Functional Dependencies

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1 BCNF and 3NF

2 Functional Dependencies Functional dependencies: modeling constraints on attributes stud-id name address course-id session-id classroom instructor Functional dependencies should be obtained from application requirement analysis Functional dependencies should be maintained for all legal relations in the application CMPT 354: Database I -- BCNF and 3NF 2

3 Closure of Functional Dependencies A set of functional dependencies may logically imply other functional dependencies If A B and B C, then A C The set of all functional dependencies logically implied by F is the closure of F We denote the closure of F by F + F + is a superset of F CMPT 354: Database I -- BCNF and 3NF 3

4 Armstrong s Axioms Finding F + (reflexivity) If β α, then α β (augmentation) If α β, then γ α γβ (transitivity) If α β, and β γ, then α γ These rules are Sound: generate only functional dependencies that actually hold Complete: generate all functional dependencies that hold CMPT 354: Database I -- BCNF and 3NF 4

5 Computing Closure F + = F repeat for each functional dependency f in F + apply reflexivity and augmentation rules on f add the resulting functional dependencies to F + for each pair of functional dependencies f 1 and f 2 in F + if f 1 and f 2 can be combined using transitivity then add the resulting functional dependency to F + until F + does not change any further CMPT 354: Database I -- BCNF and 3NF 5

6 Problem of Non-Key Dependencies bor_loan(customer_id, loan_number, amount) loan_number amount, but loan_number customer_id does not hold The amount information repeats unnecessarily Is loan_number also redundant? CMPT 354: Database I -- BCNF and 3NF 6

7 Update and Deletion Anomalies bor_loan(customer_id, loan_number, amount) Update anomalies: if the amount of a loan is changed and the loan has n customers, the n tuples have to be updated Deletion anomalies: if a loan having n customers is deleted, we have to delete n tuples The amount information has to be deleted n times Failing to update/delete all tuple leads to inconsistency CMPT 354: Database I -- BCNF and 3NF 7

8 Reducing Redundancy Make a table loan(loan_number, amount) where loan_number is the key Rename bor_loan to borrower(customer_id, loan_number) where (customer_id, loan_number) is the key No information loss by joining on loan_number Update/deletion anomalies are eliminated CMPT 354: Database I -- BCNF and 3NF 8

9 Boyce-Codd Normal Form A relation schema R is in BCNF if for all functional dependencies in F + of the form α β at least one of the following holds α β is trivial (i.e., β α) α is a superkey for R bor_loan = (customer_id, loan_number, amount) is not in BCNF loan_number amount holds on bor_loan but loan_number is not a superkey CMPT 354: Database I -- BCNF and 3NF 9

10 Decomposing into BCNF For schema R and a non-trivial dependency α β causes a violation of BCNF, decompose R into (α U β ): α is the key (R -( β - α )) bor_loan = (customer_id, loan_number, amount), loan_number amount (loan_number, amount ) (customer_id, loan_number) CMPT 354: Database I -- BCNF and 3NF 10

11 Missing Functional Dependencies cust_banker_branch(customer_id, employee_id, branch_name, type) R1: employee_id branch_name R2: customer_id, branch_name employee_id Not in BCNF, decompose into banker_branch(employee_id, branch_name) cust_banker(customer_id, employee_id, type) Functional dependency R2 is lost CMPT 354: Database I -- BCNF and 3NF 11

12 Insertion Anomalies Decompose cust_banker_branch(customer_id, employee_id, branch_name, type) into banker_branch(employee_id, branch_name) and cust_banker(customer_id, employee_id, type) Cannot insert (c1, e1, Burnaby, loan) and (c1, e2, Burnaby, loan) into cust_banker_branch R2: customer_id, branch_name employee_id is violated CMPT 354: Database I -- BCNF and 3NF 12

13 Insertion Anomalies Decompose cust_banker_branch(customer_id, employee_id, branch_name, type) into banker_branch(employee_id, branch_name) and cust_banker(customer_id, employee_id, type) Can insert (e1, Burnaby) (e2, Burnaby) into banker_branch, and (c1, e1, loan) and (c1, e2, loan) into cust_banker (c1, e1, Burnaby, loan) and (c1, e2, Burnaby, loan) are inserted into the join of banker_branch and cust_banker CMPT 354: Database I -- BCNF and 3NF 13

14 BCNF and Insertion Anomalies Decomposition to BCNF may lose some functional dependencies The lost functional dependencies lead to insertion anomalies CMPT 354: Database I -- BCNF and 3NF 14

15 BCNF and Dependency Preservation If it is sufficient to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold, then that decomposition is dependency preserving Constraints, including functional dependencies, are costly to check in practice unless they pertain to only one relation joining multiple tables is costly BCNF may not preserve functional dependencies Consider a weaker normal form if functional dependencies cannot be preserved in BCNF CMPT 354: Database I -- BCNF and 3NF 15

16 Decomposition for Preservation Banker-info-schema = (branch-name, customername, banker-name, office-number) R1:banker-name branch-name office-number R2:customer-name branch-name banker-name Decomposition preserving all functional dependencies Banker-office-schema = (banker-name, branch-name, office-number), preserving R1 Banker-schema = (customer-name, branch-name, banker-name), preserving R2 (banker-name, branch-name) is repeated in both tables CMPT 354: Database I -- BCNF and 3NF 16

17 Third Normal Form A relation schema R is in the third normal form (3NF) if for all α β in F + at least one of the following holds α β is trivial (i.e., β α) α is a superkey for R Each attribute A in β α is contained in a candidate key for R Each attribute may be in a different candidate key If a relation is in BCNF it is in 3NF In BCNF one of the first two conditions above must hold The third condition is a minimal relaxation of BCNF to ensure dependency preservation CMPT 354: Database I -- BCNF and 3NF 17

18 Comparison of BCNF and 3NF It is always possible to decompose a relation into a set of relations that are in 3NF such that the decomposition is lossless and the dependencies are preserved It is always possible to decompose a relation into a set of relations that are in BCNF such that the decomposition is lossless It may not be possible to preserve all functional dependencies CMPT 354: Database I -- BCNF and 3NF 18

19 Design Goals Goal for a relational database design is BCNF Lossless join Dependency preservation If we cannot achieve all, we have to trade off between Lack of dependency preservation Redundancy due to use of 3NF CMPT 354: Database I -- BCNF and 3NF 19

20 Functional Dependencies and SQL SQL does not provide a direct way of specifying functional dependencies other than superkeys Can specify FDs using assertions, but they are expensive to test Even if we had a dependency preserving decomposition, using SQL we would not be able to efficiently test a functional dependency whose left hand side is not a key CMPT 354: Database I -- BCNF and 3NF 20

21 Summary Non-key dependencies causes redundancy and anomalies BCNF lossless decomposition, but may not preserve all functional dependencies 3NF lossless decomposition preserving all functional dependencies, but may still contain some redundancy CMPT 354: Database I -- BCNF and 3NF 21

22 To-Do-List Read Section 7.3 in the textbook Given a relation R, a set of functional dependencies F, and a decomposition of R how can we determine whether the decomposition is in BCNF? How can we determine whether the decomposition is in 3NF? CMPT 354: Database I -- BCNF and 3NF 22

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