1. C s d liuM HNH D LIU QUAN HNguyn Thanh Tng1 C s d liu 12/04/2011

2. Ni dung Cc khi nim trong M hnh d liu quan h Ph thuc hm2 C s d liu 12/04/2011 3. Cu trc d liu quan h Cu trc d liu quan h (Relational data structure)dng lu tr cc quan h. H qun tr CSDL quan h cho php ngi s dngnhn thy CSDL di dng cc bng d liu. p dng mc ngoi (external level) v mc nim (conceptual level) trong kin trc CSDL ba lc . C s d liu quan h (relational database) l CSDL m d liu c lu tr trong cc quan h (bng). CSDL quan h bao gm nhiu quan h.3 C s d liu 12/04/2011 4. Quan h Quan h (relation) l mt bng d liu hai chiu bao gm nhiu hng (mu tin) v nhiu ct (thuc tnh hoc vng tin). Mi hng l duy nht: khng th c hai hng c cng ccgi tr tt c vng tin. Th t ca cc hng l khng quan trng. Th t ca cc ct l khng quan trng. Khng phi mi bng u l quan h. Quan h l mt bngkhng cha cc hng ging ht nhau.4 C s d liu 12/04/2011 5. Quan h (tt) V d: quan h Supplier Snum Name CityS1 Nguyn Trung Tin SFS2 Trn Th YnLAS3 Nguyn Vn An SF5 C s d liu 12/04/2011 6. Cc khi nim Thuc tnh (attribute) ca quan h l mt ct c t tn ca mt bng. Th t xut hin ca cc thuc tnh trong quan h l khng quan trng v quan h vn khng b thay i. Min tr (domain) l mt tp hp cc gi tr cho php ca mt hoc nhiu thuc tnh. Min tr l mt tp con ca m kiu d liu (data type) Da vo tp cc gi tr ca mt min tr m chng ta c th bit c ngha ca thuc tnh tng ng.6 C s d liu12/04/2011 7. Cc khi nim (tt) B (tuple) ca mt quan h l mt hng (row) ca mt bng. Cc phn t ca mt quan h l cc b hoc cc hng ca mt bng.V d: Mt b t ca quan h Employee l:t = (100, Margaret Simpson, Marketing, 48000) Thnh phn (component) ca mt b l gi tr ca mt hng. V d: Thnh phn Name ca b t l Margaret Simpson.7 C s d liu 12/04/2011 8. Cc khi nim (tt) Bc (degree) ca mt quan h l s lng cc thuc tnh ca quan h. Quan h nht phn (Unary relation): quan h c mtthuc tnh Quan h nh phn (Binary relation): quan h c hai thuctnh Quan h tam phn (Ternary relation): quan h c bathuc tnh Quan h n-phn (n-ary relation): quan h c n thuc tnhV d: Quan h Employee c 4 thuc tnh, do bc caquan h ny l 4.8 C s d liu12/04/2011 9. Cc khi nim (tt) Lng s (cardinality) ca mt quan h l s lng cc b ca quan h. Lng s ca quan h s b thay i khi thm vo hocxa b cc b ca quan h. Lng s ca quan h c xc nh t th hin caquan h ti mt thi im. V d: Quan h Employee c 5 b, do lng s caquan h ny l 5.9 C s d liu 12/04/2011 10. Lc quan h Lc quan h (Relation schema) c biu din bi mt tn i theo sau l mt tp hu hn cc thuc tnh {A1, A2,, An} c ghi trong hai du ngoc n.R(A1, A2,, An) R l tn ca lc quan h. Bc ca lc quan h l s lng cc