From: Martin Mares Date: Thu, 20 Oct 2011 21:37:15 +0000 (+0200) Subject: Dinicuv algoritmus: Oziveno z historie X-Git-Url: http://mj.ucw.cz/gitweb/?a=commitdiff_plain;h=2bb2972f62b45b749ad05b12bfda1d4262057b4e;p=ads2.git Dinicuv algoritmus: Oziveno z historie --- diff --git a/3-dinic/3-dinic.tex b/3-dinic/3-dinic.tex new file mode 100644 index 0000000..d451f01 --- /dev/null +++ b/3-dinic/3-dinic.tex @@ -0,0 +1,223 @@ +\input lecnotes.tex + +\prednaska{3}{Dinicùv algoritmus}{(zapsala Markéta Popelová)} + + +Na~minulé pøedná¹ce jsme si~ukázali Fordùv-Fulkersonùv algoritmus. Tento algoritmus hledal maximální tok tak, ¾e zaèal s~tokem nulovým a~postupnì ho zvìt¹oval. Pro~ka¾dé zvìt¹ení potøeboval v~síti najít cestu, na~které mají v¹echny hrany kladnou rezervu (po takovéto cestì mù¾eme poslat více, ne¾ po~ní aktuálnì teèe). Ukázali jsme, ¾e pokud takováto cesta existuje, jde tok vylep¹it (zvìt¹it). Zároveò pokud tok jde vylep¹it, pak takováto cesta existuje. Dokázali jsme, ¾e pro~racionální kapacity je algoritmus koneèný a~najde maximální tok. + +Fordùv-Fulkersonùv algoritmus má ov¹em znaèné nevýhody. Funguje pouze pro~racionální kapacity a~je pomìrnì pomalý. Nyní si~uká¾eme jiný algoritmus, který nevylep¹uje tok pomocí cest, ale pomocí tokù\dots Budeme k~tomu potøebovat sí» rezerv. + +\s{Definice:} {\I Sí» rezerv} k~toku~$f$ v~síti $S=(V,E,z,s,c)$ je sí» $R=(S,f)=(V,E,z,s,r)$, kde~$r(e)$ je rezerva hrany~$e$ v~toku~$f$. + +\s{Konvence:} Pro~hranu~$e$ znaèí~$\overleftarrow{e}$ hranu opaènou. Napø. pokud $e=uv$, tak $\overleftarrow{e}=vu$. + +Je dùle¾ité si~uvìdomit, ¾e sí» rezerv je závislá jak na~pùvodní síti~$S$, tak na~nìjakém toku~$f$ v~síti~$S$. Sí» rezerv~$R$ se~pak od sítì~$S$ li¹í pouze kapacitami -- sí»~$R$ má jako kapacitu hrany rezervu hrany v pùvodní síti. Pro~pøipomenutí: rezervu hrany~$e$ v~síti $S=(V,E,z,s,c)$ s~tokem~$f$ jsme si~definovali jako $r(e)=c(e) - f(e) + f(\overleftarrow{e})$. + +Ne¾ si~uká¾eme samotný algoritmus, doká¾eme si~následující lemma. + +\s{Lemma:} Pro~ka¾dý tok~$f$ v~síti~$S$ a~pro~ka¾dý tok~$g$ v~síti $R=(S,f)$ lze v~èase $\O(m)$ nalézt tok~$f'$ v~síti~$S$ takový, ¾e $\vert f' \vert = \vert f \vert + \vert g \vert$. + +\proof + +Dùkaz rozdìlíme do~tøí krokù. V~prvním kroku si~uká¾eme, jak budeme tok~$f'$ v~síti~$S$ konstruovat. V~druhém kroku doká¾eme, ¾e takto zkonstruované~$f'$ je opravdu tok. A~nakonec uká¾eme, ¾e splòuje po¾adovanou vlastnost, tedy ¾e jeho velikost je souèet velikostí tokù~$f$ a~$g$. + +\>{\it 1. konstrukce~$f'$} + +\noindent +Pro ka¾dou dvojici hran~$e, \overleftarrow{e}$ urèíme~$f'(e)$ a~$f'(\overleftarrow{e})$ následovnì: + +\itemize\ibull +\:Pokud~$g(e) = g(\overleftarrow{e}) = 0$, pak nastavme: + \itemize\ibull + \:$f'(e) := f(e)$, + \:$f'(\overleftarrow{e}) := f(\overleftarrow{e})$. + \endlist + +\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) = 0$, pak polo¾me: + \itemize\ibull + \:$\varepsilon := \min (g(e), f(\overleftarrow{e}))$, + \:$f'(e) := f(e) + g(e) - \varepsilon$, + \:$f'(\overleftarrow{e}) := f(\overleftarrow{e}) - \varepsilon$. + \endlist + +\:Pøípad~$g(e) = 0$ a~$g(\overleftarrow{e}) > 0$ vyøe¹íme obdobnì. + +\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) > 0$, pak odeèteme od toku~$g$ cirkulaci po cyklu tvoøeném hranami~$e$ a $\overleftarrow{e}$: + \itemize\ibull + \:$\delta := \min (g(e),g(\overleftarrow{e}))$, + \:$g'(e) := g(e) - \delta$, + \:$g'(\overleftarrow{e}) := g(\overleftarrow{e}) - \delta$. + \endlist + + Tok $g'$ nyní spadá pod nìkterý z~pøedchozích pøípadù, které u¾ umíme vyøe¹it. + +\endlist + +\>{\it 2. $f'$ je tok} + +\numlist{\ndotted} + +\:Nejdøíve ovìøme první podmínku: $\forall e \in E: 0 \leq f(e) \leq c(e)$. Vezmìme libovolnou hranu~$e \in E$. Podle toho, co teèe po~hranách~$e$ a~$\overleftarrow{e}$ v~toku~$g$, jsme rozdìlili konstrukci toku na~tøi pøípady: + + \numlist{\ndotted} + + \:Pokud po~hranách~$e$ a~$\overleftarrow{e}$ netekl ¾ádný tok~$g$, pak jsme nastavili $f'(e) := f(e)$ a~$f'(\overleftarrow{e}) := f(\overleftarrow{e})$. Tedy pokud~$f$ dodr¾oval kapacity, tak pro~$f'$ musí platit to samé. + + + \:Pokud po~hranì~$e$ tekl tok~$g$ nenulový a~po opaèné nulový, tak jsme zvolili: $f'(e) := f(e) + g(e) - \varepsilon$. Víme, ¾e jsme si~$\varepsilon$ vybrali tak, ¾e $\varepsilon \leq g(e)$. Proto $f'(e) \geq 0$. + + Teï ovìøme, ¾e $f'(e) \leq c(e)$. V~pøípadì, ¾e $\varepsilon = g(e)$, tak $f'(e) = f(e) \leq c(e)$. V~opaèném pøípadì platí, ¾e $\varepsilon = f(\overleftarrow{e})$. Pak ov¹em + $$f'(e) = f(e) + g(e) - f(\overleftarrow{e}) \leq $$ + $$\leq f(e) + \left[ c(e) - f(e) + f(\overleftarrow{e}) \right] - f(\overleftarrow{e}) = c(e).$$ + Vyu¾ili jsme, ¾e~$g$ je tok v~síti rezerv, tedy $g(e) \leq c(e) - f(e) + f(\overleftarrow{e})$. + + Pro tok $f'(\overleftarrow{e})$ platí, ¾e $\varepsilon \leq f(\overleftarrow{e})$. Proto $f'(\overleftarrow{e}) = f(\overleftarrow{e}) - \varepsilon \geq 0$. Zároveò $f'(\overleftarrow{e}) \leq f(\overleftarrow{e}) \leq c(\overleftarrow{e})$. + + Tím jsme dokázali, ¾e~$f'(e)$ i~$f'(\overleftarrow{e})$ dodr¾ují kapacity. + + \:V posledním pøípadì tekl po~obou hranách kladný tok~$g$. Men¹í tok z~$g(e)$ a~$g(\overleftarrow{e})$ jsme vynulovali a~od vìt¹ího odeèetli ten men¹í. Tok~$g'(e)$ a~$g'(\overleftarrow{e})$ tedy zùstal korektní a~tok~$f'$ u¾ konstruujeme podle pøedchozího pøípadu. + + \endlist + +\:Teï musíme je¹tì dokázat, ¾e nový tok neporu¹uje Kirchhoffovy zákony: $$\forall v~\in V \setminus \{z,s\}: f'^\Delta(v)=0.$$ + % neboli $$\forall v~\in V \setminus \{z,s\}: \sum_{u: uv \in E}{f'(uv)}=\sum_{u: vu \in E}{f'(vu)}.$$ + + Vezmìme si~libovolnou hranu~$e = uv \in E$. Uvìdomme si, ¾e pøi~pøechodu z~$f(e)$ na~$f'(e)$ a~z~$f(\overleftarrow{e})$ na~$f'(\overleftarrow{e})$ bylo: + \itemize\idot + \:$f^\Delta(u)$ sní¾eno o~$g(e)$ + \:$f^\Delta(v)$ zvý¹eno o~$g(e)$. + \endlist + Seèteme-li úpravy na v¹ech hranách, dostaneme: $$f'^\Delta(v) = f^\Delta(v) + \sum_{u:uv \in E} g(uv) - \sum_{u:vu \in E} g(vu) =$$ $$= f^\Delta(v) + g^+(v) - g^-(v) = f^\Delta(v) + g^\Delta(v).$$ + + Jeliko¾~$f$ byl tok, tak $f^\Delta(v) = 0$ a jeliko¾~$g$ byl tok, tak $g^\Delta(v) = 0$. Proto $f'^\Delta(v) = f^\Delta(v) + g^\Delta(v) = 0$. + +\endlist + +Tím jsme dokázali, ¾e~$f'$ je tok v~síti~$S$. + +\>{\it 3. $\vert f' \vert = \vert f \vert + \vert g \vert$} + +Pou¾ijme vztah pro souèet pøebytkù z pøedchozího kroku: +$$\vert f' \vert = f'^\Delta(s) = f^\Delta(s) + g^\Delta(s) = \vert f \vert + \vert g \vert.$$ +\qed + + +Pro algoritmus budeme potøebovat vybírat kvalitní toky~$g$ v~síti rezerv. Pokud se~nám to bude daøit, bude se~tok~$f'$ rychle zvìt¹ovat, a¾ bychom mohli dojít k~maximálnímu toku. Nejlépe by se~nám hodily co nejvìt¹í toky v~síti rezerv. Kdybychom si~dali za cíl najít v¾dy maximální tok v~síti rezerv, výsledek by byl sice krásný (dostali bychom tak rovnou i~maximální tok v~pùvodní síti), ale problém hledání maximálního toku bychom pouze pøenesli na~jinou sí». Na¹e po¾adavky na~tento tok budou tedy takové, aby byl dostateènì velký, ale abychom bìhem jeho hledání nestrávili moc èasu. Podívejme se, jak se~s~tímto problémem vyrovná {\I Dinicùv algoritmus}. Nejdøíve si~ale zadefinujme nìkolik pojmù. + +\s{Definice:} Tok~$f$ je {\I blokující}, jestli¾e pro~ka¾dou orientovanou cestu~$P$ ze~$z$ do~$s$ existuje hrana~$e \in P$ taková, ¾e $f(e) = c(e)$. + +\s{Definice:} Sí» je {\I vrstevnatá (proèi¹tìná)}, kdy¾ v¹echny vrcholy a~hrany le¾í na~nejkrat¹ích cestách ze~$z$ do~$s$. + +Dinicùv algoritmus zaèíná s~nulovým tokem. Potom v¾dy podle toku~$f$ sestrojí sí» rezerv a~v~ní vyma¾e hrany s~nulovou rezervou. Pokud v~této promazané síti rezerv neexistuje cesta ze~zdroje do~stoku, tak skonèí a~prohlásí tok~$f$ za maximální. Jinak proèistí sí» rezerv tak, aby se~z ní stala vrstevnatá sí» (rozdìlí vrcholy