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- 1. Relational Algebra 1 Fall 2001 Database Systems 1 Relational Algebra • A relation is a set of tuples. Each relational algebra operation takes as input a list of relations and produces a single relation. • General form: OperatorArguments (List of Relations) New Relation • After each operation, the remaining attributes are carried to the new relation. The attributes may be renamed, but their domain remains the same. Fall 2001 Database Systems 2 Set Theoretic Operations • Regular set operations on relations • A set operation requires two participating relations R and S to be compatible – R and S should have the same attributes • R(Name:D1, Email:D2) • S(Name:D1, Email:D2, Address:D3) • T(Name:D1, Email:D4) • V(Name:D1, Email:D2) Which relations above are union (set operation) compatible? – Union compatibility may require type conversion (casting).
- 2. Relational Algebra 2 Fall 2001 Database Systems 3 Set Operations ¢¡¤£¦¥ §©¨£ ¥¡ ! #¤$%¥(')0¥$¡1$) 243©5 673 SR ∪ 8 9 @BADC7E F©GIHQP©RDHSR©T¤F F 5H QPUFVT 5 3 8 6 T 9 W 3 H¤FVTGQFVXYH 56 V3 SR ∩ 8 9 @BADC7E F©GIHQP©RDHSR©T¤F ` 6 HDP 5 3 8aR 3Yb 9 c 5d Qd FVT¤F 3 XQF SR − 8 9 @BADC7E F©GIHQP©RDHSR©T¤F 5 3 8e`¦AQH 36 H 5 3 9 ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff gggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii Fall 2001 Database Systems 4 pYq¢rs0t%u pQvw r¤x€y¤ ‚yx0vƒ „Y… †(‡Yˆ‰ Smith@aol.com „Y‘ ’“€”€•—– Johns@geocities.com ˜™ d0e “gf)• Brown@netmail.com ˜h i1e j—k Gray@microsoft.com l0m)n)o0p0q l‚r s0tu vw‚xy o0x w0tz {%| })~ € Gray@microsoft,com {% ‚‚ƒ‚„g…‚~ † ‡ Robert432@aol.com {0ˆ ‰ƒgŠ‚‹—‡ Johns@somewhere.com {€Œ ‰—‡ ƒg‹ Jj@yahoo.com {0 Ž(g—… Duke@jazz.com‘g’¤“0”Q• ‘0–— ˜Q–— ™š‚›œ —–ž—Ÿ g¡ œ ¢g£gž š€Ÿ—–£g¤ ¥%¦ §Y¦ ¨)© ª‚«¬‚ ®€¯€ ° ¯g± ¥%² §Y² ³1´µgµ€´± ¶U®%´· ¸ ¹ ¯gº— ° ¯€± ¥0» §Y² ¼¬%«¬‚ © ¯gµ—¬ ½¤¬€¾¦¿¯€· À ¥€Á §» ‚¯gäµ—ª— ³ª‚« «ª— Ä0Å ÆÇ ÈÉ Ê€Ëg̤͂ʂΗςРɀÑÒ Ó—Ô Ô Ê ÕÖ Ó—× ×—Ø Ö Ä0Ù ÆYÚ Æ¤Û€ÑË—ÜÝÉ€ÑÛ Ö Þ Ê€Ô × Êgß à‚áâ%ã à%äå â%æ€ç è á‚äå à0é—êgá€å ë1ì¤í—égî ç ï0ðñ%ò ó0ô—õ€ö ÷%ò ïø òù ï0ðñ%ø ó0ô—õ€ö ÷gú ï%ù û0ù ï0ðñ0õ ó0ô—õ%ò ÷%ò ï€ú òó ï0ðñgú ü%ôgò ÷0ù ï%ù ý%ù ï0ðñ0ù ü%ôgò ÷%ò ïø øgö ï0ðñ0û ü%ôgø ÷gú ï€ú ý‚ö ï0ðñ‚ý ü%ôgø ÷%ø ïò ú ï0ðñ0ó ü%ô—õ ÷%ø ï%õ ù
- 3. Relational Algebra 3 Fall 2001 Database Systems 5 Assignment and Alias • Given a relation R, head(R)=A1…An is the list of attributes in R. • For each attribute, dom(A) gives the domain of A, name(A) gives the name of the attribute • Suppose B1…Bn is a list of attributes such that for each i, dom(Bi) = dom(Ai) Then assignment (:=) changes the names of some or all of the attributes in R and results in a new relation S, S(B1…Bn) := R(A1…An) and dom(B1) = dom(A1) … dom(Bn) = dom(An). Fall 2001 Database Systems 6 Set Operations • Find all people who are registered as buyers or owners ¢¡¤£¦¥¨§© ! ! # $% ')(10 243 Smith@aol.com $65 798@3¤ACB Johns@geocities.com DE% F¦GIHCP Gray@microsoft.com QRQRQ QRQ@Q QRQRQ BUYERS-2(oid,name,email) := BUYERS(buyid,name,email) RESULT := BUYERS-2 ∪ OWNERS
