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K0 @(  l  C X ^~  l  C , 3    H  0޽h ? @ ff3Ιd332z   K0 `@4(  @l @ C \\ p`    @ S b  " PpH @ 0޽h ? @ ff3Ιd332z  K0    ( q@G@ l  C Hށ p`   $  J(A Papeteria`,$  0 v(Problem Niech E bdzie dan przestrzeni, a X jej skoDczonym podzbiorem. Zbada, czy dany element x przestrzeni E nale|y do wybranego podzbioru X.   0`z ,$  0 Metoda rozwizania polega na porwnywaniu kolejnych elementw cigu, poczynajc od pierwszego, z elementem x. Je[li badany element jest rwny x, to koDczymy postpowanie; je[li nie, to rozwa|amy nastpny element cigu.  2   0XɁ  ,$  0 Specyfikacja wp ={(ei ) cig n elementw przestrzeni E , n 1} wk ={ wynik ($ i n) ei = x}i 2   % t2  0 @,$  0 ZakBadamy, |e elementy cigu s przechowywane w pewnej tablicy e[1], e[2],...,e[n].T 2T TH  0޽h ? @ ff3Ιd332z6.___PPT10+\ëD' = @B D' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*i%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*iD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*iD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*T%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*TD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*T++0+ ++0+ ++0+ ++0+ +i.  K0 (  l  C   p`   2  6hu  Vpocztek   2  6  p@ Lkoniec  2  6p 0 p 0i n and not wynik ,&2  6t p @  Mx= e[i]   6` `  m wynik := true&   6PP  wwynik := false; i := 1;&    6t#` p   W i := i+1;   ^B  6Dz z ^B  @ 6D0 0 ^B  6Dp0 0 vB  ND? @^B @ 6D@ `@` ^B  6D 0 XB @ 0D ^B  6D P P ^B @ 6D`P jB  BD``jB  BD`  b,APapeteria* ,$D  0 T x {e[1],...,e[i-1]}, wynik=false , i=14+ 2%&   bd3APapeteria`  ,$D  0 ` x {e[1],...,e[i-1]}, in+1,wynik=false, in X1 2 1l  b8<APapeteria P ,$D  0 0 x=e[i], in wynik=true , 2 &    bAAPapeteria@ @ : ,$D  0 L x {e[1],...,e[i-1]}, in, x e[i] X' 2 '  bJAPapeteria 0:,$D  0 V x {e[1],...,e[i-1]}, in+1, wynik=false B, 2&   <Q@ 0'  ITak 2   <U@ P '  INie 2 H  0޽h ? @ ff3Ιd332z___PPT10+& @D' = @B DP' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*++0+ ++0+ ++0+ ++0+ ++0+ +h  K0   r ( ,mz3 l  C h p`      tLjAA)Papeteria00 ,$D  0 Wersja 1 { i :=1; wynik := false; while ( i n and not wynik) { if x=e[i] then wynik := true else i := i+1 fi; } }8 2  )  ,   luAR|owa lignina Z ,$D  0 @wynik= true, i n , e[i]=x lub wynik=false, i=n+1, x {e[1],...,e[i-1]} BN 2)@    <~ ,$  0 d wynik = true wttw dla pewnego i n , e[i]=x:3 2$ @ T   tPAA)Papeteria  ,$D  0 |Wersja 2 { e[n+1]:= x; i:=1; while (e[i] x ) { i := i+1; } wynik := (i n ); }dh 2 (,&$?H  0޽h ? @ ff3Ιd332z___PPT10+Z0D' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(D:' =A@BBB*B0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B)barn(outHorizontal)*<3<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*3%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*3D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*3D' =%(D' =%(D<' =A@BBB B0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B+checkerboard(across)*<3<*++0+ ++0+ ++0+ ++0+ +A  K0 TL (  l  C  p`   d  bAPapeteria`Z,$D  0 XOperacja dominujca = porwnywanie elementw- 2- -  <lppj,$  0 W najlepszym razie algorytm wykona 1 porwnanie. W najgorszym razie n- porwnaD.R 2R R  bAPapeteria ,$D  0 VKoszt [redni A(n)=S d Dn p(d) * t(d)v, 2  & B  BD ,$D   0h  `pAPergaminp0,$D 0 PPrawdopodo-bieDstwo wystpienia danych d) 2)B  BDp` ,$D   01  `APergaminp` ,$D 0 m'Koszt realizacji algorytmu dla danych d( 2( (  <`  ,$   0 8Niech p(xX)= p i zaB|my, |e jest jednakowo prawdopodobne, |e x znajduje si na 1szym, 2gim czy i-tym miejscu.*p 2 f p  <D  ,$  0 ]Wtedy p(x=e[i]) = p/n. 2 $  <@ 0,$   0 >Zatem A(n) =1* p/n+ 2*p/n +...+n*p/n +n*(1-p) = n - (n-1)*p/2.? 2? ?R   LA$A)Niebieska lignina 0,$D  0 j&A koszt pamiciowy? 