Co So Du Lieu 2 (Phan Tan Va Suy Dien)

8/8/2019 Co So Du Lieu 2 (Phan Tan Va Suy Dien)

1/134

MUC LUCMUC LUC....................................................................................1Li ni u.................................................................................3PHN 1.......................................................................................6C S D LIU PHN TN.........................................................6CHNG 1. TNG QUAN V C S D LIU PHN TN............6

1.1. H CSDL phn tn ............................................................ 61.1.1. nh ngha CSDL phn tn.........................................61.1.2. Cc c im chnh ca c s d liu phn tn.........71.1.3. Mc ch ca vic s dng c s d liu phn tn...101.1.4. Kin trc c bn ca CSDL phn tn........................111.1.5. H qun tr CSDL phn tn.......................................12

1.2. Kin trc h qun tr C s d liu phn tn ..................131.2.1. Cc h khch / i l................................................131.2.2. Cc h phn tn ngang hng...................................14

CHNG 2. CC PHNG PHP PHN TN D LIU...............152.1.Thit k c s d liu phn tn .......................................15

2.1.1.Cc chin lc thit k..............................................152.2. Cc vn thit k ........................................................ 16

2.2.1. L do phn mnh.....................................................162.2.2. Cc kiu phn mnh................................................162.2.3. Phn mnh ngang....................................................18

2.3. Phn mnh dc ..............................................................372.5. Phn mnh hn hp ....................................................... 482.6. Cp pht ........................................................................ 49

2.6.1 Bi ton cp pht......................................................492.6.2 Yu cu v thng tin.................................................492.6.3. M hnh cp pht....................................................51

CHNG 3. X L VN TIN.....................................................543.1. Bi ton x l vn tin .....................................................543.2. Phn r vn tin ............................................................... 593.3. Cc b ha d liu phn tn ..........................................67

3.4. Ti u ho vn tin phn tn ...........................................743.4.1. Khng gian tm kim................................................743.4.2. Chin lc tm kim.................................................783.4.3. M hnh chi ph phn tn..........................................803.4.4. Xp th t ni trong cc vn tin theo mnh.............88

CHNG 4. QUN L GIAO DCH.............................................974.1. Cc khi nim ................................................................974. 2. M hnh kho c bn ...................................................1064.4. Thut ton iu khin tng tranh bng nhn thi gian

........................................................................................... 112PHN 1...................................................................................115C S D LIU SUY DIN.....................................................115

1


2/134

2.1. Gii thiu chung ........................................................... 1152.2- CSDL suy din ........................................................... 1152.2.1. M hnh CSDL suy din.........................................1152.2.2. L thuyt m hnh i vi CSDL quan h................1172.2.3. Nhn nhn CSDL suy din......................................120

2.2.4. Cc giao tc trn CSDL suy din...........................1202.3. CSDL da trn Logic ..................................................... 1212.3.4. Cu trc ca cu hi..............................................1262.3.5. So snh DATALOG vi i s quan h....................127

2.3.6. Cc h CSDL chuyn gia.....................................1322.4. Mt s vn khc ......................................................133

2


3/134

Li ni u

Cc h c s d liu (h CSDL) u tin c xy dng theo cc mhnh phn cp v m hnh mng, xut hin vo nhng nm 1960, cxem l th h th nht ca cc h qun tr c s d liu (h QTCSDL).

Tip theo l th h th hai, cc h QTCSDL quan h, c xy dngtheo m hnh d liu quan h do E.F. Codd xut vo nm 1970.

Cc h QTCSDL c mc tiu t chc d liu, truy cp v cp nhtnhng khi lng ln d liu mt cch thun li, an ton v hiu qu.

Hai th h u cc h QTCSDL p ng c nhu cu thu thp vt chc cc d liu ca cc c quan, x nghip v t chc kinh doanh.

Tuy nhin, vi s pht trin nhanh chng ca cng ngh truynthng v s bnh trng mnh m ca mng Internet, cng vi xu th toncu ho trong mi lnh vc, c bit l v thng mi, lm ny sinhnhiu ng dng mi trong phi qun l nhng i tng c cu trcphc tp (vn bn, m thanh, hnh nh) v ng (cc chng trnh, cc mphng). Trong nhng nm 1990 xut hin mt th h th ba cc hQTCSDL cc h hng i tng, c kh nng h tr cc ng dng aphng tin (multimedia).

Trc nhu cu v ti liu v sch gio khoa ca sinh vin chuynnghnh cng ngh thng tin, nht l cc ti liu v CSDL phn tn, CSDLsuy din, CSDL hng i tng, chng ti a ra gio trnh mn hc Cs d liu 2.

Mc ch ca gio trnh C s d liu 2 nhm trnh by cc khinim v thut ton c s ca CSDL bao gm: cc m hnh d liu v cch CSDL tng ng, cc ngn ng CSDL, t chc lu tr v tm kim, xl v ti u ho cu hi, qun l giao dch v ieukhin tng tranh, thitk cc CSDL.

Trong qu trnh bin son, chng ti da vo ni dung chngtrnh ca mn hc hin ang c ging dy ti cc trng i hc trongnc, ng thi cng c gng phn nh mt s thnh tu mi ca cngngh CSDL.

3


4/134

Gio trnh C s d liu 2 c chia thnh 2 phn

Phn 1: C s d liu phn tn

Phn 2: C s d liu suy din

Sau mi chng u c nhng phn tm tt cui chng, cu hi ntp v bi tp nhm gip sinh vin nm vng ni dung chnh ca tngchng v kim tra trnh ca chnh mnh trong vic gii cc bi tp.

Tuy rt g gng, gio trnh chc chn cn c nhng thiu st. Rtmong nhn c kin ng gp ca c gi trong ln ti bn sau, giotrnh s hon chnh hn.

Thi Nguyn thng 10 nm 2009

Cc tc gi

4


5/134

5


6/134

PHN 1

C S D LIU PHN TN

CHNG 1. TNG QUAN V C S D LIU PHN TN

Vi vic phn b ngy cng rng ri ca cc cng ty, x nghip, dliu bi ton l rt ln v khng tp trung c. Cc CSDL thuc th hmt v hai khng gii quyt c cc bi ton trong mi trng mi khngtp trung m phn tn, song song vi cc d liu v h thng khng thunnht, th h th ba ca h qun tr CSDL ra i vo nhng nm 80 trong

c CSDL phn tn p ng nhng nhu cu mi.1.1. H CSDL phn tn

1.1.1. nh ngha CSDL phn tn

Mt CSDL phn tn l mt tp hp nhiu CSDL c lin i logic vc phn b trn mt mng my tnh

- Tnh cht phn tn: Ton b d liu ca CSDL phn tn khng c

c tr mt ni m c tr ra trn nhiu trm thuc mng my tnh, iuny gip chng ta phn bit CSDL phn tn vi CSDL tp trung n l.

- Tng quan logic: Ton b d liu ca CSDL phn tn c mt s ccthuc tnh rng buc chng vi nhau, iu ny gip chng ta c th phnbit mt CSDL phn tn vi mt tp hp CSDL cc b hoc cc tp c trti cc v tr khc nhau trong mt mng my tnh.

6

Trm 1

Trm 2

Trm 3Trm 4

Trm 5

Mng truyn d liu

Hnh 1.1 Mi trng h CSDLphn tn


7/134

Trong h thng c s d liu phn tn gm nhiu trm, mi trm cth khai thc cc giao tc truy nhp d liu trn nhiu trm khc.

V d 1.1: Vi mt ngn hng c 3 chi nhnh t cc v tr khcnhau. Ti mi chi nhnh c mt my tnh iu khin mt s my k toncui cng (Teller terminal). Mi my tnh vi c s d liu thng k aphng ca n ti mi chi nhnh c t mt v tr ca c s d liuphn tn. Cc my tnh c ni vi nhau bi mt mng truyn thng.

1.1.2. Cc c im chnh ca c s d liu phn tn

(1) Chia s ti nguyn

Vic chia s ti nguyn ca h phn tn c thc hin thng quamng truyn thng. chia s ti nguyn mt cch c hiu qu th mi tinguyn cn c qun l bi mt chng trnh c giao din truyn thng,cc ti nguyn c th c truy cp, cp nht mt cch tin cy v nhtqun. Qun l ti nguyn y l lp k hoch d phng, t tn cho cclp ti nguyn, cho php ti nguyn c truy cp t ni ny n ni khc,nh x ln ti nguyn vo a ch truyn thng, ...

(2) Tnh m

Tnh m ca h thng my tnh l d dng m rng phn cng (thmcc thit b ngoi vi, b nh, cc giao din truyn thng ...) v cc phnmm (cc m hnh h iu hnh, cc giao thc truyn tin, cc dch vchung ti nguyn, ... )

Mt h phn tn c tnh m l h c th c to t nhiu loi phncng v phn mm ca nhiu nh cung cp khc nhau vi iu kin l ccthnh phn ny phi theo mt tiu chun chung.

Tnh m ca h phn tn c xem xt thao mc b sung vo ccdch v dng chung ti nguyn m khng ph hng hay nhn i cc dchv ang tn ti. Tnh m c hon thin bng cch xc nh hay phnnh r cc giao din chnh ca mt h v lm cho n tng thch vi ccnh pht trin phn mm.

7


8/134

Tnh m ca h phn tn da trn vic cung cp c ch truyn thnggia cc tin trnh v cng khai cc giao din dng truy cp cc tinguyn chung.

(3) Kh nng song song

H phn tn hot ng trn mt mng truyn thng c nhiu mytnh, mi my c th c 1 hay nhiu CPU. Trong cng mt thi im nuc N tin trnh cng tn ti, ta ni chng thc hin ng thi. Vic thchin tin trnh theo c ch phn chia thi gian (mt CPU) hay song song(nhiu CPU)

Kh nng lm vic song song trong h phn tn c thc hin dohai tnh hung sau:

- Nhiu ngi s dng ng thi ra cc lnh hay cc tng tc vi ccchng trnh ng dng

- Nhiu tin trnh Server chy ng thi, mi tin trnh p ng ccyu cu t cc tin trnh Client khc.

(4) Kh nng m rng

H phn tn c kh nng hot ng tt v hiu qu nhiu mc khc

nhau. Mt h phn tn nh nht c th hot ng ch cn hai trm lm vicv mt File Server. Cc h ln hn ti hng nghn my tnh.

Kh nng m rng c c trng bi tnh khng thay i phnmm h thng v phn mm ng dng khi h c m rng. iu ny cht c mc d no vi h phn tn hin ti. Yu cu vic m rngkhng ch l s m rng v phn cng, v mng m n tri trn cc khacnh khi thit k h phn tn.

(5) Kh nng th li

Vic thit k kh nng th li ca cc h thng my tnh da trn haigii php:

- Dng kh nng thay th m bo s hot ng lin tc v hiuqu.

- Dng cc chng trnh hi phc khi xy ra s c.

Xy dng mt h thng c th khc phc s c theo cch th nht thngi ta ni hai my tnh vi nhau thc hin cng mt chng trnh,

8


9/134

mt trong hai my chy ch Standby (khng ti hay ch). Gii phpny tn km v phi nhn i phn cng ca h thng. Mt gii php gim ph tn l cc Server ring l c cung cp cc ng dng quan trng c th thay th nhau khi c s c xut hin. Khi khng c cc s c cc

Server hot ng bnh thng, khi c s c trn mt Server no , ccng dng Clien t chuyn hng sang cc Server cn li.

Cch hai th cc phn mm hi phc c thit k sao cho trng thid liu hin thi (trng thi trc khi xy ra s c) c th c khi phc khili c pht hin.

Cc h phn tn cung cp kh nng sn sng cao i ph vi ccsai hng phn cng.

(6) Tnh trong sut

Tnh trong sut ca mt h phn tn c hiu nh l vic che khuti cc thnh phn ring bit ca h i vi ngi s dng v nhng ngi lptrnh ng dng.

Tnh trong sut v v tr: Ngi s dng khng cn bit v tr vt lca d liu. Ngi s dng c quyn truy cp ti n c s d liu nm bt

k ti v tr no. Cc thao tc ly, cp nht d liu ti mt im d liu xa c t ng thc hin bi h thng ti im a ra yu cu, ngi sdng khng cn bit n s phn tn ca c s d liu trn mng.

Tnh trong sut trong vic s dng: Vic chuyn i ca mt phnhay ton b c s d liu do thay i v t chc hay qun l, khng nhhng ti thao tc ngi s dng.

Tnh trong sut ca vic phn chia: Nu d liu c phn chia do

tng ti, n khng c nh hng ti ngi s dng.Tnh trong sut cas trng lp: Nu d liu trng lp gim chi

ph truyn thng vi c s d liu hoc nng cao tin cy, ngi s dngkhng cn bit n iu .

(7) m bo tin cy v nht qun

H thng yu cu tin cy cao: s b mt ca d liu phi c bov, cc chc nng khi phc h hng phi c m bo. Ngoi ra yu cu

ca h thng v tnh nht qun cng rt quan trng trong th hin: khng

9


10/134

c c mu thun trong ni dung d liu. Khi cc thuc tnh d liu l khcnhau th cc thao tc vn phi nht qun.

1.1.3. Mc ch ca vic s dng c s d liu phn tn

Xut pht t yu cu thc t v t chc v kinh t: Trong thc tnhiu t chc l khng tp trung, d liu ngy cng ln v phc v cho angi dng nm phn tn, v vy c s d liu phn tn l con ng thchhp vi cu trc t nhin ca cc t chc . y l mt trong nhng yut quan trng thc y vic pht trin c s d liu phn tn.

S lin kt cc c s d liu a phng ang tn ti: c s d liuphn tn l gii php t nhin khi c cc c s d liu ang tn ti v scn thit xy dng mt ng dng ton cc. Trong trng hp ny c s dliu phn tn c to t di ln da trn nn tng c s d liu ang tnti. Tin trnh ny i hi cu trc li cc c s d liu cc b mt mcnht nh. D sao, nhng sa i ny vn l nh hn rt nhiu so vi victo lp mt c s d liu tp trung hon ton mi.

