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Daa Unit 1 Notes

DAA aktu unit 1 notes

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Tukesh Kumar
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7K views67 pages

Daa Unit 1 Notes

DAA aktu unit 1 notes

Uploaded by

Tukesh Kumar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Unit-1- Introduction Algosithm : Algoorthm us dhe det off sules to obtain Yhe expected! output from +he given input: Probtesn | [Apes _—— Program Compile We omallyse omy algowrthm with sespect do wo pasiarmeters Time omct Space. Complexity of Algooithm : Time Complexity: Rules Gudsudus p= | 0 Single Statement : Time Complexity OC1) i) hoop: fr le 1 tn . S& §+ afi] ia Ten) = an = O(n) We neglect +he congtmt values chile coorting the time complexity ( genenal ) Sob dS SUID UDG UY PRB PP PBIB » m case of polynomial equations te 6m + unt we cosite dhe highest olegree team oc") Gi Nested hoop : sa le Ltom ™m eee oe a Se St+alt] bm? poe Ten) = m+am> = m* = OCm) WW) Tf-else : 2a ~ uta [ wv (R] Pe ——* OC) <——— (v) While Leep : "70 Lett n ne 2 dog n Ten) = O(n) a, THT TY » TiTPTA ( TL i// OVERS SUH HHEDHEHEUHU HHH HHHKHHUYYHEHHEEUEEEEY Wi Recursion : fact Cm) “he mét 1 wetuin (1) else vetusin : fack (n-1) mn TEN) = O(n) Example ci Sub Casn) t S= a0 1 or / L=iton obo n S= S+aCil setusin S n Ten) = atan = On) Example (ii) Froduct Ca[i---m, en} b[I---b, --p]) for L=—1 dom . Je idop do cLijl =o ke idon do c[ujl-clejl+aluki* b [kK] vretusin C[i---m1-- Example (iiiy te te ttn As A+alil B— B+ briy for 1h, fer k— 1 dom C[ujl=o B= Atc do ™ mp mp mp. 1p “PJ ' ee Ten) = mM+mp+ mp +mpn +mpn +1 = mHMpt 2mMpn +t Ton) = OCmpn) in hn oe mpm knpm 1 npm TOU = O(mpm) S22 IND HHH NAAN AAVAAVAAVYVIY IVY a sf 243325232) )°9 or dh ie Ue dr de ee eee Paul ul of Jf fel oJ of os of OS Cs eeddddddddddidddididddidididdée bal ou ave vd bob sbduus Graowth of function : to Such shat for alt integer nen, dhon EEEEEH EEF Bh+9 > En B 2 g foo = cgey eas Pee She —ll © t fou aby O2) g bs ae iy No n Theta Notation : Ctight bounds ) The owe amd upbes bemd for the function f(x) fs proviclecl by the ‘6’ notation FFP FETER Hf, ‘fos Non-nogative function foU amd gcu Yhoxe exist an integer me and positive constonty Cc amd Cc, ie Goo G70 Such 4hot for att integer m2m, shen Cigen) £ Fem) Ca cn) HTT Aa y \ OS au VASA SHH AD wD C2 gtr) fr) c. gen) Ne nm SoWing Recusnence : There ase dhsiee methods to solve wHecmwience UJ Master Method / theorem tii] Substitution Method Cilterotien Method ) Wii Recursion +ree Method wi? Master Method : Master Method is used -for Solving dhe following types of Recusurence = n Tony aT (P)+ Fou Q=1 od b>1 m Whis, dhe problam us