--- /dev/null
+input lib
+
+figname("8-geom2_");
+figtag("usecky");
+beginfig(1);
+ def drawusecka(expr p,q) = draw vertex(p); draw vertex(q); draw p--q; enddef;
+ pair A[],B[];
+ A0 := origin; B0 := (4cm,1.5cm);
+ A1 := (0.5cm,1.5cm); B1 := (1cm,0);
+ A2 := (2cm,1.3cm); B2 := (3cm, -0.9cm);
+ A3 := (1cm,-1.3cm); B3 := (3.5cm,0);
+ z0 = whatever[A0,B0]; z0 = whatever[A1,B1];
+ z1 = whatever[A0,B0]; z1 = whatever[A2,B2];
+ z2 = whatever[A2,B2]; z2 = whatever[A3,B3];
+
+ for i:=0 upto 2: fill fullcircle scaled 7pt shifted z[i]; drawemptyvertex(z[i]); endfor
+ for i:=0 upto 3: drawusecka(A[i], B[i]); endfor
+endfig;
+
+figtag("polorovina");
+beginfig(2);
+ pair A,B;
+ A := origin;
+ B := (1.7cm,0.5cm);
+ an := angle(B-A);
+ path p; p := from(.5[A,B],an+90,3cm)--from(.5[A,B],an-90,3cm);
+ fill p--reverse(p shifted (A-0.8[A,B]))--cycle withcolor 0.8white;
+ %fill p{dir (an+180)}..-0.35[A,B]..{dir an}cycle withcolor 0.8white;
+ draw p withpen boldpen;
+
+ draw A--B dashed evenly;
+ drawemptyvertex(A); drawemptyvertex(B);
+ label.llft(btex $a$ etex, A);
+ label.urt(btex $b$ etex, B);
+endfig;
+
+figtag("voroneho_diagram");
+beginfig(3);
+ u := 1.35cm;
+ pair A[],B[];
+ A0 := origin; A1 := (-u,-u); A2 := (1.3u,-u); A3 := (1.5u,0.7u); A4 := (0.6u,0.8u); A5 := (0.2u,1.9u); A6 := (-1.3u,1u);
+ def osa(expr a, b, an,l) = from(.5[a,b], angle(b-a)+an, l) enddef;
+ vardef prusecik_os(expr p,q,r) =
+ save b; pair b;
+ b = whatever[osa(p,q, 90,1cm), osa(p,q,-90,1cm)];
+ b = whatever[osa(q,r, 90,1cm), osa(q,r,-90,1cm)];
+ b
+ enddef;
+ B[0] := prusecik_os(A0,A1,A2); B[1] := prusecik_os(A0,A1,A6); B[2] := prusecik_os(A0,A4,A6); B[3] := prusecik_os(A4,A5,A6);
+ B[4] := prusecik_os(A3,A4,A5); B[5] := prusecik_os(A0,A3,A4); B[6] := prusecik_os(A0,A2,A3);
+ draw osa(A1,A2,-90,2cm)--B[0]--B[1]--osa(A1,A6,90,1.3cm) withpen boldpen;
+ draw B[1]--B[2]--B[3]--osa(A6,A5,90,2cm) withpen boldpen;
+ draw B[3]--B[4]--osa(A3,A5,-90,2.7cm) withpen boldpen;
+ draw B[4]--B[5]--B[2] withpen boldpen;
+ draw B[5]--B[6]--osa(A2,A3,-90,1.3cm) withpen boldpen;
+ draw B[6]--B[0] withpen boldpen;
+
+ draw A0--A1--A2--A0--A6--A1 dashed evenly;
+ draw A6--A5--A4--A0--A3--A2 dashed evenly;
+ draw A3--A5 dashed evenly; draw A6--A4--A3 dashed evenly;
+
+ for i:=0 upto 6: draw vertex(B[i]); endfor
+ for i:=0 upto 6: drawemptyvertex(A[i]); endfor
+endfig;
+end
--- /dev/null
+\input lecnotes.tex
+
+\prednaska{8}{Geometrie vrací úder}{(sepsal Pavel Klavík)}
+
+\>Kdy¾ s geometrickými problémy poøádnì nezametete, ony vám to vrátí! Ale kdy¾ u¾ zametat, tak urèitì ne pod koberec a místo smetáku si na nì vezmìte
+pøímku. V této pøedná¹ce nás spolu s dvìma geometrickými problémy samozøejmì èeká pokraèování pohádky o ledních medvìdech.