Lm gim tng chi ph tm kim: Vic phn tn d liu cho php ccnhm lm vic cc b c th kim sot c ton b d liu ca h. Tuyvy, ti cng thi im ngi s dng c th truy cp n d liu xa nu

cn thit. Ti cc v tr cc b, thit b phn cng c th chn sao cho phhp vi cng vic x l d liu cc b ti im .

S pht trin m rng: Cc t chc c th pht trin m rng bngcch thm cc n v mi, va c tnh t tr, va c quan h tng i vicc n v t chc khc. Khi gii php c s d liu phn tn h tr mts m rng uyn chuyn vi mt mc nh hng ti thiu ti cc nv ang tn ti

Tr li truy vn nhanh: Hu ht cc yu cu truy vn d liu t ngis dng ti bt k v tr cc b no u tho mn d liu ngay ti thi im.

tin cy v kh nng s dng nng cao: nu c mt thnh phnno ca h thng b hng, h thng vn c th duy tr hot ng.

Kh nng phc hi nhanh chng: Vic truy nhp d liu khng ph

thuc vo mt my hay mt ng ni trn mng. Nu c bt k mt lino h thng c th t ng chn ng li qua cc ng ni khc.

10


11/134

1.1.4. Kin trc c bn ca CSDL phn tn

y khng l kin trc tng minh cho tt c cc CSDL phn tn,tuy vy kin trc ny th hin t chc ca bt k mt CSDL phn tn no

- S tng th: nh ngha tt c cc d liu s c lu tr trongCSDL phn tn. Trong m hnh quan h, s tng th bao gm nhngha ca cc tp quan h tng th.

- S phn on: Mi quan h tng th c th chia thnh mt viphn khng gi ln nhau c gi l on (fragments). C nhiu cch khcnhau thc hin vic phn chia ny. nh x (mt - nhiu) gia s tng th v cc on c nh ngha trong s phn on.

- S nh v: Cc on l cc phn logic ca quan h tng thc nh v vt l trn mt hoc nhiu v tr trn mng. S nh v nhngha on no nh v ti cc v tr no. Lu rng kiu nh x c nhngha trong s nh v quyt nh CSDL phn tn l d tha hay khng.

- S nh x a phng: nh x cc nh vt l v cc i tngc lu tr ti mt trm (tt c cc on ca mt quan h tng th trncng mt v tr to ra mt nh vt l)

11


12/134

1.1.5. H qun tr CSDL phn tn

H qun tr CSDL phn tn (Distributed Database ManagementSystem-DBMS) c nh ngha l mt h thng phn mm cho phpqun l cc h CSDL (to lp v iu khin cc truy nhp cho cc h

CSDL phn tn) v lm cho vic phn tn tr nn trong sut vi ngi sdng.

c tnh v hnh mun ni n s tch bit v ng ngha cp cao ca mt h thng vi cc vn ci t cp thp. S phn tn dliu c che du vi ngi s dng lm cho ngi s dng truy nhp voCSDL phn tn nh h CSDL tp trung. S thay i vic qun tr khngnh hng ti ngi s dng.

H qun tr CSDL phn tn gm 1 tp cc phn mm (chng trnh)sau y:

Cc chng trnh qun tr cc d liu phn tn

Cha cc chng trnh qun tr vic truyn thng d liu

Cc chng trnh qun tr cc CSDL a phng.

Cc chng trnh qun tr t in d liu.

12

S tng th

S phn on

S nh v

S nh x a phng 2S nh x a phng 1

DBMS ca v tr 1

CSDL a phng ti v tr 1

Cc v tr khc

DBMS ca v tr 2

CSDL a phng ti v tr 2

Hnh1.2 Kin trc c bn ca CSDL phn tn


13/134

to ra mt h CSDL phn tn (Distributed Database System-DDBS) cc tp tin khng ch c lin i logic chng cn phi c cu trcv c truy xut qua mt giao din chung.

Mi trng h CSDL phn tn l mi trng trong d liu c

phn tn trn mt s v tr.

1.2. Kin trc h qun tr C s d liu phn tn

1.2.1. Cc h khch / i l

Cc h qun tr CSDL khch / i l xut hin vo u nhngnm 90 v c nh hng rt ln n cng ngh DBMS v phng thc xl tnh ton. tng tng qut ht sc n gin: phn bit cc chc nng

cn c cung cp v chia nhng chc nng ny thnh hai lp: chc nngi l (server function) v chc nng khch hng (client function). N cungcp kin trc hai cp, to d dng cho vic qun l mc phc tp cacc DBMS hin i v phc tp ca vic phn tn d liu.

i l thc hin phn ln cng vic qun l d liu. iu ny cngha l tt c mi vic x l v ti u ho vn tin, qun l giao dch vqun l thit b lu tr c thc hin ti i l. Khch hng, ngoi ng

dng v giao din s c modun DBMS khch chu trch nhim qun l dliu c gi n cho bn khch v i khi vic qun l cc kho cht giaodch cng c th giao cho n. Kin trc c m t bi hnh di rt thngdng trong cc h thng quan h, vic giao tip gia khch v i lnm ti mc cu lnh SQL. Ni cch khc, khch hng s chuyn cc cuvn tin SQL cho i l m khng tm hiu v ti u ho chng. i l thchin hu ht cng vic v tr quan h kt qu v cho khch hng.

C mt s loi kin trc khch/ i l khc nhau. Loi n gin nhtl trng hp c mt i l c nhiu khch hng truy xut. Chng ta giloi ny l nhiu khch mt i l. Mt kin trc khch/ i l phc tphn l kin trc c nhiu i l trong h thng (c gi l nhiu khchnhiu i l). Trong trng hp ny chng ta c hai chin lc qun l:hoc mi khch hng t qun l ni kt ca n vi i l hoc mi khchhng ch bit i l rut ca n v giao tip vi cc i l khc qua i l khi cn. Li tip cn th nht lm n gin cho cc chng trnh i lnhng li t gnh nng ln cc my khch cng vi nhiu trch nhimkhc. iu ny dn n tnh hung c gi l cc h thng khch t phc

13


14/134

v. Li tip cn sau tp trung chc nng qun l d liu ti i l. V ths v hnh ca truy xut d liu c cung cp qua giao din ca i l.

T gc tnh logc c d liu, DBMS khch/ i l cung cpcng mt hnh nh d liu nh cc h ngang hng s c tho lun

phn tip theo. Ngha l chng cho ngi s dng thy mt hnh nh vmt CSDL logic duy nht, cn ti mc vt l n c th phn tn. V th sphn bit ch yu gia cc h khch/i l v ngang hng khng phi mc v hnh c cung cp cho ngi dng v cho ng dng m mhnh kin trc c dng nhn ra mc v hnh ny.

1.2.2. Cc h phn tn ngang hng

M hnh client / server phn bit client (ni yu cu dch v) vserver (ni phc v cc yu cu). Nhng m hnh x l ngang hng, cc hthng tham gia c vai tr nh nhau. Chng c th yu cu va dch v tmt h thng khc hoc va tr thnh ni cung cp dch v. Mt cch ltng, m hnh tnh ton ngang hng cung cp cho x l hp tc gia ccng dng c th nm trn cc phn cng hoc h iu hnh khc nhau.Mc ch ca mi trng x l ngang hng l h tr cc CSDL cni mng. Nh vy ngi s dng DBMS s c th truy cp ti nhiuCSDL khng ng nht.

14


15/134

CHNG 2. CC PHNG PHP PHN TN D LIU

2.1.Thit k c s d liu phn tn

2.1.1.Cc chin lc thit k

Qu trnh thit k t trn xung (top-down)

15

Phn tch yu cu

Yu cu h thng(mc tiu)

Thit k khi nimThit k khung nhn

Lc khi

nim toncc

Thng tintruy xut nh ngha

lc ngoi

Thit k phn tn

Lc khi nim cc b

Thit k vt l

Lc vt l

Theo di v bo tr

Phn hi

Nguyn liu tngi dng

Nguyn liu

t ngi dng

Hnh 2.1. Qu trnh thit k t trnxung


16/134

Phn tch yu cu: nhm nh ngha mi trng h thng v thu thpcc nhu cu v d liu v nhu cu x l ca tt c mi ngi c s dngCSDL

Thit k khung nhn: nh ngha cc giao-din cho ngi s dngcui (end-user)

Thit k khi nim: xem xt tng th x nghip nhm xc nh ccloi thc th v mi lin h gia cc thc th.

Thit k phn tn: chia cc quan h thnh nhiu quan h nh hn gil phn mnh v cp pht chng cho cc v tr.

Thit k vt l: nh x lc khi nim cc b sang cc thit b lu

tr vt l c sn ti cc v tr tng ng. Qu trnh thit k t di ln (bottom-up)

Thit k t trn xung thch hp vi nhng CSDL c thit k tu. Tuy nhin chng ta cng hay gp trong thc t l c sn mt sCSDL, nhim v thit k l phi tch hp chng thnh mt CSDL. Tipcn t di ln s thch hp cho tnh hung ny. Khi im ca thit k tdi ln l cc lc khi nim cc b . Qu trnh ny s bao gm vic

tch hp cc lc cc b thnh khi nim lc ton cc.2.2. Cc vn thit k

2.2.1. L do phn mnh

Khung nhn ca cc ng dng thng ch l mt tp con ca quan h.V th n v truy xut khng phi l ton b quan h nhng ch l cc tpcon ca quan h. Kt qu l xem tp con ca quan h l n v phn tn s

l iu thch hp duy nht.Vic phn r mt quan h thnh nhiu mnh, mi mnh c x l

nh mt n v, s cho php thc hin nhiu giao dch ng thi. Ngoi ravic phn mnh cc quan h s cho php thc hin song song mt cu vntin bng cch chia n ra thnh mt tp cc cu vn tin con hot tc trn ccmnh. V th vic phn mnh s lm tng mc hot ng ng thi vnh th lm tng lu lng hot ng ca h thng.

2.2.2. Cc kiu phn mnh Cc quy tc phn mnh ng n

16


17/134

Chng ta s tun th ba quy tc trong khi phn mnh m chngbo m rng CSDL s khng c thay i no v ng ngha khi phnmnh.

a) Tnh y (completeness).

Nu mt th hin quan h R c phn r thnh cc mnh R1, R2,,Rn, th mi mc d liu c th gp trong R cng c th gp mt trongnhiu mnh Ri. c tnh ny ging nh tnh cht phn r ni khng mtthng tin trong chun ho, cng quan trng trong phn mnh bi v n bom rng d liu trong quan h R c nh x vo cc mnh v khng bmt. Ch rng trong trng hp phn mnh ngang mc d liu munni n l mt b, cn trong trng hp phn mnh dc, n mun ni n

mt thuc tnh.b) Tnh ti thit c (reconstruction).

Nu mt th hin quan h R c phn r thnh cc mnh R1, R2,,Rn, th cn phi nh ngha mt ton t quan h sao cho

R=Ri, Ri Fr

Ton t thay i tu theo tng loi phn mnh, tuy nhin iu

quan trng l phi xc nh c n. Kh nng ti thit mt quan h t ccmnh ca n bo m rng cc rng buc c nh ngha trn d liu didng cc ph thuc s c bo ton.

c) Tnh tch bit (disjointness).

Nu quan h R c phn r ngang thnh cc mnh R1, R2,,Rn,v mc d liu di nm trong mnh Rj, th n s khng nm trong mnh Rkkhc

(kj ). Tiu chun ny m bo cc mnh ngang s tch bit (ri nhau).Nu quan h c phn r dc, cc thuc tnh kho chnh phi c lp litrong mi mnh. V th trong trng hp phn mnh dc, tnh tch bit chc nh ngha trn cc trng khng phi l kho chnh ca mt quanh.

Cc yu cu thng tin

Mt iu cn lu trong vic thit k phn tn l qu nhiu yu t c

nh hng n mt thit k ti u. t chc logic ca CSDL, v tr cc ngdng, c tnh truy xut ca cc ng dng n CSDL, v cc c tnh ca

17


18/134

h thng my tnh ti mi v tr u c nh hng n cc quyt nh phntn. iu ny khin cho vic din t bi ton phn tn tr nn ht scphc tp.

Cc thng tin cn cho thit k phn tn c th chia thnh bn loi:

- Thng tin CSDL- Thng tin ng dng- Thng tin v mng- Thng tin v h thng my tnh

Hai loi sau c bn cht hon ton nh lng v c s dng trongcc m hnh cp pht ch khng phi trong cc thut ton phn mnh

2.2.3. Phn mnh ngang

Trong phn ny, chng ta bn n cc khi nim lin quan n phnmnh ngang (phn tn ngang). C hai chin lc phn mnh ngang c bn:

- Phn mnh nguyn thu (primary horizontal fragmentation) ca mtquan h c thc hin da trn cc v t c nh ngha trn quan h .

- Phn mnh ngang dn xut (derived horizontal fragmentation ) l phnmnh mt quan h da vo cc v t c nh trn mt quan h khc.

Hai kiu phn mnh ngang

Phn mnh ngang chia mt quan h r theo cc b, v vy mi mnh lmt tp con cc b t ca quan h r.

Phn mnh nguyn thu (primary horizontal fragmentation) ca mtquan h c thc hin da trn cc v t c nh ngha trn quan h .Ngc li phn mnh ngang dn xut (derived horizontal fragmentation )l phn mnh mt quan h da vo cc v t c nh trn mt quan hkhc. Nh vy trong phn mnh ngang tp cc v t ng vai tr quantrng.

Trong phn ny s xem xt cc thut ton thc hin cc kiu phnmnh ngang. Trc tin chng ta nu cc thng tin cn thit thc hinphn mnh ngang.