do~vrstev podle vzdálenosti od zdroje a~odstraní pøebyteèné hrany). Ve~vrstevnaté síti najde blokující tok, pomocí nìho¾ zlep¹í tok~$f$. Pak opìt pokraèuje sestrojením sítì rezerv na~tomto vylep¹eném toku~$f$ atd. + +\figure{dinic-cistasit.eps}{Proèi¹tìná sí» rozdìlená do~vrstev}{0.4\hsize} + +\s{Algoritmus (Dinicùv)} + +\algo +\:$f \leftarrow$ nulový tok. +\:Sestrojíme sí» rezerv~$R$ a~sma¾eme $e: r(e) = 0$. +\:$l \leftarrow$ délka nejkrat¹í cesty ze~$z$ do~$s$ v~$R$. +\:Pokud $l = \infty$, zastavíme se~a vrátíme~$f$. +\:Proèistíme sí» $R \rightarrow$ sí»~$C$. +\:$g \leftarrow$ blokující tok v~$C$. +\:Zlep¹íme tok~$f$ pomocí~$g$. +\:GOTO 2. +\endalgo + +\s{Pozorování:} Pokud se~algoritmus zastaví, vydá maximální tok. + +\proof +Víme, ¾e~$f$ je stále korektní tok (jediné, jak ho mìníme je pøièítání toku~$g$, co¾ je, jak jsme si~dokázali, \uv{ne¹kodná operace}). Jakmile neexistuje cesta ze~$z$ do~$s$ v~$R$, tak je $f$ maximální tok, nebo» v~tuto dobu by se~zastavil (a vydal maximální tok) i~Fordùv-Fulkersonùv algoritmus, který je korektní. +\qed + +A teï je¹tì musíme ujasnit, jak budeme èistit sí» rezerv a~vybírat blokující tok~$g$. + +\s{Algoritmus proèi¹tìní sítì rezerv} + +\algo +\:Rozdìlíme vrcholy do~vrstev podle vzdálenosti od~$z$. +\:Odstraníme vrstvy za~$s$ (tedy vrcholy, které jsou od~$z$ vzdálenìj¹í ne¾~$s$), hrany do~minulých vrstev a~hrany uvnitø vrstev. +\:Odstraníme \uv{slepé ulièky}, tedy vrcholy s~$\deg^{out}(v) = 0$, a~to opakovanì pomocí fronty. (Nejdøíve zaøadíme do~fronty v¹echny vrcholy s~ $\deg^{out}(v) = 0$. Pak dokud není fronta prázdná, v¾dy vybereme vrchol~$v$ z~fronty, odstraníme~$v$ a~v¹echny hrany~$uv$. Pro~ka¾dý takový vrchol~$u$ zkontrolujeme, zda se~tím nesní¾il výstupní stupeò vrcholu~$u$ na~nulu ($\deg^{out}(u) = 0$). Pokud sní¾il, tak ho zaøadíme do~fronty.) +\endalgo + +\figure{dinic-neprocistenasit.eps}{Neproèi¹tìná sí». Obsahuje zpìtné hrany, hrany uvnitø vrstvy a~slepé ulièky.}{0.45\hsize} + +Hledání blokujícího toku zaèneme s~tokem nulovým. Pak vezmeme v¾dy orientovanou cestu ze~zdroje do~stoku v~síti~$C$. V~této cestì najdeme hranu s~nejni¾¹í hodnotou výrazu $r(e) - g(e)$ (neboli $c(e) - f(e)$ v~pùvodní síti). Tuto hodnotu oznaèíme~$\varepsilon$. Pak ke~ v¹em hranám na~této cestì pøièteme~$\varepsilon$. Pokud tok~$g$ na~nìjaké hranì dosáhne kapacity hrany, co¾ je zde~$r(e)$, tak hranu vyma¾eme. Následnì sí» doèistíme, aby splòovala podmínky vrstevnaté sítì. A~pokud je¹tì existuje nìjaká orientovaná cesta ze~zdroje do~stoku, tak opìt pokraèujeme s~touto cestou. + +\s{Algoritmus hledání blokujícího toku} + +\algo +\:$g \leftarrow$ nulový tok. +\:Dokud existuje orientovaná cesta~$P$ ze~$z$ do~$s$ v~$C$, opakuj: +\::$\varepsilon \leftarrow \min_{e \in P} (r(e) - g(e))$. +\::Pro~$\forall e \in P: g(e) \leftarrow g(e) + \varepsilon$. +\:::Pokud $g(e) = r(e)$, sma¾eme~$e$ z~$C$. +\::Doèistíme sí» zase pomocí fronty. +\endalgo + +\s{Èasová slo¾itost} Rozeberme si~jednotlivé kroky algoritmu. + +\numlist{\ndotted} +\:Inicializace toku~$f$ \dots $\O(m)$. +\:Sestrojení sítì rezerv a~smazání hran s~nulovou rezervou \dots $\O(m + n)$. +\:Najití nejkrat¹í cesty (prohledáváním do~¹íøky) \dots $\O(m + n)$. +\:Zkontrolování délky nejkrat¹í cesty \dots $\O(1)$. +\:Proèi¹tìní sítì \dots $\O(m + n)$. + \numlist{\ndotted} + \:Rozdìlení vrcholù do~vrstev -- provedlo ji¾ prohledávání do~¹íøky \dots $\O(1)$. + \:Odstranìní nìkterých hran \dots $\O(m + n)$. + \:Odstranìní \uv{slepých ulièek} pomocí fronty -- ka¾dou hranu odstraníme nejvý¹e jedenkrát, ka¾dý vrchol se~dostane do~fronty nejvý¹e jedenkrát \dots $\O(m + n)$. + \endlist +\:Najití blokujícího toku~$g$ \dots $\O(m \cdot n)$. + \numlist{\ndotted} + \:Inicializace toku~$g$ \dots $\O(m)$. + \:Najití orientované cesty v~proèi¹tìné síti rezerv (staèí vzít libovolnou cestu ze~zdroje, nebo» ka¾dá z~nich v~této síti vede do~stoku) \dots $\O(n)$. + \:Výbìr minima