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Relational Algebra (continued)

Relational Algebra (continued). Lecture 5. Relational Algebra (Summary). Basic operations: Selection (  ) Selects a subset of rows from relation. Projection (  ) Deletes unwanted columns from relation. Cross-product (  ) Allows us to combine two relations.

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Relational Algebra (continued)

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  1. Relational Algebra(continued) Lecture 5

  2. Relational Algebra (Summary) • Basic operations: • Selection (  ) Selects a subset of rows from relation. • Projection (  ) Deletes unwanted columns from relation. • Cross-product(  ) Allows us to combine two relations. • Set-difference ( - ) Tuples in reln. 1, but not in reln. 2. • Union(  ) Tuples either in reln. 1 or in reln. 2 or in both. • Rename(  ) Changes names of the attributes • Since each operation returns a relation, operationscan be composed ! (Algebra is “closed”.) • Use of temporary relations recommended.

  3. Additional operators • We define additional operations that do not add any power to the relational algebra, but that simplify common queries. • Natural join • Conditional Join • Equi-Join • Division • Also, we’ve already seen “Set intersection”: • rs = r - (r - s) All joins are really special cases of conditional join

  4. Quick note on notation bad_customers good_customers • If we have two or more relations which feature the same attribute names, we could confuse them. To prevent this we can use dot notation. • For example • good_customers.loan-number

  5. Natural-Join Operation: Motivation borrower loan Very often, we have a query and the answer is not contained in a single relation. For example, I might wish to know where Apu banks. The classic relational algebra way to do such queries is a cross product, followed by a selection which tests for equality on some pair of fields. borrower.l-number = loan.l-number(borrower x loan))) • While this works… • it is unintuitive • it requires a lot of memory • the notation is cumbersome Note that in this example the two relations are the same size (2 by 2), this does not have to be the case. So, we have a more intuitive way of achieving the same effect, the natural join, denoted by the symbol

  6. borrower loan Natural-Join Operation: Intuition Natural join combines a cross product and a selection into one operation. It performs a selection forcing equality on those attributes that appear in both relation schemes. Duplicates are removed as in all relation operations. For the previous example (one attribute in common: “l-number”), we have… = borrower.l-number = loan.l-number(borrower x loan))) • If the two relations have no attributes in common, then their natural join is simply their cross product. • If the two relations have more than one attribute in common, then the natural join selects only the rows where all pairs of matching attributes match. (let’s see an example on the next slide).

  7. B A Both the l-name and the f-name match, so select. Only the f-names match, so don’t select. Only the l-names match, so don’t select. We remove duplicate attributes… The natural join of A and B AB = Note that this is just a way to visualize the natural join, we don’t really have to do the cross product as in this example

  8. Natural-Join Operation • Notation: rs • Let r and s be relation instances on schemas R and S respectively.The result is a relation on schema RS which is obtained by considering each pair of tuples tr from r and ts from s. • If tr and ts have the same value on each of the attributes in RS, a tuple t is added to the result, where • t has the same value as tr on r • t has the same value as ts on s • Example: • R = (A, B, C, D) • S = (E, B, D) • Result schema = (A, B, C, D, E) • rs is defined as: • r.A, r.B, r.C, r.D, s.E (r.B = s.B r.D = s.D (r x s))

  9. r s Natural Join Operation – Example B D E A B C D • Relation instances r, s: 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b r s A B C D E How did we get here? Lets do a trace over the next few slides…      1 1 1 1 2      a a a a b     

  10. A B C D B D E      1 2 4 1 2      a a b a b 1 3 1 2 3 a a a b b      r s First we note which attributes the two relations have in common…

  11. A B C D B D E      1 2 4 1 2      a a b a b 1 3 1 2 3 a a a b b      r s A B C D E   1 1   a a   There are two rows in s that match our first row in r, (in the relevant attributes) so both are joined to our first row…

  12. B D E A B C D 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b r s A B C D E   1 1   a a   …there are no rows in s that match our second row in r, so do nothing…

  13. B D E A B C D 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b r s A B C D E   1 1   a a   …there are no rows in s that match our third row in r, so do nothing…

  14. B D E A B C D 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b r s A B C D E     1 1 1 1     a a a a     There are two rows in s that match our fourth row in r, so both are joined to our fourth row…

  15. B D E A B C D 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b r s A B C D E      1 1 1 1 2      a a a a b      There is one row that matches our fifth row in r,.. so it is joined to our fifth row and we are done!

