From 73e881dcd496b63d2ef59b8fe22fea2706732d75 Mon Sep 17 00:00:00 2001
From: Martin Mares <mj@ucw.cz>
Date: Tue, 23 Jan 2007 18:08:13 +0100
Subject: [PATCH] Revize analyzy Dinicova algoritmu a obrazku okolo.

---
 2-dinic/2-dinic.tex           |  81 ++++---
 2-dinic/dinic-vrcholrez.eps   |  78 +++----
 2-dinic/dinic-zpetnahrana.eps | 426 ----------------------------------
 2-dinic/dinic.dia             | Bin 11124 -> 11036 bytes
 4 files changed, 80 insertions(+), 505 deletions(-)
 delete mode 100644 2-dinic/dinic-zpetnahrana.eps

diff --git a/2-dinic/2-dinic.tex b/2-dinic/2-dinic.tex
index 58cba61..1d3cb38 100644
--- a/2-dinic/2-dinic.tex
+++ b/2-dinic/2-dinic.tex
@@ -23,6 +23,7 @@ kdyby neuva
 \:Zaèneme s~libovolným tokem~$f$, napøíklad prázdným (v¹ude nulovým).
 \:Iterativnì vylep¹ujeme tok pomocí zlep¹ujících tokù: {\I (vnìj¹í cyklus)}
 \::Sestrojíme sí» rezerv: vrcholy a hrany jsou tyté¾, kapacity jsou urèeny rezervami v~pùvodní síti.
+Dále budeme pracovat s~ní.
 \::Najdeme nejkrat¹í $st$-cestu. Kdy¾ ¾ádná neexistuje, skonèime.
 \::Proèistime sí», tj. ponecháme v ní pouze vrcholy a hrany na nejkrat¹ích $st$-cestách.
 \::Najdeme v~proèi¹tìné síti blokující tok $f_B$:
@@ -56,30 +57,26 @@ Ozna
   \endlist
   Takto kroky 5 a 12 dohromady spotøebují èas $\O(m)$.
   \:Jeden prùchod vnìj¹ím cyklem tedy trvá $\O(mn)$.
+  \:Jak za chvíli doká¾eme, ka¾dým prùchodem vnìj¹ím cyklem $l$ vzroste, tak¾e prùchodù
+    bude maximálnì~$n$ a celý algoritmus tak pobì¾í v~èase $\O(n^2m)$.
 \endlist
 
+\s{Korektnost algoritmu:}
+Kdy¾ se Dinicùv algoritmus zastaví, nemù¾e u¾ existovat ¾ádná zlep¹ující cesta
+(viz krok~4) a tehdy, jak u¾ víme z~analýzy F-F algoritmu, je nalezený tok maximální.
+
 \s{Vìta:}
 V~ka¾dém prùchodu Dinicova algoritmu vzroste $l$ alespoò~o~1.
 
-\s{Dùsledek:}
-Vnìj¹ím cyklem projdeme $\O(n)$-krát, proto¾e nejdel¹í cesta bude mít
-délku $\leq n$. Celková slo¾itost celého algoritmu proto bude $\O(n^2m)$
-
-\s{Korektnost algoritmu}
-plyne z korektnosti F-F algoritmu. Kdy¾ se Dinicùv
-algoritmus zastaví, musí vydat maximální tok. Pokud ne, tak podle F-F algoritmu
-existuje zlep¹ující cesta, ale to je ve sporu s~krokem~4, který po¾aduje, aby ¾ádná
-taková cesta neexistovala.
-
-\s{Dùkaz vìty:}
+\proof
 Podíváme se na~prùbìh jednoho prùchodu vnìj¹ím cyklem.
 Délku aktuálnì nejkrat¹í $st$-cesty oznaème~$l$.
 V¹echny pùvodní cesty délky~$l$ se bìhem prùchodu zaruèenì nasytí, proto¾e
 tok~$f_B$ 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, mù¾e se stát, ¾e v~protismìru z~dosud nulové rezervy vyrobíme
-nenulovou. Rozmysleme si tedy, jaké hrany mohou pøibývat:
+neteklo, tak v~protismìru z~dosud nulové rezervy vyrobíme nenulovou.
+Rozmysleme si tedy, jaké hrany mohou pøibývat:
 