Yu cu thng tin ca phn mnh ngang

a) Thng tin v c s d liu

18


19/134

Thng tin v CSDL mun ni n l lc ton cc v quan hgc, cc quan h con. Trong ng cnh ny, chng ta cn bit c ccquan h s kt li vi nhau bng php ni hay bng php tnh khc. vimc ch phn mnh dn xut, cc v t c nh ngha trn quan h

khc, ta thng dng m hnh thc th - lin h (entity-relatinhip model),v trong m hnh ny cc mi lin h c biu din bng cc ng nic hng (cc cung) gia cc quan h c lin h vi nhau qua mt ni.

19


20/134

Th d 1:

Hnh 2.2. Biu din mi lin h gia cc quan h nh cc ng ni.

Hnh trn trnh by mt cch biu din cc ng ni gia cc quanh. ch rng hng ca ng ni cho bit mi lin h mt -nhiu.Chng hn vi mi chc v c nhiu nhn vin gi chc v , v th

chng ta s v mt ng ni t quan h CT (chi tr) hng n NV(nhn vin). ng thi mi lin h nhiu- nhiu gia NV v DA(d n)c biu din bng hai ng ni n quan h PC(phn cng).

Quan h nm ti u (khng mi tn ) ca ng ni c gi l chnhn (owner) ca ng ni v quan h ti cui ng ni (u mi tn)gi l thnh vin (member).

Th d 2:

Cho ng ni L1 ca hnh 2.2, cc hm owner v member c ccgi tr sau:

Owner( L1 ) = CT

Member (L1) = NV

Thng tin nh lng cn c v CSDL l lc lng (cardinality) cami quan h R, l s b c trong R, c k hiu l card (R)

b) Thng tin v ng dng

20

Chc v, Lng

MNV, tnNV, chc vMDA, tnDA, ngn sch, a im

MNV , MDA, nhim v, thi gian

CT

NVDA

PC

L1

L2 L3


21/134

phn tn ngoi thng tin nh lng Card(R) ta cn cn thng tinnh tnh c bn gm cc v t c dng trong cc cu vn tin. Lngthng tin ny ph thuc bi ton c th.

Nu khng th phn tch c ht tt c cc ng dng xc nh

nhng v t ny th t nht cng phi nghin cu c cc ng dng quantrng nht.

Vy chng ta xc nh cc v t n gin (simple predicate). Choquan h R ( A1, A2,, An ), trong Ai l mt thuc tnh c nh nghatrn mt min bin thin D(Ai) hay Di..

Mt v t n gin P c nh ngha trn R c dng:

P:Ai ValueTrong {=,, } v

value c chn t min bin thin ca Ai (value Di).

Nh vy, cho trc lc R, cc min tr Di chng ta c th xcnh c tp tt c cc v t n gin Pr trn R.

Vy Pr={P: Ai Value}. Tuy nhin trong thc t ta ch cn nhngtp con thc s ca Pr.

Th d 3: Cho quan h D n nh sau:

P1 : TnDA = thit b iu khin

P2 : Ngn sch 200000

L cc v t n gin..

Chng ta s s dng k hiu Pri biu th tp tt c cc v t ngin c nh ngha trn quan h Ri. Ccphn t ca Pri c k hiu lpij.

Cc v t n gin thng rt d x l, cc cu vn tin thng chanhiu v t phc tp hn, l t hp ca cc v t n gin. Mt t hp cnc bit ch , c gi l v t hi s cp (minterm predicate), l hi(conjunction) ca cc v t n gin. Bi v chng ta lun c th bin imt biu thc Boole thnh dng chun hi, vic s dng v t hi s cptrong mt thut ton thit k khng lm mt i tnh tng qut.

21


22/134

Cho mt tp Pri = {pi1, pi2, , pim } l cc v t n gin trn quan hRi, tp cc v t hi s cp Mi={mi1, mi2, , miz } c nh ngha l:

Mi={mij | mij= p*ik} vi 1 k m, 1 j z

Trong p*ik=pik hoc p*ik= pik . V th mi v t n gin c thxut hin trong v t hi s cp di dng t nhin hoc dng ph nh.

22


23/134

Th d 4:

Xt quan h CT:

chc v Lng

K s inPhn tch h

thng

K s c kh

Lp trnh

4000034000

27000

24000

Di y l mt s v t n gin c th nh ngha c trn PAY. p1: chc v= K s in

p2: chc v= Phn tch h thng

p3: chc v= K s c kh

p4: chc v= Lp trnh

p5: Lng 30000

p6: Lng > 30000

Di y l mt s cc v t hi s cp c nh ngha da trn ccv t n gin ny

m1: chc v= K s in Lng 30000

m2: chc v = K s in Lng > 30000

m3: (chc v= K s in ) Lng 30000

m4: (chc v= K s in ) Lng> 30000

m5: chc v= Lp trnh Lng 30000

m6: chc v= Lp trnh Lng > 30000

Ch :+ Php ly ph nh khng phi lc no cng thc hin c.Th d:xt hai v t n gin sau: Cn_di A; A Cn_trn. Tc lthuc tnh A c min tr nm trong cn di v cn trn, khi phn b

ca chng l:(Cn_di A);

23


24/134

(A Cn_trn) khng xc nh c. Gi tr ca A trongcc ph nh ny ra khi min tr ca A.

Hoc hai v t n gin trn c th c vit li l:

Cn_di A Cn_trn c phn b l: (Cn_di A Cn_trn)khng nh ngha c. V vy khi nghin cu nhng vn ny ta chxem xt cc v t ng thc n gin.

=> Khng phi tt c cc v t hi s cp u c th nh ngha c.

+ Mt s trong chng c th v ngha i vi ng ngha ca quan hChi tr.Ngoi ra cn ch rng m3 c th c vit li nh sau:

m3: chc v K s in Lng 30000

Theo nhng thng tin nh tnh v cc ng dng, chng ta cn bit haitp d liu.

tuyn hi s cp (minterm selectivity): s lng cc b ca quanh s c truy xut bi cu vn tin c c t theo mt v t his cp cho. chng hn tuyn ca m1 trong Th d 4 l zero biv khng c b no trong CT tha v t ny. tuyn ca m2 l 1.Chng ta s k hiu tuyn ca mt hi s cp mi l sel (mi).

Tn s truy xut (access frequency): tn s ng dng truy xut dliu. Nu Q={q1, q2,....,qq} l tp cc cu vn tin, acc (qi) biu th chotn s truy xut ca qi trong mt khong thi gian cho.

Ch rng mi hi s cp l mt cu vn tin. Chng ta k hiu tns truy xut ca mt hi s cp l acc(mi)

Phn mnh ngang nguyn thu

Phn mnh ngang nguyn thu c nh ngha bng mt phpton chn trn cc quan h ch nhn ca mt lc ca CSDL. V thcho bit quan h R, cc mnh ngang ca R l cc Ri:

Ri = Fi(R), 1 i z.

Trong Fi l cng thc chn c s dng c c mnh Ri.Ch rng nu Fi c dng chun hi, n l mt v t hi s cp (mj).

24


25/134

Th d 5:Xt quan h DA

MDA TnDA Ngn sch a im

P1

P2P3

P4

Thit b o c

Pht trin d liuCAD/CAM

Bo dng

150000

135000250000

310000

Montreal

New YorkNew York

Paris

Chng ta c th nh ngha cc mnh ngang da vo v tr d n. Khi cc mnh thu c, c trnh by nh sau:

DA1=a im=Montreal (DA)

DA2=a im=New York (DA)DA3=a im=Paris (DA)

DA1

MDA

TDA Ngn sch a im

P1 Thit b o c 150000 Montreal

DA2

MDA

TnDA Ngn sch a im

P2

P3

Pht trin d liu

CAD/CAM

135000

250000

NewYork

New

YorkDA3

MDA TnDA Ngn sch a im

P4 thit b o c 310000 Paris

By gi chng ta c th nh ngha mt mnh ngang cht ch v rrng hn

Mnh ngang Ri ca quan h R c cha tt c cc b R tha v t his cp mi

25


26/134


27/134

mt tp v t Pr xt tnh cc tiu chng ta c th kim tra bng cch vtb nhng v t tha c tp v t Pr l cc tiu v tt nhin Pr cng ltp y vi Pr.

Thut ton COM_MIN: Cho php tm tp cc v t y v cc

tiu Pr t Pr. Chng ta tm quy c:

Quy tc 1: Quy tc c bn v tnh y v cc tiu , n khngnh rng mt quan h hoc mt mnh c phn hoch thnh t nhthai phn v chng c truy xut khc nhau bi t nht mt ng dng.

Thut ton 1.1 COM_MIN

Input : R: quan h; Pr: tpcc v t n gin;

Output: Pr: tp cc v t cc tiu v y ;Declare

F: tp cc mnh hi s cp;

Begin

Pr= ; F = ;

For each v t p Pr if p phn hoch R theo Quy tc 1 then

Begin

Pr: = Pr p;

Pr: = Pr p;

F: = F p; {fi l mnh hi s cp theo pi }

End; {Chng ta chuyn cc v t c phn mnh R vo Pr}

Repeat

For each p Pr if p phn hoch mt mnh fk ca Prtheo quy tc 1 then

Begin

Pr: = Pr p;

Pr: = Pr p;

F: = F p;

End;

27


28/134

Until Pr y {Khng cn p no phn mnh fk ca Pr}

For each p Pr, ifp m pp then

Begin

Pr:= Pr-p;F:= F - f;

End;

End. {COM_MIN}

Thut ton bt du bng cch tm mt v t c lin i v phn hochquan h cho. Vng lp Repeat-until thm cc v t c phn hoch cc

mnh vo tp ny, bo m tnh y ca Pr. on cui kim tra tnhcc tiu ca Pr. V th cui cng ta c tp Pr l cc tiu v y .

Bc hai ca vic thit k phn mnh nguyn thy l suy dn ra tpcc v t hi s cp c th c nh ngha trn cc v t trong tp Pr. Ccv t hi s cp ny xc nh cc mnh ng c vin cho bc cp pht.Vic xc nh cc v t hi s cp l tm thng; kh khn chnh l tp ccv t hi s cp c th rt ln (thc s chng t l hm m theo s lngcc v t n gin). trong bc k tip chng ta s tm cch lm gim slng v t hi s cp cn c nh ngha trong phn mnh.

Bc ba ca qu trnh thit k l loi b mt s mnh v ngha. iuny c thc hin bng cch xc nh nhng v t mu thun vi tp ccphp ko theo (implication) I. Chng hn nu Pr={p1, p2}, trong

P1: att= value_1

P2: att=value_2

V min bin thin ca att l {value_1, value_2}, r rng I cha haiphp ko theo vi khng nh:

I1: (att=value_1) (att=value_2)

I2: (att=value_1)(att=value_2)

Bn v t hi s cp sau y c nh ngha theo Pr:

M1: (att=value_1) (att=value_2)

M2: (att=value_1)(att=value_2)

M3: (att=value_1)(att=value_2)

28


29/134

M4: (att=value_1) (att=value_2)

Trong trng hp ny cc v t hi s cp m1, m4 mu thun vi ccphp ko theo I v v th b loi ra khi M.

Thut ton phn mnh ngang nguyn thy c trnh by trong thutton 1.2.

Thut ton 1.2 PHORIZONTAL

Input: R: quan h; Pr: tp cc v t n gin;

Output: M: tp cc v t hi s cp;

Begin

Pr:= COM_MIN(R, Pr);Xc nh tp M cc v t hi s cp;

Xc nh tp I cc php ko theo gia cc piPr;

For each mi M do

Begin

IF mi mu thun vi I then

M:= M-mi

End;

End. {PHORIZONTAL}

Th d 8: Chng ta hy xt quan h DA. Gi s rng c hai ngdng. ng dng u tin c a ra ti ba v tr v cn tm tn v ngn

sch ca cc d n khi cho bit v tr. Theo k php SQL cu vn tin cvit l:

SELECT TnDA, Ngn sch

FROM DA

WHERE a im=gi tr

i vi ng dng ny, cc v t n gin c th c dng l:

P1: a im=MontrealP2: a im=New York

29


30/134

P3: a im=Paris

ng dng th hai l nhng d n c ngn sch di 200.000 lac qun l ti mt v tr, cn nhng d n c ngn sch ln hn cqun l ti mt v tr th hai. V th cc v t n gin phi c s dng

phn mnh theo ng dng th hai l:

P4: ngn sch200000

P5: ngn sch>200000

Nu kim tra bng thut ton COM_MIN, tp Pr={p1, p2, p3, p4,p5} r rng y v cc tiu

Da trn Pr chng ta c th nh ngha su v t hi s cp sau y

to ra M:M1: (a im=Montreal) (ngn sch200000)

M2: (a im=Montreal) (ngn sch>200000)

M3: (a im=New York) (ngn sch200000)

M4: (a im=New York) (ngn sch>200000)

M5: (a im=Paris) (ngn sch200000)

M6: (a im=Paris) (ngn sch>200000)

y khng phi l cc v t hi s cp duy nht c th c to ra.Chng hn vn c th nh ngha cc v t:

p1 p2 p3 p4 p5

Tuy nhin cc php ko hin nhin l:

I1: p1 p2 p3

I2: p2 p1 p3

I3: p3 p1 p2

I4: p4p5

I5: p5 p4

I6: p4p5

I7: p5p4

30


31/134

Cho php loi b nhng v t hi s cp ny v chng ta cn li m1n m6.

Cn nh rng cc php ko theo phi c nh ngha theo ng nghaca CSDL, khng phi theo cc gi tr hin ti. Mt s mnh c nh

ngha theo M={m1,,m6} c th rng nhng chng vn l cc mnh. Ktqu phn mnh nguyn thu cho DA l to ra su mnh FDA={DA1, DA2,DA3, DA4, DA5, DA6}, y c hai mnh rng l {DA2, DA5 }

DA1

MDA TnDA

Ngnsch

a im

P1 Thit b oc 150000 Montreal

31


32/134

DA3

MDA TnDA Ngnsch

a im

P2 Pht trin d liu 135000 NewYork

DA4

MDA TnDA Ngnsch

a im

P3 CAD/CAM 250000 New

YorkDA 6

MDA TnDA Ngnsch

a im

P4 bo dng 310000 Paris

Phn mnh ngang dn xutPhn mnh ngang dn xut c nh ngha trn mt quan h thnh

vin ca ng ni da theo php ton chn trn quan h ch nhn cang ni .