- 4. Relational Algebra 4 Fall 2001 Database Systems 7 Set Operations • Find all people who are owners but not buyers ¢¡¤£¦¥¨§© ! !¤ # $% ')(0 132 Smith@aol.com $4 5768293@ Johns@geocities.com $¨A B¦CED3F Gray@microsoft.com G G G BUYERS-2(oid,name,email) := BUYERS(buyid,name,email) RESULT:= OWNERS - BUYERS-2 Fall 2001 Database Systems 8 Set Operations • Find all people who are both owners and buyers HPI7QSRUTV WYX ` acb7dUe e7dUbfXhg BUYERS-2(oid,name,email) := BUYERS(buyid,name,email) RESULT := OWNERS ∩ BUYERS-2
- 5. Relational Algebra 5 Fall 2001 Database Systems 9 Projection • Select all tuples from relation R, restrict each tuple to the input attributes – written as R[A1…An] or for attributes A1…An from R – meaning nAA ...1 π for each tuple in R construct a new tuple containing only attributes A1…An place the new tuple in the result end-for remove duplicate tuples to preserve the set property Fall 2001 Database Systems 10 Projection • Find the names and email addresses of people who are both buyers and owners ¢¡¤£¦¥¨§© !# $%('0) G ra y @ m ic ro s o ft.c o m RESULT:= OWNERS[name,email] ∩ BUYERS[name,email]
- 6. Relational Algebra 6 Fall 2001 Database Systems 11 Projection • Find all the items on which at least one bid was placed ¢¡¤£¦¥¨§© ! # RESULT := BIDS[iid] Fall 2001 Database Systems 12 Selection • Select certain tuples from a relation R based on a boolean condition C – written as R where C or – meaning: RCσ for each tuple t in R if t makes C true then return t in the resulting relation end-for
- 7. Relational Algebra 7 Fall 2001 Database Systems 13 Selection • The selection condition C is built of – atomic conditions of the form Ai ∝ Aj or Ai ∝ c where • Ai, Aj are attributes and c is a constant of the same domain of values, and • ∝ is a comparison predicate of the form: =, , , = , = , etc. • If C1, C2 are selection conditions, then so are • C1 AND C2 • C1 OR C2 • NOT C1 Fall 2001 Database Systems 14 Selection • Find the bid identifiers of bids for item number “I1”: ¢¡¤£¦¥¨§© ! $# ¤% I1-BIDS := BIDS WHERE iid = “I1” RESULT := I1-BIDS[bid]
- 8. Relational Algebra 8 Fall 2001 Database Systems 15 Selection • Find the bid identifiers of bids for item number “I1” that are strictly more than $15. I1-BIDS := BIDS WHERE iid = “I1” AND amount 15 RESULT := I1-BIDS[bid] ¢¡¤£¦¥¨§© ! # Fall 2001 Database Systems 16 Selection • Find the dates in which a bid for item number “I1” or “I5” was placed I1orI5-BIDS := BIDS WHERE iid = “I1” OR iid = “i5” RESULT := I1orI5-BIDS[date] $%('#)¤021 3542687 9A@2BDC 9A@2BFE G(@HE
- 9. Relational Algebra 9 Fall 2001 Database Systems 17 Cartesian Product • Multiply the relations R and S, producing |R|x|S| many tuples with attributes head(R)+head(S) • Cartesian product combines two relations into one “big” relation • meaning for each tuple r in R do for each tuple s in S do compute a new tuple containing all attributes in r and s end-for end-for Fall 2001 Database Systems 18 R := Bids WHERE Amount 20 T := Items WHERE oid=“O2” S := T[iid,name] P := R X S ¡£¢ ¤ ¥§¦©¨ ¢ ¤ ¡¤ !