Chng I:PHN M U1.1 LI CH CA VIC NN D LIUMt trong nhng chc nng chnh ca my tnh l x l d liu v lu tr. Bn cnh vic x l nhanh, ngi ta cn quan tm n vic lu tr c nhiu d liu nhng li tit kim c vng nh v gim chi ph lu tr. V mt l thuyt th cc thit b lu tr l khng c gii hn nhng ngy nay do nhu cu x l nhiu tp tin, nhiu loi d liu trong cng mt tp do vy m kch thc tp tr nn kh ln. Trong nhiu nm gn y, mng my tnh tr nn ph bin trn th gii. S ra i ca mng thc hin c c m chinh phc khong cch gia con ngi. Nhng li ch m mng cung cp rt a dng v phong ph trn cc lnh vc khc nhau ca x hi nh cung cp, trao i thng tin gia cc my tnh, gia my tnh vi server hoc gia cc server vi nhau. iu ny dn n phi lm nh th no gim thiu thi gian, chi ph s dng trao i d liu trn mng. N cng ng ngha vi vic bn cnh nng cao cht lng ca cc thit b truyn d liu trn mng th mt khc chng ta phi ngh ra mt phng php no khng sao cho vic truyn d liu c hiu qu hn.Tt c nhng vn trn ny sinh ra khi nim nn d liu. Mt trong nhng hnh thc nn d liu u tin l h ch Braille, l mt h ch dng phng php m ho k hiu cho ngi m c th c v vit. Ngy nay, nn d liu mang li rt nhiu li ch khc nhau m in hnh l :+ i vi vic tm kim th i khi ta tm kim thng tin trn d liu nn li nhanh hn so vi vic tm kim thng tin trn d liu khng nn v do d liu lu tr t nn s php ton tm kim gim v lng thng tin cao.+ Nn d liu c bit hiu qu vi vic truyn d liu trn mng. Khi nn d liu th chi ph cho vic truyn d liu trn mng s gim mt khc tc ng truyn s tng ln bi v cng mt lng thng tin thi gian truyn d liu s gim.1Vi nhng u im trn th do , nn d liu l gii php hp l nht nhm mc ch gim chi ph cho ngi s dng. Tm li nhng li ch m nn d liu mang li xoay quanh vn tit kim ti a chi ph cho vic mua cc thit b lu tr d liu v cho vic lun chuyn thng tin trn mng.* Mt s vn gp phi khi nn d liu l :+ Cc thut ton thc hin trc ht phi gim chi ph lu tr.+ Cc thut ton c thc hin nhanh, hiu qu.Vi nhiuloi thngtinkhcnhaumtaccckthut nnkhc nhau, c hiu qu khc nhau, v d nh nn tp vn bn thng tit kim 20% - 50%, cn i vi tp tin nh phn khong 50% - 90% tuy nhin i vi cc tp tin ngu nhin th lng khng gian tit kim c rt t hoc hu nh khng tit kim c (chng hn nh tp *.exe, ...). Do cc d liu c d tha khc nhau nn mi phng php nn nu p dng ng s tr nn khng cn thit, khng c linh hot.1.2 KHI NIM V D LIU: Trong mt bi ton, d liu bao gm mt tp cc phn t c s m ta gi l d liu nguyn t (atoms). N c th l mt ch s, mt k t... nhng cng c th l mt con s, hay mt t..., iu tu thuc vo tng bi ton.1.3 NN BO TON l m hnh nn d liu m n cho php ngi s dng bo ton thng tin trong sut qu trnh nn. iu ny c gii thch nh sau:Gi s ta c d liu ngun l D v d liu nn l D'. Sau khi ta gii nn D' th c tp D'' m tp D'' hon ton ging vi tp D ban u khi c gii nn. Thng thng, k thut ny c p dng vi cc loi d liu nh vn bn v chnh xc ca vn bn. Nu hiu theo ngn ng ton hc th l mt nh x 1 - 1 t 1 tp X -> Y m mi phn t Xi X tng ng vi mt phn t Yi Y.21.4 NN KHNG BO TONTrong k thut nn, bn cnh nn bo ton th ngi ta cn a ra khi nim nn khng bo ton. Nn khng bo ton l m hnh nn d liu m tnh bo ton ca d liu khng c coi trng. N c ngha l nu ta c tp d liu D, tp nn D' th sau khi gii nn ta thu c tp D'' khc tp D ban u. K thut ny thng p dng cho vic nn d liu l cc loi tp nh v ni chung n cng khng nh hng g nhiu n hnh dng nh. Nu biu din bng ton hc, chng c m hnh sau:F(x) { y1, y2, ...}1.5 QA TRNH NN V GII NNQu trnh nn d liu l mt qu trnh gm 2 cng on: 1) Cng on nn.2) Cng on gii nn.Chng ta c th minh ho nh sau:a) Cng on nn:D liu M ho ng gi D liu nn.b) Cng on gii nn:D liu nn Gii m M ho D liu gc.Hai cng on trn l 2 in hnh tri ngc nhau. i vi tin trnh nn th module m ho thc hin vic ct vn bn ngun thnh cc on v gn cho chng k hiu xc nh chng. Ngc li i vi tin trnh gii nn th module gii m s da vo cc m m module m ho tin trnh nn sinh ra tm on tng ng. Qu trnh tm on tng ng c thc hin trn rt nhiu on trong tin trnh nn, gii nn sinh ra. Tp hp cc on chng ta gi l t in.Chng II: NHNG KHI NIM V M HNH NGUN32.1 T VN Xt qu trnh truyn bn tin tA n B.H1.1Qu trnh truyn bn tin t A n B da trn quan im mv giimngita c th phn chia h thng truyn tin thnh cc loi m hnh sau:2.2M hnh tnh (static modeling) Trong m hnh tnh,ngita coib m(encoder) v gii m (decoder) gi chung lcodebook , c ngi pht bn tin v nhn bn tin dng chung. y l m hnh tng t nh h thng truyn telegraph. Trong m hnh ny c hai nhn vin hai u c chungcodebook, nunhn vin hai u cui lm vic tin