2 H  0޽h ? @ ff3Ιd332z11___PPT101+VuD/'  = @B DW/' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*R%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*RD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*RD' =%(D|' =%(D$' =A@BBB B0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-o6Bbox(out)*<3<* DR' =%(D' =%(D' =4@BBBB%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* D' =%(D' =%(DV' =A@BBB*B0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B)barn(outHorizontal)*<3<* DR' =%(D' =%(D' =4@BBBB%(E' =4 B`BPB`B?<* %(!/%,( < +D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* D' =%(D' =%(DV' =A@BBB*B0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B)barn(outHorizontal)*<3<* D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*p%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*pD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*pD' =%(D' =%(D8' =A@BBB B0B%(E' =4 B`BPB`B?<* %(2/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*?%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*?D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*?D' =%(D' =%(D6' =A@BB BB0B%(E' =4 B`BPB`B?<* %(=/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B%slide(fromBottom)*<3<*++0+ ++0+ ++0+  ++0+  ++0+  ++0+ ++0+ ++0+ ++0+ ++  K0 0(  x  c $<-Y p`  Y   V03Y̙AA)Db0p,$D   0 pProblem Dany jest cig rosncy e[1],..,e[n] oraz element x pewnej przestrzeni liniowo uporzdkowanej <E, >. Chcemy ustali przedziaB e[i], e[i+1], w ktrym ewentualnie mie[ci si x.8 2cN <  <Y@m ,$  0 Specyfikacja : wp= {("i<n) e[i]<e[i+1] , n>0 } wk= {wynik= i wttw x nale|y do itego przedziaBu}@c 2  Mt      <Y 0,$  0 N Metoda sekwencyjna 1. Zbadamy przypadki skrajne: przedziaB 0 i przedziaB n-ty i ew. koDczymy . 2. Nastpnie porwnujemy x z kolejnymi elementami cigu poczynajc od e[2]. Je[li x jest mniejsze prawego koDca rozwa|anego i-tego przedziaBu, to znaczy , |e x znajduje si w tym przedziale, e[i] x <e[i+1]. Je[li tak nie jest, to trzeba zbada nastpny przedziaB. 6j 2E j @` z  7   7,$D  0`B  0D` @ `B  0D`B   0D`B   0D`B   0D   < XpP 7 Je[1] 2    < XP 7 Je[2] 2   <DXP7 Le[n-1] 2   < $X`P7 Je[n] 2   <-X  G... 2   <7X0 \PrzedziaB 0 2    <.X  \PrzedziaB 1 2    <4JX `PrzedziaB n-1 2   <SX 0  \PrzedziaB n 2  `B  0D@ @   <h]X P0 7 Je[3] 2 H  0޽h ? @ ff3Ιd332z0(___PPT10+DD' = @B D' = @BA?%,( < +O%,( < +D_' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*Dp' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =+4 8?RCB?BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?RCB?BCB#ppt_yB*Y3>B ppt_y<*D' =%(D|' =%(D$' =A@BBB B0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*c%(D' =-o6Bbox(out)*<3<*cD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*i%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*iD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*i++0+X ++0+X ++0+X +7  K0 P \(  x  c $|X p`  Y R  bdXAPapeteria` J,$D 0 0{ if x<e[1] then i := 0 else if x e[n] then i := n else i := 1; while x e[i+1] { i := i+1; } fi; fi; wynik := i; }B 2/)?    )  rlXA$Niebieska lignina  ,$D  0 0 e[i] x < e[i+1], i +B#style.visibility<*%(D' =-6B-randombar(horizontal)*<3<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D&' =%(D' =%(Dv' =A@BBB B0B@B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' K=-6B-randombar(horizontal)*<3<* D' =%(Do' =%(D' =4@BBB B%(E' =4 B`BPB`B?<* %(-/%,( < +D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-o6Bbox(out)*<3<* D' =%(D' =%(DG' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+v4 ,0B*O3> B<* ++0+ ++0+ ++0+ ++0+ ++0+ ++0+ ++0+  ++0+  +&  K0   p# (  x  c $ p`     <:,$  0 ZRozszerzamy dany cig o elementy a i b, tzn. przyjmujemy e[0] = a i e[n+1] =b, gdzie [a,b] jest przedziaBem, w ktrym mieszcz si elementy cigu oraz x. ZaB|my, |e prawdopodobieDstwo tego, |e x przyjmuje jak[ warto[ z przedziaBu [a,b] jest zawsze takie samo.   2    b)APapeteria``Z,$D  0 `Mamy p(x [e[i],e[i+1])) = (e[i+1]  e[i])/(b-a),1 2( 1  <,$D 0 g!Koszt oczekiwany algorytmu A(n) =" 2" "  c $A ??  