Ngan Hang Cau Hoi Csdl - Ptit

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  • 7/31/2019 Ngan Hang Cau Hoi Csdl - Ptit

    1/4

    December 13, 2011 W.K

    N.D.P 1

    HC VIN CNG NGH BU CHNH VIN THNG KHOA:CNG NGH THNG TIN I

    NGN HNG CU HI THI T LUN

    Tn hc phn:C s d liu M hc phn: CNTT1202

    Ngnho to : ........Cng nghthng tin..............................Trnh o to: ...i hc.....

    1. Ngn hng cu hi thi Cu hi loi 1im

    Cu hi 1.1: Trnh by vai tr ca m hnh d liu bc cao trong qu trnh thit kCSDLCu hi 1.2: Lit k cc tr ng h p khc nhau m vic sdng gi tr NULL l thch h pCu hi 1.3: Gii thch skhc nhau gia mt tp thc thv 1 thhin ca tp thc th Cu hi 1.4: Gii thch skhc nhau gia thuc tnh v tp gi tr Cu hi 1.5: Vai tr tham gia l g? Khi no cn phi dng cc tn vai tr khi m tcc kiu miquan h?Cu hi 1.6: Khi no chng ta xem cc mi quan hnh l thuc tnh, cc tp gi tr ca cc thuctnh ny l ci g?Cu hi 1.7 Tp mi quan h quy c ngh a l g? Hy cho vi v dvtp mi quan h quy.Cu hi 1.8: Mt tp mi quan hxcnh ca mt tp thc thyu c thc bc l n h n 2khng? Cho v v.Cu hi 1.9:nh ngh a superclass, subclass, mi quan hsuperclass/subclass, mi quan hIS-A,c bit ha, tng qut ha.Cu hi 1.10: Trnh by c chktha thuc tnh, ch l iCu hi 1.11: Trnh by l p con c xcnh b i ng i sdng v b i mnh , ch ra skhcbit gia chngCu hi 1.12 : Trnh by 2 loi rng buc chnh ca tng qut ha vc bit haCu hi 1.13 : Phn cp l p con v dn cc l p con l g?Cu hi 1.14 :c bit ha v tng qut ha khc nhau nh thno? Ti sao khng thhin s khc nhau trnhnh l c Cu hi 1.15 : M hnh EER gii quyt vn tch h p nh thnoCu hi 1.16 : Ti sao cc btrong quan hkhng c sp xpCu hi 1.17 : Ti sao khng cho php cc btrng nhau trong mt quan h Cu hi 1.18 : Kho v siu kho khc nhau nh thnoCu hi 1.19 : Trnh by cc nguyn nhn dn n sxut hin cc gi tr null trong quan h Cu hi 1.20 : Trnh by ngngh a thuc tnh nh l mt cch thc phi hnh thc nh gi chtl ng ca mt l c quan h Cu hi 1.21 : Trnh by nhng bt th ng bsung, loi b, sa i. Ti sao chng c xem lkhng ttCu hi 1.22 : Ti sao nhiu gi tr NULL trong quan h c xem l khng tt

    Mu 2

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  • 7/31/2019 Ngan Hang Cau Hoi Csdl - Ptit

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    December 13, 2011 W.K

    N.D.P 2

    Cu hi 1.23 : Ph thuc hm l g? Ai l ng i c tcc ph thuc hm trn cc thuc tnh camt l c quan h Cu hi 1.24 :nh ngh a dng chun Boyce-Codd. Dng chun Boyce-Codd khc dng chun 3nh thnoCu hi 1.25 : Bt u tmt l c quan hR={A1,A2,,An} v tp cc ph thuc hm F trn

    R. Cc thut ton phn r ca R phi m bo nhngiu gCu hi 1.26 : Ni khng mt mt thng tin l g. Ti sao li quan trng

    Cu hi loi 2imCu hi 2.1:nh ngh a cc thut ngsau: thc th, tp thc th, thuc tnh, gi tr thuc tnh,thuc tnh ghp, thuc tnha tr, thuc tnh dn xut, thuc tnh kha, tp gi tr (domain)Cu hi 2.2: Tp mi quan hl g? Gii thch skhc nhau gia mt thhin ca tp mi quan h v tp mi quan h Cu hi 2.3: M t2 cchc tcc rng buc cu trc trn cc kiu quan h. Cch no cuim h n?Cu hi 2.4: V i nhngiu kin no mt thuc tnh ca tp mi quan hnh phn c thtr thnhmt thuc tnh ca mt trong cc tp thc ththam gia?Cu hi 2.5: Khi no khi nim thc thyu c ch trong qu trnh m hnh ha d liu? nhngh a cc thut ng tp thc ths hu (owner), tp thc thyu, tp mi quan hxcnh(identifying relationship set), v kha bphnCu hi 2.6: Trnh by cc quy c thhin mt l c ER nh l mt hnh ERCu hi 2.7: Trnh by nhngiu kin mt tp mi quan hbc 3 c thbiu din b i mt s tp mi quan hnh phnCu hi 2.8:nh ngh a cc khi nim sau: min gi tr (domain), thuc tnh, n-b(n-tuple), l cquan h, thhin quan h, bc ca quan h, l c CSDL, thhin CSDL quan h