#©$¨ % '( )0213 465 %7 8£7 % '5 9@0A 47 %7 BC7 % '@8 9@02( 465 %5 B3 D EF G HPIQSR T@U VXW Y2YW `badc£W egf h@i p£q£r qstuY6q v w§x€y‚ ƒ w§x „£…†b‡ w§x ˆ ƒ w‰x€y2g‘2ˆ6ƒ w§x ’§“X”•–† —Sx ˆ ƒ —Sx ˜#…©“X‡ ™d e@f g2h6i6j kl ™m n‚m k2f oCd pp©d qgr ™d ste ™d e©l u2h©v k2m ™m w@m k2f oCd pp©d qgr ™d ste ™d en u2h©f kl ™l wj k2f oCd pp©d qgr ™d ste ™d e@f g2h6i6j kl ™m n‚m k©i x‚y@zy©{b|–}2pgy ™d e©l u2h©v k2m ™m w@m k©i x‚y@zy©{b|–}2pgy ™d en u2h©f kl ™l wj k©i x‚y@zy©{b|–}2pgy
- 10. Relational Algebra 10 Fall 2001 Database Systems 19 Relational Algebra - Join • For all pairs of tuples r and s from relations R and S where Head(R)=A1…AnB1…Bk and Head(S)=B1…BkC1…Cm such that r[B1…Bk] = s[B1…Bk], include a tuple in R ¢¡ S with attributes A1…AnB1…BkC1…Cm • The attributes in common between the two relations (B1…Bk) are called the “join attributes” • The join is a cartesian product followed by a selection on the join attributes, and a projection to remove the duplication of the join attributes [formally: you have to rename the join attributes in one relation so that you can express the selection/projection operations] Also called equijoin since the main criteria is the equality of join attributes Fall 2001 Database Systems 20 Relational Algebra - Join • Find all people with the same name and email address from the “buyers” and “owners” relations – join attributes are “Name” and “Email” from the buyer and owners relations – the joined relation will have attributes “Buyid, Name, Email, Oid” – there are 5 × 4 pairs of tuples to compare on the attributes “Name” and “Email” £¥¤§¦©¨ ¥ ¨¥ £!$#%' )(1032 ¨ 03( %54 %' 67 8@9BADC G r a y @ m ic ro s o ft,c o m E3F
- 11. Relational Algebra 11 Fall 2001 Database Systems 21 Relational Algebra - Join ¢¡¤£¦¥¢§¢¨ ¢© Gray@microsoft,com !# $% Smith@aol.com #')(10323 45 6¤798A@B Johns@geocities.com 4DC)E#F¢GIH PDQ RASUTWVYX Brown@netmail.com `badcDegf3h iqp r¤sUtvu Gray@microsoft.com ixw€ybA‚¢ƒ „¢…¤†¦‡¢ˆ¢‰ „¢ ‘’“” Gray@microsoft,com „¢…¤†¦‡¢ˆ¢‰ „¢ ‘’“” Gray@microsoft,com •¢–¤—¦˜¢™¢d •¢e fghi Gray@microsoft,com Example below shows four of the twenty pairs of tuples considered for the join Fall 2001 Database Systems 22 Relational Algebra - Join • Find the names of items with bids over $20 • Find the names of the owners of items with bids over $20 and are located in Boston Temp1 := Bids WHERE Amount 20 Temp2 := Items jlk Temp1 Result := Temp2[Name] Temp1 := Bids WHERE Amount 20 Temp3 := Items WHERE Location = “Boston” Temp2 := Temp3 jlk Temp1 Temp4 := Temp2 jlk Owners Result := Temp4[Owners.Name]
- 12. Relational Algebra 12 Fall 2001 Database Systems 23 Relational Algebra - Division • Let head(R)=A1…AnB1…Bk and head(S)= B1…Bk • R ÷ S is the “maximal” set of tuples from R[A1…An] such that R contains all tuples in (R ÷ S) x S. • Division is often used for “for-all” queries, e.g. “Find tuples in R that are in a relation with ‘all’ tuples in S.” • All the attributes in the dividing relation S must exist in the divided relation R! for all tuples r in R[A1…An] if for all tuples s in S, r+s is in R place r in R ÷ S end-if end-for Fall 2001 Database Systems 24 Relational Algebra - Division ¡ ¢ £ ¤¦¥ §¨¥ ©¥ ¤ §¨¥ ©¥ ¤¦¥ § ©¥ ¤¦¥ § © ¤ §¨¥ © ¤¦¥ § © ¤¦¥ § © ¤¦¥ §¨¥ © ! $# $% ' ( )10 230 4 5 6 798 @38 7BA @38 Find: • R ÷ S • R ÷ V • R ÷ Y C DEGFIH¦PQSR TUR VXW ÷ S
