Chng 5: M hnh d liu quan h - L thuyt thit k Phn 3: Cc bt thng trong quan h v ph thuc hm

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Mc chNm c cc vn v khi nimCc bt thng trong mt quan h (Anomalies)Ph thuc hm (Functional Dependencies)S tch lc quan h (Decomposition)*/26

Cc ni dung chnhGii thiu v cc bt thng trong mt quan hCc ph thuc hmCc nguyn nhn gy ra cc bt thngPhp tch lc QH*/26

1. Cc bt thng trong quan hHy quan st bng Student

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IDNameClassDepartmentSubjectMark1E1-001Nguyen Van AE1ElectronicsElectronic Circuit82E1-001Nguyen Van AE1ElectronicsDigital Technique73E1-002Tran Thi BE1ElectronicsDigital Technique94E1-002Tran Thi BE1ElectronicsElectronic Circuit85E2-001Nguyen Ho CE2ElectronicsDigital Technique66IT1-001Tran Thi BIT1ITElectronic Circuit107IT1-002Le Van DIT1ITDigital Technique8

1. Cc bt thng trong quan hT bng trn c th nhn thy mt s bt thng:

D tha (Redundancy): gi tr ca mt s thnh phn trong cc b b lp li khng cn thit nh : Name, Class, DepartmentBt thng khi cp nht (Update Anomalie): xut hin khi cp nht gi tr cho mt b hin c, v thc t thc hin ca thao tc c v phc tp hn rt nhiu so vi logic ca thao tc . Bt thng khi b sung (Insertion Anomalie): xut hin khi b sung thm mt b mi. Bt thng khi xa (Deletion Anomalie): xut hin khi xa mt b hin c trong quan h.

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1. Cc bt thng trong quan hHu qu ca cc bt thng trn

S khng nht qun d liu c nguy c rt caoTn ch lu tr do d thaCc thao tc c bn trn CSDL khng hiu qu

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1. Cc bt thng trong quan hVy nguyn nhn ca cc bt thng ny l g?

Khi nim Ph thuc hm c th gip chng ta hiu nguyn nhn, cng nh gip tm ra gii php cho vn trn.*/26

2. Ph thuc hmnh ngha ngha ca PTHH tin AmstrongMt s loi PTH c bitKha v thuc tnh kha*/26

2. Ph thuc hmnh ngha v PTH (Functional Dependency)Cho lc quan h R(A1,A2,An), v A = A1A2 An, v 2 tp cc thuc tnh X v Y A. Chng ta ni rng:X Y (X xc nh hm Y, hay Y ph thuc hm vo X), nu th hin r ca R, th khng tn ti 2 b t1, t2, sao cho t1[X]=t2[X] v t1[Y] t2[Y]X c gi l Quyt nh (determinant) ca PTH.

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2. Ph thuc hm v dMt s PTH t bng Student: ID Name; ID Class; Class Department; ID,Subject Mark*/26

2. Ph thuc hm ngha ngha ca mt PTH X Y trong lc R: th hin r ca R th:Hoc khng c bt k 2 b t1, t2 no, sao cho t1[X]=t2[X] Hoc nu tn ti 2 b t1, t2 m t1[X]=t2[X], th t1[Y]=t2[Y]Trong c hai trng hp, b t, vi mi gi tr ca t[X] th ch c mt gi tr ca t[Y], do nu bit trc gi tr t[X], th c th xc nh gi tr t[Y]. */26

2. Ph thuc hm ngha ngha ca PTH X Y:Th th PTH n t u?N xut pht cc quy tc nghip v (hay logic nghip v) ca c s d liu. Cc quy tc nghip v ny s quy nh cc ph thuc gia cc thuc tnh, v t s to ra cc ph thuc hm. */26

2. Ph thuc hm nghaV d: mt s quy tc nghip v trong lc Student:r1: Mi sinh vin c mt ID duy nht. r2: Mi sinh vin ch c duy nht 1 tn v ch thuc v mt lp. T r1 v r2, ta c cc PTH sau:ID Name; ID Classr3: Vi mi mn hc, mi sinh vin ch thi mt lnT r3, ta c PTH: (ID, Subject) Mark;

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2. Ph thuc hmH tin AmstrongTnh phn x: Y X th X Y (PTH tm thng)Tnh tng trng: X, Y, Z, nu X Y th XZ YZ (XZ = X Z)Tnh bc cu: X, Y, Z, nu X Y v Y Z th X ZCc h qu:Tnh hp : nu X Y v X Z th X YZTnh tch: nu X Y v Z Y th X ZTnh ta bc cu: nu X Y v YW Z th XW Z */26

2. Ph thuc hmMt s loi PTH c bitPTH b phn v PTH y : X Y c gi l PTH b phn nu X X sao cho X Y. Tri li, nu khng tn ti X nh trn th PTH ny l y .