8 $D 0  c $A ?? w8 $D 0   t@AA)Papeteria ,$D   0 Je[li dBugo[ci przedziaBw s takie same, to A(n) = n/2 +c, gdzie c<2F 2F FH  0޽h ? @ ff3Ιd332z___PPT10}+ oD' = @B D<' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D' =%(D' =%(DR' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B%slide(fromBottom)*<3<*D' =%(D' =%(DM' =4@BBB B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B-randombar(horizontal)*<3<*D' =%(D' =%(DM' =4@BBB B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B-randombar(horizontal)*<3<*DY' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<* %( /%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<*D' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<*++0+ ++0+ ++0+ ++0+ +  K0   , (  ,x , c $|e p`  e p , <,e`0,$  0 ~Idea Porwnujemy element x z co czwartym elementem danego cigu tak dBugo a| znajdziemy element wikszy od x. Wtedy szukamy x sekwencyjnie w[rd czterech poprzedzajcych go elementach cigu." 2 N , <#`` ,$D  0 Metoda dokBadniej: 1 krok: Zbada czy x [e[1],e[n]). Je[li nie, to ustali wynik zgodnie ze specyfikacj, a je[li tak to wykona krok 2.l 24  , <# `,$D  0 t$2 krok: Porwnywa x z kolejnymi elementami o indeksach 4i, tak dBugo a| (a) znajdziemy element e[4i]>x lub (b) a| przekroczymy n (a) szukamy wyniku sekwencyjnie w przedziale [e[4i- 4], e[4i]), (b) szukamy wyniku sekwencyjnie w przedziale [e[4i- 4], e[n]).   2  , Hs @U 0U 0 NomiD H , 0޽h ? @ ff3Ιd332zwo___PPT10O+-JD' = @B DF' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*,D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*,D}' =%(D%' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*,D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*,D' =+4 8?RCB?BCB#ppt_xB*Y3>B ppt_x<*,D' =+4 8?RCB?BCB#ppt_yB*Y3>B ppt_y<*,Du' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*,D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*,D' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<*,D' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<*,++0+,Y ++0+,Y ++0+,Y +=  K0  0(  0x 0 c $8GY p`  Y  0 <NY:,$D 0 ^{ if x < e[1] then i :=0 else if x e[n] then i := n else i := 4; bool := false; while (i n and not bool) do if x e[i] then i := i + 4 else bool := true fi od; i := i- 4; while x e[i+1] do i := i+1 od; fi; fi; wynik := i }n] 2-M3     !4"l 0 bhcYAPapeteriaP 0z,$D  0  e[1] x < e[n]V 2  0 bpYAPapeteria0 ,$D  0 He[j] x dla j=1,2,...,i-4, boolV% 2&  0 b}YAPapeteria  ,$D  0 Re[j] x dla j=1,2,...,i, bool, int* 2 & 0 bTYAPapeteriap j ,$D  0 Xe[j] x dla j=1,2,...,i-4, bool , in+4t- 2 &   0 bYAPapeteria 0 ,$D  0 Re[j] x dla j=1,2,...,i-4, x<e[i], bool8* 2$&%   0 bجYAPapeteria 0 ,$D  0 Fzbool oraz e[i- 4] x<e[i] lub bool oraz e[i- 4] x<e[n]t> 247  0 <Y0 @,$D   0 %Koszt pesymistyczny: W(n)= 2 +[n/4]+3B& 2 &H 0 0޽h ? @ ff3Ιd332z))___PPT10c)+kclD'' = @B DB'' = @BA?%,( < +O%,( < +D' =%(D' =%(DV' =A@BBB*B0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =-6B)barn(outHorizontal)*<3<*0D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*0D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*0D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*0D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*0D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*0D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* 0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* 0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* 0DY' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<* %(./%,( < +D' =1:Bvisible*o3>+B#style.visibility<* 0%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<* 0D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<* 0D' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<* 0D' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<* 0++0+0Y ++0+0Y ++0+0Y ++0+0Y ++0+0Y ++0+0Y ++0+ 0Y ++0+ 0Y +f!  