clevicloal into sub problems each of size * cohone a.b axe positive constants, the cost of clevicleng the problam ond combining Yho Besutts of the sub problems us ctescribe by function Foy Master theorem : Wt .Tin) be defined on the non-negative Integer by dhe wecurorence » the vunning time Ten) cam be Ten) = a T (3) +fon) cohane az1 ond b>I ‘Then Ten) can be bound asymptotically as followed Method 1: "f fory < ntogya Yhen Ton) = 9 Cn Joga }) Method 2: 4 foamy = n%9,0 4hen Tin) = @ C88. tog n ) Method 3: if feny > my 09,2 then Tony = © (fon) hy) « CD DPAEEEKEREFE FE REE} Py - 2992222222 Ge (OP UP OPO RPP Pe OP PP PUCCIO dG L lf [— [Op ™, ‘ obo ee ee Oe cx) Solve dhe Recussience using Master Theorem Ton) = 9T C2) 4% Compose qho given SECUHHENCE GH4th - un Témy = at (2) + fo) we get a arg b= 3 fend =m Now, m Log, neg, 9 mn 40g,(3)* n 2 403° 2 nv Compasie ooith value of ft) Fon en cn? Hance we cam Opbly Method 4 Coase 4) & the Solution ss Ten) = © (m403Q% ) = o(n) Exciiy Solve 4he Recussience Ten) = ast (3) +o Compasie «oth Ten) = aT CE) + Foy a=ac b= 5 $e =n? Now, m tog, % tog. as at Compare asith value of Fen) fm) = v= n* Method 2 fs applted. Teny= @ (m Jog a Jog ) = OCW. Logan ) Extil) Solve tre Recurnence Tem) = act (By an’ Com paste csi-th To = at Ch) + Fon) a=ag b=5 fo = m8 FEEFFEFFFF RRP PY AEE po [T) Novo i : mtog a : : n Log. os [> 1 5 Gena eth fy : ; Foy= 73 > n> ; Metheol 3 ts applied i 4 Ten) = ©( $0n)) ; > a Ole.) i > 4S C2) +1 Sewe at by Masten Theorem S Com posce z= Tens at CE) + fey — aon 7s a =I 12 on AP§ye 2 ny, Ue :? : -2 2 :? Now Fem) ete) : 2 Cage 2 oppdiedt L 2 Tony = OC 48% Jog m ) 2 @ C4 dog ) J Tem) = 3T C7y) + dog Com pase Tenj= aT GQ) + $e) a=3 b-4 fon) = niogn 19 po “m 0g. 3! eee Noto, Fen) = ntogn > 1s Case 37 abbtied Te = oC Ftd) = 0 Cattogn ») WI Ten) = IGT a) + 4% Compare Ten) = at GS) + For) a=I6 b= 4 f= dn 9p on 2S, | m*~ Now fone < me NN; © Ppeeerrrrerqaaqaayvvey >) wii) SOUS DT OHHH CHU UKO OU HYUUUwUd 2 vill) OP eee rke teh ba Ph hPa Cage ISt abplte d Ten) = © (mteg,2 ) = o(w) Ten) = act (F + Un? Compose - a=ar be Fe) = yn? Now, tog a on 109, aS aka Fm = yn? > n> Cage att opblied Ten) = © (fer) = O(N) = 31 Teny= at (8%) + mdogn Compasie a=y b= 4%, fen) = ntogen Now, np" on dog fy oh tog Qa n3 FOU = ndogn > n-? Cage gra Tend= (fem) “6 (nibogm ) Tens at (9m ¢ ) +7 Compare a=2 b= Im% fon m m8, 778 94 nes Now fen) HN < mos Coge ist abpliect Ten) = OCnes ) © Ten) « er (y) + 3m? Compaye a=8 b-a fon) = an2 QOQ 4 MA ‘\ aopannegaaqaqdqqgqyyy yy a 949 oe LALELIF TITTLE PET GERHRECRCECKC HKCU UYUYUYEYUUUe BRASRERH hn VOVvVSeVSY VV Now, nlog be Nn 4og,8 ms fem> = av? TEM) 4 Cr-mM41) + (mm 42) + (M43 Jt = 4-1 4) = 1 F142 43-44 ---- 4m : I+ WCm+t) 2 Saar fact 2 - 6c) ‘yh agheladabaaea ealaeliaeliaialielke/k& HEEKEKKEE} TELE PREP TT; o l AAG rrr 2 ; EEO UU EERE EE BERRHEECE VO eee EUY VY (ii) Ten) = | leeeettenel ar (jen 57> Ton) = at (Ban Ten) -ar(h)+n *a[aBjez] on 1 ar¢ 3g CB ar( Aja a ¥ 2? . at(B)en +N pe 7 aX [ar(h)+ | 4am 3 UT (43m _— . Cae = gk Ce)* Kn log n = kK =) EG) + m Logan Ten) = O(m Logan ) e "Recuseion Tee Me-+thodl- Recursion Tree Method us a pictorial representatim of an idewation method, which us fh dhe form of tree cohere at each Jlavel noolss axe expomdlct. dn genesial | ewe consiclen second tenm in Teeunnence as woot. ut us useful ushon olevide amet congo algosithm ug used Exc) Solve given vecurvrence by Fecwtsion Tree Methocl Ten = > | ica! ar(B)in iy n>! r(B)+7(2)+ are! 771 ae n n., n=2a ak log = oga® @) (2) > m= ak K= dogn logan = dog SrveTeTUTENANAUSNNNE Addi dd dd dl da/ada/a/a/a/a/a/aw/a/a/da'a«a a t PPP rp | a ee ee | eS 1 ADRARBBRAADHOD THOTT (7 aa a Yi bGbbbeddb td ddd VtCuuwuuuv’e bbb bb RbELLLEI JP PPP PPP a a le bseuu v vo ARAB Eb ) Tony = TG) +1( 2 )+n Tony= kn : 0g am z Ten) “oe 133") NOTE: We claooys Consicten the fovigest value of K because ut well give dhe Complaset'ty of worst case - here tn the above question Lyk sicte atitl give best Case cohmw ay Ry . , . bh grolo cane USE. paee Rg gv 3h \ eC) al CAL) +0 e. Solve dhe given secon by wecunsive tree oe Me+thool MBit of Map n Ger! herged of Payee m=5K cane a) “Se [« GTI ATAANIN AS \ a Qe) S DID2APIPAPAP AP PPPOOIGD Tem) = Km i. O(™ tog - ma) a dy) Ww) Ten) = a Ge) + 1(32) + 7 Tey-O(7 9 100 ") Sowe by Recusision Tree Method: Ton) = &T (CB) ant Tew TCR) + Tew Gey xX C86 &* oe 2 = ty . [rttete ~ 1 Ce & [4 ano re ® ant = yy Sowting : a Bubble Sort: Bubble Soest algosithm known as exchonge, =~ Sosk, us a simple Sosting algorithm . Ca St wovke by Compasuing doo utems at a time ci and Swapping thom Af they ase fn wrong order, a The pass “hewugh dhe st us vepectecl until a FA scoop 050 paoctod , which moos ish Tg us sorted. Lt ds algo cated compasuson gart -T Comploxtty of bubble gart = ©Cm?) © 1 Bubble Sock tigosithm : a 4- fos te— 1 do length (A) RS 2- fox ja lang th (A) clown to 14 oh © . oh 3- Tf alll < actin eho Y~ exchomge CALII, ALJ-11); Pa ia a Soskd the follocoing efament by Bubble Sot : a A= 15,16, 68,5 fe > A= 16,16,6,8,5 NS Is 16, 6,85 aw 5,6 ,16,8,5 fass-1 iY a> “e Sy 15.6.8,16, § ay Is,¢,8, ©, [fe] TH - 2a? > 1S,6,2,8, 16 S 6.18,8,5, 16 as Fass-9. ¢ fass- 5 5,6,8,18 16 fass-y 5,6,8.1S ,16 . As 5,6,8, IS, 16 mA [5.6 8 15,16 Complexity : 1424 3+U4----F TH) nN(m-1) 2 ate 2 Ten) = 8») bbobleeb LLL LEI PIPPI PTTL TITTY cs To sovt sw follocoiveg clement by Buloble Soxt az [5/2] 