 Vnìj¹í cyklus zaèíná s neproèi¹tìnou sítí. Pøíklad takové sítì je na~následujícím obrázku.
 Po~proèi¹tìní zùstanou v~síti jen èerné hrany, tedy hrany vedoucí z~$i$-té
@@ -91,9 +88,6 @@ je barevn
 
 Nové hrany mohou vznikat výhradnì jako opaèné k~èerným hranám (hrany ostatních barev
 padly za obì» proèi¹tìní). Jsou to tedy v¾dy zpìtné hrany vedoucí z~$i$-té vrstvy do~$(i-1)$-ní.
-
-\figure{dinic-zpetnahrana.eps}{Vznik nové zpìtné hrany}{0.4\hsize}
-
 Vznikem nových hran by proto mohly vzniknout nové $st$-cesty, které pou¾ívají
 zpìtné hrany. Jen¾e $st$-cesta, 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
@@ -120,15 +114,25 @@ d
 
 \h{Speciální sítì (ubíráme na obecnosti)}
 
-Dále se budeme vìnovat modifikacím Dinicova algoritmu na speciální druhy sítí,
-kde je mo¾né dostat lep¹í èasovou slo¾itost ne¾ v~obecném pøípadì.
+Pøi pøevodu rùzných úloh na~hledání maximálního toku èasto dostaneme sí» v~nìjakém
+speciálním tvaru -- tøeba s~omezenými kapacitami èi stupni vrcholù. Podíváme se
+proto podrobnìji na~chování Dinicova algoritmu v~takových pøípadech a uká¾eme,
+¾e èasto pracuje pøekvapivì efektivnì.
 
 \s{Jednotkové kapacity:}
-V¹echny rezervy jsou jen 0 nebo~1. Na $st$-cestì má v¹echno kapacitu/rezervu~1.
-Mù¾eme poslat~1~a rovnou celou cestu zahodit -- není potøeba si pamatovat minimum z~rezerv.
+Pokud sí» neobsahuje cykly délky~1 (dvojice navzájem opaèných hran), v¹echny rezervy
+jsou jen 0 nebo~1. Pokud obsahuje, mohou rezervy být i dvojky, a~proto sí» upravíme tak,
+¾e ke~ka¾dé hranì pøidáme hranu opaènou s~nulovou kapacitou a rezervu proti smìru
+toku pøiøkneme jí. Vzniknou tím sice paralelní hrany, ale to tokovým algoritmùm
+nikterak nevadí.%
+\foot{Èasto se to implementuje tak, ¾e protismìrné hrany vùbec nevytvoøíme a kdy¾
+hranu nasytíme, tak v~síti rezerv prostì obrátíme její orientaci.}
+
+Pøi hledání blokujícího toku tedy budou mít v¹echny hrany na~nalezené $st$-cestì
+stejnou, toti¾ jednotkovou, rezervu, tak¾e v¾dy z~grafu odstraníme celou cestu.
 Kdy¾ máme $m$ hran, poèet zlep¹ení po cestách délky $l$ bude maximálnì $m/l$.
 Proto slo¾itost podkrokù 9, 10 a 11 bude $m/l \cdot \O(l) = \O(m)$.%
-\foot{Nebo by ¹lo argumentovat, ¾e ka¾dou hranu pou¾ijeme jen $1\times$.}
+\foot{Nebo by ¹lo argumentovat, tím ¾e ka¾dou hranu pou¾ijeme jen $1\times$.}
 Tedy pro jednotkové kapacity dostáváme slo¾itost $\O(nm)$.
 
 \s{Jednotkové kapacity znovu a lépe:}
@@ -145,40 +149,42 @@ Ka
 nejvý¹e $\vert f_R \vert$ iterací.
 Proto bychom chtìli omezit velikost toku $f_R$. Napøíklad øezem.
 