+
+{\I Medvìdi vyøe¹ili rybí problém a hlad je ji¾ netrápí. Av¹ak na severu ne¾ijí sami, za sousedy mají Eskymáky. Proto¾e je rozhodnì lep¹í se sousedy
+dobøe vycházet, jsou medvìdi a Eskymáci velcí pøátelé. Skoro ka¾dý se svými pøáteli rád schází. Av¹ak to je musí nejprve nalézt~\dots}
+
+\h{Hledání prùseèíkù úseèek}
+
+Zkusíme nejprve Eskymákùm vyøe¹it lokalizaci ledních medvìdù.
+
+{\I Kdy¾ takový medvìd nemá co na práci, rád se prochází. Na místech, kde se trasy protínají, je zvý¹ená ¹ance, ¾e se dva medvìdi potkají a zapovídají
+-- ostatnì co byste èekali od medvìdù. To jsou ta správná místa pro Eskymáka, který chce potkat medvìda. Jenom¾e jak tato køí¾ení najít?}
+
+Pro zjednodu¹ení pøedpokládejme, ¾e medvìdi chodí po úseèkách tam a zpìt. Budeme tedy chtít nalézt v¹echny prùseèíky úseèek v rovinì.
+
+\bigskip
+\centerline{\epsfxsize=1.5in\epsfbox{8-geom2_0_bear.eps}\hskip 4em\epsfxsize2in\epsfbox{8-geom2_1_usecky.eps}}
+\smallskip
+\centerline{Problém Eskymákù: Kde v¹ude se køí¾í medvìdí trasy?}
+\bigskip
+
+Pro $n$ úseèek mù¾eme mít a¾ $\Omega(n^2)$ prùseèíkù.\foot{Zkuste takový pøíklad zkonstruovat.} Tedy optimální slo¾itosti by dosáhl i algoritmus, který
+by pro ka¾dou dvojici úseèek testoval, zda se protínají. Èasto se slo¾itost algoritmu posuzuje i vzhledem k velikosti výstupu $p$. Typické rozmístìní
+úseèek mívá toti¾ prùseèíkù spí¹e po málu. Pro tento pøípad si uká¾eme podstatnì rychlej¹í algoritmus.
+
+Pro jednodu¹¹í popis pøedpokládejme, ¾e úseèky le¾í v obecné poloze. To znamená, ¾e ¾ádné tøi úseèky se neprotínají v jednom bodì a prùnikem ka¾dých
+dvou úseèek je nejvý¹e jeden bod. Navíc pøedpokládejme, ¾e krajní bod ¾ádné úseèky nele¾í na jiné úseèce a také neexistují vodorovné úseèky. Na závìr si
+uká¾eme, jak se s tìmito pøípady vypoøádat.
+
+Algoritmus funguje na principu zametání roviny, popsaném v minulé pøedná¹ce. Budeme posouvat vodorovnou pøímku odshora dolù. V¾dy, kdy¾ narazíme na
+nový prùnik, ohlásíme jeho výskyt. Samozøejmì spojité posouvání nahradíme diskrétním a pøímku v¾dy posuneme do dal¹ího zajímavého bodu.
+
+Zajímavé události jsou {\I zaèátky úseèek}, {\I konce úseèek} a {\I prùseèíky úseèek}. Po utøídìní známe pro první dva typy událostí poøadí, v jakém
+se objeví. Výskyty prùseèíkù budeme poèítat prùbì¾nì, jinak bychom celý problém nemuseli øe¹it.
+
+V ka¾dém kroku si pamatujeme prùøez $P$ -- posloupnost úseèek aktuálnì protnutých zametací pøímkou. Tyto úseèky máme utøídìné zleva doprava. Navíc si
+udr¾ujeme kalendáø $K$ budoucích událostí. Z hlediska prùseèíkù budeme na úseèky nahlí¾et jako na polopøímky. Pro sousední dvojice úseèek si
+udr¾ujeme, zda se jejich smìry nìkde protnou. Algoritmus pro hledání prùnikù úseèek funguje následovnì:
+
+\s{Algoritmus:}
+
+\algo
+
+\:$P \leftarrow \emptyset$.
+\:Do $K$ vlo¾íme zaèátky a konce v¹ech úseèek.
+\:Dokud $K \ne \emptyset$:
+\::Odebereme nejvy¹¹í událost.
+\::Pokud je to zaèátek úseèky, zatøídíme novou úseèku do $P$.
+\::Pokud je to konec úseèky, odebereme úseèku z $P$.