Nh th nu cho trc mt ng ni L, trong owner (L)=S vmember(L)=R, v cc mnh ngang dn xut ca R c nh ngha l:

Ri=R|>< Si , 1 i w

Trong w l s lng cc mnh c nh ngha trn R, vSi= Fi(S) vi Fi l cng thc nh ngha mnh ngang nguyn thu Si

32


33/134

Th d 9: Xt ng ni

NV

MNV TnNV Chc v

E1E2

E2

E3

E3

E4

E5

E6

E7

E8

J.DoeM.Smith

M.Smith

A.Lee

A.LeeJ.Miller

B.Casey

L.Chu

R.david

J.Jones

K s inPhn tch

Phn tch

K s c kh

K s c kh

Programmer

Phn tch hthng

K s in

K s c kh

Phn tch hthng

th th chng ta c th nhm cc k s thnh hai nhm ty theolng: nhm c lng t 30.000 la tr ln v nhm c lng di30.000 la. Hai mnh Nhn vin1 v Nhn vin2 c nh ngha nhsau:

NV1=NV |>< CT1

NV2=NV |>< CT2Trong CT1= Lng 30000( CT)

CT2= Lng>30000( CT)

CT 1 CT2

Chc v Lng Chc v Lng

K s c kh

Lp trnh

27000

24000

K s in

Phn tch hthng

40000

34000

33

Chc v, Lng

MNV, TnNV, Chc v

L1NV

CT


34/134

Kt qu phn mnh ngang dn xut ca quanh NV nh sau:

34


35/134

NV1 NV2

MNV TnNV Chc v MNV TnNV Chc v

E3

E4E7

A.Lee

J.MillerR.David

K s c

khLp trnhvin

K s c kh

E1

E2E5

E6

E8

J.Doe

M.Smith

B.Casey

L.Chu

J.Jones

K s in

Phn tchPhn tch hthng

K s in

Phn tch hthng

Ch :

+ Mun thc hin phn mnh ngang dn xut, chng ta cn banguyn liu (input): 1. Tp cc phn hoch ca quan h ch nhn (Thd: CT1, CT2).

2. Quan h thnh vin

3. Tp cc v t ni na gia ch nhn v thnh vin (Chng

hn CT.Chucvu = NV.Chucvu).

+ Vn phc tp cn ch : Trong lc CSDL, chng ta haygp nhiu ng ni n mt quan h R. Nh th c th c nhiu cchphn mnh cho quan h R. Quyt nh chn cch phn mnh no cn datrn hai tiu chun sau:

1. Phn mnh c c tnh ni tt hn

2. Phn mnh c s dng trong nhiu ng dng hn.Tuy nhin, vic p dng cc tiu chun trn cn l mt vn rc

ri.

Th d 10: Chng ta tip tc vi thit k phn tn cho CSDL btu t Th d 9. V quan h NV phn mnh theo CT. By gi xt ASG. Gis c hai ng dng sau:

1. ng dng 1: Tm tn cc k s c lm vic ti mt ni no .

ng dng ny chy c ba trm v truy xut cao hn cc k s ca cc dn nhng v tr khc.

35


36/134

2. ng dng 2: Ti mi trm qun l, ni qun l cc mu tin nhnvin, ngi dng mun truy xut n cc d n ang c cc nhn vinny thc hin v cn bit xem h s lm vic vi d n trong bao lu.

Kim nh tnh ng n

By gi chng ta cn phi kim tra tnh ng ca phn mnh ngang.

a. Tnh y

+ Phn mnh ngang nguyn thu: Vi iu kin cc v t chn l y, phn mnh thu cng c m bo l y , bi v c s ca thutton phn mnh l tp cc v t cc tiu v y Pr, nn tnh y c bo m vi iu kin khng c sai st xy ra.

+ Phn mnh ngang dn xut: C khc cht t, kh khn chnh yl do v t nh ngha phn mnh c lin quan n hai quan h. Trc tinchng ta hy nh ngha qui tc y mt cch hnh thc.

R l quan h thnh vin ca mt ng ni m ch nhn l quan hS. Gi A l thuc tnh ni gia R v S, th th vi mi b t ca R, phi cmt b t ca S sao cho

t.A=t.A

Quy tc ny c gi l rng buc ton vn hay ton vn thamchiu, bo m rng mi b trong cc mnh ca quan h thnh vin u nmtrong quan h ch nhn.

b. Tnh ti thit c

Ti thit mt quan h ton cc t cc mnh c thc hin bng tont hp trong c phn mnh ngang nguyn thy ln dn xut, V th mt

quan h R vi phn mnh Fr={R1, R2,,Rm} chng ta c R = Ri , Ri FRc. Tnh tch ri

Vi phn mnh nguyn thu tnh tch ri s c bo m min lcc v t hi s cp xc nh phn mnh c tnh loi tr tng h(mutually exclusive). Vi phn mnh dn xut tnh tch ri c th bo mnu th ni thuc loi n gin.

36


37/134

2.3. Phn mnh dc

Mt phn mnh dc cho mt quan h R sinh ra cc mnh R1, R2,..,Rr,mi mnh cha mt tp con thuc tnh ca R v c kho ca R. Mc chca phn mnh dc l phn hoch mt quan h thnh mt tp cc quan h

nh hn nhiu ng dng ch cn chy trn mt mnh. Mt phn mnhti ul phn mnh sinh ra mt lc phn mnh cho php gim ti athi gian thc thi cc ng dng chy trn mnh .

Phn mnh dc tt nhin l phc tp hn so vi phn mnh ngang.iu ny l do tng s chn la c th ca mt phn hoch dc rt ln.

V vy c c cc li gii ti u cho bi ton phn hoch dcthc s rt kh khn. V th li phi dng cc phng php khm ph(heuristic). Chng ta a ra hai loi heuristic cho phn mnh dc cc quanh ton cc.

- Nhm thuc tnh: Bt u bng cch gn mi thuc tnh cho mtmnh, v ti mi bc, ni mt s mnh li cho n khi tha mt tiuchun no . K thut ny c c xut ln u cho cc CSDL tptrung v v sau c dng cho cc CSDL phn tn.

- Tch mnh: Bt u bng mt quan h v quyt nh cch phnmnh c li da trn hnh vi truy xut ca cc ng dng trn cc thuctnh.

Bi v phn hoch dc t vo mt mnh cc thuc tnh thngc truy xut chung vi nhau, chng ta cn c mt gi tr o no nh ngha chnh xc hn v khi nim chung vi nhau. S o ny gi lt lc hay lc ht (affinity) ca thuc tnh, ch ra mc lin i gia ccthuc tnh.

Yu cu d liu chnh c lin quan n cc ng dng l tn s truyxut ca chng. gi Q={q1, q2,,qq} l tp cc vn tin ca ngi dng (ccng dng) s chy trn quan h R(A1, A2,,An). Th th vi mi cu vntin qi v mi thuc tnh Aj, chng ta s a ra mt gi tr s dng thuctnh, k hiu use(qi, Aj) c nh ngha nh sau:

1 nu thuc tnh Aj c vn tin qi tham chiu

use(qi, Aj)= 0 trong trng hp ngc li

37


38/134

Cc vct use(qi, ) cho mi ng dng rt d nh ngha nu nhthit k bit c cc ng dng s chy trn CSDL.

Th d 11:

Xt quan h DA, gi s rng cc ng dng sau y chy trn ccquan h . Trong mi trng hp chng ta cng c t bng SQL.

q1: Tm ngn sch ca mt d n, cho bit m ca d n

SELECT Ngn sch

FROM DA

WHERE MDA=gi tr

q2: Tm tn v ngn sch ca tt c mi d nSELECT TnDA, ngn sch

FROM DA

q3: Tm tn ca cc d n c thc hin ti mt thnh ph cho

SELECT tnDA

FROM DA

WHERE a im=gi trq4: Tm tng ngn sch d n ca mi thnh ph

SELECT SUM (ngn sch)

FROM DA

WHERE a im=gi tr

Da theo bn ng dng ny, chng ta c th nh ngha ra cc gi tr

s dng thuc tnh. cho tin v mt k php, chng ta gi A 1=MDA,A2=TnDA, A3=Ngn sch, A4=a im. Gi tr s dng c nh nghadi dng ma trn, trong mc (i,j) biu th use(qi , Aj ).

A1 A2 A3 A4

q1 1 0 1 0q2 0 1 1 0q3 0 1 0 1q 4 0 0 1 1

38


39/134

T lc ca cc thuc tnh

Gi tr s dng thuc tnh khng lm c s cho vic tch vphn mnh. iu ny l do chng khng biu th cho ln ca tn s ngdng. S o lc ht (affinity) ca cc thuc tnh aff(A i, Aj), biu th cho

cu ni (bond) gia hai thuc tnh ca mt quan h theo cch chng ccc ng dng truy xut, s l mt i lng cn thit cho bi ton phnmnh.

Xy dng cng thc o lc ht ca hai thuc tnh Ai, Aj.

Gi k l s cc mnh ca R c phn mnh. Tc l R = R1.Rk.

Q= {q1, q2,,qm} l tp cc cu vn tin (tc l tp cc ng dng chy

trn quan h R). t Q(A, B) l tp cc ng dng q ca Q m use(q,A).use(q, B) = 1.

Ni cch khc:

Q(A, B) = {qQ: use(q, A) =use(q, B) = 1}

Th d da vo ma trn trn ta thy Q(A1,A1) = {q1}, Q(A2,A2 ) = {q2,q3}, Q(A3,A3 ) = {q1,q2, q4}, Q(A4,A4 ) = {q3, q4}, Q(A1,A2 ) = rng,Q(A1,A3 ) = {q1}, Q(A2,A3 ) = {q2},..

S o lc ht gia hai thuc tnh Ai, Aj c nh ngha l:

aff(Ai, Aj)= refl (qk)accl(qk)

qkQ(Ai, Aj) l Rl

Hoc:

aff(Ai, Aj)= refl (qk)accl(qk)

Use(qk, Ai)=1Use(qk, Aj)=1 Rl

Trong refl (qk) l s truy xut n cc thuc tnh (A i, Aj) cho ming dng qkti v tr Rl v accl(qk) l s o tn s truy xut ng dng qk ncc thuc tnh Ai, Aj ti v tr l. Chng ta cn lu rng trong cng thctnh aff (Ai, Aj) ch xut hin cc ng dng q m c A i v Aj u s dng.

Kt qu ca tnh ton ny l mt ma trn i xng n x n, mi phn tca n l mt s o c nh ngha trn. Chng ta gi n l ma trn lc

t ( lc ht hoc i lc) thuc tnh (AA) (attribute affinity matrix).

39


40/134

Th d 12: Chng ta hy tip tc vi Th d 11. cho dn ginchng ta hy gi s rng refl (qk) =1 cho tt c qk v Rl. Nu tn s ngdng l:

Acc1(q1) = 15 Acc2(q1) = 20 Acc3(q1) = 10

Acc1(q2) = 5 Acc2(q2) = 0 Acc3(q2) = 0

Acc1(q3) = 25 Acc2(q3) = 25 Acc3(q3) = 25

Acc1(q4) = 3 Acc2(q4) = 0 Acc3(q1) = 0

S o lc ht gia hai thuc tnh A1 v A3 l:

Aff(A1, A3) = 1k=1 3t=1acct(qk) = acc1(q1)+acc2(q1)+acc3(q1) = 45

Tng t tnh cho cc cp cn li ta c ma trn i lc sau:

Thut ton nng lng ni BEA (Bond Energy Algorithm)

n y ta c th phn R lm cc mnh ca cc nhm thuc tnh davo s lin i (lc ht) gia cc thuc tnh, th d t lc ca A1, A3 l 45,ca A2, A4 l 75, cn ca A1, A2 l 0, ca A3, A4 l 3 Tuy nhin, phngphp tuyn tnh s dng trc tip t ma trn ny t c mi ngi quantm v s dng. Sau y chng ta xt mt phng php dng thut ton

nng lng ni BEA ca Hoffer and Severance, 1975 v Navathe., 1984.1. N c thit k c bit xc nh cc nhm gm cc mc

tng t, khc vi mt sp xp th t tuyn tnh ca cc mc.

2. Cc kt qu t nhm khng b nh hng bi th t a cc mc vothut ton.

3. Thi gian tnh ton ca thut ton c th chp nhn c l O(n2),vi n l s lng thuc tnh.

A1 A2 A3 A4

A1 45 0 45 0A2 0 80 5 75A3 45 5 53 3A4 0 75 3 78

40


41/134

4. Mi lin h qua li gia cc nhm thuc tnh t c th xc nhc.

Thut ton BEA nhn nguyn liu l mt ma trn i lc thuc tnh(AA), hon v cc hng v ct ri sinh ra mt ma trn i lc t (CA)

(Clustered affinity matrix). Hon v c thc hin sao cho s o i lcchung AM (Global Affinity Measure) l ln nht. Trong AM l ilng:

AM= ni=1 nj=1 aff(Ai, Aj)[aff(Ai, Aj-1)+aff(Ai, Aj+1)+aff(Ai-1, Aj)+aff(Ai+1, Aj)]

Vi aff(A0, Aj)=aff(Ai, A0)=aff(An+1, Aj)=aff(Ai, An+1)=0 cho i,j

Tp cc iu kin cui cng cp n nhng trng hp mt thuctnh c t vo CA v bn tri ca thuc tnh tn tri hoc v bnphi ca thuc tnh tn phi trong cc hon v ct, v bn trn hng trncng v bn di hng cui cng trong cc hon v hng. Trong nhngtrng hp ny, chng ta cho 0 l gi tr lc ht aff gia thuc tnh angc xt v cc ln cn bn tri hoc bn phi (trn cng hoc di y )ca n hin cha c trong CA.