cy thbn tin truyn i saukhi m vbntinnhncsaukhi gii m lnh nhau .M hnh ny c nhc im , nu xut hin nhng k t mi , t mi trong bn tin , ngi ta phi b sung vo codebook c hai pha v tt nhin li phi dng dng telegrap chnh sa, cp nht.2.3M hnh bn thch nghi (semiadaptive modeling )Theo m hnh ny, trc khi gi bn tin i, nhn vin A c trc bn tin, ghi li nhng t xut hin nhiu ln, da vo tn xut xut hin m bn tin, kt qu cng vic ny to ra codebook ca chnh bn tin cn gi. Nhn vin A gibn tin cng vicodebook ca bn tin m. Nhn vin B nhn codebook v bn tin m, nhn vin B dng 4codebook gii m nhn li bn gc. nhc im ca m hnh ny l phi gi codebook mi khi gi mt bn tin.2.4M hnh thch nghi (adaptive modeling )Trong m hnh ny nhn vin A gibn tin cho B bng cch gi i ln lt tng k t mt , c hai ngi u ghi lit truyn v h thng nht nguyn tc t no truyn nhn ln hai trong cng bn text s thm n vo codebook v nh v n trong codebook, t y tr i nu gp li t ny, h c th truyn nhn bng m mi.D nhin h thng c gi nh cc nhn vin lm vic tin cy, chc chn.Nhng m hnh ny l c s cho php ngi ta thit cc thut ton nn khc nhau.2.5 Cac m hnh To ra bn tin (finite- context models)Chng ta bit rng to ra bn tin ngi ta thng da vo mt bng ch ci no . Song nu ch bit c th th cha , c th to ra bn tin, m ha v nn bn tin mt cch hiu qu ngi ta cn phibit nhiu hn, chng hn cn phi bit v cc qui tc to ra bn tin,nh cc qui tc to ra cc xu trong bn tin, quan h ca cc k t v cc nhm k t trong xu, trong cu. Chng hn nhth t trc sau ca cc k t , cc nhm k t trong xu , trong cu. Nhng thng tin nh vy cn phi c khi qut li , kt ni li dng m hnh, m hnh to ra cc bn tin ngi ta gi l m hnhnguntin(sourcemodel). Mhnhnguncvai tr quan trng ngi ta xc nh c xc sut ca cc s kin trong qu trnh to ra bn tin, nh vic m ha v nn bn tin hiu qu hn. 2.5.1 M hnh ng cnh hu hn bc 0 (order - 0 finite context model)Gi s cho bng ch ciA={a1,a2,,an}.5 M hnh ng cnh hu hn bc 0 l m hnh n gin nht, trong mi k t ca bng ch ci A c xc sut c nh pi=p(ai) vi i=1,2,n v p1+p2++pn=1; cc pikhng ph thuc vo v tr v th tca aitrong bn tin.V d A={a, b,c }p(a)=1/2 ; p(b)=1/4; p(c)=1/4;2.5.2 M hnh ng cnh hu hn bc -1 M hnh ng cnh hu hn bc -1 l m hnhtrong mi k t ca bang ch ci A c xc sut c nh v bng nhauvi mi ai; p(ai)=p=1/nvi i=1,2,nvp1+p2++pn=1; cc pikhng ph thuc vo v tr v th t ca ai trong bn tin.VdA={a, b,c,d}p(a)=1/4; p(b)=1/4; p(c)=1/4; p(d)=1/4;2.5.3 M hnh ng cnh hu hn bc n (n>=1) M hnh ng cnh hu hn bc n l m hnhtrong mi k t ca A xut hin c xc sut ph thuc vo n k t ng trc n.M hnh dng ny xut hin tng i ph bin v d trong ting vit ch u c xc sut 0,99 khixut hin ch q.2.5.4Mhnhmyhuhntrngthi (finitestate models )M hnh my hu hn trng thi FSM (finite state machine) M=Trong A l bng ch cihu hn; S tp hu hn trng thi; P l bng phn b xc sut, phn t pi k k hiu xc sut my trng thi chuyn t trng thi sisang trng thi sk 6vi a bt k thuc A. Vimii , cc piktha mn iu kin pi1+pi2++pik=1; l hm chuyn trng thi ; (a,si)=sk vi xc sut pikMy trng thi c thm t trc quan bng th c h-ng;v d 1cho my trng thi m t bi thcho bi hnh H1.31hnh H1.31

Khi xu x=abcaab sinh ra bi my trng thi vi xc sutp=0.5 x0.3x 0.2 x0.5 x 0.5 x 0.2=0.0025.v d 2 hnh H1.32Xu x=aabba c xc sut l 1/25M hnh my trng thi cn c tn gi l m hnh Markov do nh ton hc Nga a ra .Ch :7- trn chng ta ch ra m hnh trng thi tng -ng vi th c hng m trn mi cung c ghi nhn v xc sut chuyn trng thi.2.5.5 M hnhvn phm(grammar models)M hnh my trng thi c mt s hn ch nhphi din t cc xu vi vng lp ln nhxu (a+b*(c+d)-e). Ngi ta chng minh c rng cn n+2 trng thi trong m hnh trng thi hu hn din t biu thc s hc c n cp ngoc, nh vy s trng thi s rt ln khi s cp ngoc ln. y l nh-c im ca m hnh trng thi hu hn. V vy m hnh trng thi hu hn khng th dng biu din ngn ng t nhin. khc phc nhc im ca m hnh my hu hn trng thi ngi ta a ra m hnh vn phm M hnh vn phm c dng G=; N - tp hu hn cc bin hay cc nonterminal.k l phn t thuc N , ta gi l phn t bt u T tp hu hn gm ch ci hay cc terminalP tp cc qui tc dn xut dng: ----> ; , (NUT)+; (NUT)+; Ta gi L(G)={w k --* ->w ; w T+ } l tp ngn ng sinh bi vn phm G. K hiu k --* ->w ch rng bt u k sau mt s ln s dng cc qui tc trong P ta nhn c w l xu ca T; k hiu w T+ ch rng w l xu ca bng ch ci T.V d8Hnh H1.12 Xc sut vn phm G trong v d ny sinh ra xu aaaa l 3/4 *1/2*1/2*1/2 =1/16.2.5.6M hnh state : Do A.A.Markov (1856-1922) a ra.