Roadmap of This Lecture

Transcription

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2 Reltionl Model

3 Rodmp of This Lecture Structure of Reltionl Dtbses Fundmentl Reltionl-Algebr-Opertions Additionl Reltionl-Algebr-Opertions Extended Reltionl-Algebr-Opertions Null Vlues Modifiction of the Dtbse 3

4 Exmple of Reltion 4

5 Bsic Structure Formlly, given sets D, D,. D n, reltion r is subset of D x D x x D n Thus, reltion is set of n-tuples (,,, n ) where ech i D i Exmple: If customer_nme = {Jones, Smith, Curry, Lindsy, } /* Set of ll customer nmes */ customer_street = {Min, North, Prk, } /* set of ll street nmes*/ customer_city = {Hrrison, Rye, Pittsfield, } /* set of ll city nmes */ Then r = { (Jones, Min, Hrrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsy, Prk, Pittsfield) } is reltion over customer_nme x customer_street x customer_city 5

6 Attribute Types Ech ttribute of reltion hs nme The set of llowed vlues for ech ttribute is clled the domin of the ttribute Attribute vlues re (normlly) required to be tomic; tht is, indivisible E.g. the vlue of n ttribute cn be n ccount number, but cnnot be set of ccount numbers Domin is sid to be tomic if ll its members re tomic The specil vlue null is member of every domin The null vlue cuses complictions in the definition of mny opertions We shll ignore the effect of null vlues in our min presenttion nd consider their effect lter 6

7 Reltion Schem A, A,, A n re ttributes R = (A, A,, A n ) is reltion schem Exmple: Customer_schem = (customer_nme, customer_street, customer_city) r(r) denotes reltion r on the reltion schem R Exmple: customer (Customer_schem) 7

8 Reltion Instnce The current vlues (reltion instnce) of reltion re specified by tble An element t of r is tuple, represented by row in tble customer_nme customer_street customer_city ttributes (or columns) Jones Smith Curry Lindsy Min North North Prk Hrrison Rye Rye Pittsfield tuples (or rows) customer 8

9 Reltions re Unordered Order of tuples is irrelevnt (tuples my be stored in n rbitrry order) Exmple: ccount reltion with unordered tuples 9

10 Dtbse A dtbse consists of multiple reltions Informtion bout n enterprise is broken up into prts, with ech reltion storing one prt of the informtion ccount : stores informtion bout ccounts depositor : stores informtion bout which customer owns which ccount customer : stores informtion bout customers Storing ll informtion s single reltion such s bnk(ccount_number, blnce, customer_nme,..) results in repetition of informtion e.g.,if two customers own n ccount (Wht gets repeted?) the need for null vlues e.g., to represent customer without n ccount Normliztion theory (lter) dels with how to design reltionl schems 0

11 The customer Reltion

12 The depositor Reltion

13 Keys Let K R K is superkey of R if vlues for K re sufficient to identify unique tuple of ech possible reltion r(r) by possible r we men reltion r tht could exist in the enterprise we re modeling. Exmple: {customer_nme, customer_street} nd {customer_nme} re both superkeys of Customer, if no two customers cn possibly hve the sme nme In rel life, n ttribute such s customer_id would be used insted of customer_nme to uniquely identify customers, but we omit it to keep our exmples smll, nd insted ssume customer nmes re unique. 3

14 Keys (Cont.) K is cndidte key if K is miniml Exmple: {customer_nme} is cndidte key for Customer, since it is superkey nd no subset of it is superkey. Primry key: cndidte key chosen s the principl mens of identifying tuples within reltion Should choose n ttribute whose vlue never, or very rrely, chnges. E.g. emil ddress is unique, but my chnge 4

15 Foreign Keys A reltion schem my hve set of ttributes tht corresponds to the primry key of nother reltion. These ttributes is clled foreign key. E.g. customer_nme nd ccount_number ttributes of depositor re foreign keys to customer nd ccount respectively. Only vlues occurring in the primry key ttribute of the referenced reltion my occur in the foreign key ttribute of the referencing reltion. Schem digrm 5

16 Query Lnguges Lnguge in which user requests informtion from the dtbse. Ctegories of lnguges Procedurl Non-procedurl, or declrtive Pure lnguges: Reltionl lgebr Tuple reltionl clculus Domin reltionl clculus Pure lnguges form underlying bsis of query lnguges tht people use. Pure lnguges use immutble vribles only! They re functionl. 6