  16. Natural Join on Sailors Example S1 R1

  17. Last week we saw… Query: Find the name of the sailor who reserved boat 101. * Note my use of “temporary” relation Temp.

  18. Query revisited using natural join Query: Find the name of the sailor who reserved boat 101.

  19. Conditional-Join Operation: rcs = c(rxs) The conditional join is actually the most general type of join. We introduced the natural join first only because it is more intuitive and... natural! Just like natural join, conditional join combines a cross product and a selection into one operation. However instead of only selecting rows that have equality on those attributes that appear in both relation schemes, we allow selection based on any predicate. Where c is any predicate on the attributes of r and/or s Duplicate rows are removed as always, but duplicate columns are not removed!

  20. rr.age < s.age AND r.Marr-Lic = s.Marr-Lics Conditional-Join Example: We want to find all women that are younger than their husbands… s r Note we have removed ambiguity of attribute names by using “dot” notation Also note the redundant information in the marr-lic attributes

  21. Equi-Join • Equi-Join: Special case of conditional join where the conditions consist only of equalities. • Natural Join: Special case of equi-join in which equalities are specified on ALL fields having the same names in both relations.

  22. rr.Marr-Lic = s.Marr-Lics Equi-Join s r

  23. Review on Joins • All joins combine a cross product and a selection into one operation. • Conditional Join • the selection condition can be of any predicate (e.g. rating1 > rating2) • Equi-Join: • Special case of conditional join where the conditions consist only of equalities. • Natural Join • Special case of equi-join in which equalities are specified on ALL fields having the same names in both relations.

  24. Banking Examples branch (branch-name, branch-city, assets) customer (customer-name, customer-street, customer-only) account (account-number, branch-name, balance) loan (loan-number, branch-name, amount) depositor (customer-name, account-number) borrower (customer-name, loan-number) Note that I have not indicated primary keys here for simplicity.

  25. Example Queries “select from the relation loan, only the rows which have an amount greater than 1200” • Find all loans of over $1200 amount> 1200 (loan) loan amount> 1200 (loan)

  26. Example Queries • Find the loan number for each loan of an amount greater than $1200 loan-number (amount> 1200 (loan)) “select from the relation loan, only the rows which have an amount greater than 1200, then project out just the loan_number” loan amount> 1200 (loan) loan-number (amount> 1200 (loan))

  27. Example Queries • Find all loans greater than $1200 or less than $75 amount> 1200 amount< 75(loan) “select from the relation loan, only the rows which have an amount greater than 1200 or an amount less than 75 loan amount> 1200 amount< 75(loan)

  28. Example Queries • Find the names of all customers who have a loan, an account, or both, from the bank customer-name (borrower)  customer-name (depositor) borrower depositor • customer-name (depositor) customer-name (borrower)

  29. Note this example is split over two slides! Example Queries Find the names of all customers who have a loan at the Riverside branch. customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan))) borrower loan We retrieve borrower and loan… …we calculate their cross product…

  30. customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan))) …we calculate their cross product… …we select the rows where borrower.loan-number is equal to loan.loan-number… …we select the rows where branch-name is equal to “Riverside” …we project out the customer-name.

  31. Now Using Natural Join Find the names of all customers who have a loan at the Riverside branch. customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan))) => customer-name (branch-name=“Riverside ”borrower loan)) borrower loan We retrieve borrower and loan… 1234 in borrower is matched with 1234 in loan… 3421 in borrower is matched with 3421 in loan… The rest is the same.

