k$ ve~slovì $\sigma$ v~èase $\O(\sum_i \vert \iota_i \vert + \vert \sigma \vert + \<poèet výskytù>)$.
\h{Polynomy a násobení}
\>Mìjme dva polynomy definované jako:
\figure{polynom.eps}{Polynom}{2in}
\ss{Plán:}
-\>Nech» $k=2n-1$, zvolíme $x_0, \ldots, x_k$ libolná, ale rùzná a spoèteme $P(x_0), \ldots, P(x_k)$ a $Q(x_0), \ldots, Q(x_k)$.
+
+\>Nech» $k=2^{n-1}$. Zvolíme èísla $x_0, \ldots, x_k$ libovolná, ale rùzná, a spoèteme $P(x_0)$, \dots, $P(x_k)$ a $Q(x_0), \ldots, Q(x_k)$.
Poté $\forall j: y_j=P(x_j)Q(x_j)$
musíme najít polynom $R$ stupnì $\leq k: \forall j: R(x_j)=y_j$.
\>BÚNO $n=2^m$. Uva¾me polynom:
$$P(x) = p_0 x^0 + p_1 x^1 + \ldots + p_{n-1} x^{n-1}.$$
-Tento polynom si mu¾eme rozdìlit, na dvì èásti. V~levé budeme mít èleny se sudými exponenty a v~pravé budou èleny s exponenty lichými:
+Tento polynom si mu¾eme rozdìlit na dvì èasti. V levé budeme mít èleny se sudými exponenty a v~pravé budou èleny s~exponenty lichými:
$$P(x) = (p_0 + p_2 x^2 + \ldots + p_{n-2}x^{n-2}) + (p_1 x^1 + p_3 x^3 + \ldots + p_{n-1} x^{n-1}).$$
Z pravé strany mù¾eme vytknout $x$ a dostaneme:
-$$P(x) = (p_0 + p_2 x^2 + \ldots + p_{n-2}x^{n-2}) + x(p_1 + p_3 x^2 + \ldots + p_{n-1} x^{n-2})$$
+$$P(x) = (p_0 + p_2 x^2 + \ldots + p_{n-2}x^{n-2}) + x(p_1 + p_3 x^2 + \ldots + p_{n-1} x^{n-2}),$$
$$ \vdots $$
$$P(x) = L(x^2) + xN(x^2),$$
$$P(-x) = L(x^2) - xN(x^2),$$
-kde $L(x)$ a $N(x)$ jsou polynomy stupnì $n/2$. Umocnìním $x^2$ se nám poru¹í párování $x$ a $-x$, proto musíme poèítat v $\bb{C}$.
-Musíme si ale uvìdomit, ¾e tyto vztahy platí pouze, kdy¾ existuje pár $-x$ a $x$ v tìlese, nad kterým poèítáme. V~tomto pøípadì jsme z~polynomu s~$n$ koeficienty v~$n$ bodech dostali $2$ polynomy s~$n/2$ koeficienty v~$n/2$ bodech. Z~toho vyplývá èasová slo¾itost definována vztahem:
+kde $L(x)$ a $N(x)$ jsou polynomy stupnì $n/2$. Umocnìním $x^2$ se nám poru¹í párování $x$ a $-x$, proto musíme poèítat v~$\bb{C}$ místo~$\bb{R}$.
+V~tomto pøípadì jsme z~polynomu s~$n$ koeficienty v~$n$ bodech dostali $2$ polynomy s~$n/2$ koeficienty v~$n/2$ bodech. Z~toho vyplývá èasová slo¾itost definována vztahem:
$$T(n) = 2T(n/2) + \O(n).$$
Ten mù¾eme vyøe¹it s pou¾itím Master Theoremu z~ADS~I a dostaneme:
$$T(n) = \O(n \log n).$$
\bye
-
-
\>Je asi vidìt, ¾e klíèové jsou kroky 2 a 4. Vybrání bodù jistì stihneme pohodlnì v~lineárním èase a vynásobení samotných hodnot té¾ (máme $2n+2$ bodù a $C(x_{k}) = A(x_{k}) \cdot B(x_{k}), k = 0,1,2, \ldots , 2n+1$, tak¾e na to nepotøebujeme více ne¾ $2n+2$ násobení).
-Celý trik spoèívá v~chytrém vybrání onìch bodù, ve kterých budeme polynomy vyhodnocovat. Je na to potøeba vìdìt pár zajímavostí o~komplexních èíslech, na stránce Matrina Mar¹e jsou k dispozici slajdy, zde to bude zapsáno o~trochu struènìji.
