1. Schedule Today: Next And then Relational Algebra.
Read Chapter 5 to page 199.
Next
SQL Queries.
Read Sections
And then
Subqueries, Grouping and Aggregation.
Read Sections
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2. Relational Algebra
The relational algebra is a precise mathematical notation and set of rules for manipulating “relations”.
SQL is basically a more human readable form of the relational algebra.
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3. “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.
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.
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4. Relational Algebra  SELECT π PROJECT X CARTESIAN PRODUCT NATURAL JOIN
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5. Selection Example R1 = C(R2)
where C is a condition involving the attributes of relation R2.
Example
Relation Sells:
JoeMenu = bar=Joe's(Sells)
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6. Product
R = R1  R2
pairs each tuple t1 of R1 with each tuple t2 of R2 and puts in R a tuple t1t2.
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7. Natural-Join
R = R R2 is equivalent to R = C(R1  R2) where c is the condition that the the values of the attributes that R1 and R2 have in common must match.
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8. Example
Sells = Bars =
BarInfo = Sells Bars
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9. Combining Operations Algebra = Universe or domain of objects
Operators for constructing expressions.
For relational algebra:
Domain = variables standing for relations + finite, constant relations.
Expressions constructed by applying one of the operators + parentheses.
Query = expression of relational algebra.
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10. Bag Semantics
A relation (in SQL, at least) is really a bag or multiset.
It may contain the same tuple more than once, although there is no specified order (unlike a list).
Example: {1,2,1,3} is a bag and not a set.
Select, project, and join work for bags as well as sets.
Just work on a tuple-by-tuple basis, and don't eliminate duplicates.
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11. Bag Union Bag Intersection Bag Difference
Sum the times an element appears in the two bags.
Example: {1,2,1}  {1,2,3,3} = {1,1,1,2,2,3,3}.
Bag Intersection
Take the minimum of the number of occurrences in each bag.
Example: {1,2,1}  {1,2,3,3} = {1,2}.
Bag Difference
Proper-subtract the number of occurrences in the two bags.
Example: {1,2,1} – {1,2,3,3} = {1}.
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12. Duplicate Elimination
(R) = relation with one copy of each tuple that appears one or more times in R.
Example
R =
A B
1 2
3 4
(R) =
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13. Bank Database Schema Branch = (branch-name, branch-city, assets)
Customer = (customer-name, customer-street, customer-city)
Account = (branch-name, account#, balance)
Depositor = (customer-name, account#)
Loan = (branch-name, loan#, amount)
Borrower = (customer-name, loan#)
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14. The Customer Table Customer-name C-Street C-city Bob 123 Third St
San Jose
Carol
456 Main St
Santa Clara
Ted
89 Blossom Ave
Los Gatos
Alice
64 Longwalk Dr
Oakland
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15. The Account Table Branch-name Account# Balance Oakland 101 2000
SJ-Main
205
7500
207
4500
Santa Clara
311
3100
SJ-West
251
850
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16. The Loan Table Branch-name loan# Amount Oakland 2301 5000 SJ-Main 5155
700
5709
9000
Santa Clara
1541
1800
SJ-West
4321
250
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17. The Depositor Table Customer-name Account# Bob 207 Carol 311 Ted 205
Alice
101
251
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18. Queries on the Loan Table
Branch-name
Oakland
SJ-Main
Santa Clara
SJ-West
Loan = (branch-name, loan#, amount)
Find the names of all the branches in the Loan relation
select branch-name
from Loan
(branch-name) (Loan)
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19. More Queries on the Loan Table
select *
from Loan
where amount > 3000
(amount>3000) (Loan)
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20. where branch-name="Oakland" and amount>1200
Find the loan numbers for all loans made at the Oakland branch with loan amounts greater than 1200. 
select loan#
from Loan
where branch-name="Oakland" and amount>1200
(loan#) (branch-name=“Oakland”^amount>1200) (Loan)
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21. Cross Product Select A1, A2 from R1, R2 Result R = R1  R2
pairs each tuple t1 of R1 with each tuple t2 of R2 and puts in R a tuple t1t2.
(A1, A2) (R1  R2)
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22. Natural Join
Find the name of customers with an account at the Oakland branch.
select customer-name
from Depositor, Account
where
Depositor.account# = Account.account#
and branch-name = "Oakland"
(cust-name) (b-name=“Oakland”) (Depositor Account)
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