Fall 2001Arthur Keller – CS 1805–1 Schedule Today Oct. 9 (T) Multivalued Dependencies, Relational Algebra u Read Sections 3.7, Assignment 2 due. Oct. 11 (TH) SQL Queries. u Read Sections Project Part 2 due. Oct. 16 (T) Duplicates, Aggregation, Modifications. u Read Sections , Assignment 3 due.

Fall 2001Arthur Keller – CS 1805–2 Multivalued Dependencies The multivalued dependency X  Y holds in a relation R if whenever we have two tuples of R that agree in all the attributes of X, then we can swap their Y components and get two new tuples that are also in R.

Fall 2001Arthur Keller – CS 1805–3 Example Drinkers(name, addr, phones, beersLiked) with MVD Name  phones. If Drinkers has the two tuples: it must also have the same tuples with phones components swapped: Note: we must check this condition for all pairs of tuples that agree on name, not just one pair.

Fall 2001Arthur Keller – CS 1805–4 MVD Rules 1.Every FD is an MVD. u Because if X  Y, then swapping Y’s between tuples that agree on X doesn’t create new tuples.  Example, in Drinkers : name  addr. 2.Complementation: if X  Y, then X  Z, where Z is all attributes not in X or Y.  Example: since name  phones holds in Drinkers, so does Name  addr beersLiked.

Fall 2001Arthur Keller – CS 1805–5 Splitting Doesn’t Hold Sometimes you need to have several attributes on the right of an MVD. For example: Drinkers(name, areaCode, phones, beersLiked, beerManf) name  areaCode phones holds, but neither name  areaCode nor name  phones do.

Fall 2001Arthur Keller – CS 1805–6 4NF Eliminate redundancy due to multiplicative effect of MVD's. Roughly: treat MVD's as FD's for decomposition, but not for finding keys. Formally: R is in Fourth Normal Form if whenever MVD X  Y is nontrivial (Y is not a subset of X, and X  Y is not all attributes), then X is a superkey. u Remember, X  Y implies X  Y, so 4NF is more stringent than BCNF. Decompose R, using 4NF violation X  Y, into XY and X  (R—Y).

Fall 2001Arthur Keller – CS 1805–7 Example Drinkers(name, addr, phones, beersLiked) FD: name  addr Nontrivial MVD’s: name  phones and name  beersLiked. Only key: { name, phones, beersLiked } All three dependencies above violate 4NF. Successive decomposition yields 4NF relations: D1(name, addr) D2(name, phones) D3(name, beersLiked)

Fall 2001Arthur Keller – CS 1805–8 “Core” Relational Algebra A small set of operators that allow us to manipulate relations in limited but useful ways. The operators are: 1.Union, intersection, and difference: the usual set operators. u But the relation schemas must be the same. 2.Selection: Picking certain rows from a relation. 3.Projection: Picking certain columns. 4.Products and joins: Composing relations in useful ways. 5.Renaming of relations and their attributes.

Fall 2001Arthur Keller – CS 1805–9 Relational Algebra limited expressive power (subset of possible queries) good optimizer possible rich enough language to express enough useful things Finiteness  SELECT π PROJECT X CARTESIAN PRODUCT FUNDAMENTAL U UNION BINARY – SET-DIFFERENCE  SET-INTERSECTION  THETA-JOIN CAN BE DEFINED NATURAL JOIN IN TERMS OF ÷ DIVISION or QUOTIENT FUNDAMENTAL OPS UNARY

Fall 2001Arthur Keller – CS 1805–10 Extra Example Relations DEPOSIT(branch-name, acct-no,cust-name,balance) CUSTOMER(cust-name,street,cust-city) BORROW(branch-name,loan-no,cust-name,amount) BRANCH(branch-name,assets, branch-city) CLIENT(cust-name,empl-name) Borrow B-N L-# C-N AMT T1 Midtown 123 Fred 600 T2 Midtown 234 Sally 1200 T3 Midtown 235 Sally 1500 T4 Downtown 612 Tom 2000

Fall 2001Arthur Keller – CS 1805–11 Selection R 1 =  C (R 2 ) where C is a condition involving the attributes of relation R 2. Example Relation Sells : JoeMenu =  bar=Joe's ( Sells )

Fall 2001Arthur Keller – CS 1805–12 SELECT (  ) arity(  (R)) = arity(R) 0  card(  (R))  card(R)  c (R)  c (R)  (R) c is selection condition: terms of form: attr op value attr op attr op is one of  ≠ ≥ example of term: branch-name = "Midtown" terms are connected by   branch-name = "Midtown"  amount > 1000 (Borrow)  cust-name = emp-name (client)

Fall 2001Arthur Keller – CS 1805–13 Projection R 1 =  L (R 2 ) where L is a list of attributes from the schema of R 2. Example  beer,price ( Sells ) Notice elimination of duplicate tuples.