1[4[3] 7[¢ ro2 3 4 § 6 7 durgth A=7 tsi-7 j= 7-2 (v4) Now, | s vet ge7 Als] = AL7 = 6 Als) ALC] =7 ASI < ALS] Ali] < Ae] 6< 7 5 True exchange A[7] and A[6] Now away 4: [sye[' fas [el7] 12 3 4 5 6 7 Now, a na ALi] = Ale] = ¢ A [s-1] =A[5J = 3 Al¢é] < a[s] é< 3 False No exchange , [5[2 Iya 3[ 6] 7] Tee ote HAIN “HF — fy if of oP oP oP PP ee oO BO 666d GdIII Idd 7 bSS SE bu4 4 VOOR SUSE EUS Now, Novo z 3 & ven gr5 ACs] < aly] 3 °< 4 True. exchomge ALS J amd A[y] Now asnay is [s[2 1] 12 3 > ged sey AWW] < AL3l 3 <1 False . No exchange. ist 9 f= 3 A[s] < Af[e] '< 2 Tous exchonge AL3] amd A[2} Now asoray 42: BIBER $23 4 S 6 7 jr 2 Az] < A[1] 1 Note? at this point ou wuiay get sovted but deop will coctinue dilt Last citenation | Wosat case — O(n) SA ory Average case O(n) 23, 1,4,5 Best case 2(m) by Be Selechion Sort: We find sho next Lasigest (smallest) : etement fn the asyviay ond move af do atts final position un the Sostedt AsHCLY. Assume that we Saas cto sort dhe A991 ay im Mereasing oven hat ug he smallest elomert at dhe beginerg oF asciey and tho closgest elument at the last — positon. We begin by Solectirg Ha Longest rrrrhh ery E NAAAANAKA TET EEETPETPT ty? ~~ eoed LY ELID TPT T Arad PNECEnOEE: AN ee Atk etement and Moving it do Wwghest fnolex Position. We cam do shu by swappivg the | element, at dre heighest index and de Lasgest etemenk. \ "pee (feofen-, 1 Selection Sort Mago : A jn tn ci? IF Me—Length (AI py (ge itly In, FH) a for fer do mi * 9 cocin ¢atmind 3- | smallest — j 4 . 4+ mnz? 4 for Te-j+) ton \ } 5- A lest if ae J \ Le —— =1) 6- than smauest é Ci {s{sTs} Now — [ au Smallest = 4 | - aS T=5 - smodugt. 4 Afsy < Atul 5 < 4 wY al False . go q7 Sovted apsay sill be - (EErys) BELL EGEEP REFEREE E-HEEEKERE Ye) PEP PI beeebbdddd ddd due vuuv'd ‘ PREP APP PPP o bd ibibb v eo y b yee Ee ee FFI WHS Insertion Sov: Imsesction Sorting us one dat sort A Bet fvolue by senting value into am existivg Sostecl volue Algosithm : a- for ge 2 do dongtb LAT i 2- clo key < ACSI $- te grt 4- while (iro & ALi] > key ) i 5- do ALTHICALI 6- tela 5 7- ALM] key j Exomple: Sort the following element by meetion Art 9 66 0.8.2,7.1% ae fates [olel2(7 [13] 7 4 “eR E TT] Sed, Yeo! ye (ERP AG ifs] tage & GECOBEGHE ofsje/a sf2f7fi 3] Execution : 12 3 4 5 (3je]sfols J=Qt05 dea key = ALS] = API = 6 veg-=4 tro ALi] > key Ge Ge eu. Att at] apple 9 L= t-1 Fil eo = 0 A [rst] <— key A [ors] < 6 jm < rrr) ere rrrFRRRe THPTH ETT 7, euveeeeeeeeee feb ER EERE OP OPP Poe ae bbesese sues d> OTST id yA A AT Key = ALJ] = Als] =5 Te jriz2 Tso s A[eJ> key 9>5 Tour ALin] — At AB) 4] Als] <— 9 tel = a) = iso SALT > key Alt] > key G>5 , Teun ALinyJ © ALI AbI—6 tet =e Irl=o0 Falke Alot — key