thuc tnh ca lc quan h. Quan h r trn lc quan h R (k hiu r(R)) l mt tp hu hn cc nh x {t1, t2,, tp} t R vo D vi iu kin mi nh x t r th t[Ai] Di, 1 i n10 C s d liu12/04/2011 11. Lc c s d liu quan h Lc c s d liu quan h (Relational database schema) l mt tp hp cc lc quan h trong CSDL. Lc ny t b thay i theo thi gian. Th hin CSDL quan h (Relational database schema) l mt tp hp tt c cc th hin quan h trong CSDL. Th hin CSDL thng xuyn b thay i theo thi gian.11 C s d liu12/04/2011 12. Kha quan h Kha quan h (Key) l mt tp nh nht cc thuc tnh dng xc nh duy nht mt hng. Mt kha ch c mt thuc tnh c gi l kha n (simple key). Mt kha c nhiu thuc tnh c gi l kha phc hp (composite key). Kha thng c s dng lm ch mc (index) ca bng d liu lm tng tc x l cu truy vn.12 C s d liu12/04/2011 13. Kha quan h (tt) Mt quan h phi c t nht mt kha v c th c nhiu kha. Cc thuc tnh thuc mt kha c gi l thuc tnh kha (prime attribute), cc thuc tnh cn li trong lc quan h c gi l cc thuc tnh khng kha (nonprime attribute). Cc thuc tnh kha c gch di. Cc thuc tnh kha khng c c gi tr rng (null value).13 C s d liu 12/04/2011 14. Kha quan h (tt) Tt c cc kha ca mt quan h c gi l kha d tuyn (candidate key). Mt trong cc kha d tuyn c chn lm kha tiu biu, kha ny c gi l kha chnh (primary key). Mt quan h ch c mt kha chnh v c th c nhiu kha d tuyn. Trong mt quan h, mt hoc nhiu thuc tnh c gi l kha ngoi (foreign key) nu chng l kha chnh ca mt quan h khc.14 C s d liu 12/04/2011 15. Ni dung Cc khi nim trong M hnh d liu quan h Ph thuc hm15 C s d liu 12/04/2011 16. Ph thuc hm Functional Dependency Lc quan h R(U), r l mt quan h bt k trn lc quan h R, X v Y l 2 tp thuc tnh con ca U. Ph thuc hm XY trn R c gi l X xc nh hm Y hoc Y ph thuc hm vo X nu: t1, t2 r(R): t1[X]=t2[X] => t1[Y]=t2[Y] X/ Y c gi l X khng xc nh hm Y hay Y khng ph thuc hm vo X X c gi l v tri ca f, k hiu l left(f). Y c gi l v phi ca f, k hiu l right(f).16 C s d liu 12/04/2011 17. Ph thuc hm (tt) Ph thuc hm m t mi lin h, rng buc gia cc thuc tnh, l c im ng ngha ca cc thuc tnh trong lc quan h. c s dng trong qu trnh chun ha. Da vo cc ph thuc hm ny chng ta c th thit k li cc lc CSDL loi b s d tha d liu tn ti trong CSDL.17 C s d liu 12/04/2011 18. Ph thuc hm (tt) Ph thuc hm X lun lun tha mn trong mi quan h. Ph thuc hm Y tha mn trong cc quan h m mi b phn ca quan h ny u c cng gi tr Y. Ph thuc hm XY l mt ph thuc hm trn lc quan h R(U) nu X v Y u l tp con ca U F l tp ph thuc hm trn lc quan h R nu mi ph thuc hm trong F l ph thuc hm trn R.18 C s d liu12/04/2011 19. Ph thuc hm (tt) Ph thuc hm tm thng (Trivial