+\::Pokud je to prùseèík, nahlásíme ho a prohodíme úseèky v $P$.
+\::Navíc v¾dy pøepoèítáme prùseèíkové události, v¾dy maximálnì dvì odebereme a dvì nové pøidáme.
+\endalgo
+
+Zbývá rozmyslet si, jaké datové struktury pou¾ijeme, abychom prùseèíky nalezli dostateènì rychle. Pro kalendáø pou¾ijeme napøíklad haldu. Kalendáø
+obsahuje v¾dy nejvý¹e $\O(n)$ událostí. Seznam úseèek si budeme udr¾ovat ve vyva¾ovaném stromì, který obsahuje nejvý¹e $\O(n)$ prvkù. Proto jednu
+událost kalendáøe doká¾eme vyøe¹it v èase $\O(\log n)$. V¹ech událostí je $\O(n+p)$, a tedy celková slo¾itost algoritmu je $\O((n+p) \log n)$.
+
+Na závìr poznamenejme, ¾e Balaban vymyslel efektivnìj¹í algoritmus, který funguje v èase $\O(n \log n + p)$, ale je podstatnì komplikovanìj¹í.
+
+\h{Hledání nejbli¾¹ího bodu}
+
+Nyní se pokusíme vyøe¹it i problém druhé strany.
+
+{\I Eskymáci tráví vìt¹inu èasu doma, ve svém Iglù. Takový medvìd je na své toulce zasnì¾enou krajinou, kdy¾ tu se najednou rozhodne nav¹tívit nìjakého
+Eskymáka. Proto se podívá do své medvìdí mapy a nalezne nejbli¾¹í Iglù. Má to ale jeden háèek, medvìdi toti¾ neumí èíst v mapách!}
+
+Popí¹eme si nejprve, jak vypadá medvìdí mapa. Medvìdí mapa je {\I Voroného diagram}, co¾ je pro zadané body $x_1, \ldots, x_n$ rozdìlení roviny na
+oblasti $B_1, \ldots, B_n$, kde $B_i$ je mno¾ina bodù nejblí¾e k bodu $x_i$. Formálnì jsou tyto oblasti zadefinovány následovnì:
+$$B_i = \left\{y \in {\bb R}^2\ \vert\ \forall j:\rho(x_i,y) \le \rho(x_j,y)\right\}.$$
+
+Pro dva body $a$ a $b$, mno¾ina v¹ech bodù bli¾¹ích k $a$ ne¾ k $b$ tvoøí polorovinu ohranièenou osou úseèky $ab$. Ka¾dá z oblastí $B_i$ je tvoøena
+prùnikem $n-1$ polorovin, tedy je to (mo¾ná neomezený) mnohostìn.\foot{Sly¹eli jste u¾ o lineárním programování? Jak název vùbec nenapoví,
+{\I lineární programování} je teorii zabývající se øe¹ením a vlastnostmi soustav lineárních nerovnic. Lineární program je popsaný lineární funkcí, kterou chceme
+maximalizovat za podmínek popsaných soustavou lineárních nerovnic. Ka¾dá nerovnice urèuje poloprostor, ve kterém se pøípustná øe¹ení nachází.
+Proto¾e pøípustné øe¹ení splòuje v¹echny nerovnice zároveò, je mno¾ina v¹ech pøípustných øe¹ení (mo¾ná neomezený) mnohostìn. Mno¾iny $B_i$ lze
+snadno popsat jako mno¾iny v¹ech pøípustných øe¹ení lineárního programù, pomocí vý¹e ukázaných polorovin. Na závìr poznamenejme, ¾e dlouho otevøená
+otázka, zda lze nalézt optimální øe¹ení lineárního programu v polynomiálním èase, byla vyøe¹ena, je znám polynomiální algoritmu (kterému se øíká metoda
+vnitøního bodu). Na druhou stranu, pokud chceme najít pøípustné celoèíselné øe¹ení, je úloha NP-úplná a je jednoduché na ni pøevést spoustu
+optimalizaèních problémù. Dokázat v¹ak, ¾e tento problém le¾í v NP, není vùbec jednoduché.}
+Voroneho diagram je naznaèen na obrázku, oblasti $B_i$ jsou ohranièené èernými èárami.