* nh ngha :M hnh state l mt th nh hng m mi mt cnh c mt nhn v mt trng s l xc sut chuyn trng thi c theo hng , tng cc xc sut chuyntrng thi ra khi mt nh bt k ca th lun lun = 1.V d: a 0.5a 0.2a 0.2b 0.3 d 0.3c 0.5 a 0.2c 0.8Vi m hnh trn, ta c th c cc chui vn bn khc nhau nh vy vi mt chui vn bn bt k th lung thng tin sinh ra t m hnh trn l mt chu trnh no .9M hnh state c gi l xc nh nu c mt s nguyn B ln sao cho mi dy tn cc cnh do m hnh state sinh ra c di ln hn B xc nh duy nht mt dy cc nh cc cnh m n i qua. Nh th lung tin di s tng ng vi mt chu trnh no i qua dc cc nh v cc cnh. Entropycalungtinkhi cnhnghathngquaentropycachu trnh. Vy entropy l g ? Chng ta tip tc xem xt khi nim v entropy.nh ngha entropy:Entropy l kh trung bnh on nhn 1 thng tin trng thi c sinh ra t m hnh ngun.Cng thc tnhentropy :-n1 ii 2 iP log PCch tnh :1. Xc nh s trng thi.2. Tm ma trn xc sut chuyn trng thi.3. Tm vector ring4. Tnh entropy ca mi trng thi.5. Tm entropy ngun bng cch ly tng cc tch xc sut xut hin ca trng thi vi entropy ca ring n.V d : Gi s c m hnh sau :a =30b =7c =12c =26 c =26 d =7Bc 1 : S trng thi = 2 Bc 2 : Ma trn chuyn 10a11 =7549) 7 19 ( ) 12 7 30 (12 7 30+ + + ++ + ; a21 = ) 7 19 ( ) 12 7 30 (7 19+ + + ++ = 7526a12 = 2626 = 1;a22 = 0P = 1]1

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0 75 / 261 75 / 49a aa a22 2112 11Bc 3: Gi tr ring l No ca phng trnh.det 1111]1

0752617549 = 0Khai trin ra ta c phng trnh sau :0752675492 nghim l :1 = 1 ; 2 = - 7526V

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00xx0 75 / 261 75 / 4921trong X1v X2 l xc sut xut hin ca trng thi 1 v 2 nn 0 do ta chn = 1 =1.Gii h phng trnh trn vi = 1 do X1 + X2 = 1 => c No: X1 =10075 ; X2 =10126Bc 4 : Tnh entropy c tng trng thi.- Trng thi 1 :E1 =) 26 12 7 30 (73026 12 7 30log) 26 12 7 30 (302+ + +++ + ++ + +1226 7 12 30log26 12 7 3012726 12 7 30log2 2+ + ++ + +++ + +11+ 2626 12 7 30log26 12 7 30262+ + ++ + + =1.80096- Trng thi 2 :E2 =77 19log7 197197 19log7 19192 2+++++ = 0.84036Bc 5 : E = X1E1 + X2E2 =84036 . 0 x101260096 . 1 x10175+= 1.55368Kt lun : Vy entropy ca ngun l : 1.5536812Chng III: PHNG PHP V K THUT C BN V NN D LIU3.I. NH NGHA NN D LIUNn d liu thc cht l mt hnh thc m ho d liu ghi li dng d liu sao cho tn t b nh hn m li cho php chng ta khi phc li d liu ban u.3.2.MT S LOI M3.2.1 M k hiu:nh ngha : l mt h thng quy c cc m c s dng nhn ra 1 chui cc s kin khc nhau th c gi l k hiu. M k hiu l 1 hnh thc ca phng php nn d liu.M k hiuxut hin hng ngy xung quanh chng ta.V d : Trong vn bn Nh nc thng c cm t :CV cng vn ;Q quyt nh TTCP Th tng Chnh ph ...Vi cc nc pht trin,h thng chun ho k hiu cho php dng chng mt cch d dng v tit kim nh trong lnh vc in tn, telex, th in t, ... Xt v cc mt khc nhau m m k hiu c p dng th m k hiu cung cp thng tin cho chng ta y hn, n d dng gi nh tr nh ca mi ngi do vy lng thng tin t m k hiu sinh ra l rt ln.Mt s c im ca m k hiu:+ Tit kim b nh, tit kim thi gian. + Tnh nguyn thy ca m.+ Tnh tng i ca h thng m.13Nh nhng c im trn m m k hiu c s dng nhiu trong lnh vc nh qun l d liu, qun l dn s, qun l vic mua bn hng ho...* Tnh nguyn thu ca m c xem xt nh sau:Mt trong nhng lnh vc s dng m k hiu nhiu nht l qun tr d liu. D liu l c tnh ca i tng qun l v c chia ra lm 2 c tnh sau:* Qun l s vt:S vt l cc ch th m chng ta gi n l cc thc th.* Qun l s vic :S vic cc hot ng bin nhn ghi li s tng tc ca cc s vt c gi l cc form.Chnh m k hiu s l a ch dng truy cp kho d liu. ng thi m ho k hiu a ra mt s tnh cht duy nht ca mt i tng no tho mn m thi. Nh vy tnh nguyn thu ca m k hiu chnh l cch ghi nhn cc s vt hay s vic.+ Tnh tng i: tra cu mt thng tin no trong CSDL,chng ta da vo cc kho xem xt c im ca thng tin nhng vic tm kim ny khng th chng minh c rng CSDL ch c duy nht thng tin c kho m n cn c th c cc thng tin khc cng c kho nh vy. y chnh l tnh tng i ca m k hiu. Chnh v vy m tnh tng i ca m k hiu cng ny sinh ra mt s vn .V d : - H thng chun m Ting Vit trn PC nc ta do cc nh lm m qu lm dng tnh tng i ca m k hiu to ra bng ch ci chun cho my tnh nn Vit Nam c ti 3 h thng m Ting Vit khc nhau cho 3 khu vc khc nhau nh : Khu vc pha Bc - ABC ; min Trung - Vietware ; min Nam VNI.14- H thng in tn (telex) do Moorse pht minh ra. Trong h thng ny ch gm 2 k t chm v gch. Gia cch du chm v gch c ngn cch bng cc khong cch.3.2.2 M ng giLmt phngphpnndliutrongmi phngphpmbao gi cng c mt khu ng gi. Thng thng chng ta hay ghi cc byte gm 8 bit 0/1 cho nn ng gi y l cch thc to ra cc byte t dy cc bit 0/1. a s cc thut ton m chng ta s dng m phng thut ton nn d/l u cho ra dy cc bit 0/1 sau l bc ng gi cc byte 0/1 y li.V d: ng gi cc ch "abcdu" thnh u byte. a bc d u100 10001 01 001 0010111010011010101Trong hnh thc ny chng ta c cc k thut ng gi khc nhau nhng ph bin vn l k thut ng gi nh lng.