17 Reltionl Algebr Procedurl lnguge Six bsic opertors select: project: union: set difference: Crtesin product: x renme: The opertors tke one or two reltions s inputs nd produce new reltion s result. 7

18 Select Opertion Exmple Reltion r A B C D A=B ^ D > 5 (r) A B C D

19 Select Opertion Nottion: p (r) p is clled the selection predicte Defined s: p (r) = {t t r nd p(t)} Where p is formul in propositionl clculus consisting of terms connected by logicl connectives: (nd), (or), (not) Ech term is one of: <ttribute> op <ttribute> or <constnt> where op is one of: =,, >,. <. Exmple of selection: brnch_nme= Perryridge (ccount) 9

20 Project Opertion Exmple Reltion r: A B C A,C (r) A C A C = 0

21 Project Opertion Nottion: A, A,, A k ( r ) where A, A re ttribute nmes nd r is reltion nme. The result is defined s the reltion of k columns obtined by ersing the columns tht re not listed Duplicte rows removed from result, since reltions re sets Exmple: To eliminte the brnch_nme ttribute of ccount ccount_number, blnce (ccount)

22 Union Opertion Exmple Reltions r, s: A B A B 3 r s r s: A B 3

23 Union Opertion Nottion: r s Defined s: For r s to be vlid. r s = {t t r or t s}. r, s must hve the sme rity (sme number of ttributes). The ttribute domins must be comptible (exmple: nd column of r dels with the sme type of vlues s does the nd column of s) Exmple: to find ll customers with either n ccount or lon customer_nme (depositor) customer_nme (borrower) 3

24 Set Difference Opertion Exmple Reltions r, s: A B A B 3 r s r s: A B 4

25 Set Difference Opertion Nottion r s Defined s: r s = {t t r nd t s} Set differences must be tken between comptible reltions. r nd s must hve the sme rity ttribute domins of r nd s must be comptible 5

26 Crtesin-Product Opertion Exmple Reltions r, s: A B C D E r b b r x s: A B C D E s b b b b 6

27 Crtesin-Product Opertion Nottion r x s Defined s: r x s = {t q t r nd q s} Assume tht ttributes of r(r) nd s(s) re disjoint. (Tht is, R S = ). If ttributes of r(r) nd s(s) re not disjoint, then renming must be used. 7

28 Composition of Opertions Cn build expressions using multiple opertions Exmple: A=C (r x s) r x s A B C D E b b b b A=C (r x s) A B C D E b 8

29 Renme Opertion Allows us to nme, nd therefore to refer to, the results of reltionllgebr expressions. Allows us to refer to reltion by more thn one nme. Exmple: x (E) returns the expression E under the nme X If reltionl-lgebr expression E hs rity n, then r x ( ( E A, A,..., A n ) returns the result of expression E under the nme X, nd with the ttributes renmed to A, A,., A n. ) 9

30 Bnking Exmple brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) Highlighted ttributes re primry keys. 30

31 Exmple Queries Find ll lons of over $00 mount > 00 (lon) Find the lon number for ech lon of n mount greter thn $00 lon_number ( mount > 00 (lon)) Find the nmes of ll customers who hve lon, n ccount, or both, from the bnk customer_nme (borrower) customer_nme (depositor) brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) 3

32 Exmple Queries Find the nmes of ll customers who hve lon t the Perryridge brnch. customer_nme ( brnch_nme= Perryridge ( borrower.lon_number = lon.lon_number (borrower x lon))) Find the nmes of ll customers who hve lon t the Perryridge brnch but do not hve n ccount t ny brnch of the bnk. customer_nme ( brnch_nme = Perryridge ( borrower.lon_number = lon.lon_number (borrower x lon))) customer_nme (depositor) brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) 3

33 Exmple Queries Find the nmes of ll customers who hve lon t the Perryridge brnch. Query customer_nme ( brnch_nme = Perryridge ( borrower.lon_number = lon.lon_number (borrower x lon))) Wht s the benefit of the second query? Query customer_nme ( lon.lon_number = borrower.lon_number ( ( brnch_nme = Perryridge (lon)) x borrower)) brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) 33

34 Exmple Queries Find the lrgest ccount blnce Strtegy: Find those blnces tht re not the lrgest Renme ccount reltion s d so tht we cn compre ech ccount blnce with ll others Use set difference to find those ccount blnces tht were not selected in the erlier step. Tht missing blnce is the MAX mong ll ccount blnces The query is: blnce (ccount) - ccount.blnce ( ccount.blnce < d.blnce (ccount x d (ccount))) 34