    Cu hi 2.9: Trnh by cc rng buc ton vn thc th, rng buc ton vn tham chiu, khongoi. Ti sao chng quan trngCu hi 2.10: Trnh by s t ngng gicc cu trc ca m hnh ER v cc cu trc ca mhnh quan h. Mi cu trc ca m hnh ER c nh xnh thno vo m hnh quan h Cu hi 2.11:Trnh by vn vcc blv ngn chn n nh thnoCu hi 2.12: Cho F l tp cc ph thuc hm trn l c quan hR, X v Y l cc tp con ca R.nh ngh a F suy dn ra ph thuc hm XY (k hiu F X Y) vnh ngh a baong caF (F+). Pht biu tinAmstrong, gii thch tnhng vy ca tin Cu hi 2.13:nh ngh a baong ca tp thuc tnh v trnh by thut ton tnh baong.Cu hi 2.14:nh ngh a 2 tp ph thuc hm t ng ng, kim tra s t ng ng nh thnoCu hi 2.15: Tp ti thiu cc ph thuc hm l g. Ph ti thiu l g. Mi tp cc ph thuc hmc ph ti thiu hay khngCu hi 2.16 :Tnh cht bo ton ph thuc hm ca phn r l g. Ti sao li quan trngCu hi 2.17 : Trnh by khi nim vhqun tr c s d liu v hc s d liu. Nu tn mt s hqun tr c s d liu c mt trn th tr ng phn mm hin nay.

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    December 13, 2011 W.K

    N.D.P 3

    Cu hi loi 3imCu hi 3.1:nh ngh a dng chun 1, 2, 3 khi ch xem xt kho gc. nh ngh a dng chun 2

    v 3 tng qut (khi xem xt tt ccc kho). Mi quan hgia cc dng chun

    Cu hi 3.2 : Trnh by thut ton kim tra phn r bo ton ph thuc hmCu hi 3.3: Trnh by thut ton kim tra phn r c tnh cht ni khng mt mt thng tinCu hi 3.4 : Trnh by thut ton phn r ca l c thnh cc l c dng chun 3 v tho

    mn c2 tnh cht: bo ton ph thuc hm v ni khng mt mt thng tinCu hi 3.5 : Trnh by thut ton phn r ca l c thnh cc l c dng chun Boyce-Coddv thomn tnh cht ni khng mt mt thng tin. Vn phn r vdng chun Boyce-Codd vbo ton cc ph thuc hm th nh thnoCu hi 3.6 : a) Trnh by cc khnng ca mt hqun tr c s d liu th ng mi.

    b) Cc thnh phn chnh ca mt hqun tr c s d liu.

    Cu hi 3.7 : a) Trnh by khi nim vhqun tr c s d liu v hc s d liu. Nu tnmt shqun tr c s d liu c mt trn th tr ng phn mm hin nay.b) Cc thnh phn chnh ca mt hqun tr c s d liu.

    Cu hi 3.8 : a) Trnh by ba mc tru t ng ha d liu.b) Cc thnh phn chnh ca mt hqun tr c s d liu.

    Cu hi 3.9 : Trnh by c thcc b c xy dng mt hthng c s d liu vc im catng b cCu hi 3.10 : Trnh by c php tt ccc cu lnh ca ngn ng nh ngh a d liuCu hi 3.11 : Trnh by c php tt ccc cu lnh ca ngn ng thao tc d liu khng truy vnCu hi 3.12 : Trnh by c php ca cu lnh truy vn d liu tng qut va v dminh hacho tng tr ng h pCu hi 3.13 : Trnh by v tr c thc ca cc cu truy vn con, v dminh haCu hi 3.14 : Trnh bynh ngh a v v dcho cc php ton c bn v m rng ca i squanh

    Cu hi loi 4imKhng c Cu hi loi 5imKhng c

    Ghi ch : K hiu (m) cu hi c quy nh X.Y Trong : + X tng ng s im cu hi (X chy t 1 n5).

    + Y l cu hi th Y (Y chy t 1 tr i)

    2. xut cc phng n t hp cu hi thi thnh cc thi (Nu thy cn thit ) :...................

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    December 13, 2011 W.K

    N.D.P 4

    Mi thi phn ra lm hai phn: phn I cho cu hi l thuyt v phn hai cho cu hi vbi tp(mang tnh thc tin nhiu h n). Phn I schn cc cu hi c lit k trn sao cho tng s im l 5 (c th1 cu 3im + 2 cu 1im, hoc 2 cu 2im + 1 cu 1im, hoc 1 cu 3im + 1 cu 2im)

    3. Hng dn cn thit khc: Khng c

    Ngn hng cu h i thi ny c thng qua b mn v nhm cn b ging d y hc ph n.

    H Ni , ngy 07 thng 12. n m 2010 . .Tr ng khoa Tr ng bmn Ging vin ch trbin son

    Nguyn Qunh ChiNguyn Minh Hi Nguyn Minh Hi

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