+ Khi nim ng gi nh lng:nggi nhlnglcchthcchngtaqui nhsbytednh cho cc k t c ghi trong tp nn.c im ca m ng gi nh lng:+ Phng php nn khng mm do.+ Hiu qu nn cha cao.+ Thi gian chy nhanh.V d: Chng ta xt u ra ca thut ton L778 ca 1 xu k t "aaabbabaabaaabab" l:(0+a) (10 + b) (3 + a) (4 + a) (5 + a) (4 + b).15Nu ta ghi mi k t l 1 byte th s dn ti kch thc ca tp nn c th to hn kch thc cu tp trc khi nn ng thi chng ta cng s gpmt skhkhntrongvicgii nn. V vynggi cnglmt khu quan trng trong vic nn d liu.Trongphngphpnhlngchngtaphi cmt squi c nht nh ng gi cc k t sao cho tn t b nh nht. Nh v d trn nu chng ta ng gi cc con s nm trong khong 1- 2 byte v cc ktchngtacnhrt nhiudovymhiuqunncao. Vi qui nc trn, chng ta c u ra l: 0a1a3a4a5a4b.Bn cnh khi nim ng gi nh lng th ta cng c khinim ng gi t ng.*nh ngha:ng gi t ng l hnh thc chng trnh t to ra cc byte theo mt s tiu chun nht nh no . K thut ng gi t ng c p dng nhiu trong thut ton nn Huffman.Mt s c im ca m ng gi t ng:+ Phng php nn tr nn mm do+ Hiu qu nn cao+ Thi gian chy tng i chm.3.2.3 M gn ngM gnngcsdngphbintronglnhvcnhv m thanh. Mt trong nhng ng dng ph bin nht ca m gn ng l in thoi di ng. in thoi di ng truyn thng tin i sau khi nn. Nh thchngtkhngtruynmthahtingni i mtruyn"cutoca mt ci ming" n s pht li cho chng ta nghe li. Hiu qu ca phng php ny cho php chng nn tn hiu ting ni t 8000 byte/giy xung 50 byte/giy. Tng t k thut ny cng c p dng cho cc loi in thoi hnh.3.2.4 M theo di (RLE)16Trong s cc phng php dng nn d liu th phng php n gin nht l phng php m ho theo di.+ Nguyn tc m:Nguyn tc c bn ca phng php ny l pht hin mt k t c slnxut hinlintipvt quamt ngngcnhno. Trong trng hp ny dy s c thay th bng 3 k t.K t th nht l k t c bit, thng bo dy tip l dy c bit.K th th hai ch s ln lp.K t th ba ch k t lpNh vy, t tng ca phng php ny l thay th mt dy bng mt dykhcngnhntuntheomt ngngno, vthngthng ngng c gi tr l 4.+V d: Dy ngun: ..SSOOOOONNNLLLLLLAAANNN..Dy nhn c sau khi nn:.... HH CS S O NNN CS 6L AAAN...Tuy nhin phng php ny cng c nhiu khuyt im. Trc ht l k t c bit khng c php xut hin trong tp d liu ngun. Nu k t xut hin vi t cch l d liu th khi g nn n s dn n tnh trng nhp nhng.Ngoi ra ngi ta cn ch n ch s lp. Nu ch s lp c lu tr trn 1 byte th s ln lp ca mt k t khng vt qua 255.Khi s ln lp ca mt k t vt qu 255 th ch s lp s c lu tr trn 2 byte. Trong trng hp ny, ch s lp ti a s c nng ln thnh 65535. Tuy nhin trong thc t s s ln lp ca mt k t thng 255 -> qu trnh nn s chim 1 byte v ch.Da vo nhng c im trn ta thy phng php m ho theo di khng c li nhiu trong vic nn tp vn bn.17Nhng i vi nhng tp hnh nh th phng php ny li c hiu qu cao v nh en trng l dy cc s 0 (im en) v cc s 1 (im trng) an xen nhau. Trong trng hp ny vic m s ln xut hin lin tip ca cc s 0 v s 1 tng i d dng v khi m ho khng cn k t c bit cng nh khng cn phi ch r k t lp l k t no m ch cn ghi s ln xen k.+ V d: Ngun: 0000000000 11111111 00000 111111=> s c m: 1085 6N... vi hnh nh m im u tin khng phi l im en nh v d trn th phi bt u dy m bng 1 s 0.+ V d: Ngun: ... 00 .. 011111100 ...280 ln Nn:... 2550 25 52 ...i vi nh mu th mi mu sc c hiu th bng mt s nguyn. phn bit c s khc nhau trong lp mu ngi ta phi xen vo mt k t c bit v tip theo sau s l ch s lp v k t c lp. C ngha l m ho nh mu c th dng mt cp, k t u l ln lp li, k t sau l mt mu v dng mt k t c bit (v d nh #) bo hiu s xut hin cc cp nht.+V d:Ngun: 11111122223344444444Nn: #61 # 42 # 23 # 84i vi nhng mu l duy nht (k t c ch s lp l 1) th mu khng cn i vi k t c bit.