Atiie—s Slo Us J- = 4-1 =3 Atti © ALY AW <— 9 Lett = 3-te0 AtinT © AGT ABI <—6 Usd sare] A[zJ— ADI Able 5 U2t1= reo ATL kay ABICO 0 (s[é[9[8| Jes 28 Myre ¢ Time Complexity of; msextion 0sct : Complexity = comparison + movement d=Q = 1 | oe 2 + 2 a-4 = ce g=n = (M1) + (m-1) Sei) 2 Lye pyc eueb) ce NGS co suey) afn(m-1) [ment] OC) ot TITEL (LT TUA 94 » 9 TY Gaagaqaae > Fy bb bb bebe UUUHHYHUHCUUUEUEEL vos yIISSSISGS so a Fre- Requiget for Heap Sort: Binasy Tree Representation Foul binosay Tree Complete bina Tree Koop > Max hoop > Min Insertion and cletetion heap Sost algo. Binary Tree cmol Asay Robreg entation : i203 4 5¢ 73783 > best [" [sls P [ely] Grray Representation) ad t% position Le child = girindes Right Chill = att) 2 poxents of Le je 2 Fut Binasy Tree : Q oO | ‘cat ae a atl noolts have dwo chile! except demntigl foe (eof node) (\ pt © OO ® a Qo Complete Binary Tree A t @) SS ® ® ©© 6 oO A Binosuy Tree he ving ctwoo child at each twek except traf, moto C Lah, nocte com have one clulel) All Futl Binasty Tree as Complete Binasy Tree paiovity given abwoouys fo Lbt obit ® Not complete Binary Tree : ( eymee) gh s/f s\ 6 ® © (Comptete Binasuy Tree’) roy f AaAane [IEEE e > es et ~~ ~~ on ony oe oe Se. Ss ew “aus —— “Ss “Se Se om SS ~Ne VIAIAFIIIIVIIVIIFS PD dD DRS SHeKHSKHOHHUKHOHHHETEEOHO” Keab: A Binary heap dato structwie cs am aver ‘d %Y that cam be view as a compute bin asyy Tree. Fach noolo of dhe binary tree comespoals to metemant of Ihe oveoy, The aswiay ag completely filled at ot duvets exenpit - possibly Lowost (Lost) Lovet Q 6 © ©®@ OO ® Moab sree = no. of etements Koop property : Max Heob: A Heab us called max heab ox cleconeluing heap sf every mole of heap has a value greater thom er equal dhe value of every child of that nocke Al fined (J = ALI Min A is caltecl mtn hoop oF acandieing heap feap ig ene) ads of heap has a vole lss dham or equal cto yhe value of ery child of dlot node: AL Poment (1) < ADI as [se [25 31 [>> [37] (59) a) US ® @ © @ COTO) THrr IP, AfiiL} 9 99 rDNA RID D FELLECCLULLT 2 @. @ “ee @ ® t \a S<@ © . &) e@® .@ De ® “2 Ge . ee@ 2 e® ; oy i) 3 @) DAAAAMMAMIAN ADH HHMY| VA hd hdd hdd dda tTASISIILILIDI DES Oaeaaeecaencn X Heob Sov: algo : Heap Sost CA) I Builol- max- heap CA) 2- for t— slangth [A] clocon to 2 3- obo exchange AGI < afi] heap - size ca) Heap size (A)-! 5- Max - Heapify C Asi) Build -Max- Heap CArt) 1- Heop - size (A) — length (Ad & forte | SHO | own do 4 3- Max Heapity CAD) Max- Heapity CAi) f- he Ot [i] a 8 — Right C1] 3- If Lx neop-size(A) 8 ATQI> ALI tH $hen Losigest <— f S- else Aovigest — 4 . 