Hm cc i ho ch xt nhng ln cn gn nht, v th n nhm ccgi tr ln vi cc gi tr ln , gi tr nh vi gi tr nh. V ma trn lc htthuc tnh AA c tch cht i xng nn hm s va c xy dng trnthu li thnh:

AM= ni=1 nj=1 aff(Ai, Aj)[aff(Ai, Aj-1)+aff(Ai, Aj+1)]

Qu trnh sinh ra ma trn t lc (CA) c thc hin qua ba bc:

Bc 1: Khi gn:

t v c nh mt trong cc ct ca AA vo trong CA. Th d ct1, 2 c chn trong thut ton ny.

Bc 2: Thc hin lp

Ly ln lt mt trong n-i ct cn li (trong i l s ct ct vo CA) v th t chng vo trong i+1 v tr cn li trong ma trn CA.Chn ni t sao cho cho i lc chung AM ln nht. Tip tc lp n khikhng cn ct no dt.

Bc 3: Sp th t hng

41


42/134

Mt khi th t ct c xc nh, cc hng cng c t li cc v tr tng i ca chng ph hp vi cc v tr tng i ca ct.

Thut ton BEA

Input: AA - ma trn i lc thuc tnh;

Output: CA - ma trn i lc t sau khi sp xp li cc hng ccct;

Begin

{Khi gn: cn nh rng l mt ma trn n x n}

CA(, 1)AA(, 1)

CA(, 2)AA(, 2)Index:=3

while index


43/134

AM = ni=1 nj=1 [aff(Ai, Aj) aff(Ai, Aj-1)+aff(Ai, Aj) aff(Ai, Aj+1)]

= nj=1[ ni=1 aff(Ai, Aj) aff(Ai, Aj-1)+ ni=1 aff(Ai, Aj) aff(Ai,Aj+1)]

Ta nh ngha cu ni (Bond) gia hai thuc tnh Ax, v Ay l:

Bond(Ax, Ay )= nz=1aff(Az, Ax)aff(Az, Ay)

Th th c th vit li AM l:

AM = nj=1[ Bond(Ai, Aj-1)+Bond(Ai, Aj+1)]

By gi xt n thuc tnh sau:

A1 A2 Ai-1 AiAj Aj+1 An

Vi A1 A2 Ai-1 thuc nhm AM v AiAj Aj+1 An thuc nhmAM

Khi s o lc ht chung cho nhng thuc tnh ny c th vit li:

AMold = AM + AM+ bond(Ai-1, Ai) + bond(Ai, Aj) + bond(Aj, Ai)+

bond(bond(Aj+1, Aj) = nl=1[ bond(Al, Al-1)+bond(Ai, Al+1)] + nl=i+1[bond(Al, Al-1)+bond(Ai, Al+1)] + 2bond(Ai, Al))

By gi xt n vic t mt thuc tnh mi Ak gia cc thuc tnhAi v Aj trong ma trn lc ht t. S o lc ht chung mi c th c vittng t nh:

AMnew = AM + AM+ bond(Ai, Ak) + bond(Ak, Ai) + bond(Ak, Aj)+bond(Aj, Ak) = AM + AM+ 2bond(Ai, Ak) + 2bond(Ak, Aj)

V th ng gp thc (net contribution) cho s o i lc chung khit thuc tnh Ak gia Ai v Aj l:

Cont(Ai, Ak, Aj) = AMnew - AMold = 2Bond(Ai, Ak )+ 2Bond(Ak, Aj ) -2Bond(Ai, Aj )

Bond(A0, Ak)=0. Nu thuc tnh Ak t bn phi thuc tnh tn bnphi v cha c thuc tnh no c t ct k+1 ca ma trn CA nnbond(Ak, Ak+1)=0.

Th d 13: Ta xt ma trn c cho trong Th d 12 v tnh tonphn ng gp khi di chuyn thuc tnh A4 vo gia cc thuc tnh A1 v

A2, c cho bng cng thc:

Cont(A1, A4, A2)= 2bond(A1, A4)+ 2bond(A4, A2)-2bond(A1, A2)

43


44/134

Tnh mi s hng chng ta c:

Bond(A1, A4) = 4z=1aff(Az, A1)aff(Az, A4) = aff(A1,A1) aff(A1,A4)+aff(A2,A1) aff(A2,A4) + aff(A1,A3) aff(A3,A4) + aff(A1,A4) aff(A4,A4)

= 45*0 +0*75+ 45*3+0*78 = 135

Bond(A4, A2)= 11865

Bond(A1,A2) = 225

V th cont(A1, A4) = 2*135+2*11865+2*225 = 23550

Th d 14:

Chng ta hy xt qu trnh gom t cc thuc tnh ca quan h D

n v dng ma trn i lc thuc tnh AA.bc khi u chng ta chp cc ct 1 v 2 ca ma trn AA vo ma

trn CA v bt u thc hin t ct th ba. C 3 ni c th t c ct 3l: (3-1-2), (1, 3, 2) v (1, 2, 3). Chng ta hy tnh ng gp s i lcchung ca mi kh nng ny.

th t (0-3-1):

cont(A0, A3, A1) = 2bond(A0, A3)+ 2bond(A3, A1) - 2bond(A0, A1)

bond(A0, A3) = bond(A0, A1)=0

bond(A3, A1) = 45*48+5*0+53*45+3*0=4410

cont(A0, A3, A1) = 8820

th t (1-3-2)

cont (A1, A3, A2)=10150

th t (2-3-4)

cont (A2, A3, A4)=1780

Bi v ng gp ca th t (1-2-3) l ln nht, chng ta t A3 vobn phi ca A1. Tnh ton tng t cho A4 ch ra rng cn phi t n vobn phi ca A2. Cui cng cc hng c t chc vi cng th t nh ccct v cc hng c trnh by trong hnh sau:

44


45/134

A1 A2 A1 A3 A2A1 45 0 A1 45 45 0A2 0 80 A2 0 5 80

A3 45 5 A3 45 53 5

A4 0 75 A4 0 3 75

(a) (b)

A1 A3 A2 A4 A1 A3 A2 A4A1 45 45 0 0 A1 45 45 0 0A2 0 5 80 75 A3 45 53 5 3

A3 45 53 5 3 A2 0 5 80 75A4 0 3 75 78 A4 0 3 75 78

(b) (d)

trong hnh trn chng ta thy qu trnh to ra hai t: mt gc trntri cha cc gi tr i lc nh, cn t kia di gc phi cha cc gi tri lc cao. Qu trnh phn t ny ch ra cch thc tch cc thuc tnh caD n. Tuy nhin, ni chung th ranh ri cc phn tch khng hon ton r

rng. Khi ma trn CA ln, thng s c nhiu t hn c to ra v nhiuphn hoch c chn hn. Do vy cn phi tip cn bi ton mt cch ch thng hn.

Thut ton phn hoch

Mc ch ca hnh ng tch thuc tnh l tm ra cc tp thuctnh c truy xut cng nhau hoc hu nh l cc tp ng dng ring bit.Xt ma trn thuc tnh t:

A1 A2 A3 ... Ai Ai+1 ... An

A1 A1

:

Ai Ai+1 :

45

TA

B A


46/134

: An

Nu mt im nm trn ng cho c c nh, hai tp thuctnh ny c xc nh. Mt tp {A1, A2,..., Ai} nm ti gc trn tri v tp

th hai {Ai+1, Ai+2,..., An} nm ti gc bn phi v bn di im ny.Chng ta gi 2 tp ln lt l TA, BA. Tp ng dng Q={q 1, q2,...,qq} vnh ngha tp ng dng ch truy xut TA, ch truy xut BA hoc c hai,nhng tp ny c nh ngha nh sau:

AQ(qi) = {Aj |use(qi, Aj)=1}

TQ = {qi | AQ(qi) TA}

BQ = {qi | AQ(qi) BA}

OQ = Q - {TQ BQ}

y ny sinh bi ton ti u ho. Nu c n thuc tnh trong quanh th s c n-1 v tr kh hu c th l im phn chia trn ng chochnh ca ma trn thuc tnh t cho quan h . V tr tt nht phn chial v tr sinh ra tp TQ v BQ sao cho tng cc truy xut ch mt mnh lln nht cn tng truy xut c hai mnh l nh nht. V th chng ta nh

ngha cc phng trnh chi ph nh sau:CQ = refj(qi)accj(qi) qiQ Sj

CTQ = refj(qi)accj(qi) qiTQ Sj

CBQ= refj(qi)acc(qi) qiBQ Sj

COQ= refj(qi)acc(qi) qiOQ Sj

Mi phng trnh trn m tng s truy xut n cc thuc tnhbi cc ng dng trong cc lp tng ng ca chng. Da trn s liu ny,bi ton ti u ho c nh ngha l bi ton tm im x (1 x n) saocho biu thc:

Z=CTQ+CBQ-COQ2

ln nht. c trng quan trng ca biu thc ny l n nh nghahai mnh sao cho gi tr ca CTQ v CBQ cng gn bng nhau cng tt.

46


47/134

iu ny cho php cn bng ti trng x l khi cc mnh c phn tnn cc v tr khc nhau. Thut ton phn hoch c phc tp tuyn tnhtheo s thuc tnh ca quan h, ngha l O(n).

Thut ton PARTITION

Input: CA: ma trn i lc t; R: quan h; ref: ma trn s dng thuctnh;

acc: ma trn tn s truy xut;

Output: F: tp cc mnh;

Begin

{xc nh gi tr z cho ct th nht}

{cc ch mc trong phng trnh chi ph ch ra im tch}

tnh CTQn-1

tnh CBQn-1

tnh COQn-1

best CTQn-1*CBQn-1 (COQn-1)2

do {xc nh cch phn hochtt nht}

begin

for i from n-2 to 1 by -1 do

begin

tnh CTQi

tnh CBQitnh COQi

z CTQi*CBQi (COQi)2

if z > best then

begin

best z

ghi nhn im tch bn vo trong hnh ng xdch

47


48/134

end-if

end-for

gi SHIFT(CA)

end-beginuntil khng th thc hin SHIFT c na

Xy dng li ma trn theo v tr x dch

R1TA(R) K {K l tp thuc tnh kho chnhca R}

R2BA(R) K

F {R1, R2}End. {partition}

p dng cho ma trn CA t quan h d n, kt qu l nh ngha ccmnh Fd n={D n1, D n2}

Trong : D n1={A1, A3} v D n2= {A1, A2, A4}. V th

D n1={M d n, Ngn sch}

D n2={M d n, Tn d n, a im}( y M d n l thuc tnh kho ca D n)

Kim tra tnh ng n:

Tnh y : c bo m bng thut ton PARTITION v mithuc tnh ca quan h ton cc c a vo mt trong cc mnh.

Tnh ti thit c: i vi quan h R c phn mnh dc FR={R1,

R2,...., Rr} v cc thuc tnh kho KR= KRi , RiFR

Do vy nu iu kin mi Ri l y php ton ni s ti thit ling R. Mt im quan trng l mi mnh Ri phi cha cc thuc tnhkho ca R.

2.5. Phn mnh hn hp

Trong a s cc trng hp, phn mnh ngang hoc phn mnhdc n gin cho mt lc CSDL khng p ng cc yu cu t ngdng. Trong trng hp phn mnh dc c th thc hin sau mt s

48


49/134

mnh ngang hoc ngc li, sinh ra mt li phn hoch c cu trc cy.Bi v hai chin lc ny c p dng ln lt, chn la ny c gi lphn mnh hn hp.

2.6. Cp pht

2.6.1 Bi ton cp pht

Gi s c mt tp cc mnh F={F1, F2, ...,Fn} v mt mng baogm cc v tr S={S1, S2, ...,Sm} trn c mt tp cc ng dng Q={q1,q2, ...,qq} ang chy.

Bi ton cp pht l tm mt phn phi ti u ca F cho S.

Tnh ti u c th c nh ngha ng vi hai s o:

- Chi ph nh nht: Hm chi ph c chi lu mnh Fi vo v tr Sj, chiph vn tin mnh Fi vo v tr Sj, chi ph cp nht Fi ti tt c mi v tr ccha n v chi ph tryn d liu. V th bi ton cp pht c gng tm mtlc cp pht vi hm chi ph t hp nh nht.

- Hiu nng: Chin lc cp pht c thit k nhm duy tr mt hiu

qu ln l h thp thi gian p ng v tng ti a lu lng h thngti mi v tr.

Ni chung bi ton cp pht tng qut l mt bi ton phc tp v c phc tp l NP-y (NP-complete). V th cc nghin cu cdnh cho vic tm ra cc thut gii heuristec tt c li gii gn ti u.

2.6.2 Yu cu v thng tin

49

R23

R1

R2

R

HH

R11

R12

R21

R22

VV V V


50/134

giai on cp pht, chng ta cn cc thng tin nh lng vCSDL, v cc ng dng chy trn , v cu trc mng, kh nng x l vgii hn lu tr ca mi v tr trn mng.

Thng tin v CSDL

tuyn ca mt mnh Fj ng vi cu vn tin qi. y l s lngcc b ca Fj cn c truy xut x l qi. Gi tr ny k hiu l seli(Fj)

Kch thc ca mt mnh Fj c cho bi

Size (Fj) = card (Fj)* length(Fj)

Trong : Length(Fj) l chiu di (tnh theo byte) ca mt b trongmnh Fj.

Thng tin v ng dng

Hai s liu quan trng l s truy xut c do cu vn tin q i thc hintrn mnh Fj trong mi ln chy ca n (k hiu l RRij), v tng ng lcc truy xut cp nht (URij). Th d chng c th m s truy xut khi cnphi thc hin theo yu cu vn tin.

Chng ta nh ngha hai ma trn UM v RM vi cc phn t tngng uij v rij c c t tng ng nh sau:

1 nu vn tin qi c cp nht mnh Fj

uij=0 trong trng hp ngc li

1 nu vn tin qi c cp nht mnh Fjrij =

trong trng hp ngc liMt vct O gm cc gi tr o(i) cng c nh ngha, vi o(i) c

t v tr a ra cu vn tin qi .