K0   4~ (  4x 4 c $Y p`  Y  4 <Y:,$D 0 \{ if x < e[1] then i :=0 else if x e[n] then i := n else i := k; bool := false; while (i n and not bool) do if x e[i] then i := i + k else bool := true fi od; i := i-k; while x e[i+1] do i := i+1 od; fi; fi; wynik := i }n\ 2-M3     !3" 4 6X  ,$D  0 Te[j] x<e[n] dla j=1,2,...,i-k, i n+kd+ 2      +B 4 BD P ,$D  0B 4@ BD0  @ ,$D  0 4  W3fԔ?Niezmiennik: Times New Roman p ,$D 0H 4 0޽h ? @ ff3Ιd332z___PPT10+y DT' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DX' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =-6B+checkerboard(across)*<3<*4D' =%(Do' =%(D' =4@BBB B%(E' =4 B`BPB`B?<*%( /%,( < +D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =-o6Bbox(out)*<3<*4D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*4D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*4Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*4D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*4Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*4D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*4+p+0+4c ++0+4c +&  K0 "  8(  8x 8 c $c p`  c j 8 < ,$  0 Dla jakich warto[ci k algorytm ma najmniejszy koszt pesymistyczny?C 2C Cr 8 <J,$  0 Koszt pesymistyczny wyra|a si funkcj f(n) = 2 + n/k + ( k-1) B 2A B 8 < ,$  0 bSzukamy minimum tej funkcji: 2  8 <l z ,$   0 f (k) = -n/k2 + 1 f (k) = 0 dla k = n oraz f  ( n)>0XF 2   Fb 8 < ,$  0 p Koszt pesymistyczny bdzie najmniejszy, gdy k = n"9 25 9 8  W3fԔ?WniosekTimes New Roman0 dz ,$D  0H 8 0޽h ? @ ff3Ιd332z___PPT10+o+B#style.visibility<*8C%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*8CD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*8CD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*8B%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*8BD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*8BD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*8D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*8D_' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*8F%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*8FD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*8FD' =%(Do' =%(D' =4@BB BB%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =-o6Bdissolve*<3<*8D' =%(D|' =%(D$' =A@BBB B0B%(E' =4 B`BPB`B?<* %("/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*89%(D' =-o6Bbox(out)*<3<*89++0+8k ++0+8k ++0+8k ++0+8k ++0+8k +*  K0    <$ (  <x < c $\p p`  p  < <Pp0,$  0 |4Metoda  dziel i zwyci|aj  2   < <,p`Z,$  0 t.Podziel dany cig na dwie cz[ci. Sprawdz, w ktrej poBowie cigu znajduje si poszukiwany element i dalej szukaj t sam metod w tej wBa[nie poBowie. 2  < <p ,$  0 *{ i :=1; j := n; while j-i >1 do m := (i+j) div 2; if e[m] x then i := m else j := m fi od; wynik := i }, 2RB-   < <m z,$  0 Ft wp : e[1] x < e[n], n>1 wk : e[wynik] x< e[wynik+1]n; 2@ < N mA) 6 ,$D  0 V Niezmiennik e[i] x < e[j], i < jX, 2  ,P < 6 m̙`  Z ,$D 0 D1+i = j oraz e[i] x < e[j] B# 2 #B  <@ BD p ,$D 0H < 0޽h ? @ ff3Ιd332z___PPT10+_LDB' = @B D' = @BA?%,( < +O%,( < +D' =%(D|' =%(D$' =A@BBB B0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-o6Bbox(out)*<3<*<D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*<D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*<D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<;%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*<;D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*<;D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*<D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*<D' =%(D' =%(D<' =A@BBB B0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-6B+checkerboard(across)*<3<*<D' =%(D' =%(D3' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* <%(D' =-o6Bdissolve*<3<* <D' =%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-o6Bdissolve*<3<*<+P+0+<l ++0+<l ++0+<l ++0+<l ++0+<l ++0+<l +9  K0 LD@ D (  Dx D c $ l p`  l , D <<&l`,$  0 DCzy ten algorytm zatrzymuje si ?