z~výrazu $r(e) - g(e)$ pøes v¹echny hrany cesty -- ta mù¾e být dlouhá nejvý¹e~$n$ \dots $\O(n)$. + \:Pøepoèítání v¹ech hran cesty \dots $\O(n)$. + \:Smazání hran cesty, jejich¾ tok~$g(e)$ se~zvý¹il na~hodnotu~$r(e)$ \dots $\O(n)$. + \:Doèi¹»ování vyøe¹me zvlá¹». + \endlist + + Vnitøní cyklus (kroky 2 a¾ 6) provedeme nejvý¹e~$m$ krát, nebo» pøi~ka¾dém prùchodu vyma¾eme alespoò jednu hranu (tak jsme si~volili~$\varepsilon$). + + Èi¹tìní bìhem celého hledání blokujícího toku~$g$ v~proèi¹tìné síti rezerv trvá dohromady $\O(m + n)$, nebo» ka¾dou hranu a~vrchol sma¾eme nejvý¹e jedenkrát. + + Najití blokujícího toku bude tedy trvat $\O(m \cdot n + (m + n)) = \O(m \cdot n)$. + +\:Zlep¹ení toku~$f$ pomocí toku~$g$ \dots $\O(m)$. +\:Skok na~2. krok \dots $\O(1)$. +\endlist + +Zbývá nám jen urèit, kolikrát projdeme vnìj¹ím cyklem (fází). Doká¾eme si~lemma, ¾e hodnota~$l$ vzroste mezi prùchody vnìj¹ím cyklem (fázemi) alespoò o~1. Z~toho plyne, ¾e vnìj¹ím cyklem mù¾eme projít nejvý¹e $n$-krát, nebo» cesta v síti na~$n$ vrcholech mù¾e být dlouhá nejvý¹e $n$. + +Uvìdomme si, ¾e uvnitø vnìj¹ího cyklu pøevládá èlen $\O(m \cdot n)$, tak¾e celková èasová slo¾itost bude $\O(n^2 \cdot m)$. + +\s{Lemma:} Hodnota~$l$ (délka nejkrat¹í cesty ze~$z$ do~$s$ v~proèi¹tìné síti) vzroste mezi fázemi alespoò o~1. + +\proof + +Podíváme se~na~prùbìh jednoho prùchodu vnìj¹ím cyklem. Délku aktuálnì nejkrat¹í cesty ze~zdroje do~stoku oznaème~$l$. V¹echny pùvodní cesty délky~$l$ se~bìhem prùchodu zaruèenì nasytí, proto¾e tok~$g$ je blokující. Musíme v¹ak dokázat, ¾e nemohou vzniknout ¾ádné nové cesty délky~$l$ nebo men¹í. V~síti rezerv toti¾ mohou hrany nejen +ubývat, ale i~pøibývat: pokud po¹leme tok po~hranì, po~které je¹tì nic neteklo, tak v~protismìru z~dosud nulové rezervy vyrobíme nenulovou. Rozmysleme si~tedy, jaké hrany mohou pøibývat. + +Hrany mohou pøibývat jen tehdy, kdy¾ jsme po~opaèné hranì nìco poslali. Ale my nìco posíláme po~hranách pouze z~vrstvy do~té následující. Hrany tedy pøibývají do~minulé vrstvy. + +Vznikem nových hran by proto mohly vzniknout nové cesty ze~zdroje do~stoku, které pou¾ívají zpìtné hrany. Jen¾e cesta ze~zdroje do~stoku, která pou¾ije zpìtnou hranu, musí alespoò jednou skoèit o~vrstvu zpìt a~nikdy nemù¾e skoèit o~více ne¾ jednu vrstvu dopøedu, a~proto je její délka alespoò $l+2$. Pokud cesta novou zpìtnou hranu nepou¾ije, má buï délku~$> l$, co¾ je v~poøádku, nebo má délku~$= l$, pak je zablokovaná. + +Tím je lemma dokázáno. +\qed + +\figure{dinic-cestashranouzpet.eps}{Cesta u¾ívající novou zpìtnou hranu}{0.4\hsize} + +V¹echna dokázaná tvrzení mù¾eme shrnout do~následující vìty: + +\s{Vìta:} Dinicùv algoritmus najde maximální tok v~èase $\O(n^2\cdot m)$. + +\s{Poznámka:} Algoritmus se~chová hezky na~sítích s~malými celoèíselnými kapacitami, ale kupodivu i~na~rùzných jiných sítích. Èasto se~pou¾ívá, nebo» se~chová efektivnì. A~je mnoho zpùsobù, jak ho je¹tì vylep¹ovat, èi odhadovat ni¾¹í slo¾itost na~speciálních sítích. + +\s{Poznámka:} Algoritmus nevy¾aduje racionální kapacity! Dal¹í z~dùvodù, proè maximální tok existuje i~v~síti s~iracionálními kapacitami. + + +%k,s,v,na,do,ke,pro,pøi,a,u,i,po, + +\bye diff --git a/3-dinic/Makefile b/3-dinic/Makefile new file mode 100644 index 0000000..fc07c0a --- /dev/null +++ b/3-dinic/Makefile @@ -0,0 +1,3 @@ +P=3-dinic + +include ../Makerules diff --git a/3-dinic/dinic-cestashranouzpet.eps b/3-dinic/dinic-cestashranouzpet.eps new file mode 100644 index 0000000..5633e61 --- /dev/null +++ b/3-dinic/dinic-cestashranouzpet.eps @@ -0,0 +1,472 @@ +%!PS-Adobe-2.0 EPSF-2.0 +%%Title: /home/mj/texts/ga/2-dinic/Diagram1.dia +%%Creator: Dia v0.96.1 +%%CreationDate: Tue Feb 9 16:36:37 2010 +%%For: mj +%%Orientation: Portrait +%%Magnification: 1.0000 +%%BoundingBox: 0 0 546 219 +%%BeginSetup +%%EndSetup +%%EndComments +%%BeginProlog +[ /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef +/.