CS145 Lecture Notes (6) -- Relational Database Design

For all but the very simplest database applications, there are many, many different relational database "designs" (schemas) that can be used to store the relevant data.

Q: Can one schema be much better than another?
A: Often, yes

Relational database design is the topic of identifying and selecting a good relational schema for an application.

In reality, for big applications higher-level design tools are often used (we'll cover that topic soon), but sometimes designers go straight to relations.
Relational design is one of the more theoretical topics of the course, but it is still of practical use.

Motivation

Student application information:

ID and name
Applying to one or more campuses
Played zero or more sports in high school
Attended one or more high schools

(Assume students played the same sports at every high school attended.)

Proposed relation to capture this information:

   Apply(ID, name, campus, sport, HScode, HScity)

Consider "Mary" (ID 123) who is applying to Berkeley and Santa Cruz. She played tennis, basketball, and volleyball at both of her high schools, Gunn (code 26) and Menlo-Atherton (code 28).

Question: How many tuples in the relation?

Design "Anomalies"

This Apply relation exhibits three types of anomalies:

Redundancy
Update Anomaly
Deletion Anomaly

Question: What's the "best design" for this information?

Good designs:

No anomalies
Can reconstruct all original information

Question: What's the best design if different sports can be played at different high schools?

Design by Decomposition

Start with "mega" relations containing all information
Then decompose them into smaller, better relations containing the same information
Do the decomposition automatically

Overview of Automatic Decomposition

Designer specifies mega-relations plus properties of the data
System decomposes based on properties
Final set of relations satisfies normal form -- guarantees no anomalies, no lost information

Properties and Normal Forms

If designer specifies functional dependencies (FDs) => system produces Boyce-Codd Normal Form (BCNF)
If designer also specifies multivalued dependencies (MVDs) => system produces Fourth Normal Form (4NF)
4NF is stronger than BCNF: Any design that is in 4NF is also in BCNF, but not vice-versa.

FDs and BCNF: Example

Apply(ID,name,campus)
Students can apply to multiple campuses

Question: Is this a good design or should we decompose?

Can have many duplicates of each (ID,name)

Redundant: a given ID always has the same name (but not necessarily vice-versa)
Better to store each ID-name connection once

Functional dependency "A -> B" says a given A value always has the same B value.

Boyce-Codd Normal Form says if A,B are both in relation R, then A must be a key => A-B connection stored only once.

BCNF decomposition says make a separate relation (A,B) and take B out of the original.

MVDs and 4NF: Example

Consider just Apply(ID,campus,sport)

Question: Is this a good design or should we decompose?

Multiplicative effect: C campuses and S sports yields CxS tuples

Question: captured by FDs and BCNF?

Multivalued dependency "A ->> B" says a given A value always has every combination of B and C values (C is remaining attributes).

Fourth Normal Form says if A,B are both in relation R, then A must be a key => don't get multiplicative effect of B and C.

4NF decomposition says make a separate relation (A,B) and take B out of the original.

Details and Theory

Now we will formalize the intuition above.

Functional dependencies
Boyce-Codd Normal Form
BCNF decomposition algorithm
Multivalued dependencies
Fourth Normal Form
4NF decomposition algorithm

Example schema:

   Student(ID, name, address, HScode, HSname, HScity, cumGPA, priority)
   Apply(ID, campus, date, major)

Functional Dependencies

Suppose Student.priority is determined by Student.cumGPA.

(show example)

Formally:

   For every pair of tuples t and u in Student:
   if t[cumGPA] = u[cumGPA] then t[priority] = u[priority]

This is a "functional dependency" (FD): a constraint specified with the schema of a relation.

Based on knowledge of the real world
All data must adhere to it

Notation for declaring an FD for a relation R:

   A1, A2, ..., Am -> B1, B2, ..., Bn   (commas may be omitted)

states that for every pair of tuples t and u in R: if

t[A1,...,Am]
= u[A1,...,Am]

then t[B1,..,Bn] = u[B1,..,Bn].

We will abbreviate A1, A2, ..., Am as AA (or "A-bar") and B1, B2, ..., Bn as BB (or "B-bar").

(simple abstract example)

Question: What are some functional dependencies for Student besides cumGPA -> priority?

Question: What are some functional dependencies for Apply?

Functional Dependencies and Keys

Assume R has no duplicates.