+ Thut ton m theo di:* Nn: Tp ngun: fK t c bit: db18S ln lp: demdb = 255dem = 1c k t u tin trong f l ktlapWhile not eof (f) dobegin- c k t tip theo -kt;if kt := kt lp then dem = dem + 1else if dem > 3 thenbeginIn db;In dem;In ktlap;End;end;for|:= 1 to dem doIn ktlapEnd;Dem: = 1ktlap := ktEnd;* Gii nn:Tp nn:fK t c bit: dbS ln lp: demdb = 255While not eof (f) dobeginc k t trong f l kt;If kt:=db then19beginc k t tip - dem;c k t tip - ktlap;for | = 1 to dem doIn ktlapelsec k t tip;end; End ;Trong phng php ny, ngi ta s dng k t c bit vi t cch l k t khng bao gi xut hin trong tp d liu ngun. y l t tng khng chun mc bi v tp d liu ngun c th cha tt c cc k t ca bng m ASCII.Do phng php ny c s dng nhiu nht trong vic nn hnh nh.3.2.5 M ho thch nghi dn ca tn s (AFE)Phng php ny l phng php m ho thch nghidn ca tng k t theo tn s xut hin.Khi u ca vic nn, nu mt k t c m ho trn 5 bits th n cng c th c m ho bng 1 hay 9 bits tu thuc vo s xut hin ca n l nhiu hay t.Trongphngphpny, vicnnvgnnphi tinhnhsong song. Mi modul scchungbngmbanucho1bytetruyni. Nhng chng s ph thuc vo tn s xut hin ca k t v tun theo qui tc ci bin.Vi mc ch nh vy, phi lp m cho mi byte v lu tr chng trong mt bng tham kho hay l trong mt t in.Bng ny dng lm c s cho s lp m ca mi k t, c mt tng hai modul nn v g nn.20M ca mi byte bao gm 2 phn: Phn u (Header) v phn thn (Body). Phn u thng gm 3 bits v phn thn c 1 - 8 bits, tu thuc vo byte lp m v nht l tn s xut hin ca m. Phn u ch ra di ca thn m.Khigii m khng cn x l tng bit phn bit cc byte truyn nh trong phng php Huffman. gii m ch cn c 3 bits u xc nh di n ca thn m, sau c n bits tip theo khi phc li m truyn i.3.3. M HNH NNChng ta c th c thut ton nn tt khi bit c m hnh sinh ra ngun tin. Song vic tm ra m hnh sinh ra d liu l khng th. Vy c cch no nn c m khng bit m hnh sinh ngun tin. Chng ta c 2 phng php sau:- M hnh tnh: l m hnh tm c thng qua nghin cu cc c trng thng k ca vn bn ri sau s dng chng.- M hnh thch ng: M hnh thch ng c xut pht im l mt m hnh tng qut no sau hiu chnh dn. -c im ca m hnh thch ng:+ M hnh thch ng da vo thng k ca mt s rt ln cc vn bn cng loi v p dng cho vn bn mi. u im ca phng php ny l trnh nn dn chy nhanh.+ Da vo m hnh nn gi nh chng ta tnh gi tr cc trng ti v tin hnh nn.3.3.1. Nn d liu c m hnh ngunMt trong nhng c im ca vic nn d liu ny l c ngi nn cng bit ngun sinh ra tin.Nhng thut ton nn d liu c trng cho vic nn d liu c m hnh ngun ny l:+ Thut ton Fano - Shannon+ Thut ton Huffman.213.3.2. Nn d liu cha c m hnh ngunMt trong nhng v d in hnh v nn cha c m hnh ngun l ngn ng t nhin. Chng ta lun ri vo tnh trng c vn bn m khng c m hnh ngun. tm ra mt thut ton nn tt nht th chng ta phi tm ra qui lut ca chng. Qua nghin cu c th rt ra c hai cch tip cn sau:a. Cch tip cn tng th:Cch tip cn ny cn c gi l phng php thng k. Cch ny da vo nhn xt tinh t, l nhng g xy ra trong qu kh th s xy ra trong tng lai. iu ny cng c tha nhn trn thc t m c trng l nhng cu tc ng ca cha ng chng ta c c kt t nhiu i nay nh: "m thng 5 cha nm sng,ngy thng 10 cha ci ti", ...Nh vy, c m hnh lung tin, chng ta phi tm ra mt s iu kin cho lung tin. y chng ta ch xt m hnh sinh tin l ngu nhin nhn hu hn cc gi tr t mt ngn tin bt k sao cho s xut hin ca thng tin sau ph thuc vo gi tr ca lng thng tin ng trc n.Cch tip cn ny gm hai phng php sau:+ Phng n tnh.+ Phng n ng.Phng n tnh c tin hnh m theo 2 bc:1. Tm m hnh thch hp da vo thng k.2. Tin hnh m nn theo phng n c m hnh.Phng n ng:c trng chnh trong phng n ng l cc thut ton, da vo s cm nhn sau.t gi s chng ta tin hnh nn d liu trong khicha bit r k t no s xut hin tip theo. Nhng nu chng ta bit c xc sut xut hin ca mt ch tip theo l g th trn thc t chng ta c th bit c m hnh nn v s c thut ton nn d liu. Tuy nhin y 22li ny sinh ra vn l lm th no d on k t no s xut hin, xut hin t hay xut hin nhiu nu chng ta ch da vo k t xut hin.Xt v d sau y:Gischngtacmt vnbngm"aabbababab... bbaaba". Tt nhin, nu vn bn cha kt thc th ta khng th khng nh l vn bn ch gm 2 k t "a" & "b" th cha chc chn v c th c 1 k t bt k xut hinnh"s", "t"... Dovytrongphngnngtalunphi c phng n d phng. Vy xc xut xut hin ca k t cha xut hin l bao nhiu? Ta c th ni rng "xc sut ca mt k t no cha xut hin tuy rt nh nhng vn phi ln hn 0".im mu cht ny c gi l zero - frequency problem. Vn ny cho n nay vn ang c tm hiu v cha c chng minh.