Fall 2001Arthur Keller – CS 1805–14 Projection (π) 0  card (π A (R))  card (R) arity (π A (R)) = m  arity(R) = k π i 1,...,i m (R) 1  i j  k distinct  produces set of m-tuples  a 1,...,a m  such that  k-tuple  b 1,...,b k  in R where a j = b i j for j = 1,...,m π branch-name, cust-name  (Borrow) Midtown Fred Midtown Sally Downtown Tom

Fall 2001Arthur Keller – CS 1805–15 Product R = R 1  R 2 pairs each tuple t 1 of R 1 with each tuple t 2 of R 2 and puts in R a tuple t 1 t 2.

Fall 2001Arthur Keller – CS 1805–16 Cartesian Product (  ) arity(R) = k1 arity(R  S) = k1 + k2 arity(S) = k2 card(R  S) = card(R)  card(S) R  S is the set all possible (k1 + k2)-tuples whose first k1 attributes are a tuple in R last k2 attributes are a tuple in S R S R  S A B C D D E F A B C D D' E F

Fall 2001Arthur Keller – CS 1805–17 Theta-Join R = R 1 C R 2 is equivalent to R =  C (R 1  R 2 ).

Fall 2001Arthur Keller – CS 1805–18 Example Sells = Bars = BarInfo = Sells Sells.Bar=Bars.Name Bars

Fall 2001Arthur Keller – CS 1805–19 Theta-Join R arity(R) = r arity(S) = s arity (R S) = r + s 0  card(R S)  card(R)  card(S) S i  j  $i  $  r  j)  R  S) R S 1... r 1... s   can be = ≠  If equal (=), then it is an  EQUIJOIN RS =  (R  S) c c R(A B C) S(C D E) result has schema T(A B C C' D E) R.A<S.D ij R(ABC) S(CDE) T(ABCC’DE)

Fall 2001Arthur Keller – CS 1805–20 Natural Join R = R 1 R 2 calls for the theta-join of R 1 and R 2 with the condition that all attributes of the same name be equated. Then, one column for each pair of equated attributes is projected out. Example Suppose the attribute name in relation Bars was changed to bar, to match the bar name in Sells. BarInfo = Sells Bars

Fall 2001Arthur Keller – CS 1805–21 Renaming  S(A 1,…,A n ) (R) produces a relation identical to R but named S and with attributes, in order, named A 1,…,A n. Example Bars =  R( bar,addr ) ( Bars ) = The name of the second relation is R.

Fall 2001Arthur Keller – CS 1805–22 Union (R  S) arity(R) = arity(S) = arity(R  S) max(card(R),card(S))  card(R  S)  card(R) + card(S) set of tuples in R or S or bothR  R  S S  R  S Find customers of Perryridge Branch π Cust-Name (  Branch-Name = "Perryridge" (BORROW  DEPOSIT) )

Fall 2001Arthur Keller – CS 1805–23 Difference(R  S) arity(R) = arity(S) = arity(R  –  S) 0  card(R –  S)  card(R)  R – S  R is the tuples in R not in S Depositors of Perryridge who aren't borrowers of Perryridge π Cust-Name (  Branch-Name = "Perryridge" ( DEPOSIT – BORROW) ) Deposit Borrow π Cust-Name (  Branch-Name = "Perryridge" ( DEPOSIT) ) — π Cust-Name (  Branch-Name = "Perryridge" ( BORROW) ) Does  (π (D)  π (B) ) work?

Fall 2001Arthur Keller – CS 1805–24 Combining Operations Algebra = 1.Basis arguments + 2.Ways of constructing expressions. For relational algebra: 1.Arguments = variables standing for relations + finite, constant relations. 2.Expressions constructed by applying one of the operators + parentheses. Query = expression of relational algebra.