6 4 bE heop- size (A) BACB) S A ( Aasiges+) 7 Fhen Losigest < 8 Losigest # 1 8- Then exchange CATII,A [Loriges+] ) _ 1 max. heapity CA, Losi gest ) i] PUIIILL PES MELLEL LL OTST eneeeee’ iLustrote He operation of build! max hea en the A> £ 5.3.17, 0,84, 19, 6,2%,9¢ [ear | «Ld sb) Heap size A=9 So first we call max: heaprty (Ax) often max- hoa pity G4) After max hoapify CA!) Agaut max hoobify a) Q ® ® OO © @® Sost Hoo followi vg odio using hoop sost A=§4,1,3,2, 16,9, 10,1, 8,7 ¢ After max haobify (AS) , No exchange | 3 RYVYVVHVANRNAAHAMAKAHAAAK j >» % HL) PPrrrr rrr rr err % / 999979 FERC CEELCLOUOOPEL a a aa USSU U A asaganreeerrr eye HeAAs HEA often max-hoapify (A2) often max. heabify (As!) Agen marx hoabify (12) @ (9 ® O®@ © © OO After tng we sort the elements- & O® © 6060 © 0 S OW Now fe heabi ty: Jers “foe M® Novo Max heabify fits Now max heap fy Ad ®O®® © DPELERPEEEPFPPOOOD DY LEECT CLL TT EEEEREN zis HAANMAHHAAAAAAA C +44 rr r dha ae 9 yw f | 2 Ti r Noto Max~heapify : (2) Q ALO > (1) @) © @O Y O© (CROROROMC) Now Max heabify : (2 Q QS 7 § @ O®@ OF @@HOO®@ Now max heopity : sy - ge 6 G@©G@OGO®@ Now max heebify : &} > 3% OOOOH ® © © Now max heabify : ce > & © © OO0OO OH © Sosk Ho element by heap Soot: £4:1.8,2,16,9,10,14,8,7§ ]o¢ jo T>16 , Foe dasiges t = 5 (No exchange } Now, . t=4 eo a9 8410 W>2 , True Josiges+ = 8 9410 B>ly | False &#i UF Q (exchange Qe) COOEYHE ane SB, r ANHHANKNAAKS x Peet. 9 pidcccecedeeitiiiTHe PEOPLE ADA Chop t tt e Y Quick sort: Gutck gost us baged on clevide ond conqun sul, BivicleafP .-. w J into ALP..-- 4-1] and fan a A[p.--¥] l™ Aled} ala 0] Suchthat — A[P---Ft1] 4 apa] < A[4qti----3] Conqusy, shost dhe sub assiay Teawuively, Combing, excise aseny ALP. v] us moto costed . Quek Sod algosithm coosks dike paschtion- ving the aswiay do be gosded amd each postition fm terms costed vecusisively. mn pastidion, ens of cho aseiay efoment 8 Choosen OS a key value dhs Koy value can be dhe fisat efoment ef sho ascjay. Rest of drs BUI eloments wu grouped flo hoo postion Such as- dl) one portion contains eloments smatbte Yvhan Yho key value. Gy Anothes postion confatns elements Lasiges than Ye key value Algo: Fasctation) pivot ition» Volur < key key key < valuz E 4 8 LOE TT] oo Ranit-1 key Poot-2 (P--- 4-1) (qti--- 8) Firstelement asc |p tastetem ent Quick Sost CAP) 1- Tf Pes dhen a- ae 4- 4<— PARTITION CALA) Quick soot CA,P, 4-1) Quick sost CA,441,% ) PARTITION CA: P, 5) fvot | k iC hwu last element as pivot x Cpivoty key ) x= A[r] Ta pd fos GP dod alo Ff ALIS % Yhon Te ted exehamge ALiI< Ati] exchange ACH] << ALrI vetwin [44 NANHNNRANKAAAAHA’ POPP h rrr) a Tr « 6 de LLC ay a i 2% ? il LL 2p ULULLLLL e lal leleriey ev be o rr aT > \ 4 Ex: ferform Quick Sost : 3 5.1,2,9,0, 8,13? ER BEET) 3| t Prot 2.0, El 1 P [5] toe! Pp 2>h], P p [ol BIST #] 3 13] of GB, [8]. 