- Thut ton nn ng1. Thng k cc k t c trong vn bn to m hnh mi.2. Nn m hnh mi va xy dng bc mt.3. Lp li bc mt v bc hai cho n ht.Tmli, trnthctnhngkthut tipcntngth(kthut thng k) c p dng rt t do cc thut ton chy chm v cc chng trnh vit theo k thut trn phi thc hin mt s lng ln cc php tnh ton v mt khc hiu qu nn li khng cao.b. Cch tip cn i t chi tit:Cch tip cnnydavo nhngnhn xtph thuc thiim hin ti cng tng t mt thi im no trong qu kh.Ni theo mt cch khc th cch tip cn ny da vo nhng qui lut nht nh. Cc qui lut gp li thnh t in nn ta c th gi cch tip cn ny l phng php t in m chng ta c th xem xt k hn phn sau.23Chng IV: PHNG PHP M HA T IN4.1. m ha T in tnh (static dictionary Encoders)T in m ha tnh, gi tt l t in tnh. N c tn gi nh vy v t in ny c lp ra khng ph thuc vo vn bn c th. Ni chung hiu qu ca nn da vo t in tnh khng cao, nhng c u im l n gin, khi trun vn bn ingi ta khng cn truyn t in km theo.V d in hnh ca t in tnh l bng m ASCII. N dng m 256 k t, mi k t c di m nh nhau gm 8 bit hay mt byte.Vi t in tnh ngi ta c th c mt s ci tin nng cao hiu qu. Ngi ta c th ghp hai k t i lin nhau ng vi mt m, gi l digram.Thut ton m v gii m r rng l rt n gin v nhanh chng. V d trong cc k t ca bng m ASCII ngi ta phn chia thnh hailoigm 96 k t (94 k t in c, space, newline) v 160 k t cn li c m theo digram hay theo cp k t. Chng hn ta c th m cc digram {*a, *t, *s, th,he,.}.Mtcchtng qut, nu c qk t c coilquan trng hay dng, khi ta cn 256-q digram lp vo t in. Trongtrnghpnychngtacthchai gii php. Gii php th nht l kim tra cc vn bn tiu biu 24 xc nh 256-q digram chung nht. Gii php th hai l xc nh haik tS1 v S2 v xy dng digram cho cc phn t sinh ra bi hai tp ny. Kh t in s c lp y nuS1 *S2 =256-q H4.1Hnh H4.1minh ha tng nu trn, trong cc con tr, tr n cc bng khc s dng 4 bit.Cthmrngtngdigramtrongtintnh bng cch lp cc n - gram m thc chtnl ccon(fragment) gmnktlintip. Bi tonvi sn- gramtnh (static n- gram scheme) l bi ton chn cu di nht c th lp t in n ph thuc vo vn bn c th.4.2. T in m ha bnthch nghi (semiadaptive dictionary Encoders)Logictnhincaquanimn- gramtrongtin tnha ngi ta n tng xy dng t in cha m ca on vn bn, t in ny cha ng phong cch rt ring (idiosyncratic) cavnbnchotrc. iunyrt haygp trong cc vn bn khoa hc k thut. u im ca vic nn 25theo t in bn thch nghi nhiu khi rt hiu qu, song nh-c im ca n l phi gi t in i km theo bn m.Chng ta ch rng vic xc nh trc mt t in ti u khi cho trc mt vn bn l bi ton c phc tp NP y .Tuy nhin c nhiu thut ton heuristic c th tm ra gii php ti u cho bi ton.Gi s s cc icc phn t trong t in M c xc nhtrc lM , mi phnttrongtinM c m log Mbit, nh vy vic m s ti u khi cc cu trong t in c tn xut bng nhau (equifrequent). Nh vy vn ct li l thut ton xc nh cc cu sao cho chng c tn xut nh nhau. Ta xt thut ton c m t nh sau:Trc ht tnh tn xut ca tng k t n l, sau l tn xut ca digram,trigram v.vKhi to t in lc u ton cc t n.Tm tn xut cao nht trong cc digram, b sung n vo t in, gim (reduce) i tn xut ca cc cp k t, Xc nh tn xut ln th hai ca digram v thm n vo v tr tip theo trong t in, qu trnh c tip tc chonkhi tnxutlnnht catrigramlnhncctn xut ca digram cn li. Tip theo trigram ny c thm vo t in, cc tn s lin quan n hai digram v 3 k t n c rt gn (Reduce) theo. Qu trnh c lp li cho n lp y t in.Trong qu trnh trn c th hy imt s phntkhi tinnutnscangimxungqu thp. Chng hn *of* th c tn xut cao ta b xung n vo t in, nhng nu *of* the cng c, khi tn xut c ca *of* the s b gim xung do *of* th xy ra trc e.26Ch rng vic chnm cho cc phn t trong t in cn phi dung ha gia t l nn v tc m.4.3. T in m ha thch nghi(adaptive dictionary Encoders)T in mhathchnghi cnctn gi lbng m ZIV_LEMPEL.Nm 1967 bi bo m t k thut xy dng t in bn thch nghic cng b. Nm 1977 tng ny c Jacob Ziv v Abraham Lemple citin tr thnh t in m ha thch nghi. Thut ton ca Jacob Ziv v Abraham Lemple c ci tin nhiu ln, n tr thnh h thut ton ZIV_LEMPEL hay gi tt l nn LZ . tng ct li ca h thut ton LZ l mt xu s c thay bng con tr, tr nni mxu xut hintrc. Contrdng (m,l), s tr n v tr m ca dng vo x v l l di xu trong dng vo, chnh l xu x[m . . m+l-1]; u im c bn ca h thut ton ZIV_LEMPEL cho t l nn cao, tc gii nn nhanh. V d: Con tr (7,2) ch xu th 7, th 8 trong dng vo. khi xu abbaabbbabab c m abba(1,3)(3,2)(8,3) nh hnhHnh H1.8Hnh H4.327Ch :Qutrnhm lqui, tuynhinvicgii m khng nhp nhng.Cnhiuthut tonthuchthut tonLZ, ccthut ton thuc h LZ khc nhau ch yu hai yu t sau: yu t th nht l s k t con tr c th tr quay lui v dng con no trong gii hn c th l im tr ca con tr. Chng ta s tm hiu mt vi thut ton h LZ.4.3.1LZ77LZ77 l thut ton cng b sm nht ca h thut ton ton LZ (1977).Theo s ny,cc con tr biu th cc cu trong mt ca s c kch thc c nh N, ca s ny t trc v tr m. C di cc i F cho cc dng con, c th thay th bng con tr. LZ77 cho php s dngmt ca s trt. t-ng ca LZ77 nh sau: Cho ca s (window) kch thc N (thng N K t Tn xut MA 0.2 01E 0.3 00I 0.1 101O 0.2 100U 0.1 110 0.1 111Cch nhm nh sau: ME 0.3 0.3 00A 0.2 0.5 0.2 01O 0.2 