+Celý trik spoèívá v~chytrém vybrání onìch bodù, ve kterých budeme polynomy vyhodnocovat. Je na to potøeba vìdìt pár zajímavostí o~komplexních èíslech, na stránce Matrina Mare¹e jsou k dispozici slajdy, zde to bude zapsáno o~trochu struènìji.
\ss{Vyhodnocení polynomu metodou rozdìl a panuj (algoritmus FFT):}
Mìjme polynom $P$ øádu $n$ a chceme jej vyhodnotit v $n$ bodech. Vybereme si body tak, aby byly spárované, èili $\pm x_{0}, \pm x_{1}, \ldots , \pm x_{n/2-1} $. To nám výpoèet urychlí, proto¾e pak se druhé mocniny $x_{i}$ shodují s~druhými mocninami $-x_{i}$.
Máme tedy algoritmus, který \uv{pøevede} koeficienty polynomu na hodnoty tohoto polynomu v~námi zadaných bodech. Ale potøebujeme také algoritmus, který doká¾e reprezentaci polynomu pomocí hodnot pøevést zpìt na koeficienty polynomu. Tedy nìjaký inverzní algoitmus.
-Definuje me si algoritmus DFT, která vyu¾ívá maticovou reprezentaci a s~jeho¾ pomocí získáme hledaný algoritmus.
+Definujeme si algoritmus DFT, která vyu¾ívá maticovou reprezentaci a s~jeho¾ pomocí získáme hledaný algoritmus.
\s{Definice:}
\>{\I Diskretní Fourierova transformace} $(DFT)$
je funkce $f: { {\bb C} ^n} \rightarrow { {\bb C} ^n}$, kde $y=f(x) \equiv \forall j \ y_{j} = \sum \limits ^{n-1}_{k=0} x_{k} . \omega ^{jk}$.
\s{Poznámka:}
-Vezmeme polynom, který má $x_{kj}$ jako koeficienty a vyhodnotíme ho v~bodì $\omega ^{j} [y_{j} = x(\omega^{j})] \Rightarrow {f}$ je linearní $\Rightarrow$ mù¾eme napsat $f(x) = \Omega_{x} ,\ \Omega _{jk} =\omega ^{jk}$, kde $\Omega$ je matice.
+Vezmeme polynom, který má $x_{kj}$ jako koeficienty a vyhodnotíme ho v~bodì $\omega ^{j} [y_{j} = x(\omega^{j})] \Rightarrow {f}$ je linearní $\Rightarrow$ mù¾eme napsat $f(x) = \Omega x ,\ \Omega _{jk} =\omega ^{jk}$, kde $\Omega$ je matice.
\s{Jak najít inverzní matici?} Víme, ¾e $\Omega =\Omega ^{T}$ proto¾e $\omega ^{jk} = \omega ^{kj}$.
\h{Hledání konvexního obalu}
Ptáte se o co pùjde? Zkusme si to pøiblí¾it na problému ledních medvìdù :)
-{\I Lední medvìdi si po dlouhé dobì zmapovaly vody severního moøe a zjistili pøesnì místa, kde se nacházejí jejich oblíbené ryby. No a proto¾e to jsou medvìdi chytøí tak se rozhodli v¹echny tyto rybky pochytat najednou do jedné veliké sítì. A problém, který tady mají je takovýto: Jaký nejmen¹í obvod mù¾e mít taková sí», aby se dovnitø ve¹ly je¹tì v¹echny rybky?!}
+{\I Lední medvìdi si po dlouhé dobì zmapovali vody severního moøe a zjistili pøesnì místa, kde se nacházejí jejich oblíbené ryby. No a proto¾e to jsou medvìdi chytøí, rozhodli se v¹echny tyto rybky pochytat najednou do jedné velké sítì. A problém, který tady mají, je následující: jaký nejmen¹í obvod mù¾e mít taková sí», aby se dovnitø ve¹ly je¹tì v¹echny rybky?!}
Neboli budeme øe¹it, jak nìjakou zadanou mno¾inu bodù v~rovinì obalit co nejkrat¹í uzavøenou køivkou, do které by se je¹tì v¹echny body ve¹ly.
-Intuice nám napovídá ¾e výsledek bude nìjaký konvexní\foot{Mno¾ina bodù v~rovinì je konvexní útvar, pokud platí ¾e pro ka¾dé dva body této mno¾iny je úseèka spojující tyto dva body také celá v~této mno¾inì} mnohouhelník, který bude mít vrcholy v~nìkterých uvedených bodech. Ostatní vrcholy pak budou buï nìkde na hranách mnohouhelníku, nebo uvnitø.
+Intuice nám napovídá ¾e výsledek bude nìjaký konvexní\foot{Mno¾ina bodù v~rovinì je konvexní útvar, pokud platí ¾e pro ka¾dé dva body této mno¾iny je úseèka spojující tyto dva body také celá v~této mno¾inì} mnohoúhelník, který bude mít vrcholy v~nìkterých uvedených bodech. Ostatní vrcholy pak budou buï nìkde na hranách mnohoúhelníku, nebo uvnitø.