9.13 F l Ex: Perform Quick Sust: {76.10,5,9,2,1,15,7f Avot [6.5,2, L 7 [7] 9, 10, 15 t tf Fivot 5,8! [Elz 7] 9. 10,15 r P Sj + Z 1,7, 4], 8.10, 15 P 5[¢]7]7]s lof 15 AT Rs Xd letd vouve ei P [1jo2]s 4 5]6[7] Laos y oe Pis|s)7] [plsl¢) m= U-14) m= B-¥=4 m= Y 123 us ‘a3 4g L- [2[uls]7][o ® [1 ]2] se] eo tiiaa g¢aag = pox e 3.4 §S ¢ 7 es i* Oe a+ Ll2talstetste [7] 4 Counting Sovt : Coumt Sort assume thot each dee om Inout elements fg an integen in dhe -wange idek, for some integen K ushen k= 0(m) the gost sum fr O(n) dimes. The basic iotea of count ost ug co atermixe for each Prbut element oc, the number ef element les than oe. his fo com be used ato place element oo directly nto bs position fk dhe output arciay Algosithm : Counting Sort (A,B. K) Se OGeel a onsale a do CliJ—o 3- for j= 1 to dength [a] doc CACIIJ<- c [acd3]-r4 5 for i= 140k 6 do Chil O(n) —| > > Sostane given element by Rackese sort: —~ 2 dy 329 720 720 329 — us7 asc 329 356 a 657 436 436 u36 —L, 839 > 457 5 639 5 uc7 > 436 657 3s 657 > 720 323 usT = Ls oo 839 6s7 839 La viata hy Gi) M0, 46 45) BO, 802 24,2 66 —L_» — (70 170 B02 002 —4 ous ©30 002 oou Ls eae B02 Ooy ous Ls go. > 082 S our 5 086 —* ai O24 66 oS —4* oon org '70 O90 —? Oe o1s os \70 1” O66 80 802 —_”» .* —_» \ ( : Gy Cow SEA BAR BAR do EAR Cow G Ruc TAR DOm SEA DOG cea EAR Ruc BAR Row Row > EAR Dom > Rue COW BAR TAR SEA EAR Cow ee TAR TAR Rov -FEEEEW “PP rr rer l. de BUCKET sort :- Bucket Sosct wm Gn Linea tme on dhe cwenage dike cormt gost, bucket gort us fast becore cut assuma somthing obout- the input chee 8 coum cont — assume that the fnput consist ef dhe integer na gmall wonge, bucket Sovst assume Yhot dhe Frnpuct fs genes ateol by a wondom process that clistrbutes — elaments umefowmly pve dhe Prdesivat (0,1). Be sort ov Input umber, bucket sont W) posctitien trto “ny mon evetabbing tnterwal Caltect bucket td fut each Fabect mumbens into it! buckef (io Sort each — bucket using a Rimple algorithm exampls- fhgection sot ond dhen TORUIVOOTTOPFPAONNMNMMOMOOHRAKRE TCCCULLLLLL ELL LULL Le POPP PPP PR PD HLL ! Phe tt t tt WW) comcatinate dhe goxted Lad Bucket Sost CA) Ir me dength (A) 2 for te. a don 3- oo insect ALi] Mato dist BL LmAlrl] | Me for 1— 0 to N-! 5- do sort list @[t] usith frsection sort 6- concatenate yro Lust Blo], B[2]------ B[m-1] dogethen im ovclen Sost dhe geven eloment by bucket godt jo- 123 4 5 6 7g 49 og to A= [lisa |-26[- 72] « g7)0, ~ro}as)- 6a] °o ' I —7 WT je eILior a ee pe bn] 3] F-C8s7 121 ]-23[-26) 4y s|_| L338} 6 [68 r— r-|:68) ao ta 8 18 | 72 3 F317 37] 123 4 § 6 7 8.9 10 \ofrl-2[-23f-26]-39[- | ae) 72] 7]

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