Fall 2001Arthur Keller – CS 1805–25 π Cust-Name,Cust-City (  CLIENT.Banker-Name = "Johnson" (CLIENT  CUSTOMER) ) = π Cust-Name,Cust-City (CUSTOMER) Is this always true? π CLIENT.Cust-Name, CUSTOMER.Cust-City (  CLIENT.Banker-Name = "Johnson"  CLIENT.Cust-Name = CUSTOMER.Cust-Name (CLIENT  CUSTOMER) ) π CLIENT.Cust-Name, CUSTOMER  Cust-City (  CLIENT.Cust-Name =CUSTOMER.Cust-Name (CUSTOMER  π Cust-Name  CLIENT.Banker-Name="Johnson" (CLIENT) ) ) )

Fall 2001Arthur Keller – CS 1805–26 SET INTERSECTION arity(R) = arity(S) = arity (R  S) (R  S) 0  card (R  S)  min (card(R), card(S)) tuples both in R and in S R  (R  S) = R  S SR  R  S  R  R  S  S

Fall 2001Arthur Keller – CS 1805–27 Operator Precedence The normal way to group operators is: 1. Unary operators , , and  have highest precedence. 2. Next highest are the “multiplicative” operators,, C, and . 3. Lowest are the “additive” operators, , , and —. But there is no universal agreement, so we always put parentheses around the argument of a unary operator, and it is a good idea to group all binary operators with parentheses enclosing their arguments. Example Group R   S T as R  (  (S ) T ).

Fall 2001Arthur Keller – CS 1805–28 Each Expression Needs a Schema If , , — applied, schemas are the same, so use this schema. Projection: use the attributes listed in the projection. Selection: no change in schema. Product R  S: use attributes of R and S. u But if they share an attribute A, prefix it with the relation name, as R.A, S.A. Theta-join: same as product. Natural join: use attributes from each relation; common attributes are merged anyway. Renaming: whatever it says.

Fall 2001Arthur Keller – CS 1805–29 Example Find the bars that are either on Maple Street or sell Bud for less than $3. Sells(bar, beer, price) Bars(name, addr)

Fall 2001Arthur Keller – CS 1805–30 Example Find the bars that sell two different beers at the same price. Sells(bar, beer, price)

Fall 2001Arthur Keller – CS 1805–31 Linear Notation for Expressions Invent new names for intermediate relations, and assign them values that are algebraic expressions. Renaming of attributes implicit in schema of new relation. Example Find the bars that are either on Maple Street or sell Bud for less than $3. Sells(bar, beer, price) Bars(name, addr) R1(name) :=  name (  addr = Maple St. (Bars)) R2(name) :=  bar (  beer=Bud AND price<$3 (Sells)) R3(name) := R1  R2

Fall 2001Arthur Keller – CS 1805–32 Why Decomposition “Works”? What does it mean to “work”? Why can’t we just tear sets of attributes apart as we like? Answer: the decomposed relations need to represent the same information as the original. u We must be able to reconstruct the original from the decomposed relations. Projection and Join Connect the Original and Decomposed Relations Suppose R is decomposed into S and T. We project R onto S and onto T.

Fall 2001Arthur Keller – CS 1805–33 Example R = Recall we decomposed this relation as:

Fall 2001Arthur Keller – CS 1805–34 Project onto Drinkers1 ( name, addr, favoriteBeer ): Project onto Drinkers3(beersLiked, manf): Project onto Drinkers4(name, beersLiked):

Fall 2001Arthur Keller – CS 1805–35 Reconstruction of Original Can we figure out the original relation from the decomposed relations? Sometimes, if we natural join the relations. Example Drinkers3 Drinkers4 = Join of above with Drinkers1 = original R.

Fall 2001Arthur Keller – CS 1805–36 Theorem Suppose we decompose a relation with schema XYZ into XY and XZ and project the relation for XYZ onto XY and XZ. Then XY XZ is guaranteed to reconstruct XYZ if and only if X  Y (or equivalently, X  Z). Usually, the MVD is really a FD, X  Y or X  Z. BCNF: When we decompose XYZ into XY and XZ, it is because there is a FD X  Y or X  Z that violates BCNF. u Thus, we can always reconstruct XYZ from its projections onto XY and XZ. 4NF: when we decompose XYZ into XY and XZ, it is because there is an MVD X  Y or X  Z that violates 4NF. u Again, we can reconstruct XYZ from its projections onto XY and XZ.