functionaldependency) Ph thuc hm XY l PTH tm thng nu YX. Cn c gi l ph thuc hm hin nhin. nh thuc (determinant) L thuc tnh xc nh ca mt ph thuc hm. L tp thuc tnh bn tri ca ph thuc hm Trong ph thuc hm XY th X l nh thuc.19 C s d liu12/04/2011 20. H tin Armstrong Ph thuc hm XY c suy din lun l t F, k hiu l F|=XY, nu mi quan h tha mn tt c cc ph thuc hm trong F th cng tha mn XY. F|=XY cn c gi l F bao hm XY hay XY c suy din theo quan h t F. Qui tc suy din (inference rule) l qui tt nu mt quan h tha mn mt s ph thuc hm no th quan h ny cng tha mn mt s ph thuc hm khc. Cn c gi l Lut suy din hay Tin suy din(inference axiom)20 C s d liu12/04/2011 21. H tin Armstrong(tt) Cho lc quan h R(X,Y,Z,W) H tin Armstrong gm cc tin suy din sau: F1: Phn x (reflexivity): YX => XY F2: Gia tng (augmentation): XY => XZYZ F3: Bc cu (transitivity): XY v YZ => XZ T , ta c th suy ra cc qui tc suy din sau: F4: Hp (additivity): XY v XZ => XYZ F5: Chiu (projectivity): XYZ => XZ F6: Bc cu gi (pseudotransitivity):XY v YZW => XZW21 C s d liu 12/04/2011 22. Bao ng ca tp ph thuc hm Bao ng (closure) ca mt tp ph thuc hm F (k hiu F+) l mt tp ph thuc hm nh nht cha F sao cho khng th p dng h tin Armstrong trn tp ny to ra mt ph thuc hm khng c trong tp hp ny. F+ l tp hu hn. F+ = (F+)+. F+ ph thuc vo lc quan h R. Nu R(A,C) th F+ khng cha BB. Tp ph thuc hm F suy din XY nu XY thucF+.22 C s d liu 12/04/2011 23. Bao ng ca tp ph thuc hm (tt) V d: Cho F={ABC, CB} l mt tp ph thuc hmtrn r(ABC).F+ = {AA, ABA, ACA, ABCA,BB, ABB, BCB, ABCB,CC, ACC, BCC, ABCC,ABAB, ABCABC, ACAC, ABCAC,BCBC, ABCBC, ABCABC,ABC, ABAC, ABBC, ABABC,CB, CBC, ACB, ACAB,}23 C s d liu12/04/2011 24. Bao ng ca tp thuc tnh Bao ng ca tp thuc tnh X da trn mt tp phthuc hm F (closure of X under F), k hiu l X+F, lmt tp ph thuc hm Y sao cho: XY F+ XZF+ : Z Y Vi tp thuc tnh X bt k th X X+F v XX F+24 C s d liu 12/04/2011 25. Bao ng ca tp thuc tnh (tt) Gii thut tm bao ng ca tp thuc tnh: Procedure Closure( X, F, Closure_X ) { Closure_X = X; repeatOld_X = Closure_X;for mi WZ trong F do if W Closure_X thenClosure_X = Closure_X Z; until Closure_X = Old_X; }25 C s d liu 12/04/2011 26. Bao ng ca tp thuc tnh (tt) V d: Cho lc quan h R(A,B,C,D,E,F) v tp phthuc hm F={f1:DB, f2:AC, f3:ADE, f4:CF} Tm {A}+F: Trc tin {A}+F = {A}. Duyt ln th nht tp ph thuc hm F T f2: {A}+F = {AC} T f4: {A}+F = {ACF} Duyt ln hai {A}+F khng thay i. Vy {A}+F = {ACF} Tm {AD}+F: Trc tin {AD}+F = {AD}. Duyt ln thnht tp ph thuc hm F T f1: {AD}+F = {ADB}. T f2: {AD}+F = {ADBC} T f3: {AD}+F = {ADBCE}. T f4: {AD}+F = {ADBCEF} = U Vy {AD}+F = U26 C s d liu12/04/2011 27. Kim tra thnh vin trong F+ xc nh tp ph thuc hm F bao hm ph thuc XY (tc F|=XY) chng ta ch cn kim traXY F+. Tm F+ khng d dng (v n ln hn nhiu so vi F) Chng ta kim tra XY F+ m khng cn tm F+. Kim tra Y X+F, nu ng th F|=XY.27 C s d liu 12/04/2011 28. Kim tra thnh vin trong F+ (tt) V d: Cho tp ph thuc hm F = {DB, AC, ADE, CB} Kim tra F c bao hm AB hay khng Tm A+F = ??? A+F = {ACB} Do BA+F nn F bao hm AB.28 C s d liu12/04/2011 29. Thut ton tm kha quan h N=U- right(f); F Nu N+F = U th K = {N} Nu khng th D=right(f) - FF left(f); L = U N+FD; K= ; Vi mi Li L: Nu {N Li}+F = U th K = K {NLi}; Vi mi Ki, Kj K Nu Ki Kj th K = K {Kj};29 C s d liu12/04/2011 30. Thut ton tm kha quan h (tt) V d 1: Cho lc quan h R(A,B,C,D,E,F) v tp ph thuc hm F={DB, AC, ADE, CF}. Tm tt c cc kha ca R? N=U- F right(f) = {ABCDEF}-{BCEF} = {AD} N+F = {AD}+F = {ADBCEF} = U Vy lc quan h R ch c mt kha l {AD}30 C s d liu 12/04/2011 31. Thut ton tm kha quan h (tt) V d 2: Cho lc quan h R(A,B,C,D,E,F) v tp ph thuc hm F={AD, CAF, ABEC}. N=U-F right(f) = {ABCDEF}-{DAFEC} = {B} N+F = {B}+F = {B} U D= F right(f) -F left(f)={DAFEC}-{ACB}={DFE} L = U N+FD = {ABCDEF}-{BDFE} = {AC} Cc tp con ca L l {A}, {C} v {AC} {BA}+F = {BADECF} = U. Vy {BA} l kha ca R. Loi b tp cha ca {A}, l {AC} {BC}+F = {BCAFED} = U. Vy {BC} l kha ca R. Loi b tp cha ca {C}, l {AC} Lc quan h R c 2 kha l {BA} v {BC}.31 C s d liu 12/04/2011 32. Ph ti thiu ca tp ph thuc hm B1: Tch cc ph thuc hm sao cho v phi ch cn mt thuc tnh. B2: B cc thuc tnh d tha v tri. B3: Loi khi F cc ph thuc hm d tha.32 C s d liu12/04/2011 33. Ph ti thiu ca tp ph thuc hm V d: Cho lc quan h R(A,B,C,D) v tp ph thuc hm F={ABCD, BC, CD}. Tm ph ti thiu? B1: Tch cc ph thuc hm sao cho v phi ch c 1 thuc tnh. F = {ABC, ABD, BC, CD}33 C s d liu12/04/2011 34. Ph ti thiu ca tp ph thuc hm V d: Cho lc quan h R(A,B,C,D) v tp ph thuc hm F={ABCD, BC, CD}. Tm ph ti thiu? B2: B cc thuc tnh d tha v tri BC, CD khng xt v v tri ch c mt thuc tnh. Xt ABC Nu b A th B+=BCD khng cha A nn ko b A. Nu b B th A +=A khng cha B nn ko b B. Xt ABD Nu b A th B+=BCD khng cha A nn ko b A. Nu b D th A+=A khng cha D nn ko b D.34 C s d liu 12/04/2011 35. Ph ti thiu ca tp ph thuc hm V d: Cho lc quan h R(A,B,C,D) v tp ph thuc hm F={ABCD, BC, CD}. Tm ph ti thiu? B3: Loi khi F cc ph thuc hm d tha. Xt ABC: Tnh AB+ = ABCD = U nn loi b ABC Xt ABD: Tnh AB+ = ABCD = U nn loi b ABD Xt BC: Tnh B+ = B khng th b Xt CD: Tnh C+ = C khng th b Ph ti thiu l {BC, CD}35 C s d liu12/04/2011 36. Q&A36