-Najdeme v síti rezerv øez $C$. Kde ho vzít?\foot{Pøeci v øeznictví. Kdepak, spí¹e v cukrárnì.
+Najdeme v síti rezerv nìjaký dost malý øez $C$. Kde ho vzít?\foot{Pøeci v øeznictví. Kdepak, spí¹e v cukrárnì.
 Myslíte, ¾e v cukrárnì mají Dinicovy øezy? Myslím, ¾e v cukrárnì je vìt¹ina øezù minimální. {\sl (odposlechnuto na~pøedná¹ce)}}
 Poèítejme jen hrany zleva doprava. Tìch je jistì nejvý¹e $m$ a tvoøí alespoò $k$ rozhraní mezi
 vrstvami. Tedy existuje rozhraní vrstev
 s~nejvý¹e $m/k$ hranami\foot{Princip holubníku a nìjaká ta $\pm1$.}.
-Toto rozhraní je øez. Tedy existuje øez $C$, pro nìj¾ $\vert C\vert \leq m/k$,
+Toto rozhraní je øez. Tedy existuje øez~$C$, pro nìj¾ $\vert C\vert \leq m/k$,
 a~algoritmu zbývá maximálnì $m/k$ dal¹ích krokù.
 Celkový poèet krokù je nejvý¹ $k + m/k$, tak¾e staèí zvolit $k = \sqrt{m}$ a získáme
 odhad na~poèet krokù $\O(\sqrt{m})$.
 
-Celkovì slo¾itost Dinicova algoritmu pro jednotkové kapacity znovu a lépe
-je $\O(\sqrt{m} \cdot m) = \O(m^{3/2})$. Pomohli jsme si pro øídké grafy.
+Tím jsme dokázali, ¾e celková slo¾itost Dinicova algoritmu pro jednotkové
+kapacity je $\O(m^{3/2})$. Tím jsme si pomohli pro øídké grafy.
 
 \s{Jednotkové kapacity a jeden ze stupòù roven 1:}
 Úlohu hledání maximálního párování v~bipartitním grafu, pøípadnì hledání
 vrcholovì disjunktních cest v~obecném grafu lze pøevést (viz pøedchozí kapitola)
-na~hledání maximálního toku v~síti, v~ní¾ $\forall v \neq s,t: \min(\deg^-(v), \deg^+(v)) \leq 1$.
-Pro takovou sí» mù¾eme odhad je¹tì tro¹ku upravit. My¹lenka je stejná jako minule. Pokusíme
-se nalézt v síti po $k$ krocích nìjaký malý øez. Místo rozhraní budeme hledat jednu malou
-vrstvu a z malé vrstvy vytvoøíme malý øez: Pro ka¾dý vrchol z vrstvy vezmeme tu hranu,
+na~hledání maximálního toku v~síti, v~ní¾ má ka¾dý vrchol~$v\ne s,t$ buïto vstupní
+nebo výstupní stupeò roven jedné.
+Pro takovou sí» mù¾eme pøedchozí odhad je¹tì tro¹ku upravit. Pokusíme
+se nalézt v síti po~$k$~krocích nìjaký malý øez. Místo rozhraní budeme hledat jednu malou
+vrstvu a z malé vrstvy vytvoøíme malý øez tak, ¾e pro ka¾dý vrchol z vrstvy vezmeme tu hranu,
 která je ve svém smìru sama.
 
 \figure{dinic-vrcholrez.eps}{Øez podle vrcholù ve vrstvì}{0.2\hsize}
 
-Po $k$ krocích máme alespoò $k$ vrstev. Tedy existuje vrstva $\delta$ s nejvý¹e $n/k$ vrcholy.
+Po $k$ krocích máme alespoò $k$ vrstev, a~proto existuje vrstva $\delta$ s nejvý¹e $n/k$ vrcholy.
 Tedy existuje øez $C$ o~velikosti $\vert C\vert \leq n/k$ (získáme z vrstvy $\delta$ vý¹e popsaným postupem).
-Algoritmu zbývá do konce $\leq n/k$ iterací vnìj¹ího cyklu, celkem tedy udìlá $k + n/k$ krokù.
+Algoritmu zbývá do konce $\leq n/k$ iterací vnìj¹ího cyklu, celkem tedy udìlá $k + n/k$ iterací.
 Nyní staèí zvolit $k = \sqrt{n}$ a slo¾itost
-celého algoritmu vyjde $\O(\sqrt{n}\cdot m)$\foot{Poèet iterací vnìj¹ího cyklu je $\O(\sqrt{n})$ a
-vnitøní cyklus má $\O(m)$.}.
+celého algoritmu vyjde $\O(\sqrt{n}\cdot m)$.
 