# 2# # D <0l`,$  0 Niech k bdzie liczb wykonanych iteracji ptli oraz odl(k) = j - i.E 2C&5 h D <SkamieniaBa ryba * ,$D  0 ,odl(k+1) = odl(k)/2, je[li odl(k) jest liczb parzyst (odl(k)-1)/2 odl(k+1) (odl(k)+1)/2 , je[li odl(k) jest liczb nieparzystB 2L >Z Q D <Ol p ,$  0 Odl(k) jest liczb caBkowit z przedziaBu [1,n-1 ] i ze wzrostem k maleje! K 2JFn D <X`l @ ,$  0 Istnieje zatem takie k, |e odl(k)=1. A wic algorytm zatrzymuje si.E 2E E(  D <ilP p,$  0 BA jaka jest najwiksza warto[ k?" 2" ":  D <LklP P@p,$  0  k = lg n8 2&  D < P@,$D  0H D 0޽h ? @ ff3Ιd332zJ*B*___PPT10"*+Z}DF(' = @B D(' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*D#%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D#D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D#D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*DE%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*DED' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*DED{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*D0%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D0D' =%(D|' =%(D$' =A@BB BB0B%(E' =4 B`BPB`B?<* %(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*D%(D' =-o6Bdissolve*<3<*DD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*DK%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*DKD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*DKD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*DE%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*DED' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*DED{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* D"%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D"D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* D"Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* D%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* DD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* DD' =%(D|' =%(D$' =A@BBB B0B%(E' =4 B`BPB`B?<* %(4/%,( < +D' =1:Bvisible*o3>+B#style.visibility<* D %(D' =-o6Bbox(out)*<3<* D ++0+Dl ++0+Dl ++0+Dl ++0+Dl ++0+Dl ++0+Dl ++0+ Dl ++0+ Dl + 0 ^V@(  ^  S    YP  c $iX @  Y Na kolejnych slajdach poka|emy kilka rozwizaD tego samego problemu: wyszukiwania w cigu uporzdkowanym. PrzykBad ten ma t zalet, |e jest prosty i Batwo pokaza r|nic w kosztach kolejnych wersji rozwizania problemu.H  0޽h ? ̙33F 0 ` (   ^   S    X   c $ @  X *Chocia| zadanie zostaBo sformuBowane oglnie, to ograniczymy si w tej chwili do struktury liczb rzeczywistych. Oczywi[cie nie zmienia to oglno[ci postpowania, ale zamieszczony tutaj tekst u|ywa standardowego rozumienia relacji niewikszo[ci. W oglnym przypadku nale|aBoby napisa funkcj o warto[ciach booleowskich, ktra realizowaBaby porwnywanie elementw i odpowiednio jej u|y w testach programu.F ? XH   0޽h ? ̙33L 0  (  ^  S      c $$P @   LPrzedziaB zostaB podzielony na (n+1) kawaBkw. Skoro maj by takie same, to dBugo[ jednego wynosi (b-a)/(n+1). Oznaczmy to przez x. Zatem A(x)= 1+ (m-1)x/nx + Suma{ ((n-i) x)/(n x) : i=1...n-1} = 1 + 1/n{ n-1 + 1+ 2 + ... + n-1} = 1+ 1/n {(n-1) + n (n-1)/2} = n/2 + 1.5 - 1/n *H  0޽h ? ̙33  0 0@\(  @^ @ S    p @ c $Tl @  p RW tym algorytmi pominiemy przypadek, |e x nie nale|y do |adnego z przedziaBw wyznaczonych przez nasz cig e1,...en. Upro[cimy i skrcimy tym samym algorytm. Oczywi[cie, dopisujc takie same testy jak w poprzednich algorytmach Batwo przeksztaBcimy podany tu algorytm do przypadku oglnego. W zwizku z powy|szym, specyfikacja wyglda tym razem nieco inaczej. Aatwo te| wykry niezmiennik ptli. Zaproponowana formuBa jest rzeczywi[cie niezmiennikiem ptli w tym programie. Skoro x jest midzy i-tym i jty, elementem to znajduje si albo w lewej albo w prawej poBowie tego przedziaBu. Po ustaleniu (porwnanie w instrukcji if) , w ktrej cz[ci znajduje si nasz element , przesuwamy jeden z koDcw przedziaBu. Zatem po wykonaniu tre[ci ptli, znw x znajduje si miedzy i-tym i j-tym elementem. 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