0.2 100I 0.1 0.3 0.1 101U 0.1 0.5 0.1 110 0.1 0.2 0.1 111nnmt tpdliubngphngphpFano-Shannon, cng vic u tin l phi c tp ngun thng k tn s xut hin ca mi k hiu. Sau khi thng k xong, ngi ta sp xp bng tn s theo th t gim dn ca tn s xut hin.Bng m tng ng ca cc k hiu c gi ti cho chng trnh g nn nh sau: Mi k hiu dng 3 bytes,byte th nht mang k t 35chuyn m, 4 bits tip theo (4 bits u ca byte th hai) mang di ca m, phn cn li mang m ti a ca k hiu.Trongqu trnh g nnphis dng bng m nhn c t thut ton nn. ng thi gii m cn phi da vo nhn xt: khng mt m no l phn u ca m khc.nhgi: V nhngldotrnnnthut tonFano-Shannoncxp trong danh sch nhng thut ton thng k. Tuy nhin, nhn chung y l mt phng php km hiu qu v t c s dng.5.1.2 - Thut ton Huffman.a)Nguyn l:Nguyn l ca phng php Huffman l m ho cc bytes trong tp d liu ngun bng bin nh phn. N to m di bin thin l mt tp hp cc bits. y cng l mt phng php nn kiu thng k, nhng k t xut hin nhiu hn s c m ngn hn.M Huffman c mt tnh cht quan trng: m ca mt k hiu ny khng th l phn u ca m mt k hiu khc.Nu nh mt k hiu c m ho bng t hp nh phn 101 th t hp 10110 khng th l m ca mt k hiu khc trong tp ngun. Do khi gii m cn phi c ln lt cc bit cho n khi gp m ca k hiu no .b)Thut ton.Vic xy dng cy m ho Huffman c tin hnh bi mt thut tonkhcvi thut tonFano-Shannon. NunhcyFano-Shannon cxydngttrnxungdi bngcchchiai vgnchomi phn 1 bt, cng vic kt thc khi khng th tin hnh phn chia tip th cy Huffman li c thitk t diln, btu tcc l cacyv cng vic kt thc ti im gc.V d :36Cho m hnh ngun c cc trng thi v tn sut tng ng nh sau: (A, 0.2); (E, 0.3); (I, 0.1), (0. 0,2); (U, 0.1); (, 0.1) ta c:K t Tn xutM0 1A 0.2 10E 0.3 01I 0.1 001O 0.2 11U 0.1 0000 0.1 0001Bc1: Nhm 2 ch ci c tn sut nh nht to ra ch ci kp. Sau mi ln nhm s ch ci t i 1.c -> 0.3 e -> 0.3 e -> 0.3 {a, 0} -> 0.4{{{u, },:}a -> 0.2 a -> 0.2 {{u,},i} -> 0.3e -> 0.3 {a, o{ ->o -> 0.2 o -> 0.2 a -> 0.2 {{u, }, i} -> 0.3i -> 0.1 {u,} -> 0.2 o -> 0.2u -> 0.1 i -> 0.1 -> 0.1Bc2: Tocyphnnhnhngcvi qutrnhnhmtnhnh tri c m 0, nhnh phi m 1.{{{{u, }, 1}, e}, {a, o}}01 {{{u, }, i}, e}{a, o}0 101 {{{ u, }, i},ea o 01{u, }i37010 101 01eao0 1iu 01 u Vy m ca k t l:u -> 0000; e -> 01 -> 0001; a -> 10i -> 001; o -> 11Thut ton nn:Bc 1: Tm hai k t c trng s nh nht ghp li lm mt, trng s ca k t mi bng tng trng s ca hai k t em ghp.Bc 2: Trong khi s lng k t trong danh sch cn ln hn mt th thc hin bc mt, nu khng th thc hin bc ba.Bc ba: Tch k t cui cng v to cy nh phn vi qui c bn tri m 0, bn phi m 1.Thut ton gii nn:Bc 1: c ln lt tng bit trong tp tin nn v duyt cy nh phn c xc nh cho n khi ht mt l. Ly k t l ghi ra tp gii nn.Bc 2: Trong khi cha ht tp tin nn th thc hin bc mt, ngc li thc hin bc ba.Bc 3: Kt thc thut ton.Mt s nhng hn ch ca m Huffman:+ M Huffman ch thc hin c khi bit c tn xut xut hin ca cc k t.+ M Huffman ch gii quyt c d tha phn b k t.+ Huffman tnh i hi phi xy dng cy nh phn sn cha cc kh nng. iu ny i hi thi gian khng t do ta khng bit trc kiu d liu s c thc hin nn.+ Qu trnh gii nn phc tp do chiu di m khng bit trc cho n khi k t u tin c tm ra.385.1.3. Thut ton Huffman ng:Trong thut ton Huffman ng chng ta s dng hai cy nh phn cn v mt ca s .Khi nim ca s :Caslmt slngrt lnccktcthmvocynh phn Huffman trong qu trnh nn vn bn.Trong qu trnh nn d liu, ta tin hnh thng k cc k t. Nh vic thng k ny m sinh ra b nn.Thut ton nn:Bc 1: Khi to bng m Huffman(front tree) vi k t c bit, c s m bng 1.Bc 2: To m th t theo nguyn tc cn bng.Bc 3: While not eof(f) doBeginGetchar-> chIf ch thuc bng m th t thenBeginGhi m 0/1 ca k t v k t c bit ra tp chXa k t bng m th t, thm k t vo bng m Huffman(trong cy front tree) vi s m bng 1.End ElseBeginIf ch thuc bng m Huffman thenBeginTng s m ca k t ln mt n vIf s m ca k t ln hn s m ca k t ngay trn n thenBegini ch hai k t End;Ghi m 0/1 ca k t v k t c bit ra tp chEnd;39End;If s lng k t trong bng m Huffman>= thenBeginTm mt k t, gim s m ca k t i mt n vIf di k t l k t c bit thenGim s m ca k t c bit i mt n vBeginIf s m ca k t sau khi gim < s m ca k t di n thenBegini chEnd;End;If s m ca k t bng khng thenBeginLoi k t ra khi bng m Huffman.Tng s lng k t cy th t ln mt.Gim s lng k t cy Huffman i mt.Thm k t va loi ra khi bng m Huffman v bng m th t v tr u tin.End;To li bng m th t, m Huffman cho bng m Huffman theonguyn tc cn bng.End;Bc 4: Dng chng trnh.Thut ton gii nn:Bc 1: Khi to bng m th t theo nguyn