PTH bc cu v PTH trc tip: X Y c gi l PTH bc cu (hay gin tip) nu Z XY sao cho X Z v Z Y. Tri li, nu khng tn ti Z nh trn th PTH ny l trc tip.

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2. Ph thuc hmKha v thuc tnh khaKha: Cho 1 lc R. Mt hay mt tp thuc tnh K ca R c gi l Kha ca R nu n tha mn 2 iu kin:K RK R l mt PTH y Siu kha (SuperKey): K c gi l siu kha ca R nu n cha t nht mt kha ca R. Thuc tnh kha (Prime attribute): 1 thuc tnh A l thuc tnh kha nu n thuc mt kha bt k. Tri li, th n l thuc tnh khng kha (hay thuc tnh m t)*/26

2. Ph thuc hmKha v thuc tnh khaMt s thuc tnh ca KhaMt lc quan h lun c t nht mt kha, v thng l c nhiu kha. Gi tr ca mi b trn kha/siu kha l duy nht. l v sao n c gi l Kha. */26

3. Nguyn nhn gy ra cc bt thng*/26

IDNameClassDepartmentSubjectMark1E1-001Nguyen Van AE1ElectronicsElectronic Circuit82E1-001Nguyen Van AE1ElectronicsDigital Technique73E1-002Tran Thi BE1ElectronicsDigital Technique94E1-002Tran Thi BE1ElectronicsElectronic Circuit85E2-001Nguyen Ho CE2ElectronicsDigital Technique66IT1-001Tran Thi BIT1ITElectronic Circuit107IT1-002Le Van DIT1ITDigital Technique8

3. Nguyn nhn gy ra cc bt thngKha ca quan h Student:K = (ID, Subject)Nhn xt: cc thuc tnh d tha l cc thuc tnh ph thuc hm b phn hoc bc cu vo Kha, v nh: (ID, Subject) Name; PTH b phn v ID Name(ID, Subject) Department; PTH bc cu v (ID, Subject) Class v Class Department */26

3. Nguyn nhn gy ra cc bt thngL s tn ti ca cc ph thuc b phn hoc bc cu vo kha

Gii php: Cn loi b cc loi PTH ny trong quan h, nhng lm th no? Php tch lc !

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4. Php tchnh ngha: cho mt LQH R. Mt php R l s thay th n bng cc L con (sub-schema) R1, R2, , Rn sao cho R = R1R2 Rn. K hiu php tch l: (R) = (R1,R2, ..., Rn) ( called rho)

Mt s tnh cht mun c ca php tch:Tch ni khng mt thng tin (Loseless-join decomposition) Bo ton cc PTH (Preservation of FDs)

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4. Php tchTch ni khng mt thng tin: gi s cho = (R1,R2, ..., Rn) l 1 php tch R. c gi l tch ni khng mt thng tin nu n tha mn: th hin r ca R th biu thc sau lun tha mn: : r = R1(r) R2(r) ... Rn(r)Bo ton cc PTH: cho L R vi tp cc PTH F. Php tch = (R1,R2, ..., Rn) c gi l bo ton PTH nu n tha mn:F = R1(F) R2(F) Rn(F)Vi Ri(F) = {X Y | X Y and XY Ri}*/26

V d: tch bng Student thnh 3 bng S1, S2 v S3

S1S2S3*/26

IDNameClassE1-001Nguyen Van AE1E1-002Tran Thi BE1E2-001Nguyen Ho CE2IT1-001Tran Thi BIT1IT1-002Le Van DIT1

ClassDepartmentE1ElectronicsE2ElectronicsIT1IT

IDSubjectMarkE1-001Electronic Circuit8E1-001Digital Technique7E1-002Digital Technique9E1-002Electronic Circuit8E2-001Digital Technique6IT1-001Electronic Circuit10IT1-002Digital Technique8

4. Php tchMt php tch tt cn phi::Loi b tt c cc bt thngLu gi c cc tnh cht mong mun

Lm th no t c? Qu trnh chun ha v cc dng chun (Normalization and Normal Forms)*/26

Tm ttCc bt thng trong lc quan hPh thuc hm v nguyn nhn ca cc bt thngPhp tch v cc tnh cht ca n*/26

Thank you!*/26

*Anomaly: Bt thngFunctional Dependency: Ph thuc hm*