Thng tin v v tr

Vi mi v tr (trm) chng ta cn bit v kh nng lu tr v x lca n. Hin nhin l nhng gi tr ny c th tnh c bng cc hmthch hp hoc bng phng php nh gi n gin.

50


51/134

+ Chi ph n v tnh lu d liu ti v tr Sk s c k hiu lUSCk.

+ c t s o chi ph LPCk, l chi ph x l mt n v cng vic tiv tr Sk. n v cng vic cn phi ging vi n v ca RR v UR.

Thng tin v mng

Chng ta gi s tn ti mt mng n gin, g ij biu th cho chi phtruyn mi b gia hai v tr Si v Sj. c th tnh c s lng thngbo, chng ta dng fsize lm kch thc (tnh theo byte) ca mt b dliu.

2.6.3. M hnh cp pht

M hnh cp pht c mc tiu lm gim thiu tng chi ph x l vlu tr d liu trong khi vn c gng p ng c cc i hi v thi gianp ng. M hnh ca chng ta c hnh thi nh sau:

Min (Total Cost)

ng vi rng buc thi gian p ng, rng buc lu tr, rng buc xl.

Bin quyt nh xij c nh ngha l

1 nu mnh Fi c lu ti v tr Sj

xij= 0 trong trng hp ngc li

Tng chi ph

Hm tng chi ph c hai thnh phn: phn x l vn tin v phn lutr. V th n c th c biu din l:

TOC= QPCi + STCjk

qi Q SkS FjF

vi QPCi l chi ph x l cu vn tin ng dng q i, v STCjk l chi phlu mnh Fj ti v tr Sk.

Chng ta hy xt chi ph lu tr trc. N c cho bi

STCjk = USCk * size(Fj) *xjk

Chi ph x l vn tin kh xc nh hn. Hu ht cc m hnh cho bi

ton cp pht tp tin FAP tch n thnh hai phn: Chi ph x l ch c vchi ph x l ch cp nht. y chng ti chn mt hng tip cn

51


52/134

khc trong m hnh cho bi ton DAP v xc nh n nh l chi ph x lvn tin bao gm chi ph x l l PC v chi ph truyn l TC. V th chi phx l vn tin QPC cho ng dng qi l

QPCi=PCi+TCi

Thnh phn x l PC gm c ba h s chi ph, chi ph truy xut AC,chi ph duy tr ton vn IE v chi ph iu khin ng thi CC:

PCi=ACi+IEi+CCi

M t chi tit cho mi h s chi ph ph thuc vo thut ton cdng hon tt cc tc v . Tuy nhin minh ho chng ti s m tchi tit v AC:

ACi= (uij*URij+rij*RRij)* xjk*LPCk SkS FjFHai s hng u trong cng thc trn tnh s truy xut ca vn tin q i

n mnh Fj. Ch rng (URij+RRij) l tng s cc truy xut c v cpnht. Chng ta gi thit rng cc chi ph x l chng l nh nhau. K hiutng cho bit tng s cc truy xut cho tt c mi mnh c qi thamchiu. Nhn vi LPCk cho ra chi ph ca truy xut ny ti v tr Sk. Chng

ta li dng xjk ch chn cc gi tr chi ph cho cc v tr c lu cc mnh.

Mt vn rt quan trng cn cp y. Hm chi ph truy xutgi s rng vic x l mt cu vn tin c bao gm c vic phn r nthnh mt tp cc vn tin con hot tc trn mt mnh c lu ti v tr, theo sau l truyn kt qu tr li v v tr a ra vn tin.

H s chi ph duy tr tnh ton vn c th c m t rt ging thnhphn x l ngoi tr chi ph x l cc b mt n v cn thay i nhm

phn nh chi ph thc s duy tr tnh ton vn.

Hm chi ph truyn c th c a ra ging nh cch ca hm chiph truy xut. Tuy nhin tng chi ph truyn d liu cho cp nht v choyu cu ch c s khc nhau hon ton. Trong cc vn tin cp nht, chngta cn cho tt c mi v tr bit ni c cc bn sao cn trong vn tin ch cth ch cn truy xut mt trong cc bn sao l . Ngoi ra vo lc kt thcyu cu cp nht th khng cn phi truyn d liu v ngc li, cho v tr

a ra vn tin ngoi mt thng bo xc nhn, cn trong vn tin ch c cth phi c nhiu thng bo tryn d liu.

52


53/134

Thnh phn cp nht ca hm truyn d liu l:

TCUi = uj*xjk*go(i),k + uj*xjk*g k,o(i)SkS FjF SkS FjF

S hng th nht gi thng bo cp nht t v tr gc o(i) ca qi

n tt c bn sao cp nht. S hng th hai dnh cho thng bo xc nhn.

Thnh phn chi ph ch c c th c t l:

TCRi= min (uij * xjk * go(i), k+rij * xjk * (seli(Fj)* length (Fj)/fsize) *gk, o(i))

Fj F SkSS hng th nht trong TCR biu th chi ph truyn yu cu ch c

n nhng v tr c bn sao ca mnh cn truy xut. S hng th hai truyn cc kt qu t nhng v tr ny n nhng v tr yu cu. Phngtrnh ny khng nh rng trong s cc v tr c bn sao ca cng mtmnh, ch v tr sinh ra tng chi ph truyn thp nht mi c chn thc hin thao tc ny.

By gi hm chi ph tnh cho vn tin qi c th c tnh l:

TCi=TCUi+TCRi

Rng bucRng buc thi gian p ng cn c c t l thi gian thc thi

ca qi thi gian p ng ln nht ca qiqiQ

Ngi ta thch c t s o chi ph ca hm theo thi gian bi v nn gin ho c t v rng buc thi gian thc thi.

Rng buc lu tr l: STCjk kh nng lu tr ti v tr Sk,SkS

FjF

Trong rng buc x l l:

ti trng x l ca q i ti v tr Sk kh nng x l ca Sk,SkS.

qi Q

53


54/134

CHNG 3. X L VN TIN

Ng cnh c chn y l php tnh quan h v i s quan h.

Nh chng ta thy cc quan h phn tn c ci t qua cc mnh.Thit k CSDL c vai tr ht sc quan trng i vi vic x l vn tin vnh ngha cc mnh c mc ch lm tng tnh cc b tham chiu, v ikhi tng kh nng thc hin song song i vi nhng cu vn tin quantrng nht. Vai tr ca th x l vn tin phn tn l nh x cu vn tin cpcao trn mt CSDL phn tn vo mt chui cc thao tc ca i s quan htrn cc mnh. Mt s chc nng quan trng biu trng cho nh x ny.Trc tin cu vn tin phi c phn r thnh mt chui cc php ton

quan h c gi l vn tin i s. Th hai, d liu cn truy xut phi ccc b ha cc thao tc trn cc quan h c chuyn thnh cc thao tctrn d liu cc b (cc mnh). Cui cng cu vn tin i s trn cc mnhphi c m rng bao gm cc thao tc truyn thng v c ti uha hm chi ph l thp nht. Hm chi ph mun ni n cc tnh tonnh thao tc xut nhp a, ti nguyn CPU, v mng truyn thng.

3.1. Bi ton x l vn tin

C hai phng php ti u ha c bn c s dng trong cc b xl vn tin: phng php bin i i s v chin lc c lng chi ph.

Phng php bin i i s n gin ha cc cu vn tin nh ccphp bin i i s nhm h thp chi ph tr li cu vn tin, c lp vid liu thc v cu trc vt l ca d liu.

Nhim v chnh ca th x l vn tin quan h l bin i cu vn tin

cp cao thnh mt cu vn tin tng ng cp thp hn c din tbng i s quan h. Cu vn tin cp thp thc s s ci t chin lcthc thi vn tin. Vic bin i ny phi t c c tnh ng n ln tnhhiu qu. Mt bin i c xem l ng n nu cu vn tin cp thp ccng ng ngha vi cu vn tin gc, ngha l c hai cng cho ra mt ktqu. Mt cu vn tin c th c nhiu cch bin i tng ng thnh is quan h. Bi v mi chin lc thc thi tng ng u s dng tinguyn my tnh rt khc nhau, kh khn chnh l chn ra c mt chin

lc h thp ti a vic tiu dng ti nguyn.

Th d 3.1:

54


55/134

Chng ta hy xt mt tp con ca lc CSDL c cho

NV( MNV, TnNV, Chc v)

PC (MNV, MDA, Nhim v, Thi gian)

V mt cu vn tin n gin sau:Cho bit tn ca cc nhn vin hin ang qun l mt d n

Biu thc vn tin bng php tnh quan h theo c php ca SQL l:

SELECT TnNV

FROM NV, PC

WHERE NV.MNV=PC.MNV

AND Nhimv=Qunl

Hai biu thc tng ng trong i s quan h do bin i chnhxc t cu vn tin trn l:

TnNV( Nhimv=Qunl NV.MNV=PC.MNV(NV x PC))

v

TnNV(NV|>


56/134

cc chin lc thc thi tng ln, lm cho vic x l vn tin phn tn tngln rt nhiu.

Th d 3.2:

Th d ny minh ha tm quan trng ca vic chn la v tr vcch truyn d liu ca mt cu vn tin i s. Chng ta xt cu vn tinca th d trn:

TnNV(NV|> E3(NV)

PC1= MNV E3(PC)

PC2= MNV E3(PC)

Cc mnh PC1, PC2, NV1, NV2 theo th t c lu ti cc v tr 1, 2,3 v 4 v kt qu c lu ti v tr 5

56

Hnh 4.1a) Chin lc a

Kt qu = NV1

NV2

NV1= NV|>


57/134

Mi tn t v tr i n v tr j c nhn R ch ra rng quan h R cchuyn t v tr i n v tr j. Chin lc A s dng s kin l cc quan hEMP v ASG c phn mnh theo cng mt cch thc hin song songcc php ton chn v ni. chin lc B tp trung tt c cc d liu ti vtr lu kt qu trc khi x l cu vn tin.

nh gi vic tiu dng ti nguyn ca hai chin lc ny,chng ta s dng mt m hnh chi ph n gin sau. Chng ta gi s rng

thao tc truy xut mt b (tuple access) c k hiu l tupacc, l mt nv v thao tc truyn mt b (tuple transfer) tuptrans l 10 n v. ngthi chng ta cng gi s l cc quan h NV v PC tng ng c 400 v1000 b, v c 20 gim c d n thng nht cho cc v tr. Cui cngchng ta gi s rng cc quan h PC v NV c gom t cc b tng ngtheo cc thuc tnh Nhimv v MNV. V vy c th truy xut trc tipn cc b ca PC da trn gi tr ca thuc tnh Nhimv (tng ng lMNV cho NV)

* Tng chi ph ca chin lc A c th c tnh nh sau:

1.To ra PC bng cch chn trn PC cn (10+10)* tupacc= 20

2. Truyn PC n v tr ca NV cn (10+10)*tuptrans = 200

3. To NV bng cch ni PC v NV cn (10+10)*tupacc*2= 40

4. Truyn NV n v tr nhn kt qu cn (10+10)*tuptrans = 200

Tng chi ph 460* Tng chi ph cho chin lc B c th c tnh nh sau:

1. Truyn NV n v tr 5 cn 400*tuptrans = 4.000

2. truyn PC n v tr 5 cn 1000*tuptrans =10.000

3. To ra PC bng cch chn trn PC cn 1000*tupacc= 1.000

4. Ni NVv PC cn 400*20*tupacc = 8.000

Tng chi ph l 23.000

57


58/134

Trong chin lc B, chng ta gi s rng cc phng php truy xutcc quan h NV v PC da trn cc thuc tnh Nhimv v MNV b mttc dng do vic truyn d liu. y l mt gi thit hp l trong thc t.Chin lc A tt hn vi h s 50 rt c ngha. Hn na n a ra mt

cch phn phi cng vic gia cc v tr. Khc bit cn cao hn na nuchng ta gi thit l tc truyn chm hn v/ hoc mc phn mnhcao hn.

Ch s nh gi tiu dng ti nguyn l tng chi ph (total cost)phi tr khi x l vn tin. Tng chi ph l tng thi gian cn x l ccphp ton vn tin ti cc v tr khc nhau v truyn d liu gia cc v tr.Mt cng c khc l thi gian p ng ca cu vn tin, l thi gian cn

thit chy cu vn tin. V cc php ton c th c thc hin song songti cc v tr khc nhau, thi gian p ng c th nh hn nhiu so vi tngchi ph ca n. Trong mi trng CSDL phn tn, tng chi ph cn phigim thiu l chi ph CPU, chi ph xut nhp v chi ph truyn. Chi phCPU l chi ph phi tr khi thc hin cc thao tc trn d liu trong b nhchnh. Chi ph xut nhp (I/O) l thi gian cn thit cho cc thao tc xutnhp a. Chi ph truyn tin l thi gian cn trao i d liu gia cc vr tham gia vo trong qu trnh thc thi cu vn tin. Chi ph ny phi tr khiphi x l cc thng bo (nh dng/ gii nh dng) v khi truyn d liutrn mng. Chi ph truyn c l l yu t quan trng nht c xt ntrong CSDL phn tn.

phc tp ca cc php ton quan h: cc php ton chn, Chiu(Khng loi b trng lp) c phc tp l O(n); Cc php chiu(C loib trng lp), trng lp, ni, ni na, chia c phc tp l O(n*logn); TchDescartes c phc tp l O(n2) ( N biu th lc lng ca quan h nu

cc b thu c c lp vi nhau)

nh gi:

+ Cc thao tc c tnh chn la lm gim i lc lng cn phi thchin trc tin.

+ Cc php ton cn phi c sp xp trnh thc hin tchDescartes hoc li thc hin sau.