  32. Note this example is split over three slides! Example Queries Find the largest account balance ...we will need to rename account relation as d... balance(account) - account.balance(account.balance < d.balance(account xr(d,account))) d account We do a rename to get a “copy” of account which we call d… … next we will do a cross product…

  33. balance(account) - account.balance(account.balance < d.balance(account xr(d,account))) … do a cross product… …select out all rows where account.balance is less than d.balance… .. next we project…

  34. balance(account) - account.balance(account.balance < d.balance(account xr(d,account))) .. next we project out account.balance… …then we do a set difference between it and the original account.balance from the account relation… … the set difference leaves us with one number, the largest value! account

  35. Now Using Conditional Join Find the largest account balance ...we will need to rename account relation as d... balance(account) - account.balance(account.balance < d.balance(account xr(d,account))) • r(d,account) balance(account) - account.balance(account account.balance < d.balance d)

  36. More Examples on Sailors Relations Sailors(sid, sname, rating, age) Boats(bid, bname, color) Reserves(sid, bid, day)

  37. Find names of sailors who’ve reserved boat #103 • Solution 1: • Solution 2: • Solution 3:

  38. Find names of sailors who’ve reserved a red boat • Information about boat color only available in Boats; so need an extra join: • A more efficient solution: • A query optimizer can find this given the first solution!

  39. What happens if is replaced by in this query? Find sailors who’ve reserved a red or a green boat • Can identify all red or green boats, then find sailors who’ve reserved one of these boats: • Can also define Tempboats using union! (How?)

  40. Find sailors who’ve reserved a red and a green boat • Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors):

  41. Consider yet another query • Find the sailor(s) (sid) who reserved all the red boats. R1 B

  42. Start an attempt • who reserved what boat: • All the red boats:

  43. Division Operation r / s • Suited to queries that include the phrase “for all”, e.g. Find sailors who have reserved allred boats. • Let S1 have 2 fields, x and y; S2 have only field y: • S1/S2 = • i.e., S1/S2 contains all x tuples (sailors) such that for everyy tuple (redboat) in S2, there is an xy tuple in S1 (i.e, x reserved y). • In general, x and y can be any lists of fields; y is the list of fields in S2, and xy is the list of fields of S1. • Let r and s be relations on schemas R and S respectively where • R = (A1, …, Am, B1, …, Bn) • S = (B1, …, Bn) The result of r / s is a relation on schema R – S = (A1, …, Am)

  44. Division Operation – Example B A B Relations r, s: 1 2            1 2 3 1 1 1 3 4 6 1 2 s A   r/ s: r •  occurs in the presence of both 1 and 2, so it is returned. •  occurs in the presence of both 1 and 2, so it is returned. • does not occur in the presence of both 1 and 2, so is ignored. ...

  45. Another Division Example Relations r, s: A B C D E D E         a a a a a a a a         a a b a b a b b 1 1 1 1 3 1 1 1 a b 1 1 s A B C r/s:   a a   r <, a , > occurs in the presence of both <a,1> and <b,1>, so it is returned. < , a , > occurs in the presence of both <a,1> and <b,1>, so it is returned. <, a , > does not occur in the presence of both <a,1> and <b,1>, so it is ignored.

  46. Examples of Division A/B B1 B2 B3 A/B1 A/B2 A/B3 A

  47. Find the sailor(s) who reserved ALL red boats • who reserved what boat: • All the red boats: => S1/S2

  48. Find the names of sailors who’ve reserved all boats • Uses division; schemas of the input relations must be carefully chosen: • To find sailors who’ve reserved all ‘Interlake’ boats: .....

  49. ( ) ( ) ( ) s º s s R . . . R (Cascade) Ù Ù c 1 . . . cn c 1 cn ( ) ( ) ( ) ( ) (Commute) s s º s s R R 1 2 2 1 c c c c ( ) ( ) ( ) ( ) (Cascade) p º p p R . . . R l1 ln l1 Some Properties • Selection • Projection where li is a subset of attributes in R, li li+1 • A projection commutes with a selection that only uses attributes retained by the projection: where c(a1,…,an) is a condition on attributes a1,…,an a1,..,an(c(a1,..,an) (R))= c(a1,..,an) (a1,..,an( R ))

  50. R c1 (S c2T) (R c1S) c2T (R cS) (S cR) c(R c S)c(R) c S Properties of Joins Joins where c(a1,…,an) is a condition on attributes only in R (Associative) (Commute)

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