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef +/.notdef /.notdef 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a/old/2-dinic/2-dinic.tex b/old/2-dinic/2-dinic.tex deleted file mode 100644 index 244427c..0000000 --- a/old/2-dinic/2-dinic.tex +++ /dev/null @@ -1,223 +0,0 @@ -\input lecnotes.tex - -\prednaska{2}{Dinicùv algoritmus}{(zapsala Markéta Popelová)} - - -Na~minulé pøedná¹ce jsme si~ukázali Fordùv-Fulkersonùv algoritmus. Tento algoritmus hledal maximální tok tak, ¾e zaèal s~tokem nulovým a~postupnì ho zvìt¹oval. Pro~ka¾dé zvìt¹ení potøeboval v~síti najít cestu, na~které mají v¹echny hrany kladnou rezervu (po takovéto cestì mù¾eme poslat více, ne¾ po~ní aktuálnì teèe). Ukázali jsme, ¾e pokud takováto cesta existuje, jde tok vylep¹it (zvìt¹it). Zároveò pokud tok jde vylep¹it, pak takováto cesta existuje. Dokázali jsme, ¾e pro~racionální kapacity je algoritmus koneèný a~najde maximální tok. - -Fordùv-Fulkersonùv algoritmus má ov¹em znaèné nevýhody. Funguje pouze pro~racionální kapacity a~je pomìrnì pomalý. Nyní si~uká¾eme jiný algoritmus, který nevylep¹uje tok pomocí cest, ale pomocí tokù\dots Budeme k~tomu potøebovat sí» rezerv. - -\s{Definice:} {\I Sí» rezerv} k~toku~$f$ v~síti $S=(V,E,z,s,c)$ je sí» $R=(S,f)=(V,E,z,s,r)$, kde~$r(e)$ je rezerva hrany~$e$ v~toku~$f$. - -\s{Konvence:} Pro~hranu~$e$ znaèí~$\overleftarrow{e}$ hranu opaènou. Napø. pokud $e=uv$, tak $\overleftarrow{e}=vu$. - -Je dùle¾ité si~uvìdomit, ¾e sí» rezerv je závislá jak na~pùvodní síti~$S$, tak na~nìjakém toku~$f$ v~síti~$S$. Sí» rezerv~$R$ se~pak od sítì~$S$ li¹í pouze kapacitami -- sí»~$R$ má jako kapacitu hrany rezervu hrany v pùvodní síti. Pro~pøipomenutí: rezervu hrany~$e$ v~síti $S=(V,E,z,s,c)$ s~tokem~$f$ jsme si~definovali jako $r(e)=c(e) - f(e) + f(\overleftarrow{e})$. - -Ne¾ si~uká¾eme samotný algoritmus, doká¾eme si~následující lemma. - -\s{Lemma:} Pro~ka¾dý tok~$f$ v~síti~$S$ a~pro~ka¾dý tok~$g$ v~síti $R=(S,f)$ lze v~èase $\O(m)$ nalézt tok~$f'$ v~síti~$S$ takový, ¾e $\vert f' \vert = \vert f \vert + \vert g \vert$. - -\proof - -Dùkaz rozdìlíme do~tøí krokù. V~prvním kroku si~uká¾eme, jak budeme tok~$f'$ v~síti~$S$ konstruovat. V~druhém kroku doká¾eme, ¾e takto zkonstruované~$f'$ je opravdu tok. A~nakonec uká¾eme, ¾e splòuje po¾adovanou vlastnost, tedy ¾e jeho velikost je souèet velikostí tokù~$f$ a~$g$. - -\>{\it 1. konstrukce~$f'$} - -\noindent -Pro ka¾dou dvojici hran~$e, \overleftarrow{e}$ urèíme~$f'(e)$ a~$f'(\overleftarrow{e})$ následovnì: - -\itemize\ibull -\:Pokud~$g(e) = g(\overleftarrow{e}) = 0$, pak nastavme: - \itemize\ibull - \:$f'(e) := f(e)$, - \:$f'(\overleftarrow{e}) := f(\overleftarrow{e})$. - \endlist - -\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) = 0$, pak polo¾me: - \itemize\ibull - \:$\varepsilon := \min (g(e), f(\overleftarrow{e}))$, - \:$f'(e) := f(e) + g(e) - \varepsilon$, - \:$f'(\overleftarrow{e}) := f(\overleftarrow{e}) - \varepsilon$. - \endlist - -\:Pøípad~$g(e) = 0$ a~$g(\overleftarrow{e}) > 0$ vyøe¹íme obdobnì. - -\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) > 0$, pak odeèteme od toku~$g$ cirkulaci po cyklu tvoøeném hranami~$e$ a $\overleftarrow{e}$: - \itemize\ibull - \:$\delta := \min (g(e),g(\overleftarrow{e}))$, - \:$g'(e) := g(e) - \delta$, - \:$g'(\overleftarrow{e}) := g(\overleftarrow{e}) - \delta$. - \endlist - - Tok $g'$ nyní spadá pod nìkterý z~pøedchozích pøípadù, které u¾ umíme vyøe¹it. - -\endlist - -\>{\it 2. $f'$ je tok} - -\numlist{\ndotted} - -\:Nejdøíve ovìøme první podmínku: $\forall e \in E: 0 \leq f(e) \leq c(e)$. Vezmìme libovolnou hranu~$e \in E$. Podle toho, co teèe po~hranách~$e$ a~$\overleftarrow{e}$ v~toku~$g$, jsme rozdìlili konstrukci toku na~tøi pøípady: - - \numlist{\ndotted} - - \:Pokud po~hranách~$e$ a~$\overleftarrow{e}$ netekl ¾ádný tok~$g$, pak jsme nastavili $f'(e) := f(e)$ a~$f'(\overleftarrow{e}) := f(\overleftarrow{e})$. Tedy