+
+\twofigures{8-geom2_2_polorovina.eps}{Body bli¾¹í k $a$ ne¾ $b$.}{1.25in}
+ {8-geom2_3_voroneho_diagram.eps}{Voroneho diagram.}{2.5in}
+
+Prùnik dvou oblastí mù¾e být, v závislosti na jejich sousedìní, prázdný, bod, úseèka, nebo polopøímka. V na¹em uva¾ování uzavøeme celý Voroneho diagram do
+dostateènì velkého obdélníku. Proto¾e okraje mno¾in tvoøí rovinný graf, je podle Eulerovy formule (poèet hran je nejvý¹e $3n-6$) slo¾itost diagramu
+$\O(n)$. Voroneho diagram lze zkonstruovat v èase $\O(n \log n)$. Tím se zabývat nebudeme,\foot{Pro zvídavé, kteøí nemají zkou¹ku druhý den ráno:
+Detaily naleznete v zápiscích z loòského ADSka.} ale uká¾eme si, jak zadaný Voroneho diagram pou¾ít k rychlému nalezení nejbli¾¹ího bodu.
+
+Máme zadaný rozklad roviny na konvexní mnohoúhelníky. Chceme pro jednotlivé body rychle rozhodovat, do kterého mnohoúhelníku patøí. Na¹e øe¹ení budeme
+optimalizovat pro jeden pevný rozklad a obrovské mno¾ství rùzných dotazù, které chceme co nejrychleji odpovìdìt.\foot{Pøedstavujme si to tøeba tak, ¾e
+medvìdùm zprovozníme server. Ten jednou schroustá celou mapu a potom co nejrychleji odpovídá na jejich dotazy. Medvìdi sice neumí èíst v mapách, ale
+pøipojit se na server hravì zvládnou!} Nejprve pøedzpracujeme Voroneho diagram a vytvoøíme strukturu, která nám umo¾ní rychlé dotazy na jednotlivé
+body.
+
+Uka¾me si nejprve øe¹ení bez pøedzpracování. Rovinu budeme zametat pøímkou shora dolù. Podobnì jako pøi hledání prùseèíkù úseèek, udr¾ujeme si prùøez
+pøímky. V¹imnìte si, ¾e tento prùøez se mìní jenom ve vrcholech mnohoúhelníkù. Ve chvíli, kdy narazíme na hledaný bod, podíváme se, do kterého
+intervalu v prùøezu patøí, a ten nahlásíme. Prùøez budeme uchovávat ve vyhledávacím stromì. Takové øe¹ení má slo¾itost $\O(n \log n)$ na dotaz, co¾ je
+hroznì pomalé.
+
+Pøedzpracování bude fungovat následovnì. Rozøe¾eme si celou rovinu na pásy, bìhem kterých se prùøez pøímkou nemìní. Pro ka¾dý z nich si budeme
+pamatovat stav stromu, ve kterém se nacházel prùøez pøi procházení tímto pásem. Kdy¾ chceme lokalizovat nìjaký bod, nejprve pùlením nalezneme pás, v
+kterém le¾í. Poté polo¾íme dotaz na pøíslu¹ný strom. Dotaz doká¾eme zodpovìdìt v èase $\O(\log n)$.
+
+Jenom¾e na¹e øe¹ení má jeden háèek: Jak zkonstruovat jednotlivé verze stromu dostateènì rychle? K tomu napomohou {\I èásteènì perzistentní} datové
+struktury. Pod perzistencí se myslí, ¾e struktura si umo¾òuje uchovávat svoji historii. Èásteènì perzistentní struktury nemohou svoji historii
+modifikovat.
+
+Popí¹eme si, jak vytvoøit perzistentní strom s pamìtí $\O(\log n)$ na zmìnu. Pokud provádíme operaci na stromì, mìní se jenom malá èást stromu.
+Napøíklad pøi vkládání do stromu se mìní jenom prvky na jedné cestièce z koøene do listu (a pøípadnì rotací jejím nejbli¾¹ím okolí). Proto si ulo¾íme
+upravenou cestièku a zbytek stromu budeme sdílet s pøedchozí verzí. Mimochodem zmìny ka¾dé operace se slo¾itostí $\O(k)$ lze zapsat v pamìti
+$\O(k)$, prostì operace nemá tolik èasu, aby mohla pozmìnit pøíli¹ velikou èást stromu.