58


59/134

3.2. Phn r vn tin

Phn r vn tin l giai on u tin ca qu trnh x l cu vntin. N bin i cu vn tin dng php tnh quan h thnh cu vn tin is quan h. Cc vn tin nhp xut u tham chiu cc quan h ton cc v

khng dng n cc thng tin phn b d liu. V th phn r vn tin uging nhau trong c h thng tp trung ln phn tn, cu vn tin xut sng v ng ngha v t cht lng theo ngha l loi b cc hnhng khng cn thit. Phn r vn tin c th xem nh bn bc lin tipnhau: Chun ho, phn tch, loi b d tha, vit li cu vn tin

Chun ho

Mc ch ca chun ho (normalization) l bin i cu vn tinthnh mt dng chun x l tip. Chun ho mt vn tin ni chung gmc t cc lng t v lng t ho vn tin bng cch p dng u tinca cc ton t logic.

Vi cc ngn ng quan h nh SQL, bin i quan trng nht llng t ho vn tin (mnh Where), c th l mt v t phi lng tvi phc tp no vi tt c cc lng t cn thit ( hoc ) ct pha trc. C hai dng chun c th cho v t, mt c th bc cao cho

AND() v loi cn li cho th bc cao OR (). Dng chun hi l hi (vt ) ca cc tuyn v t (cc v t ):

(p11p12.p1n) ..(pm1pm2.pmn)

trong pij l mt v t n gin. Ngc li, mt lng t ho dng chun tuyn nh sau:

(p11p12.p1n) .(pm1pm2.pmn)

Bin i cc v t phi lng t l tm thng bng cch s ccquy tc tng ng cho cc php ton logic (, , ): 9

1. p1p2p2p1

2. p1p2p2p1

3. p1( p2p3) (p1p2 )p3

4. p1( p2p3) (p1p2 )p3

5. p1( p2p3) (p1p2 )(p1p3 )6. p1( p2p3) (p1p2 ) (p1p3 )

59


60/134

7. (p1p2 )p1p2

8. (p1 p2 )p1p2

9. (p)p

Trong dng chun tc tuyn, cu vn tin c th c x l nh cccu vn tin con hi c lp, c ni bng php hp (tng ng vi cctuyn mnh ).

Nhn xt: Dng chun tuyn t c dng v dn n cc v t ni vchn trng nhau. Dng chun hi hay dng trong thc t

Th d 3.3:

Tm tn cc nhn vin ang lm vic d n P1 trong 12 thng hoc

24 thng.

Cu vn tin c din t bng SQL nh sau:

SELECT TnNV

FROM NV, PC

WHERE NV.MNV=PC.MNV

AND PC.MDA= P1

AND Thi gian=12 OR Thi gian=24

Lng t ho dng chun hi l:

NV.MNV=PC.MNV PC.MDA= P1 (Thi gian=12 Thigian=24)

Cn lng t ho dng chun tuyn l

(NV.MNV=PC.MNV PC.MDA=P1 Thi gian=12)

(NV.MNV=PC.MNV PC.MDA=P1 Thi gian=24)

dng sau, x l hai hi c lp c th l mt cng vic tha nucc biu thc con chung khng c loi b.

Phn tch

Phn tch cu vn tin cho php ph b cc cu vn tin chunho nhng khng th tip tc x l c hoc khng cn thit, nhng l do

chnh l do chng sai kiu hoc sai ng ngha.

60


61/134


62/134

p1: PC.MaNV = NV.MaNV p2: PC.MaDA=DA.MaDA p3: TnDA=CAD/CAM p4: CV =TP p5: tgian >36

th vn tin th ni

Th d 3.5:

Select TnNV

From PC, NV,DA

Where PC.MaNV = NV.MaNVand TnDA=CAD/CAM and CV =TP and tgian >36

=> th khng lin thng

Nhn xt: Cu vn tin sai ng ngha nu th vn tin ca n khnglin thng: 1 hoc nhiu th con b tch ri vi th kt qu.

Loi b d tha

Mt cu vn tin ca ngi s dng thng c din t trn mtkhung nhn c th c b sung thm nhiu v t c c s tng ng

khung nhn - quan h, bo m c tnh ton vn ng ngha v bo mt.Th nhng lng t ho vn tin c sa i ny c th cha cc v t

62

PC

NV

DA

KQ

P1 P2

P3

P5

P4

P

C

NV

DA

PC

NV

DA

KQ

P1

P3

P5

P4

PC

NV

DA


63/134

d tha, c th phi khin lp li mt s cng vic. Mt cch lm n ginvn tin l loi b cc v t tha

- Loi b v t d tha bng qui tc lu ng: 10

1. p p p 6. p v True True2. p v p p 7. p p False

3. p Truep 8. p v p True

4. p v Falsep 9. p1 (p1v p2) p1

5. p FalseFalse 10. p1 v (p1 p2) p1

Th d 3.6 :

Select CV

From NV

Where (Not (CV =TP) and (CV=TP or CV=PP) and not(CV=PP)) or TnNV=Mai

p1: CV =TP Lng t ho:

p2: CV=PP ( p1 (p1 v p2) p2 ) v p3

p3: TnNV=Mai

p dng : ( p1 ((p1 p2 ) v (p2 p2 ))) v p3

p dng 3: ( p1 p1 p2 ) v ( p1 p2 p2 ) v p3

p dng 7: (False p2) v ( p1 False) v p3

p dng 5: False v False v p3 = p3

Vit li: Select CV

From NV

where TnNV=Mai

Vit li cu vn tin

Bc ny c chia thnh hai bc nh:

(1)Bin i cu vn tin t php tnh quan h thnh i s quan h

(2) Cu trc li cu vn tin i s nhm ci thin hiu nng.

63


64/134

cho d hiu, chng ta s trnh by cu vn tin i s quan hmt cch hnh nh bng cy ton t. Mt cy ton t l mt cy vi mint l biu th cho mt quan h c lu trong CSDL v cc nt khngphi l nt l biu th cho mt quan h trung gian c sinh ra bi cc

php ton quan h. Chui cc php ton i theo hng t l n gc biuth cho kt qu vn tin.

Bin i cu vn tin php tnh quan h b thnh mt cy ton tc th thu c d dng bng cch sau. Trong SQL, cc nt l c sn trongmnh FROM. th hai nt gc c to ra nh mt php chiu cha ccthuc tnh kt qu. Cc thuc tnh ny nm trong mnh SELECT cacu vn tin SQL. Th ba, lng t ho (mnh Where ca SQL) c

dch thnh chui cc php ton quan h thch hp (php chn, ni, hp, ..)i t cc nt l n nt gc. Chui ny c th c cho trc tip qua th txut hin ca cc v t v ton t.

Th d 3.7:

Cu vn tin: tm tn cc nhn vin tr J.Doe lm cho d nCAD/CAM trong mt hoc hai nm.

Biu thc SQL l:

SELECT TnNV

FROM DA, PC, NV

WHERE PC.MNV=NV.MNV

AND PC.MDA=DA.MDA

AND TnNV J.Doe

AND DA.TnDA=CAD/CAMAND (Thi gian=12 OR Thi gian=24)

Cc th c nh x thnh cy trong hnh di.

64


65/134

Bng cch p dng cc quy tc bin i, nhiu cy c th c thyrng tng ng vi cy c to ra bng phng php c m t trn. Su quy tc tng ng hu ch nht v c xem l cc php ton

i s quan h c bn :R, S, T l nhng quan h, trong R c nh ngha trn cc thuc

tnh A={A1, A2,,An} v quan h S c nh ngha trn cc thuc tnhB={B1, B2,,Bn}.

1. Tnh giao hon ca php ton hai ngi

R x S S x R

R SS RQuy tc ny cng p dng c cho hp nhng khng p dng cho

hiu tp hp hay ni na.

2. Tnh kt hp ca cc php ton hai ngi

(R x S)x TR x (Sx T)

(R S) TR (S T)

3- Tnh ly ng ca cc php ton n ngi

65

TnNV

Thi gian=12 Thi gian=24

TnDA =CAD/CAM

TnNV J.Doe

MDA

ENO

PC NVDA

Chiu

Chn

Ni


66/134

Nu R c nh ngha trn tp thuc tnh A v A A, A A vA A th A A( A (R)) A(R)

p1(A1)( p2(A2)(R)) p1(A1)p2(A2)(R)

trong pi l mt v t c p dng cho thuc tnh Ai

4. Giao hon php chn vi php chiu

A1An( p(Ap)(R)) A1An( p(Ap)( A1An,Ap(R)))

Ch rng nu Ap l phn t ca {A1, A2,,An} th php chiu cuicng trn {A1, A2,,An} v phi ca h thc khng c tc dng.

5. Giao hon php chn vi php ton hai ngi

p(Ai)(R x S) ( p(Ai)(R)) x S p(Ai)(R p(j, Bk) S) ( p(Ai)(R)) p(j, Bk) S

p(Ai)(R T) p(Ai)(R) p(Ai)(T)

6-Giao hon php chiu vi php ton hai ngi

Nu C=A B, trong AA, B B, v A, B l cc tp thuctnh tng ng ca quan h R v S, chng ta c

C(R x S) A(R) B(S) C(R p(i, Bj) S) A(R) p(i, Bj) B(S)

C(R S) A(R) B(S)

Cc quy tc trn c th c s dng cu trc li cy mt cchc h thng nhm loi b cc cy xu. Mt thut ton ti cu trc ngin s dng heuristic trong c p dng cc php ton n ngi (chn/chiu ) cng sm cng tt nhm gim bt kch thc ca quan h trung

gian.

Ti cu trc cy trong hnh trn sinh ra cy trong hnh sau. Kt quc xem l t cht lng theo ngha l n trnh truy xut nhiu ln ncng mt quan h v cc php ton chn la nhiu nht c thc hintrc tin.

66


67/134

3.3. Cc b ha d liu phn tn

Tng cc b ha d liu chu trch nhim dch cu vn tin i strn quan h ton cc sang cu vn tin i s trn cc mnh vt l. Cc bha c s dng cc thng tin c lu trong mt lc phn mnh.

Tng ny xc nh xem nhng mnh no cn cho cu vn tin v bini cu vn tin phn tn thnh cu vn tin trn cc mnh. To ra cu vntin theo mnh c thc hin qua hai bc. Trc tin vn tin phn tnc nh x thnh vn tin theo mnh bng cch thay i mi quan h phn

tn bng chng trnh ti thit ca n. Th hai vn tin theo mnh c ngin ho v ti cu trc to ra mt cu vn tin c cht lng. Qu

67

TenDA J.Doe

MDA, MNV

MNV, TenNV

MDA

TenDA=CAD/CAM

Thoigian=12 Thoigian=24

DA PC NV

TnNV

MDA, TenNV

MNV


68/134

trnh n gin ho v ti cu trc c th c thc hin theo nhng quy tcc s dng trong tng phn r. Ging nh trong tng , cu vn tintheo mnh cui cng ni chung cha t n ti u bi v thng tin linquan n cc mnh cha c s dng.

- Cc b ho d liu s xc nh cc mnh no cn cho cu vn tin.Bin i cu vn tin phn tn thnh cc cu vn tin theo mnh.

- Trong phn ny i vi mi kiu phn mnh ta s trnh by cc kthut rt gn to cc cu vn tin ti u v n gin hn.

- Ta s s dng cc qui tn bin i v cc khm ph, chng hn ycc php ton n ngi xung thp nh c th.

Rt gn phn mnh ngang nguyn thu- Vic phn mnh ngang phn tn 1 quan h da trn cc v t chn

Th d 3.8:

NV (MaNV, TenNV, CV)

NV1 = MaNV E3(NV)

NV2 = E3 < MaNV E6(NV) NV = NV1 NV2 NV3

NV3 = MaNV > E6(NV)

Cch lm:

+ Xc nh sau khi ti cu trc li cy con, xem cy no to ra ccquan h rng th loi b chng i.

+ Phn mnh ngang c th c dng n gin ho php chn vphp ni

Rt gn vi php chn: Chn trn cc mnh c lng t mu thun vilng t ho ca qui tc phn mnh s sinh ra quan h rng ta loi bchng.

rj = nu t thuc r : ( t(pi) t(pj) )

Trong pi, pj l cc v t chn, t biu th cho 1 b, t(p) biu th v t png vi t.

V d: NV1 = MaNV E3(NV)

NV2 = E3 < MaNV E6(NV)

68


69/134

NV3 = MaNV > E6(NV)

Select MaNV

From NV

Where MaNV=E5

Rt gn vi php ni

- Ni trn cc quan h phn mnh ngang c th c n gin khi cc

quan h ni c phn mnh theo thuc tnh ni.- n gin ho gm c phn phi cc ni trn cc hp ri b i cc ni

v dng.

( r1 r2 ) |> E6 (NV)PC1 = MaNV E3 (PC)

69

Bng cch hon v php chn vi php hp tas pht hin ra v t chn >< v t ca NV1,

NV3. => to ra cc quan h rng => loi b

MaNV

MaNV=E5

NV1 NV2 NV3

MaNV

MaNV=E5

NV2


70/134

PC (MaNV, MaDA, NV, Tg) PC2 = MaNV > E3 (NV)Cu hi:

Select *From NV, PC

Where NV.MaNV = PC.MaNV

Rt gn cho phn mnh dc

Phn mnh dc phn tn mt quan h da trn cc thuc tnh chiu.Chng trnh cc b ho cho mt quan h phn mnh dc gm c ni cacc mnh theo thuc tnh chung.

Th d 3.10:

NV(MaNV, TnNV, CV)

NV1 = MaNV, TnNV(NV)

NV2 = MaNV, CV(NV)

Chng trnh cc b ho: NV = NV1 |>


71/134

NV(MaNV, TnNV, CV)

NV1 = MaNV, TnNV(NV)

NV2 = MaNV, CV(NV)

Select TnNVFrom NV

Rt gn cho phn mnh ngang dn xut

- Phn mnh ngang dn xut l mt cch phn phi hai quan hm nh c th ci thin kh nng x l cc im giao nhau gia php

chn v php ni.- Nu quan h r phi phn mnh dn xut theo quan h s, cc mnh

ca r v s ging nhau thuc tnh ni s nm cng v tr. Ngoi ra s c thc phn mnh theo v t chn.