- 13. Relational Algebra 13 Fall 2001 Database Systems 25 Relational Algebra - Division ¡ ¢ £ ¤¦¥ §¨¥ ©¥ ¤ §¨¥ ©¥ ¤¦¥ § ©¥ ¤¦¥ § © ¤ §¨¥ © ¤¦¥ § © ¤¦¥ § © ¤¦¥ §¨¥ © ! $# $% ' ( )10 230 4 5 6 798 @38 7BA @38 Find: • R ÷ S • R ÷ V • R ÷ Y Fall 2001 Database Systems 26 Relational Algebra - Division • Find identifiers of buyers who placed a bid on “all” the items • Find the identifiers of items that have a bid from “all” the buyers Temp := Items[Iid] Temp1:= Bids[Buyid,Iid] Result := Temp1 ÷ Temp A := Buyers[BuyId] B := Bids[Iid, BuyId] Result := B ÷ A
- 14. Relational Algebra 14 Fall 2001 Database Systems 27 Relational Algebra Operations – Union R ∪ S – Intersection R ∩ S – Difference R - S – Assignment S(B1…Bn) := R(A1…An) – Selection R where C – Projection R[A1…An] – Cartesian Product R x S – Join R ¢¡ S – Division R ÷ S Fall 2001 Database Systems 28 £¥¤§¦¥¨© £¥ ¦ !!# $¥% (')021 Smith@aol.com $¥3 465718@9 Johns@geocities.com $A BC@5EDF8 Brown@netmail.com $ G H(CPIRQ Gray@microsoft.com S!TFU(VWX SY2`bac d¥ef g Vfe!ah ibp qsrut2v Gray@microsoft,com iw xy7€7!r@‚Pƒ Robert432@aol.com i„ …6y7†‡Pƒ Johns@somewhere.com i7ˆ …6tRƒRy!‡ Jj@yahoo.com i‰ (‘E’@ Duke@jazz.com “R”F•–— “˜™ de˜™ f gEh(i j2kRl2g2m ˜ kon p7q req sbtuuvwRx yEz@x@{Pz7| p7} re} ~6€R€o|u y‚„ƒ … zo†PxP{Pz7| pE‡ re} ˆ!wvwox2t@zR€@w ‰FwoŠ ‹zR‚@Œ 2Ž ‘7’o“s”@•2–P– —!•˜˜•@– E™ FŽ š› 2’Rœ2 ž’oŸR 7¡ ›R¢£u•2–P–@’ ¤!¥ •@¦P¦o˜¥ o§ e¨ ›R¢” ¥ © ’@–@¦P’7ª «!¬(® «b¯° (±E²@³ ¬!¯° «´2µ¬E° ¶s· ¸7´E¹2² º»¼b½ ¾!¿7À7Á ½ ºÃ ½uÄ º»¼bà ¾!¿7À7Á ÂRÅ ºÄ ÆÄ º»¼À ¾!¿7À½ ½ º7Å ½„¾ º»¼Å Ç!¿½ ÂEÆ ºÄ ÈbÄ º»¼Ä Ç!¿½ ½ ºÃ ÃEÁ º»¼Æ Ç!¿Ã ÂRÅ º7Å È!Á º»¼!È Ç!¿Ã Âà º½ Å º»¼¾ Ç!¿7À Âà ºÀ Ä Auction Database
- 15. Relational Algebra 15 Fall 2001 Database Systems 29 Problem 1 Find the name and email of owners of items located in “Boston” A := Items WHERE Location = “Boston” B := A ¢¡ Owners Result := B[Owners.Name, Owners.Email] Fall 2001 Database Systems 30 Problem 2 Find the identifiers and amount of bids placed by buyer “Roberts” A := Buyers WHERE Name = “Roberts” B := A ¢¡ Bids Result := B[Bid, Amount]
- 16. Relational Algebra 16 Fall 2001 Database Systems 31 Problem 3 Find the names of buyers who placed a bid on an item owned by “Brown” A := Owners WHERE Name = “Brown” B := (A ¢¡ Items) [Iid] C := B ¢¡ Bids Result := (C ¢¡ Buyers) [Buyers.Name] Fall 2001 Database Systems 32 Problem 4 Find the identifier of items with more than one bid A := Bids[Bid, Iid] B := A x Bids[Bid,Iid] C := B WHERE (A.Bid Bids.Bid) AND (A.Iid = Bids.Iid) Result := C[Iid]
- 17. Relational Algebra 17 Fall 2001 Database Systems 33 Problem 5 Find the names of items all buyers placed a bid on A := (Bids[Iid, BuyId]) ÷ (Buyers[Buyid]) B := A ¢¡ Items Result := B[Name]