 \s{Tøetí pokus pro jednotkové kapacity bez omezení na stupnì vrcholù v síti:}
 Hlavní my¹lenkou je opìt po $k$ krocích najít nìjaký malý øez. Najdeme dvì malé
 sousední vrstvy a v¹echny hrany mezi nimi budou tvoøit námi hledaný malý øez.
+Budeme tentokrát pøedpokládat, ¾e na¹e sí» není multigraf, pøípadnì ¾e
+násobnost hran je alespoò omezena konstantou.
 
 Oznaème $s_i$ poèet vrcholù v $i$-té vrstvì. Souèet poètu vrcholù ve dvou
 sousedních vrstvách oznaèíme $t_i = s_i + s_{i+1}$. Bude tedy platit nerovnost:
@@ -196,10 +202,10 @@ maxim
 Za jednu iteraci velkého cyklu projdeme malým cyklem maximálnì tolikrát,
 o kolik se v nìm zvedl tok, proto¾e ka¾dá zlep¹ující cesta ho zvedne alespoò o $1$.
 Zlep¹ující cesta se tedy hledá maximálnì  $\vert f\vert$-krát za celou dobu bìhu algoritmu.
-Cestu najedeme v èase $\O(n)$. Celkem na hledání
+Cestu najedeme v èase $\O(n)$. Celkem na~hledání
 cest spotøebujeme $\O(\vert f\vert\cdot n)$ za celou dobu bìhu algoritmu.
 
-Nesmíme ale zapomenout na cleanupy. V jedné iteraci velkého cyklu stojí cleanupy $\O(m)$ a velkých
+Nesmíme ale zapomenout na èi¹tìní. V jedné iteraci velkého cyklu nás stojí èi¹tìní $\O(m)$ a velkých
 iterací je $\leq n$. Proto celková slo¾itost algoritmu èiní $\O(\vert f\vert n + nm)$
 
 Pokud navíc budeme pøedpokládat, ¾e kapacita hran je nejvý¹e~$C$ a $G$ není multigraf,
@@ -209,9 +215,8 @@ $\O(Cn^2 + nm)$.
 \h{Scaling kapacit}
 
 Základní my¹lenka je podobná, jako u algoritmu pro tøídìní dlouhých èísel postupnì po øádech pomocí
-radix-sortu. Pro jistotu si ho pøipomeòme. Algoritmus nejprve setøídí èísla podle poslední%
-\foot{Poslední cifrou myslíme nejménì významnou cifru.}
-cifry, poté podle pøedposlední, pøedpøedposlední a tak dále.
+radix-sortu. Pro jistotu si ho pøipomeòme. Algoritmus nejprve setøídí èísla podle poslední
+(nejménì významné) cifry, poté podle pøedposlední, pøedpøedposlední a tak dále.
 
 \figure{dinic-sort.eps}{Kroky postupného tøídìní podle øádù}{0.4\hsize}
 
diff --git a/2-dinic/dinic-vrcholrez.eps b/2-dinic/dinic-vrcholrez.eps
index 4458215..a0d5bee 100644
--- a/2-dinic/dinic-vrcholrez.eps
+++ b/2-dinic/dinic-vrcholrez.eps
@@ -1,11 +1,11 @@
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+%%Title: Diagram2.dia
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-%%CreationDate: Sun Mar 26 18:47:25 2006
-%%For: bernard
+%%CreationDate: Tue Jan 23 17:57:45 2007
+%%For: mares
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+%%BoundingBox: 0 0 196 210
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-- 
2.39.2