tc cn bngBc 2: While not eof(f) doBeginLy ra ln lt tng k t v gn chui cho n khi gp k t cbit.If k t c bit ch ti bng m th t then40BeginTra trong bng m th t v ghi k t ra tp ch.Thm k t vo bng m Huffman vi s m bng 1.End;If k t c bit ch ti bng m Huffman thenBeginTra trong bng m Huffman v ghi k t ra tp ch.Tng s m ca k t ln 1 n v.If s m ca k t ln hn s m ca k t trn n thenBegini ch.End;End;If s lng k t trong bng m Huffman>= thenBeginTm k t no gim i 1 n v.If s m ca k t < s m ca k t ng trc n thenBeginIf bn di n l k t c bit thenGim s m ca k t c bit i 1.i ch hai k t .End;If s m ca k t sau khi gim i 1 bng khng thenBeginTng s lng k t cy th t ln mt.Gim s lng k t cy Huffman i mt.Loi k t ra khi bng m Huffman.Thm k t va loi ra vo bng m th t.End;To li m cho bng m th t, bng m Huffman theo nguyn l cnbng.End;Bc 4: Kt thc.41nh gi: Qu trnh m v gii m tng i chm do phi xy dng cy nh phn ng vi d liu nhp.Thut ton nn Huffman thng c dng nn cc tp dng vn bn, cc tp c kch thc ln.5.2.1. Thut ton LZ78:Thay v thng bo v tr oan vn lp li trong qu kh, m LZ78 nh s tt c cc on vn sao cho mi on ghi nhn s hiu on vn lp li trong qu kh cng vi mt k t m n lm cho on khc vi on trong qu kh.Nh vy mi on mi l mt on k t trong qu kh cng vi mt k t trong qu kh. Chnh v th m on mi khc vi on c trong qu kh.V d:Gi s chng ta c mt on vn bn sau: aaabbabaabaaabab.Theo thut ton LZ78 th chng c phn thnh cc on nh sau:Inputaaa bbabaa baaa babon1234 5 67Output0+a 1+a0+b3+a 4+a 5+a4+bNh vy bn nn ca chng ta l: (0,a);(1,a);(0,b);(3,a);(4,a);(5,a);(4,b).Tip theo chng ta s s dng tp phn tch m cc con s cn dng, m c di c nh mt byte cho cc k t.Biu din qu trnh thc hin ca thut ton LZ78 bng cy: a ba a a b420325 741aQua v d trn, khi thut ton LZ78 c biu din di cu trc d liu dng cy s lm cho vic x l n gin hn.Thut ton nn:Bc 1: c mt k t -> ch, on ch gn bng 1, kt np k t vo t in, w:=ch;Bc 2: While not eof(f) doBeginc tip k t tip theo w:=ww+ch;If w thuc t in then ww:=w;Else BeginCode(w,j);Ghi j v ch vo tp nn.Thm w vo t in.End;End;Bc 3: Dng chng trnh.Thut ton gii nnBc 1: c thng tin v t in c lu trong tp nn, tl:=false;Bc 2: While not eof(f) doBeginc byte tip theo ->bDecode(b,s,t);If tl=false then w:=w+sElse w:=ww+s;TIMCHU(w,t);If t=false thenBegin436Ghi s ra tp gii nnThm s vo t inEnd ElseBeginww:=s;End;End;Bc 3: Dng chng trnh.nh gi:Ni chung thut ton LZ78 l mt thut ton nn vn bn tt, c thi gian chy chng trnh tng i nhanh tuy nhin kh nng tit kim cha c khai thc ti a.5.2.2. Thut ton LZW:Thut ton ny l s chuyn giao ca thut ton LZ78. Nh chng ta bit thut ton LZ78, vic lu tr cc k t theo sau mi on thng gy lng ph v b nh nn hiu qu nn khng cao.Thut ton LZW qun l bng cch loi b k t sau mi on do u ra ca mi on ch cha con tr m thi.Thut ton ny lu tr bng vic chun b mt danh sch cc on bao gm rt nhiu k t trong u vo l mt bng ch ci no , n thc hin mt qu trnh m rng cc bng ch ci hay ni cch khc l n dng k t b sung biu din li cc chui ca k t chnh quy. nn LZW trn m ASCII 8 bits ta cn m rng bng ch ci bng cch dng 9 bits hay nhiu hn 256 k t b sung m m 9 bits cung cp c dng lu tr cc chui m c quyt nh t cc chui trong ngun tin.Thut ton s khng t hiu qu nn cao nu c nhng iu kin sau:+ Ngun tin khng ng nht v c tnh d tha ca n thay i trong sut tp tin.+Nguntindi mt cchngkvt qutmgii hncabng chui.Thut ton nn:Bc 1: Thng k to ra t din, ghi vo tp nn, t:=false44c k t u tin -> wBc 2: While not eof(f) doBeginc mt k t -> chif t=false then w:=w+chelsebeginw:=ww+ch;t:=false;end;TIMCHU(w,tl);If tl=false thenBeginCode(,j);Ghi j ra tp nn.Thm w vo t in.w:=ch;end elsebeginww:=w;t:=true;end;end;Bc 3: Code(ch), Dng chng trnh.Thut ton gii nn:Bc 1: c thng tin t in trong tp nn, c byte tip theo v giinn gn vo w, t=false;Bc 2: While not eof(f) doBeginc byte tip theo -> bDecode(b,s,t);If t=true thenBeginFor i:=1 to length(s) do45BeginIf t=false then w:=w+s(i)Else beginw:=ww+s(i); t:=false; end;TIMCHU(w,t);If t=false thenBeginThm vo t in.Ghi ra tp gii nn.W:=s(i);End;end;EndElseBeginGhi ra tp gii nn;w:=w+w(i);thm w vo t inEnd;End;Bc 3: Decode(b,s,t); ghi s ra tp gii nn. Dng chng trnh.nh gi:Thut ton LZW khc phc c s lng ph v b nh m cc thut tontrckhngtndngcht. Thut toncthi gianchy chng trnh nhanh nu chng ta s dng cu trc cy( nh phn, tam phn). Mt trong nhng u im na l thut ton c hiu qu nn cao.

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