\>Mo¾ná by se teï hodilo pøedvést názornì jak vypadají nejmen¹í konvexní obaly:
\:konvexní obal prázdné mno¾iny je prázdná mno¾ina
\:konvexní obal 1 bodu je bod samotný
\:konvexní obal 2 bodù je úseèka spojující tyto body
-\:konvexní obal 3 bodù je trojuhleník s vrcholy v~tìchto bodech
+\:konvexní obal 3 bodù je trojúhleník s vrcholy v~tìchto bodech
\:konvexní obal 4 bodù \dots to u¾ je slo¾itìj¹í
\endlist
-Konvexní obaly 4 a více bodù, jak si mù¾em v¹imnout, u¾ nejsou jednoznaèné. Pro $N$-prvkovou mno¾inu bude konvexní obal mnohouhelník se tøema a¾ $N$ vrcholy.
+Konvexní obaly 4 a více bodù, jak si mù¾em v¹imnout, u¾ nejsou jednoznaèné. Pro $N$-prvkovou mno¾inu bude konvexní obal mnohoúhelník se tøemi a¾ $N$ vrcholy.
Jeden dobrý zpùsob jak tento konvexní obal sestrojit se nazývá {\I Zametání roviny.}
Budeme takto potkávat body které le¾í v~na¹í mno¾inì.
v~ka¾dém okam¾iku budeme chtít, aby v~té èásti kterou jsme ji¾ zametli, jsme u¾ mìli spoèítaný konvexní obal. V¾dycky kdy¾ pak zametací pøímkou narazíme na nový bod, tak si u¾ jen rozmyslíme jak ho do toho konvexního obalu pøidat.
-BÚNO tady v¹ude bereme body v~obecné poloze, tedy takové, ¾e ¾ádné tøi nele¾í na jedné pøímce. Dál taky budem pøedpokládat ¾e budeme zametat ve smìru $X$-ové osy a ¾e v¹echny body mají rùznou $X$-ovou souøadnici.
+BÚNO tady v¹ude bereme body v~obecné poloze, tedy takové, ¾e ¾ádné tøi nele¾í na~jedné pøímce. Dál taky budem pøedpokládat ¾e budeme zametat ve smìru $X$-ové osy a ¾e v¹echny body mají rùznou $X$-ovou souøadnici.
Jde také vidìt ¾e bod s nejmen¹í a nejvìt¹í $X$-ovou souøadnicí bude le¾et v~konvexním obalu.
\s{Algoritmus:}
\algo
\:Setøídíme body podle jejich $X$-ové souøadnice.
-\:Vezmem první tøi body a sestojíme jejich konvexní obal.
+\:Vezmem první tøi body a sestrojíme jejich konvexní obal.
\:Opakuj: Najdeme dal¹í bod a podíváme se jestli ho mù¾eme do konvexního obalu rovnou pøidat:
\::Pokud jej mù¾eme rovnou pøidat, tak jej pøidáme.
\::Pokud jej pøidat rovnou nemù¾eme, pak je potøeba nejdøív~nìjaké body odzadu odebrat a pak teprv~pøipojit ná¹ nový bod .
\h{Voroného diagramy}
-Pøed tím, ne¾ vás vystra¹ím nìjakou definicí, si øekneme, co jsi pod tímto, na první pohled ne zøejmým pojmem, pøedstavit. Mìjme mno¾inu teèek T rozmístìných náhodnì po papíru. Ke ka¾dému bodu nakreslíme okraje tak, aby vniklá plo¹ka obsahovala body, které jsou nejblí¾e právì té na¹í vybrané teèce. Samozøejmì "sousední" teèky budou mít tyto hranice spoleèné. Výsledkem na¹eho dlouhého sna¾ení pak bude právì Voroného diagram. V dal¹ích odstavcích se budeme zajímat o to, jak takový útvar správnì popsat, jak ho sestrojit a jaké datové struktury k tomu pou¾ít.
+Pøed tím, ne¾ vás vystra¹ím nìjakou definicí, si øekneme, co jsi pod tímto, na první pohled ne zøejmým pojmem, pøedstavit. Mìjme mno¾inu teèek $T$ rozmístìných náhodnì po papíru. Ke ka¾dému bodu nakreslíme okraje tak, aby vniklá plo¹ka obsahovala body, které jsou nejblí¾e právì té na¹í vybrané teèce. Samozøejmì "sousední" teèky budou mít tyto hranice spoleèné. Výsledkem na¹eho dlouhého sna¾ení pak bude právì Voroného diagram. V dal¹ích odstavcích se budeme zajímat o to, jak takový útvar správnì popsat, jak ho sestrojit a jaké datové struktury k tomu pou¾ít.