pokud~$f$ dodr¾oval kapacity, tak pro~$f'$ musí platit to samé. - - - \:Pokud po~hranì~$e$ tekl tok~$g$ nenulový a~po opaèné nulový, tak jsme zvolili: $f'(e) := f(e) + g(e) - \varepsilon$. Víme, ¾e jsme si~$\varepsilon$ vybrali tak, ¾e $\varepsilon \leq g(e)$. Proto $f'(e) \geq 0$. - - Teï ovìøme, ¾e $f'(e) \leq c(e)$. V~pøípadì, ¾e $\varepsilon = g(e)$, tak $f'(e) = f(e) \leq c(e)$. V~opaèném pøípadì platí, ¾e $\varepsilon = f(\overleftarrow{e})$. Pak ov¹em - $$f'(e) = f(e) + g(e) - f(\overleftarrow{e}) \leq $$ - $$\leq f(e) + \left[ c(e) - f(e) + f(\overleftarrow{e}) \right] - f(\overleftarrow{e}) = c(e).$$ - Vyu¾ili jsme, ¾e~$g$ je tok v~síti rezerv, tedy $g(e) \leq c(e) - f(e) + f(\overleftarrow{e})$. - - Pro tok $f'(\overleftarrow{e})$ platí, ¾e $\varepsilon \leq f(\overleftarrow{e})$. Proto $f'(\overleftarrow{e}) = f(\overleftarrow{e}) - \varepsilon \geq 0$. Zároveò $f'(\overleftarrow{e}) \leq f(\overleftarrow{e}) \leq c(\overleftarrow{e})$. - - Tím jsme dokázali, ¾e~$f'(e)$ i~$f'(\overleftarrow{e})$ dodr¾ují kapacity. - - \:V posledním pøípadì tekl po~obou hranách kladný tok~$g$. Men¹í tok z~$g(e)$ a~$g(\overleftarrow{e})$ jsme vynulovali a~od vìt¹ího odeèetli ten men¹í. Tok~$g'(e)$ a~$g'(\overleftarrow{e})$ tedy zùstal korektní a~tok~$f'$ u¾ konstruujeme podle pøedchozího pøípadu. - - \endlist - -\:Teï musíme je¹tì dokázat, ¾e nový tok neporu¹uje Kirchhoffovy zákony: $$\forall v~\in V \setminus \{z,s\}: f'^\Delta(v)=0.$$ - % neboli $$\forall v~\in V \setminus \{z,s\}: \sum_{u: uv \in E}{f'(uv)}=\sum_{u: vu \in E}{f'(vu)}.$$ - - Vezmìme si~libovolnou hranu~$e = uv \in E$. Uvìdomme si, ¾e pøi~pøechodu z~$f(e)$ na~$f'(e)$ a~z~$f(\overleftarrow{e})$ na~$f'(\overleftarrow{e})$ bylo: - \itemize\idot - \:$f^\Delta(u)$ sní¾eno o~$g(e)$ - \:$f^\Delta(v)$ zvý¹eno o~$g(e)$. - \endlist - Seèteme-li úpravy na v¹ech hranách, dostaneme: $$f'^\Delta(v) = f^\Delta(v) + \sum_{u:uv \in E} g(uv) - \sum_{u:vu \in E} g(vu) =$$ $$= f^\Delta(v) + g^+(v) - g^-(v) = f^\Delta(v) + g^\Delta(v).$$ - - Jeliko¾~$f$ byl tok, tak $f^\Delta(v) = 0$ a jeliko¾~$g$ byl tok, tak $g^\Delta(v) = 0$. Proto $f'^\Delta(v) = f^\Delta(v) + g^\Delta(v) = 0$. - -\endlist - -Tím jsme dokázali, ¾e~$f'$ je tok v~síti~$S$. - -\>{\it 3. $\vert f' \vert = \vert f \vert + \vert g \vert$} - -Pou¾ijme vztah pro souèet pøebytkù z pøedchozího kroku: -$$\vert f' \vert = f'^\Delta(s) = f^\Delta(s) + g^\Delta(s) = \vert f \vert + \vert g \vert.$$ -\qed - - -Pro algoritmus budeme potøebovat vybírat kvalitní toky~$g$ v~síti rezerv. Pokud se~nám to bude daøit, bude se~tok~$f'$ rychle zvìt¹ovat, a¾ bychom mohli dojít k~maximálnímu toku. Nejlépe by se~nám hodily co nejvìt¹í toky v~síti rezerv. Kdybychom si~dali za cíl najít v¾dy maximální tok v~síti rezerv, výsledek by byl sice krásný (dostali bychom tak rovnou i~maximální tok v~pùvodní síti), ale problém hledání maximálního toku bychom pouze pøenesli na~jinou sí». Na¹e po¾adavky na~tento tok budou tedy takové, aby byl dostateènì velký, ale abychom bìhem jeho hledání nestrávili moc èasu. Podívejme se, jak se~s~tímto problémem vyrovná {\I Dinicùv algoritmus}. Nejdøíve si~ale zadefinujme nìkolik pojmù. - -\s{Definice:} Tok~$f$ je {\I blokující}, jestli¾e pro~ka¾dou orientovanou cestu~$P$ ze~$z$ do~$s$ existuje hrana~$e \in P$ taková, ¾e $f(e) = c(e)$. - -\s{Definice:} Sí» je {\I vrstevnatá (proèi¹tìná)}, kdy¾ v¹echny vrcholy a~hrany le¾í na~nejkrat¹ích cestách ze~$z$ do~$s$. - -Dinicùv algoritmus zaèíná s~nulovým tokem. Potom v¾dy podle toku~$f$ sestrojí sí» rezerv a~v~ní vyma¾e hrany s~nulovou rezervou. Pokud v~této promazané síti rezerv neexistuje cesta ze~zdroje do~stoku, tak skonèí a~prohlásí tok~$f$ za maximální. Jinak proèistí sí» rezerv tak, aby se~z ní stala vrstevnatá sí» (rozdìlí vrcholy do~vrstev podle vzdálenosti od zdroje a~odstraní pøebyteèné hrany). Ve~vrstevnaté síti najde blokující tok, pomocí nìho¾ zlep¹í tok~$f$. Pak