35 Forml Definition A bsic expression in the reltionl lgebr consists of either one of the following: A reltion in the dtbse A constnt reltion (shown lter) Let E nd E be reltionl-lgebr expressions; the following re ll reltionl-lgebr expressions: E E E E E x E p (E ), P is predicte on ttributes in E s (E ), S is list consisting of some of the ttributes in E x (E ), x is the new nme for the result of E 35

36 Additionl Opertions We define dditionl opertions tht do not dd ny power to the reltionl lgebr, but tht simplify common queries. Set intersection Nturl join Division Assignment 36

37 Set-Intersection Opertion Nottion: r s Defined s: r s = { t t r nd t s } Assume: r, s hve the sme rity ttributes of r nd s re comptible Note: r s = r (r s) 37

38 Set-Intersection Opertion Exmple Reltion r, s: A B A B 3 r s r s A B 38

39 Nturl-Join Opertion Nottion: r s Let r nd s be reltions on schems R nd S respectively. Then, r s is reltion on schem R S obtined s follows: Consider ech pir of tuples t r from r nd t s from s. If t r nd t s hve the sme vlue on ech of the ttributes in R S, dd tuple t to the result, where Exmple: t hs the sme vlue s t r on r t hs the sme vlue s t s on s R = (A, B, C, D) S = (E, B, D) Result schem = (A, B, C, D, E) r s is defined s: r.a, r.b, r.c, r.d, s.e ( r.b = s.b r.d = s.d (r x s)) 39

40 Nturl Join Opertion Exmple Reltions r, s: A B 4 C D b b B 3 3 D b b E r A B C D b E s r s 40

41 Division Opertion Nottion: r s Suited to queries tht include the phrse for ll. Let r nd s be reltions on schems R nd S respectively where R = (A,, A m, B,, B n ) S = (B,, B n ) The result of r s is reltion on schem R S = (A,, A m ) r s = { t t R-S (r) u s ( tu r ) } Where tu mens the conctention of tuples t nd u to produce single tuple 4

42 Division Opertion Exmple Reltions r, s: A B B r s: A r s 4

43 Another Division Exmple A B C D b b b b E 3 Reltions r, s: r s: D b E A B C r s 43

44 Division Opertion (Cont.) Property Let q = r s Then q is the lrgest reltion stisfying q x s r Definition in terms of the bsic lgebr opertion Let r(r) nd s(s) be reltions, nd let S R r s = R-S (r ) R-S ( ( R-S (r ) x s ) R-S,S (r )) To see why R-S,S (r) simply reorders ttributes of r R-S ( R-S (r ) x s ) R-S,S (r)) gives those tuples t in R-S (r ) such tht for some tuple u s, tu r. 44

45 Assignment Opertion The ssignment opertion () provides convenient wy to express complex queries. Write query s sequentil progrm consisting of series of ssignments followed by n expression whose vlue is displyed s result of the query. Assignment must lwys be mde to temporry reltion vrible. Exmple: Write r s s temp R-S (r ) temp R-S ((temp x s ) R-S,S (r )) temp temp The result to the right of the is ssigned to the reltion vrible on the left of the. My use vrible in subsequent expressions. 45

46 Bnk Exmple Queries Find the nmes of ll customers who hve lon nd n ccount t bnk. customer_nme (borrower) customer_nme (depositor) Find the nme of ll customers who hve lon t the bnk, the lon number nd the lon mount customer_nme, lon_number, mount (borrower lon) brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) 46

47 Bnk Exmple Queries Find ll customers who hve n ccount from both the Downtown nd the Uptown brnches. Query customer_nme ( brnch_nme = Downtown (depositor customer_nme ( brnch_nme = Uptown (depositor ccount )) ccount)) Query customer_nme, brnch_nme (depositor ccount) temp(brnch_nme ) ({( Downtown ), ( Uptown )}) Note tht Query uses constnt reltion. brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) 47

48 Bnk Exmple Queries Find ll customers who hve n ccount t ll brnches locted in Brooklyn city. customer_nme, brnch_nme (depositor ccount) brnch_nme ( brnch_city = Brooklyn (brnch)) brnch (brnch_nme, brnch_city, ssets) customer (customer_nme, customer_street, customer_city) ccount (ccount_number, brnch_nme, blnce) lon (lon_number, brnch_nme, mount) depositor (customer_nme, ccount_number) borrower (customer_nme, lon_number) 48