Then AA is a key for R if and only if AA -> BB, where BB is all attributes of R

(show abstract example)

Note:

If R can have duplicates, FD <> key
Book requires "keys" to be minimal (introduces "superkeys")

Trivial FD

AA -> BB where BB is a subset of AA

(example)

Nontrivial FD

AA -> BB where BB is not a subset of AA

(example)

Completely nontrivial FD

AA -> BB with no overlap between AA and BB

(example)

=> Most of the time we're interested in completely nontrivial FD's.

Rules for Functional Dependencies

Splitting Rule

If AA -> B1, B2, ..., Bn then AA -> B1, AA -> B2, ..., AA -> Bn

Question: Can we also split the left-hand side?

Combining Rule

If AA -> B1, AA -> B2, ..., AA -> Bn then AA -> B1, B2, ..., Bn

Trivial Dependency Rules

If AA -> BB then AA -> (BB - AA)
If AA -> BB then AA -> (BB U AA)

Transitive Rule

If AA -> BB and BB -> CC then AA -> CC

Closure of Attributes

Given a relation R, a set of FD's for R, and a set of attributes {A1, A2, ..., Am} of R:

   Find all attributes B in R such that A1, A2, ..., Am -> B

This set of attributes is the "closure" and is denoted {A1, A2, ..., Am}+

Algorithm for computing closure:

   start with {A1, A2, ..., Am}
   repeat until no change:
      if current set of attributes includes LHS of a dependency,
      add RHS attributes to the set

(Effectively applies combining and transitive rules until there's no more change.)

Example: closure of {ID, HScode} in Student given FD's:

   ID -> name, address, cumGPA
   cumGPA -> priority
   HScode -> HSname, HScity

Question: How can we exploit closure to test whether a set of attributes is a key?

Related question: How can we find all keys given a set of FD's?

Specifying FD's for a Relation

A set S2 of FD's "follows" from another set S1 of FD's if all relation instances that satisfy S1 also satisfy S2.

Example: apply rules above

When specifying FD's for a relation, we would like to specify a minimal set of completely nontrivial FD's such that all FD's that hold on the relation follow from the dependencies in this set. Sounds hard, but in practice this approach comes naturally.

Question: How can we tell if one FD follows from others?

Example: Does AA -> BB follow from a set S of FD's?

Decomposition and Boyce-Codd Normal Form

Reminder:

Designer specifies mega-relations plus properties of the data: functional dependencies
System decomposes based on FDs
Final set of relations satisfies normal form (BCNF) -- guarantees no anomalies, no lost information

Decomposition

Formally: decompose R(C1, ..., Ck) into R1(A1, ..., Am) and R2(B1, ..., Bn), where {C1, ..., Ck} = {A1, ..., Am} U {B1, ..., Bn}

(diagram)

Idea of decomposition:

R1 = PROJECT[A1, ..., Am] (R), eliminating duplicates
R2 = PROJECT[B1, ..., Bn] (R), eliminating duplicates
R1 NATURAL-JOIN R2 = R

Note: R is never actually created, it's just a step in the design process.

Example decomposition:

   Student(ID, name, address, HScode, HSname, HScity, cumGPA, priority)

decomposed to

   Student1(ID, name, address, HScode, cumGPA, priority)
   Student2(HScode, HSname, HScity)

(check criteria for decomposition)

Another example decomposition:

   Student1(ID, name, address, HScode, HSname, HScity)
   Student2(name, HSname, cumGPA, priority)

(check criteria for decomposition)

Boyce-Codd Normal Form (BCNF)

Defines which decompositions are "good" ones, based on specified functional dependencies

Given: relation R and set of FD's for R

Definition: R is in BCNF with respect to its FD's if for every nontrivial FD AA -> BB, AA contains a key.

Question: Why does violating this requirement produce a "bad" relation?

Example:

   Student(ID, name, address, HScode, HSname, HScity, cumGPA, priority)

   FD's: ID -> name, address, cumGPA
         cumGPA -> priority
         HScode -> HSname, HScity

   Key:  ID, HScode

Question: Is the relation in BCNF?

Each violation produces anomalies.

Example:

   Apply(ID, campus, date, major)

Can apply to campus multiple times for different majors, but can only apply to a campus once per day

   FD's: ID, campus, date -> major
   Key:  ID, campus, date

Question: Is the relation in BCNF?

BCNF Decomposition Algorithm

   Input: relation schema R and set of FD's for R

   (1) compute keys for R based on FD's
   (2) repeat until no more BCNF violations:
         (2a) pick any R' with AA -> BB that violates BCNF
         (2b) decompose R' into R1(AA,BB) and R2(AA,CC)
              where CC is all attributes in R' except (AA U BB)
         (2c) compute FD's for R1 and R2
         (2d) compute keys for R1 and R2 based on FD's

(diagram)

Question: How can we compute keys in steps (1) and (2d)?