- Phn mnh dn xut ch c s dng cho mi lin h 1 N (phncp) t s n r: trong 1 b ca s c th khp vi nhiu b ca r

Th d 3.12:

NV(MaNV, TnNV, CV) NV1= Cv=TP(NV)

NV2 = Cv TP(NV)

PC(MaNV, MaDA, NV, Tg) PC1 = PC |>< NV1

PC2 = PC |>< NV2

a ra tt c cc thuc tnh ca NV, PC vi NV =PP

Select *From NV, PC

71

TnNV

|>


72/134

Where NV.MaNV = PC.MaNV and NV.CV=PP

Cu vn tin trn cc mnh NV1, NV2, PC1, PC2 c nh ngha.y php chn xung cc mnh NV1. NV2 cu vn tin rt gn li do muthun vi v t chn ca NV1 = > loi b NV1

72

*

|>


73/134

Rt gn cho phn mnh ngang hn hp

Mc tiu: H tr hiu qu cc cu vn tin c cha php chiu, chn vni

Cu vn tin trn cc mnh hn hp c th c rt gn bng cch t hpcc qui tc tng ng uc dng trong cc phn mnh ngang nguynthu, phn mnh dc, phn mnh ngang dn xut.

Qui tc:

1/ Loi b cc quan h rng c to ra bi cc php ton chn mu thuntrn cc mnh ngang.

2/ Loi b cc quan h v dng c to ra bi cc php chiu trn cc mnh

dc3/ Phn phi cc ni cho cc hp nm c lp v loi b cc ni v dng.

Th d 3.13: NV1 = MaNV E4 ( MaNV, TnNV(NV) )

MaNV, TnNV, CV) NV2 = MaNV > E4 ( MaNV, TnNV (NV) )

NV3 = MaNV, CV (NV)

Chng trnh cc b ho NV = (NV1 NV2 ) |>


74/134

3.4. Ti u ho vn tin phn tn

Trong phn ny chng ta s gii thiu v qu trnh ti u ha nichung, bt k mi trng l phn tn hay tp chung. Vn tin cn ti u githit l c din t bng i s quan h trn cc quan h CSDL (c th lcc mnh) sau khi vit li vn tin t biu thc php tnh quan h.

Ti u ha vn tin mun ni n qu trnh sinh ra mt hoch nhthc thi vn tin (query execution plan, QEP) biu th cho chin lc thcthi vn tin. Hoch nh c chn phi h thp ti a hm chi ph. Th tiu ha vn tin, l mt n th phn mm chu trch nhim thc hin ti uha, thng c xem l cu to bi ba thnh phn: mt khng gian tmkim (search space), mt m hnh chi ph(cost model) v mt chin lctm kim (search strstegy) (xem hnh 1.4.4). Khng gian tm kim l tp cchoch nh thc thi biu din cho cu vn tin. Nhng hoch nh ny ltng ng, theo ngha l chng sinh ra cng mt kt qu nhng khcnhau th t thc hin cc thao tc v cch thc ci t nhng thao tcny, v th khc nhau v hiu nng. Khng gian tm kim thu c bngcch p dng cc quy tc bin i, chng hn nhng qui tc cho i squan h m t trong phn vit li cu vn tin. M hnh chi ph tin onchi ph ca mt hoch nh thc thi cho. cho chnh xc, m hnh chiph phi c thng tin cn thit v mi trng thc thi phn tn. Chin

lc tm kim s khm ph khng gian tm kim v chn ra hoch nh ttnht da theo m hnh chi ph. N nh ngha xem cc hoch nh no cnc kim tra v theo th t no. Chi tit v mi trng (tp trung hayphn tn) c ghi nhn trong khng gian v m hnh chi ph.

3.4.1. Khng gian tm kimCc hoch nh thc thi vn tin thng c tru tng ha qua cy

ton t), trn nh ngha th t thc hin cc php ton. Chng ta bsung thm cc thng tin nh thut ton tt nht c chn cho mi phpton. i vi mt cu vn tin cho, khng gian tm kim c th cnh ngha nh mt tp cc cy ton t tng ng, c c bng cch p

74


75/134

dng cc qui tc bin i . nu bt cc c trng ca th ti u ha vntin , chng ta thng tp trung cc cy ni (join tree), l cy ton t vi ccphp ton ni hoc tch Descartes. L do l cc hon v th t ni cc tcdng quan trng nht n hiu nng ca cc vn tin quan h.

75


76/134

CU VN TIN

TO RA KHNGGIAN TM KIM

Th d 3.14:

Xt cu vn tin sau:

SELECT ENAME

FROM EMP, ASG, PROJ

WHERE EMP, ENO=ASG.ENO

76

CU VN TIN

QEP TNG NG

QEP TT NHT

TO RA KHNGGIAN TM KIM QUY TC BIN I

CHIN LCTM KIM

M HNH CHIPH


77/134

AND ASG, PNO=PROJ . PNO

Hnh sau minh ha ba cy ni tng ng cho vn tin , thu c

bng cch s dng tnh cht kt hp ca cc ton t hai ngi. Mi cy nyc th c gn mt chi ph da trn chi ph ca mi ton t. Cy ni ( c )

bt u vi mt tch Des-cartes c th c chi ph cao hn rt nhiu so vi

cy cn li.

PNO ENO

ENO PROJ PNO

EMP

EMP ASG ASG PROJ

(a) (b)

ENO.PNO

X ASG

PROJ EMP

(c)

Vi mt cu vn tin phc tp (c gm nhiu quan h v nhiu ton t),

s caaytoans t tng ng c th rt nhiu. Th d s cy ni c th thu

c t vic p dng tnh giao hon v kt hp l O(N!) cho N quan h.

Vic nh gi mt khng gian tm kim ln c th mt qu nhiu thi gianti u ha, i khi cn tn hn c thi gian thc thi thc s. V th, th ti

77


78/134

u ha thng hn ch kch thc cn xem xt ca khng gian tm kim .

Hn ch th nht l dng cc heuristic. Mt heuristic thng dng nht l

thc hin php chn v chiu khi truy xut n quan h c s. Mt

heuristic thng dng khc l trnh ly cc tch Descartes khng c chnhcu vn tin yu cu. Th d trong hnh trn cy ton t (c ) khng phi l

phn c th ti u ha xem xt trong khng gian tm kim.

a) Cy ni tuyn tnh b) Cy ni xum xuMt hn ch quan trng khc ng vi hnh dng ca cy ni. Hai loi

cy ni thng c phn bit Cy ni tuyn tnh v cy ni xum xu

(xem Hnh 9.3). Mt cy tuyn tnh (linear tree) l cy vi mi nt ton t

c t nht mt ton hng l mt quan h c s. Mt cy xum xu (bushy

tree) th tng qut hn v c th c cc ton t khng c quan h c s lm

ton hng (ngha l c hai ton hng u l cc quan h trung gian). Nu

ch xt cc cy tuyn tnh, kch thc ca khng gian tm kim c rt

gn li thnh O(2N

). Tuy nhin trong mi trng phn tn, cy xum xu rtc li cho vic thc hin song song.

3.4.2. Chin lc tm kim

Chin lc tm kim hay c cc th ti u ha vn tin s dng nht

l quy hoch ng (dynamic programming) ci tnh cht n nh

(deterministic). Cc chin lc n nh tin hnh bng cch xy dng cc

hoch nh , bt u t cc quan h c s, ni thm nhiu quan h ti mibc cho n khi thu c tt c mi hoch nh kh hu nh trong Hnh

78

R1

R2

R3

R4

R1 R2 R3 R4


79/134

9.4.. Quy hoch ng xy dng tt c mi hoch nh kh hu theo hng

ngang (breadth-first) trc khi n chn ra hoch nh tt nht. h

thp chi ph ti u ha, cc hoch nh tng phn rt c kh nng khng

dn n mt hoch nh ti u u c xn b ngay khi c th. Ngcli, mt chin lc n nh khc l thut ton thin cn ch xy dng mt

hoch nh theo hng su (depth-first).

Bc 1 Bc 2 Bc3

Quy hoch ng hu nh c bn cht vt cn v bo m tm ra ccc hoch nh. N phi tr mt chi ph c th chp nhn c (theo thi

gian v khng gian) khi s quan h trong cu vn tin kh nh. Tuy nhin

li tip cn ny c chi ph qu cao khi s quan h ln hn 5 hoc 6. V l

do ny m cc ch gn y ang tp trung vo cc chin lc ngu nhin

ha (randomized strategy) lm gim phc tp ca ti u ha nhng

khng bo m tm c hoch nh tt nht. Khng ging nh cc chin

lc n nh, cc chin lc ngu nhin ha cho php th ti u ha nh

i thi gian ti u ha v thi gian thc thi.

Chin lc ngu nhin ha chng hn nh tp trung vo vic tm kim

li gii ti u xung quanh mt s im c bit no . Chung khng m

bo s thu c mt li gii tt nht nhng trnh c chi ph qu cao ca

ti u ha tnh theo vic tiu dng b nh v thi gian. Trc tin mt

79

R1

R2R

1R

2R

1

R2

R3R4R4

R3


80/134

hoc nhiu hoch nh khi u c xy dng bng mt chin lc thin

cn . Sau thut ton tm cch ci thin hoch nh ny bng cch thm

cc ln cn (neighbor) ca n. Mt ln cn thu c bng cch p dng

mt bin i ngu nhin cho mt hoch nh. Th d v mt bin i inhnh gm c hon i hai quan h ton hng c chn ngu nhin ca

hoch nh nh trong chng t bng thc nghim rng cc chin lc

ngu nhin ha c hiu nng tt hn cc chin lc n nh khi vn tin c

cha kh nhiu quan h.

3.4.3. M hnh chi ph phn tn

M hnh chi ph ca th ti u ha gm c cc hm chi ph d onchi ph ca cc ton t, s liu thng k, d liu c s v cc cng thc c lng kch thc cc kt qu trung gian.

Hm chi ph

Chi ph ca mt chin lc thc thi phn tn c th c din t ngvi tng thi gian hoc vi thi gian p ng. Tng thi gian (total time)l tng tt c cc thnh phn thi gian (cn c gi l chi ph), cn thi

gian p ng( response time) l thi gian tnh t khi khi hot n lchon thnh cu vn tin. Cng thc tng qut xc nh tng chi ph cm t nh sau:

Total_time = TCPU * #insts + TI/O * #I/Os + TMSG * #msgs + TTR *#bytcs

Hai thnh phn u tin l thi gian x l cc b, trong TCPU l thigian ca mt ch th CPU v TI/O l thi gian cho mt thao tc xut nhpa. Thi gian truyn c biu th qua hai thnh phn cui cng. TMSG lthi gian c nh cn khi hot v nhn mt thng bo, cn TTR l thigian cn truyn mt n v d liu t v tr ny n v tr khc. n v

80

R1

R2

R1

R3

R3 R2


81/134

d liu y tnh theo byte (#byte l tng kch thc ca tt c cc thngbo), nhng cng c th tnh theo nhng n v khc (th d theo gi).Thng thng chng ta gi thit TTR l mt gi tr khng i. iu ny cth khng ng trong cc mng WAN, trong mt s v tr nm xa hnso vi mt s khc. Tuy nhin gi thit ny lm n gin qu trnh ti u

ha rt nhiu. V th thi gian truyn #byte d liu t v tr ny n v trkhc c gi thuyt l mt hm tuyn tnh theo #bytes:

CT(#bytes) = TMSG + TTR * #bytes

Cc chi ph ni chung c din t theo n v thi gian, v t cth chuyn thnh cc n v khc (th d nh la).

Gi tr tng i ca cc h s chi ph c trng cho mi trng

CSDL phn tn. Topo mng c nh hng rt ln n t s gia cc thnhphn ny. Trong mng WAN nh Internet, thi gian truyn thng l h schim a phn. Tuy nhin trong cc mng LAN th cc h s thnh phncn bng hn. Nhng nghin cu ban u ch ra rng t s gia thigian truyn v thi gian xut nhp mt trang vo khong 20:1 i vimng WAN, i vi cc mng Ethernet in hnh (10Mbds) th vokhong 1:1,6. V th phn ln cc h DBMS phn tn c thit k trncc mng WAN u b qua chi ph x l cc b v tp trung vo vn cc tiu ha chi ph truyn. Ngc li cc DBMS phn tn c thit k

cho mng LAN u xt n c ba thnh phn chi ph ny. Cc mng nhanhhn c mng WAN ln mng LAN ci thin cc t l nu trn thin vchi ph truyn khi tt c mi th khc u nh nhau. Tuy nhin thi giantruyn vn l mt yu t chin a phn trong cc mng WAN nh Internetbi v d liu cn phi c di chuyn i n cc v tr xa hn.

Khi thi gian p ng vn tin l hm mc tiu ca th ti u ha,chng ta cn phi xt n vn x l cc b song song v truyn songsong. Cng thc tng qut ca thi gian p ng l:

Response_time = TCPU * seq_ #insts + TI/O * seg_ #I/Os

+ TMSG * seg_ #msgs + TTR * seg_ #bytes

Trong seq_ #x, vi x c th l cc ch th (insts), cc xut nhp I/O,cc thng bo (msgs) hoc bytes, l s lng x ti a phi c thc hinmt cch tun t khi thc hin vn tin. V vy mi x l v truyn d liu thchin song song u c b qua.

Th d 3.15:

81


82/134

Chng ta minh ha s khc bit gia tng chi ph v thi gian p ngqua th d trong Hnh 6, trong kt qu tr li c tnh ti v tr 3, dliu c ly t v tr 1 v 2. n gin, chng ta phi gi s rng ch xtn chi ph truyn.

82


83/134


84/134

cc kt qu. D nhin l c nhng c mt gia tnh chnh xc ca cc sliu thng k v chi ph qun l chng: s liu c

Documents

Co So Du Lieu 2 (Phan Tan Va Suy Dien)