\s{Definice:} Voronoi diagram pro koneènou mno¾inu $M = \{m_1, \dots, m_n\} \in {\bb R}^2$ míst je systém mno¾in $1..M_n$ takových, ¾e pro v¹echny $i$ a $j$ a pro v¹echny $x \in M_i$ je vzdálenost $x$ a $m_i$ men¹í nebo rovna vzdálenosti $x$ a $m_j$ a zároveò sjednocení $M_i$ pøes v¹echna $i$ je celý prostor ${\bb R}^2$, neboli:
\s{Pozorování:}
\itemize\ibull
-\:Pro v¹echny $i$ je $M_i$ ohranièena konvexní lomenou èarou, tak¾e oblasti mají tvar konvexních mnohoúhelníkù, ale je mo¾né, ¾e jsou oteveøené do nekneèna.
+\:Pro v¹echny $i$ je $M_i$ ohranièena konvexní lomenou èarou, tak¾e oblasti mají tvar konvexních mnohoúhelníkù, ale je mo¾né, ¾e jsou otevøené do nekoneèna.
\:Pro ka¾dou hranu $h$ ve Voroného diagramu existuje $i$ a $j$ takové, ¾e kdy¾ $x \in H$, pak vzdálenost $d(x,m_i) = d(x,m_j)$.
Øekneme, ¾e vrchol je takové místo, kde se potkávají alespoò dvì hrany.
Shrneme-li na¹e úvahy, mù¾ou se dít celkem tøi vìci. Jedna z nich je posun pøímky.
To se vlastnì dìje poøád. Nemìní se nám rif pobøe¾í a prùseèíky parabol nám kreslí
hrany. To mù¾eme poèítat najednou. Navíc nejen ¾e bychom mohli s pøímkou skoèít
-o nìjaké epsilon, ale mi dokonce mù¾eme skoèit o poøádný kus a prostì pouze dopoèítat,
+o nìjaké epsilon, ale my dokonce mù¾eme skoèit o poøádný kus a prostì pouze dopoèítat,
jak se nám zmìnilo pobøe¾í a co se nám vykreslilo.
Pokud tedy narazíme na bod, tak musíme najít místo, kde to na¹e pobøe¾í rozetnou
a kam vklínit dal¹í výbì¾ek (parabolu). Takovéto události budeme øíkat místní událost.
Pokud se pohybujeme v obecné poloze, tak se nám nestane, ¾e bychom narazili na prùseèík.
-\figure{mistni.eps}{Místní událost.}{3in}
+\figure{mistni.eps}{Místní událost -- èervená kolmice se bude dále rozevírat a vytvoøí dal¹í parabolu.}{3in}
Poslední situace, která tedy mù¾e nastat je, ¾e nám nìjaký parabola schová za jiné.
Kouknìme se na obrázek, fialový bod le¾í na ka¾dé parabole. Speciálnì tedy ta tøi
\s{Slo¾itost:}
Poèet místních událostí je roven $n$ a poèet kru¾nicových událostí není vìt¹í ne¾
$n$ (s ka¾dou místní událostí zanikne kru¾nicová), neboli velikost $P$ není vìt¹í
-ne¾ $2n$ a tedy je v¾dy lineární. Zároveò velikost $H$ není vìt¹í ne¾ $2n$,
+ne¾ $2n$, a tedy je v¾dy lineární. Zároveò velikost $H$ není vìt¹í ne¾ $2n$,
proto¾e aèkoliv~poèet místních událostí je $n$, tak v~$H$ je záznam pro ka¾dou
-trojici a tedy v~$H$ je maximálnì $2n$ událostí najednou. Zbývá nám tedy zjistit
+trojici, a tedy v~$H$ je maximálnì $2n$ událostí najednou. Zbývá nám tedy zjistit
velikost diagramu $D$, ale ten se urèitì vejde do $\O(n)$ pamìti.
Pokud tedy shrneme v¹echny na¹e odhady, pak èasová slo¾itost algoritmu je
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all:
clean:
- for a in *-* all ; do ( cd $$a && make clean ) ; done
+ for a in *-* all manual ; do ( cd $$a && make clean ) ; done
rm -f *~
.SECONDARY:
P=ads2
-X:=$(shell for a in 1 2 3 4 5 6 7 8 9 10 12 ; do echo ../$$a-*/$$a-*.tex ; done)
+X:=$(shell for a in 1 2 3 4 5 6 7 8 9 10 11 12 ; do echo ../$$a-*/$$a-*.tex ; done)
%universe: all ChangeLog
projects / ads2.git / commitdiff