Question: How do we compute FD's in step (2c)?

(run algorithm on Student example)

=> Final decomposed relations may be different depending on which violating FD is chosen in each iteration (step 2(a)), but all decompositions will be in BCNF.

Does BCNF guarantee a good decomposition?

Removes anomalies? Yes
Reconstructs original relation?
- Can we get too many tuples?
  Consider R(A,B,C) with tuples (1,2,3) and (4,2,5)
  Decompose into R1(A,B) with tuples (1,2) and (4,2) R2(B,C) with tuples (2,3) and (2,5)
  "Reassembled" relation has four tuples => wrong
  But BCNF would not decompose this way unless B -> C or B -> A
- Can we get too few tuples? No

With BCNF get:

Set of relations without anomalies
Can reconstruct original relation
But some (relatively uncommon) shortcomings to be discussed

Multivalued Dependencies

Example:

   Apply(ID, campus, major)  // removed date from earlier example

Suppose students can apply to several campuses and different majors, but must apply to the same set of majors at every campus.

Question: What are FD's?

Question: What is key?

Question: Is it in BCNF?

Question: Is it a "good" design?

MVD Definition:

   AA ->> BB is an MVD for relation R if:
   For all tuples t,u in R:
      If t[AA] = u[AA] then there exists a v in R such that:
         (1) v[AA] = t[AA]
         (2) v[BB] = t[BB]
         (3) v[CC] = u[CC] where CC is all attributes in R except (AA U BB)

' (show with picture; show implied fourth tuple)

MVD's are also called "tuple-generating dependencies."

Back to example: What are MVD's?

(show example data to verify)

Intuition: MVD's uncover situations where independent facts related to a certain object are being squished together in one relation.

Trivial MVD

AA ->> BB where BB is a subset of AA or (AA U BB) contains all attributes of R

Nontrivial MVD

AA ->> BB where BB is not a subset of AA and (AA U BB) does not contain all attributes of R

Rules for Multivalued Dependencies

FD-is-an-MVD Rule

If AA -> BB then AA ->> BB

Prove by showing (1), (2), (3) in MVD definition.

Transitive Rule

If AA ->> BB and BB ->> CC then AA ->> CC

Complementation Rule

If AA ->> BB then AA ->> CC where CC is all attributes in R except (AA U BB)

Question: Are there any rules for FD's that do not apply for MVD's?

Fourth Normal Form (4NF)

Defines which decompositions are "good" ones, based on specified functional dependencies and multivalued dependencies

Given: relation R and set of MVD's for R

Definition: R is in 4NF with respect to its MVD's if for every nontrivial MVD AA ->> BB, AA contains a key.

Note: Since every FD is also an MVD, 4NF implies BCNF

Question: What happens in the MVD definition if AA contains a key?

4NF Decomposition Algorithm

Very similar to BCNF:

   Input: relation schema R and set of FD's and MVD's for R

   (1) compute keys for R based on FD's
   (2) repeat until no more 4NF violations:
         (2a) pick any R' with AA ->> BB that violates 4NF
         (2b) decompose R' into R1(AA,BB) and R2(AA,CC)
              where CC is all attributes in R' except (AA U BB)
         (2c) compute FD's and MVD's for R1 and R2
         (2d) compute keys for R1 and R2 based on FD's

Question: How do we compute MVD's in step (2c)?

(run algorithm on Apply example)

Shortcomings of BCNF/4NF

Example 1

Apply(ID, campus, date, major)

A student can apply to each campus only once

Campuses have specific non-overlapping application dates

Question: What are FDs and keys?

Question: What is BCNF decomposition?

Question: Is this decomposition "good"?

Third Normal Form (3NF) is weaker than BCNF (and therefore weaker than 4NF). It permits Apply as above to remain undecomposed.

=> Material on 3NF (Section 3.6.6) is required reading for the course, not covered in lecture.

Example 2

Student(ID, HScode, cumGPA, priority)

  ID -> cumGPA
  cumGPA -> priority
  ID -> priority

Note: ID is not a key.

Apply BCNF decomposition splitting first on ID -> cumGPA

Question: Is the resulting decomposition "good"?

Heuristic: "close" each FD before beginning decomposition

=> Overall, BCNF/4NF decomposition does not guarantee that all of the original FDs can be enforced on the individual decomposed relations.