opìt pokraèuje sestrojením sítì rezerv na~tomto vylep¹eném toku~$f$ atd. - -\figure{dinic-cistasit.eps}{Proèi¹tìná sí» rozdìlená do~vrstev}{0.4\hsize} - -\s{Algoritmus (Dinicùv)} - -\algo -\:$f \leftarrow$ nulový tok. -\:Sestrojíme sí» rezerv~$R$ a~sma¾eme $e: r(e) = 0$. -\:$l \leftarrow$ délka nejkrat¹í cesty ze~$z$ do~$s$ v~$R$. -\:Pokud $l = \infty$, zastavíme se~a vrátíme~$f$. -\:Proèistíme sí» $R \rightarrow$ sí»~$C$. -\:$g \leftarrow$ blokující tok v~$C$. -\:Zlep¹íme tok~$f$ pomocí~$g$. -\:GOTO 2. -\endalgo - -\s{Pozorování:} Pokud se~algoritmus zastaví, vydá maximální tok. - -\proof -Víme, ¾e~$f$ je stále korektní tok (jediné, jak ho mìníme je pøièítání toku~$g$, co¾ je, jak jsme si~dokázali, \uv{ne¹kodná operace}). Jakmile neexistuje cesta ze~$z$ do~$s$ v~$R$, tak je $f$ maximální tok, nebo» v~tuto dobu by se~zastavil (a vydal maximální tok) i~Fordùv-Fulkersonùv algoritmus, který je korektní. -\qed - -A teï je¹tì musíme ujasnit, jak budeme èistit sí» rezerv a~vybírat blokující tok~$g$. - -\s{Algoritmus proèi¹tìní sítì rezerv} - -\algo -\:Rozdìlíme vrcholy do~vrstev podle vzdálenosti od~$z$. -\:Odstraníme vrstvy za~$s$ (tedy vrcholy, které jsou od~$z$ vzdálenìj¹í ne¾~$s$), hrany do~minulých vrstev a~hrany uvnitø vrstev. -\:Odstraníme \uv{slepé ulièky}, tedy vrcholy s~$\deg^{out}(v) = 0$, a~to opakovanì pomocí fronty. (Nejdøíve zaøadíme do~fronty v¹echny vrcholy s~ $\deg^{out}(v) = 0$. Pak dokud není fronta prázdná, v¾dy vybereme vrchol~$v$ z~fronty, odstraníme~$v$ a~v¹echny hrany~$uv$. Pro~ka¾dý takový vrchol~$u$ zkontrolujeme, zda se~tím nesní¾il výstupní stupeò vrcholu~$u$ na~nulu ($\deg^{out}(u) = 0$). Pokud sní¾il, tak ho zaøadíme do~fronty.) -\endalgo - -\figure{dinic-neprocistenasit.eps}{Neproèi¹tìná sí». Obsahuje zpìtné hrany, hrany uvnitø vrstvy a~slepé ulièky.}{0.45\hsize} - -Hledání blokujícího toku zaèneme s~tokem nulovým. Pak vezmeme v¾dy orientovanou cestu ze~zdroje do~stoku v~síti~$C$. V~této cestì najdeme hranu s~nejni¾¹í hodnotou výrazu $r(e) - g(e)$ (neboli $c(e) - f(e)$ v~pùvodní síti). Tuto hodnotu oznaèíme~$\varepsilon$. Pak ke~ v¹em hranám na~této cestì pøièteme~$\varepsilon$. Pokud tok~$g$ na~nìjaké hranì dosáhne kapacity hrany, co¾ je zde~$r(e)$, tak hranu vyma¾eme. Následnì sí» doèistíme, aby splòovala podmínky vrstevnaté sítì. A~pokud je¹tì existuje nìjaká orientovaná cesta ze~zdroje do~stoku, tak opìt pokraèujeme s~touto cestou. - -\s{Algoritmus hledání blokujícího toku} - -\algo -\:$g \leftarrow$ nulový tok. -\:Dokud existuje orientovaná cesta~$P$ ze~$z$ do~$s$ v~$C$, opakuj: -\::$\varepsilon \leftarrow \min_{e \in P} (r(e) - g(e))$. -\::Pro~$\forall e \in P: g(e) \leftarrow g(e) + \varepsilon$. -\:::Pokud $g(e) = r(e)$, sma¾eme~$e$ z~$C$. -\::Doèistíme sí» zase pomocí fronty. -\endalgo - -\s{Èasová slo¾itost} Rozeberme si~jednotlivé kroky algoritmu. - -\numlist{\ndotted} -\:Inicializace toku~$f$ \dots $\O(m)$. -\:Sestrojení sítì rezerv a~smazání hran s~nulovou rezervou \dots $\O(m + n)$. -\:Najití nejkrat¹í cesty (prohledáváním do~¹íøky) \dots $\O(m + n)$. -\:Zkontrolování délky nejkrat¹í cesty \dots $\O(1)$. -\:Proèi¹tìní sítì \dots $\O(m + n)$. - \numlist{\ndotted} - \:Rozdìlení vrcholù do~vrstev -- provedlo ji¾ prohledávání do~¹íøky \dots $\O(1)$. - \:Odstranìní nìkterých hran \dots $\O(m + n)$. - \:Odstranìní \uv{slepých ulièek} pomocí fronty -- ka¾dou hranu odstraníme nejvý¹e jedenkrát, ka¾dý vrchol se~dostane do~fronty nejvý¹e jedenkrát \dots $\O(m + n)$. - \endlist -\:Najití blokujícího toku~$g$ \dots $\O(m \cdot n)$. - \numlist{\ndotted} - \:Inicializace toku~$g$ \dots $\O(m)$. - \:Najití orientované cesty v~proèi¹tìné síti rezerv (staèí vzít libovolnou cestu ze~zdroje, nebo» ka¾dá z~nich v~této síti vede do~stoku) \dots $\O(n)$. - \:Výbìr minima z~výrazu $r(e) - g(e)$ pøes v¹echny hrany cesty -- ta mù¾e být dlouhá nejvý¹e~$n$ \dots $\O(n)$. - \:Pøepoèítání v¹ech hran cesty \dots $\O(n)$. - \:Smazání hran cesty, jejich¾ tok~$g(e)$ se~zvý¹il na~hodnotu~$r(e)$ \dots $\O(n)$. - \:Doèi¹»ování vyøe¹me zvlá¹». - \endlist - - Vnitøní cyklus (kroky 2 a¾ 6) provedeme nejvý¹e~$m$ krát, nebo» pøi~ka¾dém prùchodu vyma¾eme alespoò jednu hranu (tak jsme si~volili~$\varepsilon$). - - Èi¹tìní bìhem celého hledání blokujícího toku~$g$ v~proèi¹tìné síti rezerv trvá dohromady $\O(m + n)$, nebo» ka¾dou hranu a~vrchol sma¾eme nejvý¹e jedenkrát. - - Najití blokujícího toku bude tedy trvat $\O(m \cdot n + (m + n)) = \O(m \cdot n)$. - -\:Zlep¹ení toku~$f$ pomocí toku~$g$ \dots $\O(m)$. -\:Skok na~2. krok \dots $\O(1)$. -\endlist - -Zbývá nám jen urèit, kolikrát projdeme vnìj¹ím cyklem (fází). Doká¾eme si~lemma, ¾e hodnota~$l$ vzroste mezi prùchody vnìj¹ím cyklem (fázemi) alespoò o~1. Z~toho plyne, ¾e vnìj¹ím cyklem mù¾eme projít nejvý¹e $n$-krát, nebo» cesta v síti na~$n$ vrcholech mù¾e být dlouhá nejvý¹e $n$. - -Uvìdomme si, ¾e uvnitø vnìj¹ího cyklu pøevládá èlen $\O(m \cdot n)$, tak¾e celková èasová slo¾itost bude $\O(n^2 \cdot m)$. - -\s{Lemma:} Hodnota~$l$ (délka nejkrat¹í cesty ze~$z$ do~$s$ v~proèi¹tìné síti) vzroste mezi fázemi alespoò o~1. - -\proof - -Podíváme se~na~prùbìh jednoho prùchodu vnìj¹ím cyklem. Délku aktuálnì nejkrat¹í cesty ze~zdroje do~stoku oznaème~$l$. V¹echny pùvodní cesty délky~$l$ se~bìhem prùchodu zaruèenì nasytí, proto¾e tok~$g$ je blokující. Musíme v¹ak dokázat, ¾e nemohou vzniknout ¾ádné nové cesty délky~$l$ nebo men¹í. V~síti rezerv toti¾ mohou hrany nejen -ubývat, ale i~pøibývat: pokud po¹leme tok po~hranì, po~které je¹tì nic neteklo, tak v~protismìru z~dosud nulové rezervy vyrobíme nenulovou. Rozmysleme si~tedy, jaké hrany mohou pøibývat. - -Hrany mohou pøibývat jen tehdy, kdy¾ jsme po~opaèné hranì nìco poslali. Ale my nìco posíláme po~hranách pouze z~vrstvy do~té následující. Hrany tedy pøibývají do~minulé vrstvy. - -Vznikem nových hran by proto mohly vzniknout nové cesty ze~zdroje do~stoku, které pou¾ívají zpìtné hrany. Jen¾e cesta ze~zdroje do~stoku, která pou¾ije zpìtnou hranu, musí alespoò jednou skoèit o~vrstvu zpìt a~nikdy nemù¾e skoèit o~více ne¾ jednu vrstvu dopøedu, a~proto je její délka alespoò $l+2$. Pokud cesta novou zpìtnou hranu nepou¾ije, má buï délku~$> l$, co¾ je v~poøádku, nebo má délku~$= l$, pak je zablokovaná. - -Tím je lemma dokázáno. -\qed - -\figure{dinic-cestashranouzpet.eps}{Cesta u¾ívající novou zpìtnou hranu}{0.4\hsize} - -V¹echna dokázaná tvrzení mù¾eme shrnout do~následující vìty: - -\s{Vìta:} Dinicùv algoritmus najde maximální tok v~èase $\O(n^2\cdot m)$. - -\s{Poznámka:} Algoritmus se~chová hezky na~sítích s~malými celoèíselnými kapacitami, ale kupodivu i~na~rùzných jiných sítích. Èasto se~pou¾ívá, nebo» se~chová efektivnì. A~je mnoho zpùsobù, jak ho je¹tì vylep¹ovat, èi odhadovat ni¾¹í slo¾itost na~speciálních sítích. - -\s{Poznámka:} Algoritmus nevy¾aduje racionální kapacity! Dal¹í z~dùvodù, proè maximální tok existuje i~v~síti s~iracionálními kapacitami. - - -%k,s,v,na,do,ke,pro,pøi,a,u,i,po, - -\bye diff --git a/old/2-dinic/Makefile b/old/2-dinic/Makefile deleted file mode 100644 index c183fb0..0000000 --- a/old/2-dinic/Makefile +++ /dev/null @@ -1,3 +0,0 @@ -P=2-dinic - -include ../Makerules diff --git a/old/2-dinic/dinic-cestashranouzpet.eps b/old/2-dinic/dinic-cestashranouzpet.eps deleted file mode 100644 index 5633e61..0000000 --- a/old/2-dinic/dinic-cestashranouzpet.eps +++ /dev/null @@ -1,472 +0,0 @@ -%!PS-Adobe-2.0 EPSF-2.0 -%%Title: /home/mj/texts/ga/2-dinic/Diagram1.dia -%%Creator: Dia v0.96.1 -%%CreationDate: Tue Feb 9 16:36:37 2010 -%%For: mj -%%Orientation: Portrait -%%Magnification: 1.0000 -%%BoundingBox: 0 0 546 219 -%%BeginSetup -%%EndSetup -%%EndComments -%%BeginProlog -[ /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef -/.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 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