+
+Celková èasová slo¾itost je tedy $\O(n \log n)$ na pøedzpracování Voroneho diagramu a vytvoøení persistentního stromu. Kvùli persistenci potøebuje
+toto pøedzpracování pamì» $\O(n \log n)$. Na dotaz spotøebujeme èas $\O(\log n)$, nebo» nejprve vyhledáme pùlením pøíslu¹ný pás a poté polo¾íme dotaz
+na pøíslu¹nou verzi stromu. Rychleji ani celou operaci provést nepùjde, proto¾e potøebujeme utøídit souøadnice bodù. Na závìr poznamenejme, ¾e se umí
+provést vý¹e uvedená persistence vyhledávacího stromu v amortizované pamìti $\O(1)$ na zmìnu. My¹lenka je, ¾e pou¾ijeme stromy, které pøi insertu a
+deletu provádí amortizovanì jenom konstantnì mnoho úprav své struktury. To nám napøíklad zaruèí 2-4 stromy z pøedná¹ky a podobnou vlastnost lze
+dokázat i o èervenoèerných stromech. Pøi zmìnì potom nebudeme upravovat celou cestu, ale upravíme jenom jednotlivé vrcholy, kterých se zmìna týká.
+
+\bye
--- /dev/null
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--- /dev/null
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--- /dev/null
+P=8-geom2
+
+include ../Makerules
--- /dev/null
+% implementation of figure naming and figure transparency
+string name,tag; name := ""; tag := "";
+def updatefigname =
+ if tag="": filenametemplate (name & "%c.eps");
+ else: filenametemplate (name & "%c_" & tag & ".eps");
+ fi;
+enddef;
+
+def figname(expr n) = name := n; updatefigname; enddef;
+def figtag(expr t) = tag := t; updatefigname; enddef;
+
+picture transparent_picture;
+color transparent_color; transparent_color := 0.9white;
+def drawtransparent(expr num) =
+ transparent_picture := currentpicture;
+endfig;
+
+if tag="": filenametemplate (name & "%c_transparent.eps");
+else: filenametemplate (name & "%c_" & tag & "_transparent.eps");
+fi;
+beginfig(num);
+ draw transparent_picture withcolor transparent_color;
+endfig;
+
+updatefigname;
+enddef;
+
+def from(expr p,d,len) =
+ p+dir(d)*len
+enddef;
+
+def dirs(expr p,d,len) =
+ p--from(p,d,len)
+enddef;
+
+def drawvertices(expr s,n) =
+ for i:=s upto n:
+ draw vertex(PQ[i]);
+ endfor
+enddef;
+
+def drawfvertices(expr s,n,flags) =
+ for i:=s upto n:
+ draw vertex(PQ[i]) flags;
+ endfor
+enddef;
+
+def vertex(expr p) = p withpen pencircle scaled 4pt enddef;
+def drawemptyvertex(expr p) = unfill fullcircle scaled 4pt shifted p; draw fullcircle scaled 4pt shifted p; enddef;
+def drawendpointvertex(expr p) = draw vertex(p) withcolor red; draw fullcircle scaled 6pt shifted p; enddef;
+def createpath(expr p) = shakepath(p, 0.015cm,0.1cm) enddef;
+vardef shakepath(expr p,d,l) =
+ save r,b;
+ path r; r := point(arctime 0 of p) of p;
+ b := -1;
+ for i:=l step l until arclength(p):
+ r := r--(point(arctime i of p) of p)+dir(angle(direction(arctime i of p) of p rotated 90))*d*b;
+ b := -b;
+ endfor
+ r--point(arctime arclength(p) of p) of p
+enddef;
+
+pen normalpen; normalpen := pencircle scaled 0.6pt;
+pen boldpen; boldpen := pencircle scaled 1.5pt;
+pen bolderpen; bolderpen := pencircle scaled 2pt;
+def dotline = withdots scaled 0.82 withpen boldpen enddef;
+
+vardef unclosedbubblec(expr p,c) =
+ bubblec((p..reverse p..cycle),c)
+enddef;
+
+vardef bubblec(expr p,c) =
+ save r;
+ path r; r := (point(arctime 0 of p) of p)+dir(angle(direction(arctime 0 of p) of p rotated 90))*c;
+ for i:=0.01cm step 0.025cm until arclength(p):
+ r := r..(point(arctime i of p) of p)+dir(angle(direction(arctime i of p) of p rotated 90))*c;
+ endfor
+ r..(point(arctime arclength(p) of p) of p)+dir(angle(direction(arctime arclength(p) of p) of p rotated 90))*c..cycle
+enddef;
+
+vardef bubble(expr p) = bubblec(p,0.12cm) enddef;
+vardef unclosedbubble(expr p) = unclosedbubblec(p,0.12cm) enddef;
projects / ads2.git / commitdiff